NCERT Class 12 Maths Ex 6.3 Solutions

Q1

LONG ANSWER4 marks

Find the maximum and minimum values, if any, of the following functions:

\[ f(x)=(2x-1)^2+3 \]
\[ f(x)=9x^2+12x+3 \]
\[ f(x)=-(x-1)^2+10 \]
\[ g(x)=x^3+1 \]

📘 Concept & Theory

Theory and Concept

›

To find the maximum or minimum value of a differentiable function, we first locate its critical points. A critical point is generally obtained by solving

\[ f'(x)=0 \]

or by identifying points where $f'(x)$ does not exist.

Once a critical point is obtained, the second derivative test can be used:

If $f''(a)>0$, then $f(x)$ has a local minimum at $x=a$.
If $f''(a)<0$, then $f(x)$ has a local maximum at $x=a$.
If $f''(a)=0$, the second derivative test is inconclusive. Another method, such as the first derivative test or direct analysis of the function, must be used.

For polynomial functions defined on all real numbers, we should also distinguish between a local extremum and an absolute maximum or minimum. In this question, the quadratic functions can be analysed directly because their squared terms are always non-negative. For the cubic function $g(x)=x^3+1$, the derivative analysis shows that the function is increasing throughout its domain.

🗺️ Solution Roadmap

Step-by-step Plan

›

Differentiate the given function.
Set the first derivative equal to zero to obtain the critical point.
Find the second derivative.
Apply the second derivative test wherever it gives a definite conclusion.
If the second derivative is zero at a critical point, do not conclude that the function has neither maximum nor minimum. Instead, analyse the sign of the first derivative or the function itself.
Substitute the critical point into the original function to obtain the corresponding extremum value.

✏️ Solution

(i) $f(x)=(2x-1)^2+3$

›

Step-by-step Solution · 10 steps

Given —
\[f(x)=(2x-1)^2+3\]
Find the first derivative
Using the chain rule,
\[f'(x)=2(2x-1)\frac{d}{dx}(2x-1)+\frac{d}{dx}(3)\]

\[f'(x)=2(2x-1)(2)+0\]

\[f'(x)=4(2x-1)\]
Find the critical point
For a possible maximum or minimum,
\[f'(x)=0\]
Therefore,
\[4(2x-1)=0\]
Since $4\neq0$,
\[2x-1=0\]

\[2x=1\]

\[x=\frac{1}{2}\]
Find the second derivative
\[f'(x)=4(2x-1)\]
Differentiating again,
\[ f''(x)=\frac{d}{dx}\left[4(2x-1)\right] \]

\[f''(x)=4\frac{d}{dx}(2x-1) \]

\[f''(x)=4(2) \]

\[f''(x)=8 \]
Therefore,
\[f''\left(\frac{1}{2}\right)=8>0\]
Since the second derivative is positive at the critical point, the function has a minimum at
\[ x=\frac{1}{2}. \]
Find the minimum value
Therefore, the minimum value is $3$, attained at $x=\frac12$.
There is no maximum value because
\[ (2x-1)^2\to\infty \]
as $|x|\to\infty$. Hence,
\[ f(x)\to\infty. \]

✏️ Solution

(ii) $f(x)=9x^2+12x+3$

›

Step-by-step Solution · 10 steps

Given —
\[f(x)=9x^2+12x+3\]
Find the first derivative
Differentiating term by term,
\[f'(x)=18x+12+0\]

\[f'(x)=18x+12\]
Find the critical point
Set the first derivative equal to zero:
\[f'(x)=0\]

\[18x+12=0\]

\[18x=-12\]

\[x=-\frac{12}{18}\]

\[x=-\frac{2}{3}\]
Thus, the critical point occurs at
\[x=-\frac23\]
Find the second derivative
\[f'(x)=18x+12\]
Therefore,
\[f''(x)=18\]
Hence,
\[f''\left(-\frac23\right)=18>0\]
Since the second derivative is positive, the function has a minimum at
\[ x=-\frac23 \]
Find the minimum value
\[f\left(-\frac23\right)=9\left(-\frac23\right)^2+12\left(-\frac23\right)+3\]

\[=9\left(\frac49\right)-8+3\]

\[=4-8+3\]

\[=-1\]
Therefore, the minimum value is $-1$.
There is no maximum value because the coefficient of $x^2$ is positive and
\[ f(x)\to\infty \]
as $|x|\to\infty$.

✏️ Solution

(iii) $f(x)=-(x-1)^2+10$

›

Step-by-step Solution · 13 steps

Given —
\[f(x)=-(x-1)^2+10\]
Find the first derivative
Differentiating using the chain rule,
\[f'(x)=-2(x-1)\frac{d}{dx}(x-1)+0\]

\[f'(x)=-2(x-1)\]
Expanding,
\[f'(x)=-2x+2\]
Find the critical point
\[f'(x)=0\]
Therefore,
\[-2(x-1)=0\]
Since $-2\neq0$,
\[x-1=0\]

\[x=1\]
Find the second derivative
\[f'(x)=-2x+2\]
Differentiating,
\[f''(x)=-2\]
Therefore,
\[f''(1)=-2<0\]
Since the second derivative is negative at the critical point, the function has a maximum at
\[ x=1. \]
Find the maximum value
\[f(1)=-(1-1)^2+10\]

\[=-(0)^2+10\]

\[=0+10\]

\[=10\]
Therefore, the maximum value is $10$.
There is no minimum value because
\[ -(x-1)^2\to-\infty \]
as $|x|\to\infty$. Consequently,
\[ f(x)\to-\infty. \]

✏️ Solution

(iv) $g(x)=x^3+1$

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Step-by-step Solution · 14 steps

Given —
\[g(x)=x^3+1\]
Find the first derivative
Differentiating,
\[g'(x)=3x^2\]
Find the critical point
Set the first derivative equal to zero:
\[g'(x)=0\]

\[3x^2=0\]
Since $3\neq0$,
\[x^2=0\]

\[x=0\]
Thus, $x=0$ is a critical point.
Apply the second derivative test
\[g'(x)=3x^2\]
Differentiating again,
\[g''(x)=6x\]
At the critical point $x=0$,
\[g''(0)=6(0)\]

\[g''(0)=0\]
Here,
\[ g''(0)=0. \]
Therefore, the second derivative test is inconclusive. It is not correct to conclude directly that the function has neither a maximum nor a minimum merely because the second derivative is zero.
Use the first derivative to determine the nature of the critical point
We have
\[g'(x)=3x^2\]
For every real $x$,
\[x^2\geq0\]
Therefore,
\[3x^2\geq0\]
For every $x\neq0$,
\[3x^2>0\]
Thus,
\[g'(x)>0\]
on both sides of $x=0$, except at $x=0$ itself.
Hence, $g(x)$ is increasing before $x=0$ and continues to be increasing after $x=0$. The derivative does not change sign from positive to negative or from negative to positive at $x=0$.
Therefore, $x=0$ is a stationary point of inflection, not a maximum or minimum point.

💡 Answer

Final Answer

›

Function	Critical Point	Second Derivative Test	Result
$f(x)=(2x-1)^2+3$	$x=\frac12$	$f''(x)=8>0$	Minimum $=3$
$f(x)=9x^2+12x+3$	$x=-\frac23$	$f''(x)=18>0$	Minimum $=-1$
$f(x)=-(x-1)^2+10$	$x=1$	$f''(x)=-2<0$	Maximum $=10$
$g(x)=x^3+1$	$x=0$	$g''(0)=0$	No maximum or minimum

🎯 Exam Significance

Exam Significance

›

This question tests one of the fundamental applications of derivatives: using the first and second derivatives to identify maximum and minimum values. For board examinations, students should clearly show the sequence

\[ f'(x)=0\rightarrow\text{critical point}\rightarrow f''(x) \rightarrow\text{nature of extremum}\rightarrow\text{extreme value}. \]

A particularly important board-level concept appears in part (iv). When

\[ f''(a)=0, \]

students must not automatically state that there is no maximum or minimum. The second derivative test is simply inconclusive. Further analysis is required. This distinction is frequently important in conceptual and reasoning-based questions.

Significance for Competitive Entrance Examinations

For JEE Main, JEE Advanced and other entrance examinations, the problem develops the ability to distinguish between a critical point, a local extremum, an absolute extremum, and a stationary point of inflection. Competitive questions may deliberately provide a point where

\[ f'(x)=0 \]

and

\[ f''(x)=0 \]

to test whether the student incorrectly applies the second derivative test.

In particular, for

\[ g(x)=x^3+1, \]

the derivative

\[ g'(x)=3x^2\geq0 \]

shows that the function is increasing throughout its domain. This is a useful pattern to recognise quickly in objective questions: a zero derivative at an isolated point does not necessarily imply a maximum or minimum.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 10 points

For a differentiable function, possible extrema are generally investigated by solving $f'(x)=0$.
A positive second derivative at a critical point indicates a local minimum.
A negative second derivative at a critical point indicates a local maximum.
If $f''(a)=0$, the second derivative test is inconclusive.
The condition $f'(a)=0$ alone does not guarantee a maximum or minimum.
For $g(x)=x^3+1$, $g'(x)=3x^2\geq0$, so the function is increasing throughout its domain.
The critical point $(0,1)$ for $g(x)=x^3+1$ is a stationary point of inflection, not an extremum.
Always calculate the value of the original function at the critical point to obtain the corresponding maximum or minimum value.
For global maximum or minimum questions, also examine the behaviour of the function at the boundaries of its domain and as $x\to\pm\infty$, when applicable.
For quadratic functions, completing the square provides an independent and useful verification of the extremum.

📘 Concept & Theory

Theory and Concept

›

Maximum and minimum values are not always found simply by solving

\[ f'(x)=0. \]

We must first examine the nature and domain of the function.

This question contains absolute-value functions, a trigonometric function, and a function defined on an open interval. These cases illustrate several important ideas:

An absolute-value function may not be differentiable at the point where its inside expression is zero.
A function can have a maximum or minimum at a point where its derivative does not exist.
For trigonometric functions, the standard ranges of $\sin x$ and $\cos x$ can often determine extrema directly.
For an absolute value, the identity $|u|\geq0$ is often more useful than differentiation.
For a function defined on an open interval, endpoint values are not included unless the endpoints belong to the domain.

Useful Results

For every real number $u$,

\[ |u|\geq0. \]

Also,

\[ -1\leq\sin\theta\leq1. \]

and

\[ -1\leq\cos\theta\leq1. \]

For an absolute-value expression,

\[ |u|= \begin{cases} u,&u\geq0,\\ -u,&u<0. \end{cases} \]

🗺️ Solution Roadmap

Step-by-step Plan

›

Identify the domain of each function.
For absolute-value functions, identify the point where the expression inside the modulus becomes zero.
Check differentiability at that point and analyse the function on either side.
For trigonometric functions, use the known range of sine or cosine.
For a function on an open interval, determine whether the endpoint values are actually attained.
Clearly distinguish between a maximum/minimum value and a supremum/infimum that is approached but never attained.

✏️ Solution

(i) $f(x)=|x+2|-1$

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Step-by-step Solution · 16 steps

Identify the critical point caused by the modulus
The expression inside the modulus is
\[x+2\]
It becomes zero when
\[x+2=0\]

\[x=-2\]
Therefore, $x=-2$ is the point at which the absolute-value function changes its algebraic form.
Write the function piecewise
For $x < -2$, we have $x+2 < 0$. Therefore,
\[|x+2|=-(x+2)\]
Hence,
\[f(x)=-(x+2)-1\]

\[f(x)=-x-2-1\]

\[f(x)=-x-3\]
Differentiating,
\[f'(x)=-1\]
Thus,
\[f'(x) < 0\quad\text{for }x < -2\]
Therefore, $f(x)$ is decreasing on $(-\infty,-2)$
For $x>-2$, we have $x+2>0$. Therefore,
\[|x+2|=x+2\]
Hence,
\[f(x)=x+2-1\]

\[f(x)=x+1\]
Differentiating,
\[f'(x)=1\]
Thus,
\[f'(x)>0\quad\text{for }x>-2\]
Therefore, $f(x)$ is increasing on $(-2,\infty)$
Examine $x=-2$
At $x=-2$,
\[f(-2)=|-2+2|-1\]

\[=|0|-1\]

\[=-1\]
The function decreases up to $x=-2$ and increases after $x=-2$. Therefore, $x=-2$ gives a minimum.
Check whether a maximum exists
As $x\to\infty$,
\[f(x)=x+1\to\infty\]
Hence, the function is unbounded above and therefore has no maximum value.

✏️ Solution

(ii) $g(x)=-|x+1|+3$

›

Step-by-step Solution · 16 steps

Identify the breakpoint
The expression inside the modulus is
\[x+1\]
Setting it equal to zero,
\[x+1=0\]

\[x=-1\]
Thus, $x=-1$ is the point where the function changes its form.
Analyse the function for $x < -1$
When $x < -1$,
\[x+1 < 0\]
Therefore,
\[|x+1|=-(x+1)\]
Hence,
\[g(x)=-[-(x+1)]+3\]

\[=x+1+3\]

\[=x+4\]
Differentiating,
\[g'(x)=1\]
Therefore,
\[g'(x)>0\quad\text{for }x < -1.\]
Hence, $g(x)$ is increasing on $(-\infty,-1)$.
Analyse the function for $x>-1$
When $x>-1$,
\[x+1>0\]
Therefore,
\[|x+1|=x+1\]
Hence,
\[g(x)=-(x+1)+3\]

\[=-x-1+3\]

\[=-x+2\]
Differentiating,
\[g'(x)=-1\]
Therefore,
\[g'(x) < 0\quad\text{for }x>-1\]
Hence, $g(x)$ is decreasing on $(-1,\infty)$.
Evaluate the function at $x=-1$
\[g(-1)=-|-1+1|+3\]

\[=-|0|+3\]

\[=3\]
The function increases up to $x=-1$ and decreases after $x=-1$. Therefore, $x=-1$ gives a maximum.
Check whether a minimum exists
As $x\to\infty$,
\[g(x)=-x+2\to-\infty\]
As $x\to-\infty$,
\[g(x)=x+4\to-\infty\]
Hence, the function is unbounded below and has no minimum value.

✏️ Solution

(iii) $h(x)=\sin(2x)+5$

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Step-by-step Solution · 23 steps

Use the range of the sine function
For every real $x$,
\[-1\leq\sin(2x)\leq1\]
Adding $5$ throughout the inequality gives
\[-1+5\leq\sin(2x)+5\leq1+5\]
Therefore,
\[4\leq h(x)\leq6\]
Find when the minimum is attained
The minimum possible value occurs when
\[\sin(2x)=-1\]
We know
\[\sin\theta=-1\]
when
\[\theta=\frac{3\pi}{2}+2n\pi, \quad n\in\mathbb Z\]
Therefore,
\[2x=\frac{3\pi}{2}+2n\pi\]
Dividing by $2$,
\[ x=\frac{3\pi}{4}+n\pi, \quad n\in\mathbb Z \]
At these values,
\[h(x)=4\]
Find when the maximum is attained
The maximum possible value occurs when
\[\sin(2x)=1\]
We know
\[\sin\theta=1\]
when
\[ \theta=\frac{\pi}{2}+2n\pi, \qquad n\in\mathbb Z \]
Therefore,
\[2x=\frac{\pi}{2}+2n\pi\]
Dividing by $2$
\[ x=\frac{\pi}{4}+n\pi, \qquad n\in\mathbb Z. \]
At these values,
\[h(x)=6\]
Verification using derivatives
Although the range method is the quickest approach, the result can also be verified using derivatives.
\[h(x)=\sin(2x)+5\]

\[h'(x)=2\cos(2x)\]
For stationary points,
\[h'(x)=0\]
Therefore,
\[2\cos(2x)=0\]

\[\cos(2x)=0\]
Hence,
\[ 2x=\frac{\pi}{2}+n\pi, \quad n\in\mathbb Z \]
Thus,
\[x=\frac{\pi}{4}+\frac{n\pi}{2}\]
The second derivative is
\[h''(x)=-4\sin(2x)\]
When $\sin(2x)=1$,
\[h''(x)=-4<0\]
so the function has a maximum.
When $\sin(2x)=-1$,
\[h''(x)=4>0,\]
so the function has a minimum.

✏️ Solution

(iv) $f(x)=|\sin(4x)+3|$

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Step-by-step Solution · 16 steps

Determine the range of the expression inside the modulus
We know that
\[-1\leq\sin(4x)\leq1\]
Adding $3$ throughout,
\[2\leq\sin(4x)+3\leq4\]
Therefore,
\[\sin(4x)+3>0\]
for every real $x$.
Remove the modulus correctly
Since $\sin(4x)+3$ is always positive,
\[|\sin(4x)+3|=\sin(4x)+3\]
Hence,
\[f(x)=\sin(4x)+3\]
Determine the range
Since
\[-1\leq\sin(4x)\leq1,\]
adding $3$ gives
\[2\leq\sin(4x)+3\leq4\]
Therefore,
\[2\leq f(x)\leq4\]
Find the maximum value
The maximum occurs when
\[\sin(4x)=1\]
Therefore,
\[ 4x=\frac{\pi}{2}+2n\pi, \quad n\in\mathbb Z \]
Dividing by $4$,
\[ x=\frac{\pi}{8}+\frac{n\pi}{2}, \quad n\in\mathbb Z \]
At these values,
\[f(x)=|1+3|\]

\[=4\]
Find the minimum value
The minimum occurs when
\[\sin(4x)=-1\]
Therefore,
\[ 4x=\frac{3\pi}{2}+2n\pi, \quad n\in\mathbb Z \]
Dividing by $4$,
\[ x=\frac{3\pi}{8}+\frac{n\pi}{2}, \quad n\in\mathbb Z \]
At these values,
\[f(x)=|-1+3|\]

\[=|2|\]

\[=2\]

✏️ Solution

(v) $h(x)=x+1,\quad x\in(-1,1)$

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Step-by-step Solution · 11 steps

Understand the domain
The function is
\[h(x)=x+1\]
with domain
\[-1 < x < 1\]
The interval is open. Therefore, $x=-1$ and $x=1$ are not included in the domain.
Examine the derivative
\[h'(x)=1\]
for every $x\in(-1,1)$, the function is strictly increasing throughout its domain.
Examine the values approached near the endpoints
As $x\to-1^+$,
\[h(x)=x+1\to0\]
As $x\to1^-$,
\[h(x)=x+1\to2\]
Thus, the range of $h$ on the given domain is
\[(0,2)\]
Determine whether the minimum and maximum are attained
The value $0$ would occur only if
\[ x+1=0, \]
giving
\[ x=-1. \]
But $x=-1\notin(-1,1)$. Therefore, $0$ is not attained.
Similarly, the value $2$ would occur only if
\[ x+1=2, \]
giving
\[ x=1. \]
But $x=1\notin(-1,1)$. Therefore, $2$ is not attained.
Consequently, the function has neither a maximum nor a minimum value on the given open interval.
However, its infimum and supremum are
\[ \inf h(x)=0 \]
and
\[ \sup h(x)=2. \]
These are not minimum and maximum values because they are not attained.

💡 Answer

Final Answer

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Part	Function	Maximum	Minimum
(i)	$\|x+2\|-1$	Does not exist	$-1$ at $x=-2$
(ii)	$-\|x+1\|+3$	$3$ at $x=-1$	Does not exist
(iii)	$\sin(2x)+5$	$6$	$4$
(iv)	$\|\sin(4x)+3\|$	$4$	$2$
(v)	$x+1,\;x\in(-1,1)$	Does not exist	Does not exist

🎯 Exam Significance

Exam Significance

›

This question is important because it tests whether students understand that maximum and minimum values depend on both the function and its domain. In particular, parts (i) and (ii) demonstrate that an absolute-value function can attain an extremum at a point where the derivative does not exist.

Part (iii) tests the direct use of the standard trigonometric range

\[ -1\leq\sin x\leq1, \]

while part (iv) tests whether students correctly analyse the expression inside an absolute-value sign before applying the modulus.

Part (v) is especially important for board examinations because it tests the difference between an endpoint being approached and an endpoint actually belonging to the domain. The interval

\[ (-1,1) \]

excludes both endpoints, so the limiting values $0$ and $2$ are not attained.

Significance for Competitive Entrance Examinations

For JEE Main, JEE Advanced and other competitive examinations, this question develops several high-value concepts: piecewise functions, non-differentiability, range analysis, periodicity, monotonicity, and open-domain extrema.

A common competitive-exam trap is to assume that every point satisfying

\[ f'(x)=0 \]

must correspond to a maximum or minimum. Absolute-value functions show why this is insufficient. Another common trap is to treat the supremum or infimum as a maximum or minimum even when the corresponding value is not attained.

For trigonometric functions, recognising the range immediately is often faster than differentiation. For example,

\[ -1\leq\sin(4x)\leq1 \]

instantly gives the range of

\[ |\sin(4x)+3|. \]

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 11 points

For $|x+2|$, the important breakpoint is $x=-2$, not $x=0$.
An absolute-value function may have an extremum at a point where its derivative does not exist.
For $|x+2|-1$, the minimum is $-1$ at $x=-2$, while no maximum exists.
For $-|x+1|+3$, the maximum is $3$ at $x=-1$, while no minimum exists.
The range $-1\leq\sin\theta\leq1$ can often determine maximum and minimum values directly.
$\sin(2x)+5$ has maximum $6$ and minimum $4$.
Since $-1\leq\sin(4x)\leq1$, we have $2\leq\sin(4x)+3\leq4$. Therefore, the modulus in part (iv) does not change the expression's sign.
$|\sin(4x)+3|$ has minimum $2$ and maximum $4$.
A function defined on an open interval need not attain its limiting endpoint values.
For $h(x)=x+1$ on $(-1,1)$, the range is $(0,2)$, so there is neither a maximum nor a minimum.
Do not confuse $\sup f(x)$ with a maximum or $\inf f(x)$ with a minimum. A maximum or minimum must actually be attained by the function.

Q3

LONG ANSWER4 marks

Find the local maxima and local minima, if any, of the following functions. Also find the corresponding local maximum and local minimum values.

$\quad f(x)=x^2$
$\quad g(x)=x^3-3x$
$\quad h(x)=\sin x+\cos x,\qquad 0 < x < \frac{\pi}{2}$
$\quad f(x)=\sin x-\cos x,\qquad 0 < x < 2\pi$
$\quad f(x)=x^3-6x^2+9x+15$
$\quad g(x)=\frac{x}{2}+\frac{2}{x},\qquad x>0$
$\quad g(x)=\frac{1}{x^2+2}$
$\quad f(x)=x\sqrt{1-x},\qquad 0 < x < 1$

📘 Concept & Theory

Theory and Concept

›

A function $f(x)$ has a local maximum at $x=a$ if its value at $a$ is greater than or equal to the values of the function at all sufficiently nearby points. Similarly, $f(x)$ has a local minimum at $x=a$ if its value at $a$ is less than or equal to the values at all sufficiently nearby points.

For a differentiable function, a local maximum or minimum can occur at a point where

\[ f'(a)=0. \]

Such a point is called a stationary point.

The second derivative test provides a convenient way of determining the nature of a stationary point:

If $f'(a)=0$ and $f''(a)>0$, then $f$ has a local minimum at $x=a$.
If $f'(a)=0$ and $f''(a)<0$, then $f$ has a local maximum at $x=a$.
If $f'(a)=0$ and $f''(a)=0$, the second derivative test is inconclusive.

It is important to remember that $f'(a)=0$ alone does not establish a maximum or minimum. The nature of the stationary point must be determined.

🗺️ Solution Roadmap

Step-by-step Plan

›

Find the first derivative $f'(x)$.
Set $f'(x)=0$ to obtain the stationary points.
Check that the stationary points belong to the domain.
Find the second derivative $f''(x)$.
Use the sign of $f''(a)$ to determine whether the stationary point is a local maximum or local minimum.
Substitute the corresponding $x$-value into the original function to obtain the local extremum value.
For functions restricted to an interval, pay particular attention to the stated domain.

✏️ Solution

(i) $f(x)=x^2$

›

Step-by-step Solution · 8 steps

Find the first derivative
\[f(x)=x^2\]

\[f'(x)=2x\]
Find the stationary point
\[f'(x)=0\]
Therefore,
\[2x=0\]

\[x=0\]
Thus, $x=0$ is the stationary point.
Find the second derivative
\[f'(x)=2x\]

\[f''(x)=2\]
Hence,
\[f''(0)=2>0\]
Since the second derivative is positive, the function has a local minimum at $x=0$.
Find the local minimum value
\[f(0)=0^2\]

\[f(0)=0\]
Therefore,
\[\boxed{\text{Local minimum value}=0\text{ at }x=0}\]
There is no local maximum.

✏️ Solution

(ii) $g(x)=x^3-3x$

›

Step-by-step Solution · 12 steps

Find the first derivative
\[g(x)=x^3-3x\]

\[g'(x)=3x^2-3\]
Find the stationary points
\[g'(x)=0\]

\[3x^2-3=0\]
Dividing by $3$,
\[x^2-1=0\]

\[(x-1)(x+1)=0\]
Therefore,
\[x=1\]
or
\[x=-1\]
Find the second derivative
\[g'(x)=3x^2-3\]

\[g''(x)=6x\]
At $x=1$,
\[g''(1)=6(1)=6>0\]
Therefore, $x=1$ gives a local minimum.
Find the local minimum value
\[g(1)=1^3-3(1)\]

\[=1-3\]

\[=-2\]
Hence,
\[\boxed{\text{Local minimum value}=-2\text{ at }x=1}\]
Classify the stationary point $x=-1$
\[g''(-1)=6(-1)\]

\[=-6<0\]
Therefore, $x=-1$ gives a local maximum.
Find the local maximum value
\[g(-1)=(-1)^3-3(-1)\]

\[=-1+3\]

\[=2\]
Hence,
\[\boxed{\text{Local maximum value}=2\text{ at }x=-1}\]

✏️ Solution

(iii) $h(x)=\sin x+\cos x,\quad 0 < x < \frac{\pi}{2}$

›

Step-by-step Solution · 13 steps

Find the first derivative
\[h(x)=\sin x+\cos x\]

\[h'(x)=\cos x-\sin x\]
Find the stationary point
\[h'(x)=0\]
Therefore,
\[\cos x-\sin x=0\]

\[\cos x=\sin x\]
Dividing by $\cos x$, which is positive on
\[0 < x < \frac{\pi}{2},\]
we get
\[\tan x=1\]
Hence,
\[x=\frac{\pi}{4}\]
This point lies within the given domain.
Find the second derivative
\[h'(x)=\cos x-\sin x\]

\[h''(x)=-\sin x-\cos x\]
At
\[x=\frac{\pi}{4},\]
we have
\[\sin\frac{\pi}{4}=\frac{1}{\sqrt2}\]
and
\[\cos\frac{\pi}{4}=\frac{1}{\sqrt2}\]
Therefore,
\[h''\left(\frac{\pi}{4}\right)=-\frac{1}{\sqrt2}-\frac{1}{\sqrt2}\]

\[=-\frac{2}{\sqrt2}\]

\[=-\sqrt2 < 0\]
Hence, $x=\frac{\pi}{4}$ gives a local maximum.
Find the local maximum value
\[h\left(\frac{\pi}{4}\right)=\sin\frac{\pi}{4}+\cos\frac{\pi}{4}\]

\[=\frac{1}{\sqrt2}+\frac{1}{\sqrt2}\]

\[=\frac{2}{\sqrt2}\]

\[=\sqrt2\]
Therefore,
\[\boxed{\text{Local maximum value}=\sqrt2\text{ at }x=\frac{\pi}{4}}\]
There is no local minimum in the given open interval. In fact, the function is increasing on
\[ \left(0,\frac{\pi}{4}\right) \]
and decreasing on
\[ \left(\frac{\pi}{4},\frac{\pi}{2}\right). \]

✏️ Solution

(iv) $f(x)=\sin x-\cos x,\quad 0 < x < 2\pi$

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Step-by-step Solution · 17 steps

Find the first derivative
\[f(x)=\sin x-\cos x\]

\[f'(x)=\cos x+\sin x\]
Find all stationary points
\[f'(x)=0\]
Therefore,
\[\cos x+\sin x=0\]

\[\sin x=-\cos x\]
Hence,
\[\tan x=-1\]
In the interval $0 < x < 2 \pi\;$ the solutions are
\[x=\frac{3\pi}{4}\]
and
\[x=\frac{7\pi}{4}\]
Find the second derivative
\[f'(x)=\cos x+\sin x\]

\[f''(x)=-\sin x+\cos x\]
Classify $x=\frac{3\pi}{4}$
\[f''\left(\frac{3\pi}{4}\right)=-\sin\frac{3\pi}{4}+\cos\frac{3\pi}{4}\]
Since
\[\sin\frac{3\pi}{4}=\frac{1}{\sqrt2}\]
and
\[\cos\frac{3\pi}{4}=-\frac{1}{\sqrt2},\]
we get
\[f''\left(\frac{3\pi}{4}\right)=-\frac{1}{\sqrt2}-\frac{1}{\sqrt2}\]

\[=-\sqrt2 < 0\]
Therefore,
\[x=\frac{3\pi}{4}\]
gives a local maximum.
Find the local maximum value
\[f\left(\frac{3\pi}{4}\right)=\sin\frac{3\pi}{4}-\cos\frac{3\pi}{4}\]

\[=\frac{1}{\sqrt2}-\left(-\frac{1}{\sqrt2}\right)\]

\[=\frac{2}{\sqrt2}\]

\[=\sqrt2\]
Hence,
\[\boxed{\text{Local maximum value}=\sqrt2\text{ at }x=\frac{3\pi}{4}}\]
Classify $x=\frac{7\pi}{4}$
\[f''\left(\frac{7\pi}{4}\right)=-\sin\frac{7\pi}{4}+\cos\frac{7\pi}{4}\]
Since
\[\sin\frac{7\pi}{4}=-\frac{1}{\sqrt2}\]
and
\[\cos\frac{7\pi}{4}=\frac{1}{\sqrt2},\]
we get
\[f''\left(\frac{7\pi}{4}\right)=-\left(-\frac{1}{\sqrt2}\right)+\frac{1}{\sqrt2}\]

\[=\frac{2}{\sqrt2}\]

\[=\sqrt2>0\]
Therefore,
\[x=\frac{7\pi}{4}\]
gives a local minimum.
Find the local minimum value
\[f\left(\frac{7\pi}{4}\right)=\sin\frac{7\pi}{4}-\cos\frac{7\pi}{4}\]

\[=-\frac{1}{\sqrt2}-\frac{1}{\sqrt2}\]

\[=-\frac{2}{\sqrt2}\]

\[=-\sqrt2\]
Therefore,
\[\boxed{\text{Local minimum value}=-\sqrt2\text{ at }x=\frac{7\pi}{4}}\]

✏️ Solution

(v) $f(x)=x^3-6x^2+9x+15$

›

Step-by-step Solution · 14 steps

Find the first derivative
\[f'(x)=3x^2-12x+9\]
Find the stationary points
Set
\[f'(x)=0\]
Therefore,
\[3x^2-12x+9=0\]
Dividing by $3$,
\[x^2-4x+3=0\]
Factorising,
\[x^2-3x-x+3=0\]

\[x(x-3)-1(x-3)=0\]

\[(x-1)(x-3)=0\]
Hence,
\[x=1\]
or
\[x=3\]
These are stationary points. They should not be called points of inflection until their nature has been established.
Find the second derivative
\[f'(x)=3x^2-12x+9\]

\[f''(x)=6x-12\]
Classify $x=1$
\[f''(1)=6(1)-12\]

\[=6-12\]

\[=-6 < 0\]
Therefore, $x=1$ gives a local maximum.
Find the local maximum value
\[f(1)=1^3-6(1)^2+9(1)+15\]

\[=1-6+9+15\]

\[=19\]
Hence,
\[\boxed{\text{Local maximum value}=19\text{ at }x=1}\]
Classify $x=3$
\[f''(3)=6(3)-12\]

\[=18-12\]

\[=6>0\]
Therefore, $x=3$ gives a local minimum.
Find the local minimum value
\[f(3)=3^3-6(3)^2+9(3)+15\]

\[=27-6(9)+27+15\]

\[=27-54+27+15\]

\[=15\]
Therefore,
\[\boxed{\text{Local minimum value}=15\text{ at }x=3}\]

✏️ Solution

(vi) $g(x)=\frac{x}{2}+\frac{2}{x},\quad x>0$

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Step-by-step Solution · 16 steps

Find the first derivative
\[g(x)=\frac{x}{2}+\frac{2}{x}\]
Writing the second term as a power,
\[g(x)=\frac{x}{2}+2x^{-1}\]
Therefore,
\[g'(x)=\frac12-2x^{-2}\]

\[g'(x)=\frac12-\frac{2}{x^2}\]
Find the stationary point
\[g'(x)=0\]

\[\frac12-\frac{2}{x^2}=0\]
Therefore,
\[\frac12=\frac{2}{x^2}\]
Multiplying by $2x^2$,
\[x^2=4\]
Hence,
\[x=\pm2\]
But the given domain is
\[x>0\]
Therefore,
\[x=2\]
is the only stationary point in the domain.
Find the second derivative
\[g'(x)=\frac12-2x^{-2}\]
Differentiating,
\[g''(x)=4x^{-3}\]

\[g''(x)=\frac{4}{x^3}\]
At $x=2$,
\[g''(2)=\frac{4}{2^3}\]

\[=\frac48\]

\[=\frac12>0\]
Therefore, $x=2$ gives a local minimum.
Find the local minimum value
\[g(2)=\frac{2}{2}+\frac{2}{2}\]

\[=1+1\]

\[=2\]
Hence,
\[\boxed{\text{Local minimum value}=2\text{ at }x=2}\]
There is no local maximum for $x>0$. In fact,
\[g'(x)<0\]
for $0 < x < 2$, while
\[g'(x)>0\]
for $x>2$, so the function decreases and then increases.

✏️ Solution

(vii) $g(x)=\frac{1}{x^2+2}$

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Step-by-step Solution · 16 steps

Find the first derivative
\[g(x)=(x^2+2)^{-1}\]
Using the chain rule,
\[g'(x)=-(x^2+2)^{-2}(2x)\]

\[g'(x)=-\frac{2x}{(x^2+2)^2}\]
Find the stationary point
\[g'(x)=0\]

\[-\frac{2x}{(x^2+2)^2}=0\]
Since
\[x^2+2>0\]
for every real $x$, the denominator is never zero. Therefore, the numerator must be zero:
\[-2x=0\]
Hence,
\[x=0\]
Find the second derivative
Starting from
\[g'(x)=-2x(x^2+2)^{-2}\]
apply the product rule:
\[g''(x)=-2(x^2+2)^{-2}+(-2x)\left[-2(x^2+2)^{-3}(2x)\right]\]
Therefore,
\[g''(x)=-\frac{2}{(x^2+2)^2}+\frac{8x^2}{(x^2+2)^3}\]
Taking $(x^2+2)^3$ as the common denominator,
\[g''(x)=\frac{-2(x^2+2)+8x^2}{(x^2+2)^3}\]
Simplifying the numerator,
\[-2x^2-4+8x^2=6x^2-4\]
Hence,
\[g''(x)=\frac{6x^2-4}{(x^2+2)^3}\]
At $x=0$,
\[g''(0)=\frac{6(0)^2-4}{(0^2+2)^3}\]

\[=\frac{-4}{8}\]

\[=-\frac12 < 0\]
Therefore, $x=0$ gives a local maximum.
Find the local maximum value
\[g(0)=\frac{1}{0^2+2}\]

\[=\frac12\]
Hence,
\[\boxed{\text{Local maximum value}=\frac12\text{ at }x=0}\]
There is no local minimum. In fact, $g(x)>0$ for every real $x$, and
\[ g(x)\to0 \]
as $|x|\to\infty$, but $g(x)$ never becomes equal to $0$

✏️ Solution

(viii) $f(x)=x\sqrt{1-x},\quad 0 < x < 1$

›

Step-by-step Solution · 29 steps

Rewrite the function
\[f(x)=x(1-x)^{1/2}\]
Find the first derivative
Using the product rule,
\[f'(x)=x'(1-x)^{1/2}+x\frac{d}{dx}(1-x)^{1/2}\]
Since
\[x'=1\]
and
\[\frac{d}{dx}(1-x)^{1/2}=\frac12(1-x)^{-1/2}(-1)\]
we obtain
\[f'(x)=\sqrt{1-x}-\frac{x}{2\sqrt{1-x}}\]
Taking the common denominator $2\sqrt{1-x}$,
\[f'(x)=\frac{2(1-x)-x}{2\sqrt{1-x}}\]
Therefore,
\[f'(x)=\frac{2-3x}{2\sqrt{1-x}}\]
Find the stationary point
For $0 < x < 1$,
\[\sqrt{1-x}>0\]
Therefore, $f'(x)=0$ requires
\[2-3x=0\]
Hence,
\[3x=2\]

\[x=\frac23\]
Since
\[0 < \frac23 < 1,\]
the stationary point lies within the given domain.
Find the second derivative
From
\[f'(x)=\frac{2-3x}{2\sqrt{1-x}}\]
write
\[f'(x)=\frac12(2-3x)(1-x)^{-1/2}\]
Differentiate using the product rule:
\[f''(x)=\frac12\left[(-3)(1-x)^{-1/2}+(2-3x)\frac{d}{dx}(1-x)^{-1/2}\right]\]
Now,
\[\frac{d}{dx}(1-x)^{-1/2}=-\frac12(1-x)^{-3/2}(-1)\]

\[=\frac12(1-x)^{-3/2}\]
Therefore,
\[f''(x)=\frac12\left[-3(1-x)^{-1/2}+\frac12(2-3x)(1-x)^{-3/2}\right]\]
Taking $(1-x)^{-3/2}$ common,
\[f''(x)=\frac12(1-x)^{-3/2}\left[-3(1-x)+\frac12(2-3x)\right]\]
Simplifying the expression in brackets,
\[-3(1-x)+\frac12(2-3x)=-3+3x+1-\frac32x\]

\[=-2+\frac32x\]

\[=\frac{3x-4}{2}\]
Therefore,
\[f''(x)=\frac{3x-4}{4(1-x)^{3/2}}\]
Evaluate the second derivative at $x=\frac23$
\[f''\left(\frac23\right)=\frac{3\left(\frac23\right)-4}{4\left(1-\frac23\right)^{3/2}}\]
Simplifying the numerator,
\[3\left(\frac23\right)-4=2-4=-2\]
Also,
\[1-\frac23=\frac13\]
Therefore,
\[f''\left(\frac23\right)=\frac{-2}{4\left(\frac13\right)^{3/2}}\]
Since the denominator is positive,
\[f''\left(\frac23\right) < 0\]
More explicitly,
\[f''\left(\frac23\right)=-\frac{3\sqrt3}{2} < 0\]
Hence, $x=\frac23$ gives a local maximum, not a local minimum.
Find the local maximum value
\[f\left(\frac23\right)=\frac23\sqrt{1-\frac23}\]
Therefore,
\[f\left(\frac23\right)=\frac23\sqrt{\frac13}\]

\[=\frac{2}{3\sqrt3}\]
Rationalising the denominator,
\[\frac{2}{3\sqrt3}\times\frac{\sqrt3}{\sqrt3}=\frac{2\sqrt3}{9}\]
Hence,
\[\boxed{\text{Local maximum value}=\frac{2}{3\sqrt3}=\frac{2\sqrt3}{9}\text{ at }x=\frac23}\]
There is no local minimum in the open interval $0 < x < 1$. Notice that
\[f(x)\to0\]
as $x\to0^+$ and also as $x\to1^-$, but neither endpoint belongs to the given domain.

💡 Answer

Final Answer

›

Part	Function	Local Maximum	Local Minimum
(i)	$x^2$	None	$0$ at $x=0$
(ii)	$x^3-3x$	$2$ at $x=-1$	$-2$ at $x=1$
(iii)	\(\sin x+\cos x,\;0	$\sqrt2$ at $x=\frac{\pi}{4}$	None
(iv)	\(\sin x-\cos x,\;0	$\sqrt2$ at $x=\frac{3\pi}{4}$	$-\sqrt2$ at $x=\frac{7\pi}{4}$
(v)	$x^3-6x^2+9x+15$	$19$ at $x=1$	$15$ at $x=3$
(vi)	$\frac{x}{2}+\frac{2}{x},\;x>0$	None	$2$ at $x=2$
(vii)	$\frac{1}{x^2+2}$	$\frac12$ at $x=0$	None
(viii)	$x\sqrt{1-x},\;0 < x < 1$	$\frac{2}{3\sqrt3}=\frac{2\sqrt3}{9}$ at $x=\frac23$	None

🎯 Exam Significance

Exam Significance

›

This question is important for board examinations because it brings together the central procedure for identifying local extrema:

\[ f'(x)=0 \]

followed by the second derivative test.

Students should clearly distinguish between a stationary point and a local extremum. A solution should not stop after obtaining

\[ f'(x)=0. \]

The nature of every stationary point must be established.

Trigonometric functions in parts (iii) and (iv) also require careful consideration of the stated domain. In part (iv), the equation

\[ \tan x=-1 \]

has two solutions in

\[ 0 < x < 2\pi, \]

not just one. Missing $x=\frac{7\pi}{4}$ would result in an incomplete answer.

Part (viii) is particularly useful for board preparation because it requires the product rule, chain rule, careful simplification of the first derivative, and the second derivative test.

Significance for Competitive Entrance Examinations

For JEE Main, JEE Advanced and other competitive entrance examinations, this problem covers several recurring concepts: stationary points, second derivative classification, domain restrictions, trigonometric equations, rational functions and radical functions.

Competitive examinations frequently include distractors based on incomplete solutions. For example, obtaining

\[ x=\pm2 \]

in part (vi) is not sufficient because the domain is

\[ x>0. \]

Therefore, only $x=2$ is admissible.

Similarly, in part (vii), the denominator

\[ x^2+2 \]

is always positive, which makes it straightforward to identify the only stationary point. Such observations can substantially reduce calculation time in objective questions.

Part (viii) is a useful competitive-exam example because the stationary point

\[ x=\frac23 \]

is a local maximum. A common error is to incorrectly infer a minimum from the positivity of the function itself. The sign of the derivative or second derivative, rather than the fact that $f(x)>0$, determines the nature of the extremum.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 12 points

A stationary point is obtained by solving $f'(x)=0$, but every stationary point need not be a maximum or minimum.
If $f''(a)>0$, the function has a local minimum at $x=a$.
If $f''(a)<0$, the function has a local maximum at $x=a$.
If $f''(a)=0$, the second derivative test is inconclusive.
In part (ii), $x=-1$ gives a local maximum of $2$, while $x=1$ gives a local minimum of $-2$.
In trigonometric problems, all solutions within the specified domain must be considered.
For $\sin x-\cos x$ on $(0,2\pi)$, there are two stationary points: $\frac{3\pi}{4}$ and $\frac{7\pi}{4}$.
Domain restrictions can eliminate algebraically obtained stationary points, as seen in part (vi).
The function $\frac{1}{x^2+2}$ has a local maximum of $\frac12$ at $x=0$.
For $x\sqrt{1-x}$ on $(0,1)$, the stationary point $x=\frac23$ is a local maximum, not a local minimum.
For an open interval, an endpoint value may be approached without being attained; therefore, it cannot automatically be called a local or absolute extremum.
When possible, the first derivative sign test provides an excellent independent verification of the second derivative test.

Q4

LONG ANSWER4 marks

Prove that the following functions do not have maxima or minima:

\[ f(x)=e^x \]
\[ g(x)=\log x \]
\[ h(x)=x^3+x^2+x+1 \]

📘 Concept & Theory

Theory and Concept

›

A function may fail to have a maximum or minimum because it is strictly increasing or strictly decreasing throughout its domain. If the derivative maintains one sign over the entire domain, the function is monotonic and therefore cannot have an interior local maximum or minimum.

For a differentiable function:

If $f'(x)>0$ throughout an interval, then $f(x)$ is strictly increasing on that interval.
If $f'(x)<0$ throughout an interval, then $f(x)$ is strictly decreasing on that interval.
A strictly increasing or strictly decreasing function on an open interval cannot have a local maximum or local minimum at an interior point.

In this question, all three functions can be shown to be strictly increasing on their respective domains. We will prove this carefully using derivatives.

🗺️ Solution Roadmap

Step-by-step Plan

›

Determine the domain of the function.
Find the first derivative.
Determine the sign of the derivative throughout the domain.
Use the derivative sign to establish monotonicity.
Conclude that the function cannot have a local maximum or local minimum.
Where useful, examine the limiting behaviour to establish that the function also has no absolute maximum or minimum.

✏️ Solution

(i) $\displaystyle f(x)=e^x$

›

Step-by-step Solution · 12 steps

Determine the domain
The exponential function $e^x$ is defined for every real number $x$. Therefore,
\[ \text{Domain of }f=(-\infty,\infty). \]
Find the first derivative
\[f(x)=e^x\]
Differentiating,
\[f'(x)=e^x\]
Since
\[e^x>0\]
for every real $x$, we have
\[f'(x)>0\quad\forall x\in\mathbb R.\]
Establish monotonicity
Since
\[ f'(x)>0 \]
throughout its domain, $f(x)=e^x$ is strictly increasing on
\[ (-\infty,\infty). \]
Therefore, as $x$ increases, $e^x$ continuously increases. There is no point at which the function changes from increasing to decreasing, so there is no local maximum or local minimum.
Prove that there is no absolute maximum
Consider the behaviour as $x\to\infty$:
\[\lim_{x\to\infty}e^x=\infty\]
Thus, $e^x$ is unbounded above. Hence, no real number can be the maximum value of $e^x$.
Prove that there is no absolute minimum
As $x\to-\infty$,
\[\lim_{x\to-\infty}e^x=0\]
However,
\[e^x>0\]
for every finite real $x$. Therefore, $e^x$ never actually equals $0$.
Hence $0$ is the infimum of the range but is not attained. Therefore, there is no minimum value.
Conclusion
\[\boxed{f(x)=e^x\text{ has neither a maximum nor a minimum.}}\]

✏️ Solution

(ii) $\displaystyle g(x)=\log x$

›

Step-by-step Solution · 10 steps

Determine the domain
The logarithm $\log x$ is defined for positive real numbers only. Therefore,
\[x>0\]
Hence,
\[\text{Domain of }g=(0,\infty)\]
Find the first derivative
For the natural logarithm,
\[g(x)=\log x\]
gives
\[g'(x)=\frac1x\]
Since the domain requires $x>0$,
\[\frac1x>0\]
Therefore,
\[g'(x)>0\qquad\forall x>0\]
Establish monotonicity
Since
\[g'(x)>0\]
throughout its domain, $g(x)=\log x$ is strictly increasing on
\[(0,\infty)\]
Therefore, there is no point in its domain at which the function changes from increasing to decreasing or from decreasing to increasing. Hence, it has no local maximum and no local minimum.
Prove that there is no absolute maximum
As $x\to\infty$,
\[\lim_{x\to\infty}\log x=\infty\]
Therefore, $\log x$ is unbounded above and cannot have a maximum value.
Prove that there is no absolute minimum
As $x\to0^+$,
\[\lim_{x\to0^+}\log x=-\infty\]
Thus, $\log x$ is unbounded below and cannot have a minimum value.
Conclusion
\[\boxed{g(x)=\log x\text{ has neither a maximum nor a minimum.}}\]

✏️ Solution

(iii) $\displaystyle h(x)=x^3+x^2+x+1$

›

Step-by-step Solution · 15 steps

Determine the domain
Since $h(x)$ is a polynomial, it is defined for every real number. Therefore,
\[\text{Domain of }h=(-\infty,\infty).\]
Find the first derivative
\[h(x)=x^3+x^2+x+1\]
Differentiating term by term,
\[h'(x)=3x^2+2x+1\]
Determine the sign of $h'(x)$
We need to show that
\[3x^2+2x+1>0\]
for every real $x$.
Complete the square:
\[3x^2+2x+1=3\left(x^2+\frac23x\right)+1\]
Now,
\[x^2+\frac23x=\left(x+\frac13\right)^2-\frac19\]
Therefore,
\[3x^2+2x+1=3\left[\left(x+\frac13\right)^2-\frac19\right]+1.\]

\[=3\left(x+\frac13\right)^2-\frac13+1.\]

\[=3\left(x+\frac13\right)^2+\frac23.\]
Since
\[\left(x+\frac13\right)^2\geq0,\]
we have
\[3\left(x+\frac13\right)^2\geq0\]
Hence,
\[3\left(x+\frac13\right)^2+\frac23\geq\frac23>0\]
Therefore,
\[\boxed{h'(x)>0\quad\forall x\in\mathbb R}\]
Establish monotonicity
Since
\[h'(x)>0\]
for every real $x$, the function
\[h(x)=x^3+x^2+x+1\]
is strictly increasing on the entire real line.
Hence, there is no point at which the function changes from increasing to decreasing or from decreasing to increasing. Therefore, the function has neither a local maximum nor a local minimum.
Prove that there is no absolute maximum
Consider the behaviour as $x\to\infty$:
\[\lim_{x\to\infty}\left(x^3+x^2+x+1\right)=\infty.\]
Therefore, $h(x)$ is unbounded above and has no maximum value.
Prove that there is no absolute minimum
Consider the behaviour as $x\to-\infty$:
\[\lim_{x\to-\infty}\left(x^3+x^2+x+1\right)=-\infty.\]
Therefore, $h(x)$ is unbounded below and has no minimum value.
Conclusion
\[\boxed{h(x)=x^3+x^2+x+1\text{ has neither a maximum nor a minimum.}}\]

💡 Answer

Final Answer

›

Function	Domain	Derivative	Nature	Maximum	Minimum
$\displaystyle e^x$	$\mathbb R$	$\displaystyle e^x>0$	Strictly increasing	None	None
$\displaystyle \log x$	$(0,\infty)$	$\displaystyle \frac1x>0$	Strictly increasing	None	None
$\displaystyle x^3+x^2+x+1$	$\mathbb R$	$\displaystyle 3x^2+2x+1>0$	Strictly increasing	None	None

🎯 Exam Significance

Exam Significance

›

This question is important because it demonstrates that proving the absence of maxima and minima is often easier than explicitly solving

\[ f'(x)=0. \]

The key is to establish that the derivative has a constant sign throughout the domain.

For $e^x$,

\[ f'(x)=e^x>0, \]

and for $\log x$,

\[ g'(x)=\frac1x>0\quad(x>0). \]

Thus, both functions are strictly increasing.

The third function is particularly useful because its derivative is a quadratic:

\[ h'(x)=3x^2+2x+1. \]

Showing that this quadratic is always positive proves that the original polynomial is strictly increasing. This is a standard board-level application of derivatives.

Significance for Competitive Entrance Examinations

For JEE Main, JEE Advanced and other competitive examinations, the important skill is recognising the derivative's sign without unnecessary calculations.

In part (iii), instead of attempting to solve

\[ 3x^2+2x+1=0, \]

one can observe that its discriminant is

\[ \Delta=2^2-4(3)(1)=4-12=-8<0. \]

Since the leading coefficient is positive, the quadratic is positive for every real $x$. Therefore,

\[ h'(x)>0 \]

everywhere.

This discriminant-based observation is often the fastest method in objective questions.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 11 points

A function whose derivative is strictly positive throughout its domain is strictly increasing.
A strictly increasing function has neither a local maximum nor a local minimum at an interior point of its domain.
$\displaystyle e^x$ is strictly increasing because $\displaystyle e^x>0$ for every real $x$.
$\displaystyle \log x$ is strictly increasing on $(0,\infty)$ because $\displaystyle \frac1x>0$ there.
$\displaystyle x^3+x^2+x+1$ is strictly increasing because $\displaystyle 3x^2+2x+1>0$ for every real $x$.
For $\displaystyle 3x^2+2x+1$, completing the square gives
\[ 3x^2+2x+1=3\left(x+\frac13\right)^2+\frac23>0. \]
For a quadratic derivative, a negative discriminant together with a positive leading coefficient is a quick way to prove that the derivative is always positive.
For absolute extrema, monotonicity alone establishes the absence of an interior extremum, while limiting behaviour can establish whether the function is unbounded above or below.
$\displaystyle e^x$ is unbounded above and approaches $0$ without attaining it, so it has neither maximum nor minimum.
$\displaystyle \log x$ is unbounded both above and below, so it has neither maximum nor minimum.
$\displaystyle x^3+x^2+x+1$ is unbounded above and below, so it has neither maximum nor minimum.

Q5

LONG ANSWER4 marks

Find the absolute maximum value and the absolute minimum value of the following functions in the given intervals.

$\quad f(x)=x^3,\qquad x\in[-2,2] $
$\quad f(x)=\sin x+\cos x,\qquad x\in[0,\pi] $
$\quad f(x)=4x-\frac12x^2,\qquad x\in\left[-2,\frac92\right] $
$\quad f(x)=(x-1)^2+3,\qquad x\in[-3,1] $

📘 Concept & Theory

Theory and Concept

›

When a function is continuous on a closed and bounded interval

\[ [a,b], \]

it attains both an absolute maximum and an absolute minimum on that interval. This is the Extreme Value Theorem.

To find the absolute maximum and minimum of a differentiable function on

\[ [a,b], \]

we use the following procedure:

Find the derivative $f'(x)$.
Find all critical points in the open interval $(a,b)$ by solving $f'(x)=0$ or identifying points where $f'(x)$ does not exist.
Discard any critical points that do not lie in the given interval.
Evaluate the original function at every critical point lying inside the interval.
Evaluate the original function at both endpoints $a$ and $b$.
Compare all obtained function values.
The greatest value is the absolute maximum and the least value is the absolute minimum.

Important: The second derivative test is useful for classifying local maxima and minima, but for an absolute maximum/minimum on a closed interval, the endpoint values must always be checked. An endpoint can provide the absolute maximum or minimum even though it is not a stationary point.

🗺️ Solution Roadmap

Step-by-step Plan

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For each part, we will use the critical points plus endpoints method:
\[ \boxed{\text{Critical points in }(a,b)\;+\;\text{endpoints }a,b} \]
We then compare the corresponding values of the original function.

✏️ Solution

(i) $\displaystyle f(x)=x^3,\quad x\in[-2,2]$

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Step-by-step Solution · 12 steps

Find the first derivative
Differentiating,
\[f'(x)=3x^2\]
Find the critical point
\[f'(x)=0\]
Therefore,
\[3x^2=0\]
Since $3\neq0$,
\[x^2=0\]
Hence,
\[x=0\]
Since
\[-2<0<2\]
the critical point $x=0$ lies inside the given interval.
Evaluate the function at the critical point
\[f(0)=0^3\]

\[f(0)=0\]
Evaluate the function at the endpoints
At the left endpoint $x=-2$,
\[f(-2)=(-2)^3\]

\[=-8.\]
At the right endpoint $x=2$,
\[f(2)=2^3\]

\[=8\]
Compare all values

Point	Function value
$x=-2$	$-8$
$x=0$	$0$
$x=2$	$8$

The greatest value is $8$, and the least value is $-8$.
The second derivative is
\[ f''(x)=6x, \]
so
\[ f''(0)=0. \]
The second derivative test is therefore inconclusive at $x=0$. More importantly, for an absolute-extrema problem on a closed interval, we must compare the critical point with both endpoints. The point $x=0$ does not give either absolute extremum.

✏️ Solution

(ii) $\displaystyle f(x)=\sin x+\cos x,\quad x\in[0,\pi]$

›

Step-by-step Solution · 13 steps

Find the first derivative
\[f(x)=\sin x+\cos x\]

\[f'(x)=\cos x-\sin x\]
Find the critical points
Set
\[f'(x)=0\]
Therefore,
\[\cos x-\sin x=0\]
Hence,
\[\cos x=\sin x\]
Since $\cos x\neq0$ at a solution of this equation,
\[\tan x=1\]
Therefore,
\[x=\frac{\pi}{4}+n\pi,\qquad n\in\mathbb Z.\]
In the interval $[0,\pi],$ the only solution is
\[x=\frac{\pi}{4}\]
The value
\[x=\frac{3\pi}{4}\]
is not a critical point because
\[\tan\frac{3\pi}{4}=-1\neq1.\]
Equivalently,
\[f'\left(\frac{3\pi}{4}\right)=\cos\frac{3\pi}{4}-\sin\frac{3\pi}{4}=-\frac{1}{\sqrt2}-\frac{1}{\sqrt2}\neq0.\]
Evaluate the function at the critical point
\[f\left(\frac{\pi}{4}\right)=\sin\frac{\pi}{4}+\cos\frac{\pi}{4}\]

\[=\frac{1}{\sqrt2}+\frac{1}{\sqrt2}\]

\[=\frac{2}{\sqrt2}\]

\[=\sqrt2\]
Evaluate the function at the endpoints
At $x=0$,
\[f(0)=\sin0+\cos0\]

\[=0+1\]

\[=1.\]
At $x=\pi$,
\[f(\pi)=\sin\pi+\cos\pi\]

\[=0-1\]

\[=-1\]
Compare all candidate values

Point	Function value
$x=0$	$1$
$\displaystyle x=\frac{\pi}{4}$	$\sqrt2$
$x=\pi$	$-1$

Since
\[ \sqrt2>1>-1, \]
the greatest value is $\sqrt2$ and the least value is $-1$.

✏️ Solution

(iii) $\displaystyle f(x)=4x-\frac12x^2,\quad x\in\left[-2,\frac92\right]$

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Step-by-step Solution · 16 steps

Find the first derivative
\[ f(x)=4x-\frac12x^2 \]
Differentiating,
\[ f'(x)=4-\frac12(2x) \]

\[ f'(x)=4-x \]
Find the critical point
Set
\[f'(x)=0\]
Therefore,
\[4-x=0\]
Hence,
\[x=4\]
Since
\[-2<4<\frac92,\]
the critical point lies inside the given interval.
Evaluate the function at the critical point
\[f(4)=4(4)-\frac12(4)^2\]

\[=16-\frac12(16)\]

\[=16-8\]
\[=8\]
Evaluate the function at $x=-2$
\[f(-2)=4(-2)-\frac12(-2)^2\]

\[=-8-\frac12(4)\]

\[=-8-2\]

\[=-10\]
Evaluate the function at $x=\frac92$
\[f\left(\frac92\right)=4\left(\frac92\right)-\frac12\left(\frac92\right)^2\]
Simplifying,
\[=18-\frac12\left(\frac{81}{4}\right)\]

\[=18-\frac{81}{8}\]
Writing $18$ with denominator $8$,
\[18=\frac{144}{8}\]
Therefore,
\[f\left(\frac92\right)=\frac{144}{8}-\frac{81}{8}\]

\[=\frac{63}{8}\]
Thus,
\[f\left(\frac92\right)=\frac{63}{8}\]
Compare all candidate values

Point	Function value
$x=-2$	$-10$
$x=4$	$8$
$\displaystyle x=\frac92$	$\displaystyle \frac{63}{8}=7.875$

Therefore,
\[8>\frac{63}{8}>-10\]
Hence, the absolute maximum is $8$, and the absolute minimum is $-10$.

✏️ Solution

(iv) $\displaystyle f(x)=(x-1)^2+3,\quad x\in[-3,1]$

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Step-by-step Solution · 11 steps

Find the first derivative
\[f(x)=(x-1)^2+3\]
Differentiating,
\[f'(x)=2(x-1)\]
Find the critical point
Set
\[f'(x)=0\]
Therefore,
\[2(x-1)=0\]
Since $2\neq0$,
\[x-1=0\]

\[x=1\]
Notice that $x=1$ is not an interior point of the interval. It is the right endpoint of
\[ [-3,1]. \]
Therefore, it must still be checked as an endpoint, but it should not be described as an interior stationary point for this interval.
Evaluate the function at the endpoints
At $x=-3$,
\[f(-3)=(-3-1)^2+3\]

\[=(-4)^2+3\]

\[=16+3\]

\[=19\]
At $x=1$,
\[f(1)=(1-1)^2+3\]

\[=0^2+3\]

\[=3\]
Compare the endpoint values
\[19>3\]
Therefore,
\[\boxed{\text{Absolute maximum value}=19\text{ at }x=-3}\]
and
\[\boxed{\text{Absolute minimum value}=3\text{ at }x=1}\]
Final Comparison Table

Part	Function	Absolute Maximum	Point of Maximum	Absolute Minimum	Point of Minimum
(i)	$\displaystyle x^3,\;[-2,2]$	$8$	$x=2$	$-8$	$x=-2$
(ii)	$\displaystyle \sin x+\cos x,\;[0,\pi]$	$\sqrt2$	$\displaystyle x=\frac{\pi}{4}$	$-1$	$x=\pi$
(iii)	$\displaystyle 4x-\frac12x^2,\;\left[-2,\frac92\right]$	$8$	$x=4$	$-10$	$x=-2$
(iv)	$\displaystyle (x-1)^2+3,\;[-3,1]$	$19$	$x=-3$	$3$	$x=1$

🎯 Exam Significance

Exam Significance

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This question is a direct application of the method for finding absolute extrema on a closed interval. A very important examination rule is:

\[ \boxed{\text{Check all critical points and both endpoints.}} \]

Even if a critical point produces a local maximum or minimum, it cannot automatically be declared the absolute maximum or minimum. Its value must be compared with the values at the endpoints.

For example, in part (ii),

\[ f\left(\frac{\pi}{4}\right)=\sqrt2 \]

is the local and absolute maximum, but the absolute minimum occurs at the endpoint $x=\pi$, where

\[ f(\pi)=-1. \]

Part (iv) is also important because the critical point $x=1$ happens to coincide with an endpoint of the specified interval. It must therefore be included in the endpoint comparison.

Significance for Competitive Entrance Examinations

For JEE Main, JEE Advanced and other competitive entrance examinations, the key skill is to identify the complete set of candidates for absolute extrema quickly:

\[ \boxed{\text{Interior critical points}+\text{endpoints}} \]

A common trap is to find $f'(x)=0$, classify that point, and stop. That method can miss an absolute extremum at an endpoint.

In part (ii), solving

\[ \cos x-\sin x=0 \]

gives

\[ \tan x=1. \]

Within $[0,\pi]$, the only solution is

\[ x=\frac{\pi}{4}. \]

The value $\frac{3\pi}{4}$ must not be included because

\[ \tan\frac{3\pi}{4}=-1. \]

Part (iii) illustrates another useful competitive-exam pattern: a quadratic function with negative leading coefficient is concave downward. Once the vertex

\[ x=4 \]

is identified, its value is a strong candidate for the maximum, but the endpoints must still be evaluated to establish the absolute minimum and confirm the absolute maximum.

🔑 Key Takeaways

Key Takeaways

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Key Takeaways · 11 points

For a continuous function on a closed interval, absolute maximum and minimum values exist.
To find them, evaluate the function at every critical point inside the interval and at both endpoints.
Never omit endpoint values in an absolute-extrema problem.
In part (i), $\displaystyle f(x)=x^3$ has absolute minimum $-8$ at $x=-2$ and absolute maximum $8$ at $x=2$.
In part (ii), the only critical point in $[0,\pi]$ is $\displaystyle x=\frac{\pi}{4}$.
For part (ii), the absolute maximum is $\sqrt2$, while the absolute minimum is the endpoint value $-1$ at $x=\pi$.
In part (iii), the critical point $x=4$ gives the absolute maximum $8$, while the left endpoint gives the absolute minimum $-10$.
In part (iv), $x=1$ is an endpoint, not an interior stationary point. It gives the absolute minimum $3$.
The absolute maximum in part (iv) occurs at the opposite endpoint $x=-3$, with value $19$.
A local maximum or minimum need not be an absolute maximum or minimum on a restricted interval.
The second derivative test is useful for classifying stationary points, but endpoint comparison is essential for absolute extrema.

📘 Concept & Theory

Theory and Concept

›

To find the maximum or minimum value of a differentiable function, we first locate its critical points by solving

\[ p'(x)=0 \]

After obtaining the critical point, we classify it using the second derivative test:

\[ p''(x)<0 \Rightarrow \text{local maximum} \]

\[ p''(x)>0 \Rightarrow \text{local minimum} \]

Here, the profit function is a quadratic function with a negative coefficient of $x^2$. Therefore, its graph is a downward-opening parabola, so its stationary point is expected to give the maximum profit.

🗺️ Solution Roadmap

Step-by-step Plan

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Differentiate the profit function $p(x)$.
Set $p'(x)=0$ to find the critical point.
Use the second derivative to verify that the critical point gives maximum profit.
Substitute the critical value of $x$ into $p(x)$.

✏️ Solution

Complete Solution

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Step-by-step Solution · 10 steps

Write the profit function
\[p(x)=41-72x-18x^2\]
Find the first derivative
\[p'(x)=\frac{d}{dx}\left(41-72x-18x^2\right)\]

\[p'(x)=0-72-36x\]

\[p'(x)=-72-36x\]
Find the critical point
For a possible maximum or minimum, set the first derivative equal to zero:
\[p'(x)=0\]

\[-72-36x=0\]

\[-36x=72\]

\[x=-2\]
Thus, the critical point occurs at
\[\boxed{x=-2}\]
Verify whether the critical point gives maximum profit
Find the second derivative:
\[p''(x)=\frac{d}{dx}(-72-36x)\]

\[p''(x)=-36\]
Since,
\[p''(-2)=-36<0\]
the function has a local maximum at $x=-2$.
In fact, because $p(x)$ is a downward-opening quadratic and has no other critical point, this is also the absolute maximum of the profit function over its mathematical domain.
Find the maximum profit
Substitute $x=-2$ into the profit function:
\[p(-2)=41-72(-2)-18(-2)^2\]

\[=41+144-18(4)\]

\[=41+144-72\]

\[=185-72\]

\[=113\]
Therefore,
\[ \boxed{p_{\max}=113} \]

💡 Answer

Final Answer

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\[ \boxed{\text{Maximum Profit}=113} \]

🎯 Exam Significance

Exam Significance

›

Always find $p'(x)$ and solve $p'(x)=0$ to locate stationary points.
Do not stop after finding a critical point; verify whether it gives a maximum or minimum.
The second derivative test provides a quick classification:
\[ p''(x)<0 \Rightarrow \text{maximum} \]
Substitute the critical value back into the original function, not its derivative.

Significance for Competitive Entrance Examinations

This is a standard application of derivatives problem. Competitive examinations frequently combine optimization with quadratic functions. A useful shortcut is to recognize that

\[ p(x)=41-72x-18x^2 \]

is a downward-opening parabola because the coefficient of $x^2$ is negative. Therefore, its vertex gives the maximum value. The vertex occurs at

\[ x=-\frac{b}{2a} \]

where $a=-18$ and $b=-72$:

\[ x=-\frac{-72}{2(-18)} \]

\[ x=\frac{72}{-36}=-2 \]

Then

\[ p(-2)=113 \]

This shortcut can save time in objective-type examinations, although the derivative method is the more general approach.

🔑 Key Takeaways

Key Takeaways

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Key Takeaways · 5 points

For optimization, first find the stationary points using $f'(x)=0$.
Use $f''(x)<0$ to confirm a maximum.
For a downward-opening quadratic, the vertex gives the maximum value.
For this problem, the maximum occurs at $x=-2$.
The maximum profit is $\boxed{113}$.

📘 Concept & Theory

Theory and Concept

›

Since the function is a polynomial, it is continuous on the closed interval $[0,3]$. Therefore, by the Extreme Value Theorem, it must attain both an absolute maximum and an absolute minimum on this interval.

To find the absolute maximum and minimum of a differentiable function on a closed interval, we must evaluate the function at:

All critical points lying inside the interval.
The two endpoints of the interval.

The critical points are obtained by solving

\[ f'(x)=0 \]

🗺️ Solution Roadmap

Step-by-step Plan

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Differentiate $f(x)$.
Solve $f'(x)=0$ to obtain all critical points.
Retain only those critical points that lie in $[0,3]$.
Evaluate $f(x)$ at all critical points and both endpoints $x=0$ and $x=3$.
Compare all obtained values to determine the absolute maximum and minimum.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 21 steps

Define the function
\[f(x)=3x^4-8x^3+12x^2-48x+25\]
Find the first derivative
\[f'(x)=\frac{d}{dx}\left(3x^4-8x^3+12x^2-48x+25\right)\]

\[f'(x)=12x^3-24x^2+24x-48\]
Factor out $12$:
\[f'(x)=12\left(x^3-2x^2+2x-4\right)\]
For a critical point,
\[f'(x)=0\]
Therefore,
\[12\left(x^3-2x^2+2x-4\right)=0\]

\[x^3-2x^2+2x-4=0\]
Factorise the equation
Group the terms:
\[x^3-2x^2+2x-4=0\]

\[x^2(x-2)+2(x-2)=0\]
\[(x-2)(x^2+2)=0\]
Hence,
\[x-2=0\]
or
\[x^2+2=0\]
From the first equation,
\[x=2\]
The second equation give
\[x^2=-2\]
which has no real solution.
Therefore, the only real critical point is
\[\boxed{x=2}\]
Include the endpoints
Since the function is defined on the closed interval $[0,3]$, we must also examine the endpoints:
\[x=0,\quad x=3\]
Thus, the complete set of points that must be checked is
\[\boxed{x=0,\;2,\;3}\]
Evaluate $f(x)$ at $x=0$
\[f(0)=3(0)^4-8(0)^3+12(0)^2-48(0)+25\]

\[\boxed{f(0)=25}\]
Evaluate $f(x)$ at $x=2$
\[f(2)=3(2)^4-8(2)^3+12(2)^2-48(2)+25\]

\[=3(16)-8(8)+12(4)-96+25\]

\[=48-64+48-96+25\]

\[=48-64=-16\]

\[-16+48=32\]

\[32-96=-64\]

\[-64+25=-39\]
Therefore,
\[\boxed{f(2)=-39}\]
Evaluate $f(x)$ at $x=3$
\[f(3)=3(3)^4-8(3)^3+12(3)^2-48(3)+25\]

\[=3(81)-8(27)+12(9)-144+25\]

\[=243-216+108-144+25\]

\[=27+108-144+25\]

\[=135-144+25\]

\[=-9+25\]

\[\boxed{f(3)=16}\]
Compare all values

Point	Function value
$x=0$	$f(0)=25$
$x=2$	$f(2)=-39$
$x=3$	$f(3)=16$

Clearly,
\[-39<16<25\]
Therefore, the absolute minimum value is
\[\boxed{-39}\]
at
\[\boxed{x=2}\]
and the absolute maximum value is
\[\boxed{25}\]
at
\[\boxed{x=0}\]

💡 Answer

Final Answer

›

\[ \boxed{\text{Absolute Maximum Value}=25\text{ at }x=0} \]

\[ \boxed{\text{Absolute Minimum Value}=-39\text{ at }x=2} \]

🎯 Exam Significance

Exam Significance

›

For absolute extrema on a closed interval, always check both endpoints.
Find every real critical point from $f'(x)=0$.
Do not include complex roots when finding extrema over a real interval.
Compare the function values at all eligible points before stating the final answer.
A factorisation error can introduce a false critical point, so verify the factorisation by multiplication.

Significance for Competitive Entrance Examinations

This problem tests a particularly important optimization technique: finding absolute extrema on a closed interval. The key competitive-exam step is to simplify the derivative before solving it.

\[ f'(x)=12(x-2)(x^2+2) \]

Since

\[ x^2+2>0\qquad\text{for all }x\in\mathbb{R}, \]

the derivative can vanish only at $x=2$. This immediately eliminates unnecessary calculations involving $\sqrt{2}$.

Such factorisation and sign-analysis techniques are useful in JEE Main, JEE Advanced and other entrance examinations where optimization questions are often designed to test speed as well as accuracy.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 7 points

The correct derivative is
\[ f'(x)=12(x-2)(x^2+2). \]
The only real critical point is $x=2$.
$\sqrt{2}$ is not a critical point for this function.
For a closed interval, both endpoints must always be checked.
The values to compare are
\[ f(0)=25,\qquad f(2)=-39,\qquad f(3)=16. \]
The absolute maximum is
\[ \boxed{25\text{ at }x=0}. \]
The absolute minimum is
\[ \boxed{-39\text{ at }x=2}. \]

📘 Concept & Theory

Theory and Concept

›

To find the absolute maximum of a differentiable function on a closed interval, we evaluate the function at all critical points lying inside the interval and at both endpoints.

Here,

\[ f(x)=\sin 2x,\qquad x\in[0,2\pi] \]

The maximum value of the sine function is $1$. Therefore, $\sin 2x$ will attain its maximum whenever

\[ \sin 2x=1. \]

We can obtain all such points either directly from the standard sine condition or by using derivatives.

🗺️ Solution Roadmap

Step-by-step Plan

›

Differentiate $f(x)=\sin 2x$.
Set $f'(x)=0$ to find all critical points in $[0,2\pi]$.
Evaluate $f(x)$ at the critical points and endpoints.
Compare the values to identify where the maximum occurs.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 19 steps

Define the function
\[f(x)=\sin 2x\]
Find the first derivative
Using the chain rule,
\[f'(x)=\frac{d}{dx}(\sin 2x)\]

\[f'(x)=2\cos 2x\]
Find the critical points
For a critical point,
\[f'(x)=0\]
Therefore,
\[2\cos 2x=0\]

\[\cos 2x=0\]
We know that
\[\cos\theta=0\]
when
\[\theta=\frac{\pi}{2}+n\pi,\qquad n\in\mathbb Z\]
Hence,
\[2x=\frac{\pi}{2}+n\pi\]
Dividing by $2$,
\[x=\frac{\pi}{4}+\frac{n\pi}{2}\]
Now we select the values lying in
\[0\leq x\leq2\pi\]
This gives
\[x=\frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}\]
Evaluate the function at the critical points and endpoints
At the left endpoint $x=0$:
\[f(0)=\sin 0=0\]
At $x=\frac{\pi}{4}$:
\[f\left(\frac{\pi}{4}\right)=\sin\left(2\cdot\frac{\pi}{4}\right)\]

\[=\sin\frac{\pi}{2}\]

\[=1\]
At $x=\frac{3\pi}{4}$:
\[f\left(\frac{3\pi}{4}\right)=\sin\left(\frac{3\pi}{2}\right)\]

\[=-1\]
At $x=\frac{5\pi}{4}$:
\[f\left(\frac{5\pi}{4}\right)=\sin\left(\frac{5\pi}{2}\right)\]

\[=1\]
At $x=\frac{7\pi}{4}$:
\[f\left(\frac{7\pi}{4}\right)=\sin\left(\frac{7\pi}{2}\right)\]

\[=-1\]
At the right endpoint $x=2\pi$:
\[f(2\pi)=\sin(4\pi)=0\]
Compare the values

Point	Function value
$x=0$	$0$
$x=\frac{\pi}{4}$	$1$
$x=\frac{3\pi}{4}$	$-1$
$x=\frac{5\pi}{4}$	$1$
$x=\frac{7\pi}{4}$	$-1$
$x=2\pi$	$0$

The largest value is $1$, which occurs at two points in the interval:
\[\boxed{x=\frac{\pi}{4}\quad\text{and}\quad x=\frac{5\pi}{4}}\]
Thus, the maximum value is
\[\boxed{1}\]

💡 Answer

Final Answer

›

The function $\sin 2x$ attains its maximum value $1$ at

\[ \boxed{x=\frac{\pi}{4},\frac{5\pi}{4}} \]

🎯 Exam Significance

Exam Significance

›

When solving trigonometric equations, always write the general solution before restricting it to the given interval.
For absolute extrema on a closed interval, check all critical points and both endpoints.
Do not assume that the first critical point found is the only one.
The maximum value of $\sin 2x$ is $1$, and it can occur at more than one point within a sufficiently large interval.

Significance for Competitive Entrance Examinations

This problem tests the ability to combine differentiation with periodic trigonometric equations. The key step is to retain the general solution:

\[ x=\frac{\pi}{4}+\frac{n\pi}{2}. \]

Restricting this expression to $[0,2\pi]$ immediately produces all four stationary points. This avoids the common error of considering only the principal solution.

A useful direct observation is that $\sin\theta=1$ when

\[ \theta=\frac{\pi}{2}+2n\pi. \]

Hence,

\[ 2x=\frac{\pi}{2}+2n\pi \]

gives

\[ x=\frac{\pi}{4}+n\pi. \]

Within $[0,2\pi]$, this again gives

\[ x=\frac{\pi}{4},\frac{5\pi}{4}. \]

This direct approach is faster in an objective examination.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 5 points

The derivative is
\[ f'(x)=2\cos 2x. \]
The complete set of critical points in $[0,2\pi]$ is
\[ \frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}. \]
The maximum value of $\sin 2x$ is
\[ \boxed{1}. \]
This maximum is attained at two points:
\[ \boxed{x=\frac{\pi}{4},\frac{5\pi}{4}}. \]
The original solution is incomplete because it considers only $x=\frac{\pi}{4}$.

📘 Concept & Theory

Theory and Concept

›

To find the maximum value of a differentiable function using derivatives, we first find its critical points by solving

\[ f'(x)=0. \]

We then determine whether the critical point corresponds to a maximum. For this function, we can also use the standard identity

\[ \sin x+\cos x=\sqrt{2}\sin\left(x+\frac{\pi}{4}\right). \]

Since the maximum value of $\sin\theta$ is $1$, the maximum possible value is immediately $\sqrt{2}$. The derivative method confirms this result.

🗺️ Solution Roadmap

Step-by-step Plan

›

Differentiate $f(x)$.
Set $f'(x)=0$ to find the stationary points.
Verify that the stationary point gives a maximum.
Substitute the critical value into the original function.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 14 steps

Define the function
\[f(x)=\sin x+\cos x\]
Find the first derivative
\[ f'(x)=\frac{d}{dx}(\sin x+\cos x) \]

\[ f'(x)=\cos x-\sin x \]
Find the critical point
For a stationary point,
\[f'(x)=0\]
Therefore,
\[\cos x-\sin x=0\]

\[\cos x=\sin x\]
Dividing by $\cos x$, where applicable,
\[\tan x=1\]
Hence, the general solution is
\[x=\frac{\pi}{4}+n\pi,\qquad n\in\mathbb Z\]
Thus, $x=\frac{\pi}{4}$ is one stationary point, but it is important to note that it is not the only solution of $f'(x)=0$.
Verify that the stationary point gives a maximum
Find the second derivative:
\[f''(x)=\frac{d}{dx}(\cos x-\sin x)\]

\[f''(x)=-\sin x-\cos x\]
At $x=\frac{\pi}{4}$
we get
\[f''\left(\frac{\pi}{4}\right)=-\sin\frac{\pi}{4}-\cos\frac{\pi}{4}\]

\[=-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}\]

\[=-\sqrt{2} < 0\]
Therefore,
\[\boxed{x=\frac{\pi}{4}}\]
gives a local maximum.
Find the maximum value
\[f\left(\frac{\pi}{4}\right)=\sin\frac{\pi}{4}+\cos\frac{\pi}{4}\]
Since
\[\sin\frac{\pi}{4}=\cos\frac{\pi}{4}=\frac{1}{\sqrt{2}},\]
we obtain
\[f\left(\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\]

\[=\frac{2}{\sqrt{2}}\]

\[=\sqrt{2}\]
Therefore,
\[\boxed{\text{Maximum value of }f(x)=\sqrt{2}}\]

🎯 Exam Significance

Exam Significance

›

Always distinguish between a stationary point and a maximum point.
Use the second derivative test to classify a stationary point when required.
For trigonometric optimization, rewriting the expression in the form $R\sin(x+\alpha)$ or $R\cos(x-\alpha)$ can provide a very quick solution.
Remember that trigonometric equations generally have infinitely many solutions unless a specific interval is given.

Significance for Competitive Entrance Examinations

For objective-type examinations, the identity method is particularly efficient:

\[ \sin x+\cos x =\sqrt{2}\sin\left(x+\frac{\pi}{4}\right) \]

Therefore, immediately,

\[ f(x)\leq\sqrt{2}. \]

Equality occurs when

\[ \sin\left(x+\frac{\pi}{4}\right)=1, \]

giving

\[ x=\frac{\pi}{4}+2n\pi. \]

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 5 points

\[ f'(x)=\cos x-\sin x. \]
The stationary points satisfy
\[ x=\frac{\pi}{4}+n\pi. \]
The maximum value is
\[ \boxed{\sqrt{2}}. \]
The maximum occurs at
\[ \boxed{x=\frac{\pi}{4}+2n\pi}. \]
The corresponding minimum value is $-\sqrt{2}$, attained at
\[ x=\frac{5\pi}{4}+2n\pi. \]

📘 Concept & Theory

Theory and Concept

›

To find the absolute maximum value of a continuous function on a closed interval, we evaluate the function at all critical points lying inside the interval and at both endpoints.

A critical point of a differentiable function is obtained by solving

\[ f'(x)=0. \]

The important point in this question is that the same function is being considered on two different intervals. Therefore, the relevant critical points and endpoints must be considered separately for each interval.

🗺️ Solution Roadmap

Step-by-step Plan

›

Find $f'(x)$.
Solve $f'(x)=0$ to obtain all critical points.
For $[1,3]$, retain only the critical points lying in $[1,3]$, and check the endpoints.
For $[-3,1]$, retain only the critical points lying in $[-3,1]$, and check the endpoints.
Compare the corresponding function values in each interval.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 24 steps

Given —
\[f(x)=2x^3-24x+107\]
Differentiate:
\[f'(x)=\frac{d}{dx}(2x^3-24x+107)\]

\[f'(x)=6x^2-24\]
For a critical point,
\[f'(x)=0\]
Therefore,
\[6x^2-24=0\]

\[6x^2=24\]

\[x^2=4\]

\[x=\pm2\]
Hence, the two critical points are
\[\boxed{x=-2,\;2}\]
Part I: Maximum Value in $[1,3]$
Identify the Points to be Considered
The interval is
\[[1,3]\]
Of the critical points $x=-2$ and $x=2$, only $x=2$ lies inside $[1,3]$.
Therefore, we must consider
\[\boxed{x=1,\;2,\;3}\]
Evaluate $f(x)$ at $x=1$
\[f(1)=2(1)^3-24(1)+107\]

\[=2-24+107\]

\[\boxed{f(1)=85}\]
Evaluate $f(x)$ at $x=2$
\[f(2)=2(2)^3-24(2)+107\]

\[=2(8)-48+107\]

\[=16-48+107\]

\[=-32+107\]

\[\boxed{f(2)=75}\]
Evaluate $f(x)$ at $x=3$
\[f(3)=2(3)^3-24(3)+107\]

\[=2(27)-72+107\]

\[=54-72+107\]

\[=-18+107\]

\[\boxed{f(3)=89}\]
Compare the Values

Point	Function Value
$x=1$	$85$
$x=2$	$75$
$x=3$	$89$

Since
\[89>85>75\]
the maximum value in $[1,3]$ is
\[\boxed{89}\]
and it occurs at
\[\boxed{x=3}\]
Part II: Maximum Value in $[-3,1]$
Identify the Points to be Considered
The interval is
\[[-3,1]\]
Of the critical points $x=-2$ and $x=2$, only $x=-2$ lies inside $[-3,1]$.
Therefore, the points to be considered are
\[\boxed{x=-3,\;-2,\;1}\]
Evaluate $f(x)$ at $x=-3$
\[f(-3)=2(-3)^3-24(-3)+107\]

\[=2(-27)+72+107\]

\[=-54+72+107\]

\[=18+107\]

\[\boxed{f(-3)=125}\]
Evaluate $f(x)$ at $x=-2$
\[f(-2)=2(-2)^3-24(-2)+107\]

\[=2(-8)+48+107\]

\[=-16+48+107\]

\[=32+107\]

\[\boxed{f(-2)=139}\]
Evaluate $f(x)$ at $x=1$
\[f(1)=2(1)^3-24(1)+107\]

\[=2-24+107\]

\[\boxed{f(1)=85}\]

Compare the Values

Point	Function Value
$x=-3$	$125$
$x=-2$	$139$
$x=1$	$85$

Since
\[139>125>85,\]
the maximum value in $[-3,1]$ is
\[\boxed{139}\]
and it occurs at
\[\boxed{x=-2}\]

💡 Answer

Final Answer

›

Interval	Maximum Value	Point Where Maximum Occurs
$[1,3]$	$\boxed{89}$	$\boxed{x=3}$
$[-3,1]$	$\boxed{139}$	$\boxed{x=-2}$

🎯 Exam Significance

Exam Significance

›

Always use the exact interval stated in the question.
Find all critical points before restricting them to the required interval.
For a closed interval, evaluate the function at every interior critical point and both endpoints.
A critical point outside the specified interval must not be included in the comparison.
For $[1,3]$, $x=-2$ is discarded because it lies outside the interval.
For $[-3,1]$, $x=2$ is discarded because it lies outside the interval.

Significance for Competitive Entrance Examinations

This question is particularly useful for understanding how the same function can have different absolute maxima on different intervals. The derivative gives the same stationary points,

\[ x=-2,\;2, \]

but the interval determines which stationary point is relevant.

The derivative can also be factored as

\[ f'(x)=6x^2-24 \]

\[ =6(x^2-4) \]

\[ =6(x-2)(x+2). \]

Therefore, the sign of $f'(x)$ changes around the critical points. In particular, $x=-2$ corresponds to a local maximum and $x=2$ corresponds to a local minimum. This explains the results obtained on the two intervals.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 6 points

The critical points of
\[ f(x)=2x^3-24x+107 \]
are
\[ \boxed{x=-2,\;2}. \]
On $[1,3]$, only $x=2$ is an interior critical point.
The maximum on $[1,3]$ is
\[ \boxed{89\text{ at }x=3}. \]
On $[-3,1]$, only $x=-2$ is an interior critical point.
The maximum on $[-3,1]$ is
\[ \boxed{139\text{ at }x=-2}. \]
Different intervals can produce different absolute maximum values for the same function.

📘 Concept & Theory

Theory and Concept

›

Since $x=1$ lies inside the interval $[0,2]$, it is an interior point. If a differentiable function attains a maximum at an interior point, then its first derivative must vanish there.

\[ f'(1)=0 \]

Therefore, we do not need to compare $f(1)$ with another unknown point. The condition that $x=1$ is the point of maximum directly gives the value of $a$.

🗺️ Solution Roadmap

Step-by-step Plan

›

Differentiate the given function.
Since $x=1$ is an interior maximum point, put $f'(1)=0$.
Solve the resulting equation for $a$.
Verify that $x=1$ actually gives a maximum using the second derivative.
Since the question says maximum on the entire interval $[0,2]$, verify the endpoint values as well.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 14 steps

Write the function
\[f(x)=x^4-62x^2+ax+9\]
Find the first derivative
\[f'(x)=\frac{d}{dx}\left(x^4-62x^2+ax+9\right)\]

\[f'(x)=4x^3-124x+a\]
Use the maximum condition at $x=1$
Since $x=1$ is an interior point of $[0,2]$ and the function attains its maximum there,
\[f'(1)=0\]
Substituting $x=1$,
\[4(1)^3-124(1)+a=0\]

\[4-124+a=0\]

\[a-120=0\]
Therefore,
\[\boxed{a=120}\]
Verify that $x=1$ gives a maximum
The second derivative is
\[f''(x)=\frac{d}{dx}(4x^3-124x+a)\]

\[f''(x)=12x^2-124\]
At $x=1$,
\[f''(1)=12(1)^2-124\]

\[=12-124\]

\[=-112\]
Since
\[f''(1)=-112<0,\]
$x=1$ is a point of local maximum.
Verify the maximum on the complete interval $[0,2]$
Since the question specifies that the function attains its maximum on the entire closed interval $[0,2]$, we can verify the endpoints after putting $a=120$. The function becomes
\[f(x)=x^4-62x^2+120x+9\]
At $x=0$,
\[f(0)=9\]
At $x=1$,
\[f(1)=1-62+120+9=68\]
At $x=2$,
\[f(2)=2^4-62(2)^2+120(2)+9\]

\[=16-248+240+9\]

\[=17\]
Thus,
\[f(0)=9,\quad f(1)=68,\quad f(2)=17\]
and hence
\[68>17>9\]
Therefore, $x=1$ indeed gives the absolute maximum on $[0,2]$.

🎯 Exam Significance

Exam Significance

›

If a differentiable function has a maximum or minimum at an interior point, immediately use
\[ f'(x)=0. \]
If the point is explicitly given, substitute that point directly into the derivative.
Do not introduce unnecessary unknown critical points when the problem already specifies the extremum point.
For confirmation of a maximum, use
\[ f''(1)<0. \]
When the question refers to the maximum on a closed interval, endpoint verification provides a complete check.

Significance for Competitive Entrance Examinations

This is a classic parameter-based optimization problem. The fastest approach is to recognize that the given point $x=1$ is an interior maximum. Therefore,

\[ f'(1)=0. \]

Thus,

\[ 4(1)^3-124(1)+a=0 \]

\[ a=120. \]

This avoids solving a cubic equation or constructing unnecessary equations involving $f(1)$ and another critical point.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 6 points

Given
\[ f(x)=x^4-62x^2+ax+9, \]
we have
\[ f'(x)=4x^3-124x+a. \]
Since $x=1$ is an interior maximum point,
\[ f'(1)=0. \]
This gives
\[ \boxed{a=120}. \]
The second derivative confirms the maximum:
\[ f''(1)=-112<0. \]
With $a=120$,
\[ f(0)=9,\qquad f(1)=68,\qquad f(2)=17. \]
Therefore, the absolute maximum on $[0,2]$ is $68$ at $x=1$.

📘 Concept & Theory

Theory and Concept

›

Since $f(x)=x+\sin 2x$ is continuous on the closed interval $[0,2\pi]$, it attains both an absolute maximum and an absolute minimum on this interval.

To determine these values, we must evaluate the function at:

All critical points lying inside the interval.
The two endpoints $x=0$ and $x=2\pi$.

The critical points are obtained by solving

\[ f'(x)=0. \]

🗺️ Solution Roadmap

Step-by-step Plan

›

Find $f'(x)$.
Solve $f'(x)=0$ completely within $[0,2\pi]$.
Evaluate $f(x)$ at every critical point and at both endpoints.
Compare all values to determine the absolute maximum and minimum.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 30 steps

Define the function
\[f(x)=x+\sin 2x\]
Find the first derivative
Using the chain rule,
\[f'(x)=\frac{d}{dx}(x+\sin 2x)\]

\[f'(x)=1+2\cos 2x\]
Find all critical points
For a critical point,
\[f'(x)=0\]
Therefore,
\[1+2\cos 2x=0\]

\[2\cos 2x=-1\]

\[\cos 2x=-\frac{1}{2}\]
We must now solve this equation completely for
\[ 0\leq x\leq2\pi. \]
Since
\[0\leq x\leq2\pi,\]
multiplying by $2$ gives
\[0\leq2x\leq4\pi.\]
The general solutions of
\[\cos\theta=-\frac{1}{2}\]
are
\[\theta=\frac{2\pi}{3}+2n\pi\]
or
\[\theta=\frac{4\pi}{3}+2n\pi,\quad n\in\mathbb Z\]
Putting $\theta=2x$, we get
\[2x=\frac{2\pi}{3}+2n\pi\]

\[x=\frac{\pi}{3}+n\pi\]
and
\[2x=\frac{4\pi}{3}+2n\pi\]

\[x=\frac{2\pi}{3}+n\pi\]
Restricting these solutions to $[0,2\pi]$, we obtain
\[\boxed{x=\frac{\pi}{3},\frac{2\pi}{3},\frac{4\pi}{3},\frac{5\pi}{3}}\]
Include the endpoints
\[x=0,\quad x=2\pi\]
Therefore, the complete set of points to be considered is
\[\boxed{0,\frac{\pi}{3},\frac{2\pi}{3},\frac{4\pi}{3},\frac{5\pi}{3},2\pi}\]
Evaluate $f(x)$ at $x=0$
\[f(0)=0+\sin 0\]

\[\boxed{f(0)=0}\]
Evaluate $f(x)$ at $x=\frac{\pi}{3}$
\[f\left(\frac{\pi}{3}\right)=\frac{\pi}{3}+\sin\left(\frac{2\pi}{3}\right)\]
Since
\[\sin\frac{2\pi}{3}=\frac{\sqrt3}{2},\]
we get
\[\boxed{f\left(\frac{\pi}{3}\right)=\frac{\pi}{3}+\frac{\sqrt3}{2}}\]
Evaluate $f(x)$ at $x=\frac{2\pi}{3}$
\[f\left(\frac{2\pi}{3}\right)=\frac{2\pi}{3}+\sin\left(\frac{4\pi}{3}\right)\]
Since
\[\sin\frac{4\pi}{3}=-\frac{\sqrt3}{2},\]
we get
\[\boxed{f\left(\frac{2\pi}{3}\right)=\frac{2\pi}{3}-\frac{\sqrt3}{2}}\]
Evaluate $f(x)$ at $x=\frac{4\pi}{3}$
\[f\left(\frac{4\pi}{3}\right)=\frac{4\pi}{3}+\sin\left(\frac{8\pi}{3}\right)\]
Since
\[\sin\frac{8\pi}{3}=\sin\left(2\pi+\frac{2\pi}{3}\right)=\frac{\sqrt3}{2},\]
we get
\[\boxed{f\left(\frac{4\pi}{3}\right)=\frac{4\pi}{3}+\frac{\sqrt3}{2}}\]
Evaluate $f(x)$ at $x=\frac{5\pi}{3}$
\[f\left(\frac{5\pi}{3}\right)=\frac{5\pi}{3}+\sin\left(\frac{10\pi}{3}\right)\]
Since
\[\sin\frac{10\pi}{3}=\sin\left(2\pi+\frac{4\pi}{3}\right)=-\frac{\sqrt3}{2},\]
we obtain
\[\boxed{f\left(\frac{5\pi}{3}\right)=\frac{5\pi}{3}-\frac{\sqrt3}{2}}\]
Evaluate $f(x)$ at $x=2\pi$
\[f(2\pi)=2\pi+\sin(4\pi)\]

\[=2\pi+0\]

\[\boxed{f(2\pi)=2\pi}\]

Compare all values

Point	Function Value
$x=0$	$0$
$x=\frac{\pi}{3}$	$\frac{\pi}{3}+\frac{\sqrt3}{2}$
$x=\frac{2\pi}{3}$	$\frac{2\pi}{3}-\frac{\sqrt3}{2}$
$x=\frac{4\pi}{3}$	$\frac{4\pi}{3}+\frac{\sqrt3}{2}$
$x=\frac{5\pi}{3}$	$\frac{5\pi}{3}-\frac{\sqrt3}{2}$
$x=2\pi$	$2\pi$

We now compare these values. Clearly,
\[f(0)=0\]
is the smallest because for $0 < x \leq2\pi$, the function satisfies
\[f(x)=x+\sin2x\geq x-1\]
More directly, among the six values above, the smallest is $0$
For the maximum, the largest value is
\[2\pi\]
Indeed,
\[\frac{5\pi}{3}-\frac{\sqrt3}{2}<2\pi\]
and all the remaining values are smaller still.

🎯 Exam Significance

Exam Significance

›

Always solve trigonometric equations completely within the given interval.
Do not use only one principal solution when $\cos\theta=c$, $\sin\theta=c$, or $\tan\theta=c$.
For a closed interval, include both endpoints in the comparison.
A correct final answer obtained from incomplete critical-point analysis is not a complete mathematical solution.
Pay particular attention to the range of $2x$, which here is $[0,4\pi]$, not $[0,2\pi]$.

Significance for Competitive Entrance Examinations

The main challenge in this problem is solving

\[ \cos2x=-\frac12 \]

over a range spanning two complete periods. A reliable approach is to write the general solutions first:

\[ 2x=\frac{2\pi}{3}+2n\pi \]

or

\[ 2x=\frac{4\pi}{3}+2n\pi. \]

This prevents missing critical points and is particularly useful in JEE-type questions involving trigonometric optimization.

There is also a useful bounding observation:

\[ -1\leq\sin2x\leq1. \]

Therefore,

\[ x-1\leq x+\sin2x\leq x+1. \]

At $x=0$, the function is exactly $0$, and at $x=2\pi$, it is exactly $2\pi$. The complete critical-point comparison confirms that these are the absolute minimum and maximum, respectively.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 6 points

\[ f'(x)=1+2\cos2x. \]
The critical-point equation is
\[ \cos2x=-\frac12. \]
The complete set of critical points in $[0,2\pi]$ is
\[ \boxed{ \frac{\pi}{3},\frac{2\pi}{3}, \frac{4\pi}{3},\frac{5\pi}{3} }. \]
The absolute minimum is
\[ \boxed{0\text{ at }x=0}. \]
The absolute maximum is
\[ \boxed{2\pi\text{ at }x=2\pi}. \]
The original solution's final answer is correct, but its critical-point analysis is incomplete and contains an incorrect trigonometric angle.

📘 Concept & Theory

Theory and Concept

›

This is a standard optimization problem. We first express one number in terms of the other using the given condition. Then we form the product as a function of one variable and maximize that function using derivatives.

Let the two numbers be $x$ and $y$. Since their sum is $24$,

\[ x+y=24 \]

Therefore,

\[ y=24-x. \]

The product is then

\[ P(x)=xy=x(24-x). \]

We need to find the value of $x$ for which $P(x)$ is maximum.

🗺️ Solution Roadmap

Step-by-step Plan

›

Let one number be $x$.
Express the second number using the sum condition.
Form the product function $P(x)$.
Differentiate and set $P'(x)=0$.
Use the second derivative to verify that the product is maximum.
Find the two numbers and, if required, calculate their maximum product.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 14 steps

Let the first number be $x$
Since the sum of the two numbers is $24$, let the second number be $y$. Then
\[x+y=24\]
Therefore,
\[y=24-x\]
The product of the two numbers is
\[P=xy\]
Substituting $y=24-x$, we get
\[P(x)=x(24-x)\]
Expanding,
\[P(x)=24x-x^2\]
Differentiate the product
\[P'(x)=\frac{d}{dx}(24x-x^2)\]

\[P'(x)=24-2x\]
For the maximum or minimum value, set
\[P'(x)=0\]
Therefore,
\[24-2x=0\]

\[2x=24\]

\[\boxed{x=12}\]
Find the second derivative:
\[P''(x)=\frac{d}{dx}(24-2x)\]

\[P''(x)=-2\]
Since
\[P''(12)=-2<0,\]
the product function has a maximum at
\[x=12\]
Find the second number
\[y=24-x\]
Substituting $x=12$,
\[y=24-12\]

\[\boxed{y=12}\]
Therefore, the two numbers are
\[\boxed{12\text{ and }12}\]

💡 Answer

Final Answer

›

The two numbers whose sum is $24$ and whose product is as large as possible are

\[\boxed{12\text{ and }12}\]

🎯 Exam Significance

Exam Significance

›

Convert the given condition into an equation involving one variable.
Express the quantity to be optimized as a function of that variable.
Use $P'(x)=0$ to locate the stationary point.
Use $P''(x)<0$ to confirm a maximum.
Clearly state both the required numbers and the maximum product.

Significance for Competitive Entrance Examinations

This is one of the most common optimization patterns. For two numbers with fixed sum $S$, if the numbers are $x$ and $S-x$, then

\[ P(x)=x(S-x). \]

Differentiation gives

\[ P'(x)=S-2x. \]

Hence the maximum occurs at

\[ x=\frac{S}{2}. \]

Thus, for a fixed sum, the product is maximum when the two numbers are equal. For $S=24$,

\[ x=y=\frac{24}{2}=12. \]

This general result is useful for quickly solving optimization questions in JEE Main, JEE Advanced and other entrance examinations.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 6 points

The second number is
\[ 24-x. \]
The product function is
\[ P(x)=24x-x^2. \]
The critical point is
\[ x=12. \]
Since
\[ P''(x)=-2<0, \]
the product is maximum at $x=12$.
The two numbers are
\[ \boxed{12\text{ and }12}. \]
The maximum product is
\[ \boxed{144}. \]

📘 Concept & Theory

Theory and Concept

›

This is an optimization problem with a constraint. Since $x$ and $y$ are positive and their sum is fixed, we can express one variable in terms of the other and then maximize the required expression.

From

\[ x+y=60, \]

we obtain

\[ x=60-y. \]

The expression to be maximized is

\[ xy^3. \]

Therefore, after substituting $x=60-y$, it becomes a function of $y$.

🗺️ Solution Roadmap

Step-by-step Plan

›

Use $x+y=60$ to express $x$ in terms of $y$.
Form the function $f(y)=xy^3$.
Differentiate $f(y)$ and set $f'(y)=0$.
Use the positivity condition $x>0,\ y>0$ to identify the admissible critical point.
Verify that the critical point gives a maximum.
Find the corresponding values of $x$ and $y$.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 25 steps

Use the given constraint
\[x+y=60\]
Therefore,
\[x=60-y\]
Since $x$ and $y$ are positive,
\[x>0,\quad y>0\]
Since $x=60-y>0$,
\[60-y>0\]

\[y < 60\]
Therefore, the relevant domain is
\[\boxed{0 < y < 60}\]
We need to maximize $xy^3$
\[f(y)=xy^3\]
Substituting $x=60-y$,
\[f(y)=(60-y)y^3\]

\[f(y)=60y^3-y^4\]
Find the derivative
\[f'(y)=\frac{d}{dy}(60y^3-y^4)\]

\[f'(y)=180y^2-4y^3\]
Factorising,
\[f'(y)=4y^2(45-y)\]
For a critical point,
\[f'(y)=0\]
Therefore,
\[4y^2(45-y)=0\]
Hence,
\[y^2=0\]
or
\[45-y=0\]
Thus,
\[y=0\]
or
\[y=45\]
However, the problem states that $x$ and $y$ are positive. Therefore,
\[y=0\]
is not admissible
Hence, the only admissible critical point is
\[\boxed{y=45}\]
Verify that $y=45$ gives a maximum
Find the second derivative:
\[f''(y)=\frac{d}{dy}(180y^2-4y^3)\]

\[f''(y)=360y-12y^2\]
At $y=45$,
\[f''(45)=360(45)-12(45)^2\]

\[=16200-12(2025)\]

\[=16200-24300\]

\[=-8100\]
Since
\[ f''(45)=-8100<0, \]
the function has a maximum at
\[\boxed{y=45}\]
Find $x$
From
\[x=60-y,\]
substituting $y=45$,
\[x=60-45\]

\[x=15\]
Therefore, the required two positive numbers are
\[\boxed{x=15,\qquad y=45}\]
Find the maximum value of $xy^3$
Although the question asks for the numbers, we can also calculate the maximum value:
\[xy^3=15(45)^3\]
Since
\[45^3=91125\]
we get
\[xy^3=15(91125)\]

\[\boxed{xy^3=1,366,875}\]

💡 Answer

Final Answer

›

The two positive numbers satisfying $x+y=60$ for which $xy^3$ is maximum are

\[ \boxed{x=15,\qquad y=45}. \]

The maximum value of $xy^3$ is

\[ \boxed{1,366,875}. \]

🎯 Exam Significance

Exam Significance

›

In constrained optimization, first eliminate one variable using the given condition.
Once the function is expressed in one variable, differentiate with respect to that variable.
Respect restrictions such as $x>0$ and $y>0$.
A root such as $y=0$ may arise mathematically but must be rejected if the problem requires positive numbers.
Always justify why the stationary point gives a maximum or minimum.

Significance for Competitive Entrance Examinations

This problem demonstrates a standard constrained-optimization technique. The expression

\[ xy^3 \]

can be viewed as

\[ (60-y)y^3. \]

Differentiation gives

\[ f'(y)=4y^2(45-y). \]

Since $y>0$, the factor $y^2$ is positive, so the sign of $f'(y)$ is determined entirely by $45-y$. This makes the sign analysis particularly quick:

\[ f'(y)>0\quad\text{for }0 < y < 45 \]

\[ f'(y)<0\quad\text{for }45 < y < 60. \]

Hence $y=45$ is immediately identified as the maximum point.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 7 points

From
\[ x+y=60, \]
we get
\[ x=60-y. \]
The quantity to maximize is
\[ f(y)=60y^3-y^4. \]
The derivative is
\[ f'(y)=4y^2(45-y). \]
The admissible critical point is
\[ \boxed{y=45}. \]
Consequently,
\[ \boxed{x=15}. \]
Therefore, the required numbers are
\[ \boxed{15\text{ and }45}. \]
The maximum value of $xy^3$ is
\[ \boxed{1,366,875}. \]

📘 Concept & Theory

Theory and Concept

›

This is a constrained optimization problem. Since the sum of the two positive numbers is fixed, we can express one variable in terms of the other and then maximize the required product.

Given

\[ x+y=35. \]

Therefore,

\[ x=35-y. \]

Since $x$ and $y$ are positive,

\[ x>0,\quad y>0. \]

From $x=35-y>0$, we get

\[ 35-y>0 \]

\[ 0 < y < 35. \]

The quantity to be maximized is

\[ P=x^2y^5. \]

🗺️ Solution Roadmap

Step-by-step Plan

›

Use $x+y=35$ to express $x$ in terms of $y$.
Form $P(y)=x^2y^5$.
Differentiate $P(y)$.
Find the critical points within the admissible domain \(0
Reject any value that does not give positive $x$ and $y$.
Verify that the admissible critical point gives a maximum.
Find the corresponding values of $x$ and $y$.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 24 steps

Express $x$ in terms of $y$
\[x+y=35\]
Therefore,
\[\boxed{x=35-y}\]
Since $x,y>0$,
\[\boxed{0 < y < 35}\]
Form the product function
We have to maximize
\[P=x^2y^5\]
Substituting $x=35-y$,
\[P(y)=(35-y)^2y^5\]
Expanding the square,
\[(35-y)^2=35^2-70y+y^2\]
Therefore,
\[P(y)=(1225-70y+y^2)y^5\]

\[P(y)=1225y^5-70y^6+y^7\]
Differentiate
\[P'(y)=\frac{d}{dy}\left(1225y^5-70y^6+y^7\right)\]

\[P'(y)=6125y^4-420y^5+7y^6\]
Factorising $7y^4$,
\[P'(y)=7y^4\left(875-60y+y^2\right)\]
Rearranging the terms,
\[P'(y)=7y^4\left(y^2-60y+875\right)\]
Find the critical points
For a critical point,
\[P'(y)=0\]
Therefore,
\[7y^4\left(y^2-60y+875\right)=0\]
Hence,
\[y^4=0\]
or
\[y^2-60y+875=0\]
From $y^4=0$,
\[y=0\]
However, $y=0$ is not allowed because $y$ must be positive.
Solve the quadratic equation
\[y^2-60y+875=0\]
We factorise:
\[y^2-25y-35y+875=0\]

\[y(y-25)-35(y-25)=0\]

\[(y-25)(y-35)=0\]
Therefore,
\[y=25\]
or
\[y=35\]
Apply the positivity condition
We established that
\[0 < y < 35\]
Therefore,
\[y=35\]
is not admissible because it would give
\[x=35-35=0\]
whereas the question requires $x>0$
Thus, the only admissible critical point is
\[\boxed{y=25}\]
Find $x$
\[x=35-y\]
Substituting $y=25$,
\[x=35-25=10\]
Therefore, the two positive numbers are
\[\boxed{x=10,\qquad y=25}\]

💡 Answer

Final Answer

›

The two positive numbers are

\[\boxed{x=10,\qquad y=25\]

🎯 Exam Significance

Exam Significance

›

Use the constraint $x+y=35$ to eliminate one variable.
After elimination, differentiate with respect to the remaining variable.
Always apply the restrictions $x>0$ and $y>0$.
A solution of $P'(y)=0$ is not automatically an admissible solution.
Verify that the admissible critical point corresponds to a maximum.

Significance for Competitive Entrance Examinations

There is a useful shortcut for this type of problem. Instead of expanding $P(y)=(35-y)^2y^5$, logarithmic differentiation can be used because $x,y>0$, so $P>0$:

\[ \ln P=2\ln(35-y)+5\ln y. \]

Differentiating,

\[ \frac{P'}{P} = -\frac{2}{35-y}+\frac{5}{y}. \]

For a stationary value,

\[ -\frac{2}{35-y}+\frac{5}{y}=0. \]

\[ \frac{5}{y}=\frac{2}{35-y}. \]

Cross-multiplying,

\[ 5(35-y)=2y. \]

\[ 175-5y=2y. \]

\[ 7y=175. \]

\[ \boxed{y=25}. \]

Hence,

\[ x=35-25=10. \]

This method is especially efficient when the objective function contains high powers such as $x^2y^5$.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 7 points

From
\[ x+y=35, \]
we get
\[ x=35-y. \]
The product function is
\[ P(y)=(35-y)^2y^5. \]
Its derivative factors as
\[ P'(y)=7y^4(y-25)(y-35). \]
The admissible critical point is
\[ \boxed{y=25}. \]
Therefore,
\[ \boxed{x=10}. \]
The required positive numbers are
\[ \boxed{10\text{ and }25}. \]
The maximum value of $x^2y^5$ is
\[ \boxed{976,562,500}. \]

📘 Concept & Theory

Theory and Concept

›

This is a constrained optimization problem. Let the two positive numbers be $x$ and $y$. The given condition is

\[ x+y=16. \]

We need to minimize

\[ x^3+y^3. \]

Using the constraint, we can express one number in terms of the other:

\[ y=16-x. \]

Since both numbers must be positive,

\[ 0 < x < 16. \]

🗺️ Solution Roadmap

Step-by-step Plan

›

Let one number be $x$ and express the other as $16-x$.
Form the function representing the sum of cubes.
Differentiate the function.
Set the derivative equal to zero to find the critical point.
Verify that the critical point gives a minimum.
Find the two numbers and the minimum sum of their cubes.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 16 steps

Let the first number be $x$.
Since their sum is $16$, the second number is
\[y=16-x\]
Because both numbers are positive,
\[x>0\quad 16-x>0\]
Therefore,
\[\boxed{0 < x < 16}\]
The sum of the cubes of the two numbers is
\[P=x^3+y^3\]
Substituting $y=16-x$,
\[P(x)=x^3+(16-x)^3\]
Differentiate
\[P'(x)=\frac{d}{dx}\left[x^3+(16-x)^3\right]\]
Using the chain rule,
\[P'(x)=3x^2+3(16-x)^2(-1)\]

\[P'(x)=3x^2-3(16-x)^2\]
Factorising,
\[P'(x)=3\left[x^2-(16-x)^2\right]\]
Using the difference of squares,
\[a^2-b^2=(a-b)(a+b)\]
we get
\[P'(x)=3\left[x-(16-x)\right]\left[x+(16-x)\right]\]

\[P'(x)=3(2x-16)(16)\]

\[P'(x)=48(2x-16)\]

\[P'(x)=96x-768\]
Find the critical point
For a stationary point,
\[P'(x)=0\]
Therefore,
\[96x-768=0\]

\[96x=768\]

\[x=\frac{768}{96}\]

\[\boxed{x=8}\]
The second number is
\[y=16-x\]

\[y=16-8\]

\[\boxed{y=8}\]
Verify that the sum of cubes is minimum
Find the second derivative:
\[ P''(x)=\frac{d}{dx}(96x-768) \]

\[ P''(x)=96 \]
Since
\[P''(8)=96>0\]
the function has a minimum at $x=8$
Therefore, the corresponding two numbers are
\[\boxed{8\text{ and }8}\]

💡 Answer

Final Answer

›

The two positive numbers whose sum is $16$ and whose sum of cubes is minimum are

\[\boxed{8\text{ and }8}\]

🎯 Exam Significance

Exam Significance

›

Use the fixed-sum condition to eliminate one variable.
Form the quantity that needs to be minimized.
Find the stationary point using $P'(x)=0$.
Use $P''(x)>0$ to confirm a minimum.
Clearly state both the numbers and the minimum value when appropriate.

Significance for Competitive Entrance Examinations

This problem illustrates a general result: for positive numbers with a fixed sum, the sum of their cubes is minimized when the numbers are equal.

More generally, for positive $x,y$ with $x+y=S$,

\[ x^3+y^3 \]

is minimum at

\[ x=y=\frac{S}{2}. \]

Therefore, for $S=16$,

\[ x=y=\frac{16}{2}=8. \]

Recognising this pattern can make such optimization questions significantly faster in JEE Main, JEE Advanced and other entrance examinations.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 6 points

The second number is
\[ 16-x. \]
The objective function is
\[ P(x)=x^3+(16-x)^3. \]
The critical point is
\[ \boxed{x=8}. \]
The second derivative is
\[ P''(x)=96>0, \]
confirming a minimum.
The two numbers are
\[ \boxed{8\text{ and }8}. \]
The minimum sum of cubes is
\[ \boxed{1024}. \]

📘 Concept & Theory

Theory and Concept

›

This is a standard application of derivatives involving optimization of the volume of a box. When a square of side $x$ cm is removed from each corner of the $18$ cm square sheet, folding the remaining flaps produces an open-top box.

If $x$ cm is cut from each corner, then:

The height of the box becomes $x$ cm.
The length becomes $18-2x$ cm.
The breadth becomes $18-2x$ cm.

Therefore, the volume is

\[ V=x(18-2x)^2. \]

Since the dimensions of the box must be positive,

\[ x>0 \]

and

\[ 18-2x>0. \]

Therefore,

\[ 0 < x < 9. \]

🗺️ Solution Roadmap

Step-by-step Plan

›

Let the side of each square cut from the corners be $x$ cm.
Express the length, breadth and height of the resulting box.
Form the volume function $V(x)$.
Differentiate $V(x)$ and solve $V'(x)=0$.
Reject any critical point that is not physically admissible.
Verify that the admissible critical point gives maximum volume.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 21 steps

Let the side of the square cut off be $x$ cm
Let a square of side $x$ cm be cut from each of the four corners.
After cutting and folding:
\[\text{Height}=x\]
Two squares of side $x$ are removed from each side of the original $18$ cm square. Therefore,
\[\text{Length}=18-2x\]
and
\[\text{Breadth}=18-2x\]
Form the volume function
The volume of a rectangular box is
\[V=\text{Length}\times\text{Breadth}\times\text{Height}\]
Therefore,
\[V(x)=(18-2x)(18-2x)x\]

\[V(x)=x(18-2x)^2\]
Expanding,
\[V(x)=x(324-72x+4x^2)\]

\[V(x)=324x-72x^2+4x^3\]
Find the first derivative
\[V'(x)=\frac{d}{dx}(324x-72x^2+4x^3)\]

\[V'(x)=324-144x+12x^2\]
Therefore,
\[V'(x)=12x^2-144x+324\]
Find the critical points
For a maximum or minimum, set
\[V'(x)=0\]
Thus,
\[12x^2-144x+324=0\]
Dividing throughout by $12$,
\[x^2-12x+27=0\]
Factorising,
\[x^2-9x-3x+27=0\]

\[x(x-9)-3(x-9)=0\]

\[(x-3)(x-9)=0\]
Hence,
\[x=3\]
or
\[x=9\]
Apply the physical restriction
We established that
\[0 < x < 9\]
Therefore,
\[x=9\]
is not admissible because it gives
\[18-2(9)=0\]
meaning that the length and breadth of the box would both be zero. Thus, $x=9$ cannot produce a box with positive volume.
The only admissible critical point is
\[\boxed{x=3}\]
Verify that $x=3$ gives maximum volume
Find the second derivative:
\[V''(x)=\frac{d}{dx}(12x^2-144x+324)\]

\[V''(x)=24x-144.\]
At $x=3$,
\[V''(3)=24(3)-144\]

\[=72-144\]

\[=-72\]
Since
\[V''(3)=-72<0,\]
the volume is maximum at
\[\boxed{x=3}\]

💡 Answer

Final Answer

›

The side of each square that should be cut from the four corners is

\[\boxed{3\text{ cm}}\]

🎯 Exam Significance

Exam Significance

›

Translate the geometry of the problem into a mathematical volume function.
Always determine the physical domain before optimizing.
After solving $V'(x)=0$, discard values that violate the dimensions of the box.
Use the second derivative test to confirm maximum volume.
Check units in the final answer: the cut length is measured in centimetres and volume in cubic centimetres.

Significance for Competitive Entrance Examinations

This is a standard geometric optimization problem. The key modelling step is recognizing that cutting $x$ cm from both sides of each dimension leaves

\[ 18-2x. \]

Thus,

\[ V=x(18-2x)^2. \]

An efficient derivative calculation is

\[ V=x(18-2x)^2 \]

\[ V'= (18-2x)^2-4x(18-2x) \]

\[ V'=(18-2x)\left[(18-2x)-4x\right] \]

\[ V'=(18-2x)(18-6x). \]

Hence,

\[ V'=0 \]

gives

\[ x=9\quad\text{or}\quad x=3. \]

The physical restriction \(0

\[ \boxed{x=3}. \]

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 7 points

If $x$ cm is cut from each corner, the box dimensions are
\[ (18-2x)\times(18-2x)\times x. \]
The volume function is
\[ V(x)=x(18-2x)^2. \]
The physically meaningful domain is
\[ \boxed{0 < x < 9}. \]
The critical points are $x=3$ and $x=9$.
$x=9$ is inadmissible because it produces zero length and breadth.
The maximum occurs at
\[ \boxed{x=3\text{ cm}}. \]
The maximum volume is
\[ \boxed{432\text{ cm}^3}. \]

📘 Concept & Theory

Theory and Concept

›

This is a geometric optimization problem. If a square of side $x$ cm is removed from each corner of the rectangular sheet and the remaining flaps are folded upward, the resulting open-top box has:

Length $=45-2x$ cm
Breadth $=24-2x$ cm
Height $=x$ cm

Hence, the volume of the box is

\[ V(x)=x(45-2x)(24-2x). \]

For a physically meaningful box, all dimensions must be positive. Therefore,

\[ x>0, \]

\[ 45-2x>0, \]

\[ 24-2x>0. \]

The most restrictive condition is

\[ 24-2x>0 \]

\[ x < 12. \]

Thus, the physical domain is

\[ \boxed{0 < x < 12}. \]

🗺️ Solution Roadmap

Step-by-step Plan

›

Let the side of each corner square be $x$ cm.
Determine the dimensions of the resulting box.
Form the volume function.
Differentiate and solve $V'(x)=0$.
Retain only the critical points lying in the physical domain \(0
Verify that the admissible critical point gives maximum volume.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 19 steps

Let the side of each square removed from the four corners be $x$ cm.
After folding:
\[\text{Length}=45-2x\]

\[\text{Breadth}=24-2x\]

\[\text{Height}=x\]
Determine the allowable range of $x$
Since the breadth must remain positive,
\[24-2x>0\]

\[2x < 24\]

\[x < 12\]
Since $x > 0$,
\[\boxed{0 < x < 12}\]
Form the volume function
\[V(x)=(45-2x)(24-2x)x\]
First expand the two-dimensional factor:
\[(45-2x)(24-2x)\]

\[=45(24)-45(2x)-2x(24)+4x^2\]

\[=1080-90x-48x+4x^2\]

\[=1080-138x+4x^2\]
Therefore,
\[V(x)=(1080-138x+4x^2)x\]

\[V(x)=1080x-138x^2+4x^3\]
Find the first derivative
\[V'(x)=\frac{d}{dx}(1080x-138x^2+4x^3)\]

\[V'(x)=1080-276x+12x^2\]
Therefore,
\[V'(x)=12x^2-276x+1080\]
Find the critical points
For a stationary point,
\[V'(x)=0\]
Thus,
\[12x^2-276x+1080=0\]
Dividing throughout by $12$,
\[x^2-23x+90=0\]
Factorising,
\[x^2-18x-5x+90=0\]

\[x(x-18)-5(x-18)=0\]

\[(x-5)(x-18)=0\]
Hence,
\[x=5\]
or
\[x=18.\]
Apply the physical restriction
We established that
\[ 0 < x < 12\]
Therefore, $x=18$ is not physically admissible because
\[24-2(18)=-12\]
which would give a negative breadth. Thus, $x=18$ must be rejected.
The only admissible critical point is
\[\boxed{x=5}\]
Verify that $x=5$ gives maximum volume<
Find the second derivative:
\[V''(x)=\frac{d}{dx}(12x^2-276x+1080)\]

\[V''(x)=24x-276.\]
At $x=5$,
\[V''(5)=24(5)-276\]

\[=120-276\]

\[=-156.\]
Since
\[V''(5)=-156<0,\]
the volume is maximum at
\[\boxed{x=5}\]

💡 Answer

Final Answer

›

The side of each square that should be cut from the four corners is

\[\boxed{5\text{ cm}}\]

🎯 Exam Significance

Exam Significance

›

Translate the geometry into length, breadth and height before differentiating.
Determine the physical domain of the variable before selecting a critical point.
For this problem, $0 < x < 12$, not merely $x>0$.
Check the second derivative to establish that the admissible critical point gives a maximum.
Include units in the final answer: $5$ cm for the cut and $2450\text{ cm}^3$ for the volume.

Significance for Competitive Entrance Examinations

A useful way to differentiate the volume without fully expanding it is to use the product rule:

\[ V=x(45-2x)(24-2x). \]

Let

\[ g(x)=(45-2x)(24-2x). \]

Then

\[ V'=g(x)+xg'(x). \]

Since

\[ g(x)=1080-138x+4x^2 \]

and

\[ g'(x)=-138+8x, \]

we obtain

\[ V'=1080-138x+4x^2+x(-138+8x) \]

\[ =1080-276x+12x^2. \]

This leads to

\[ 12x^2-276x+1080=0 \]

\[ (x-5)(x-18)=0. \]

The domain restriction \(0

\[ \boxed{x=5}. \]

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 7 points

The box dimensions are
\[ (45-2x)\times(24-2x)\times x. \]
The volume is
\[ V(x)=x(45-2x)(24-2x). \]
The physical domain is
\[ \boxed{0 < x < 12}. \]
The critical points are $x=5$ and $x=18$.
$x=18$ is rejected because it produces an invalid negative breadth.
The maximum occurs at
\[ \boxed{x=5\text{ cm}}. \]
The maximum volume is
\[ \boxed{2450\text{ cm}^3}. \]

📘 Concept & Theory

Theory and Concept

›

This is an optimization problem in which the circle is fixed, so its radius remains constant. We need to determine which inscribed rectangle has the greatest possible area.

Let the radius of the given circle be $r$, and let the length and breadth of an inscribed rectangle be $x$ and $y$, respectively.

Since the rectangle is inscribed in the circle, its diagonal is equal to the diameter of the circle. Therefore,

\[ x^2+y^2=(2r)^2=4r^2. \]

The area of the rectangle is

\[ A=xy. \]

Thus, the problem reduces to maximizing $xy$, subject to the constraint

\[ x^2+y^2=4r^2. \]

🗺️ Solution Roadmap

Step-by-step Plan

›

Let the sides of the rectangle be $x$ and $y$.
Use the fact that the rectangle's diagonal is the diameter of the circle.
Express one side in terms of the other.
Form the area function $A(x)$.
Differentiate and find the critical point.
Use the second derivative to verify that the critical point gives maximum area.
Show that at the maximum, the two sides are equal, so the rectangle is a square.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 28 steps

Let the dimensions of the rectangle be $x$ and $y$
Let $ABCD$ be a rectangle inscribed in a circle of radius $r$. Let
\[AB=CD=x\]

\[BC=AD=y\]
Since the rectangle is inscribed in the circle, each diagonal of the rectangle passes through the centre of the circle. Hence, the diagonal of the rectangle is equal to the diameter of the circle.
Therefore,
\[AC=2r\]
Obtain the relation between $x$ and $y$
In right triangle $ABC$, by Pythagoras' theorem,
\[AC^2=AB^2+BC^2\]
Hence,
\[(2r)^2=x^2+y^2\]
Therefore,
\[\boxed{x^2+y^2=4r^2}\]
Express $y$ in terms of $x$
\[y^2=4r^2-x^2.\]
Since $y>0$,
\[y=\sqrt{4r^2-x^2}.\]
Also, because both sides of the rectangle are positive,
\[0 < x < 2r\]
Form the area function
The area of the rectangle is
\[A=xy\]
Substituting the value of $y$,
\[A(x)=x\sqrt{4r^2-x^2}\]
Differentiate the area
Using the product rule,
\[A'(x)=\frac{d}{dx}\left[x(4r^2-x^2)^{1/2}\right]\]

\[A'(x)=(4r^2-x^2)^{1/2}+x\left(\frac{1}{2}\right)(4r^2-x^2)^{-1/2}(-2x)\]
Therefore,
\[A'(x)=\sqrt{4r^2-x^2}-\frac{x^2}{\sqrt{4r^2-x^2}}\]
Taking the common denominator,
\[A'(x)=\frac{4r^2-x^2-x^2}{\sqrt{4r^2-x^2}}\]
Hence,
\[A'(x)=\frac{4r^2-2x^2}{\sqrt{4r^2-x^2}}\]
Find the critical point
For a stationary point,
\[A'(x)=0\]
Therefore,
\[\frac{4r^2-2x^2}{\sqrt{4r^2-x^2}}=0\]
Since the denominator is positive for $0 < x < 2r$, we require
\[4r^2-2x^2=0.\]
Thus,
\[2x^2=4r^2\]

\[x^2=2r^2\]

\[x=\sqrt{2}\,r.\]
Find the other side
Using
\[x^2+y^2=4r^2,\]
and $x^2=2r^2$
we get
\[2r^2+y^2=4r^2\]
Therefore,
\[y^2=2r^2\]

\[y=\sqrt{2}\,r\]
Hence,
\[\boxed{x=y=\sqrt{2}\,r}\]
Thus, at the critical point, the two sides of the rectangle are equal. Therefore, the rectangle is a square.
Verify that the area is maximum
We have
\[A'(x)=\frac{4r^2-2x^2}{\sqrt{4r^2-x^2}}\]
For $0 < x < 2r$, the denominator is positive. Therefore, the sign of $A'(x)$ is determined by $4r^2-2x^2$.
If
\[0 < x < \sqrt{2}r,\]
then
\[4r^2-2x^2>0,\]
so
\[A'(x)>0.\]
Thus, the area is increasing.
If
\[\sqrt{2}r < x < 2r,\]
then
\[4r^2-2x^2 < 0,\]
so
\[A'(x) < 0.\]
Thus, the area is decreasing.
Therefore, the area changes from increasing to decreasing at
\[ x=\sqrt{2}r. \]
Hence, the area is maximum at this point.
Find the maximum area
At the maximum,
\[x=y=\sqrt{2}r\]
Therefore,
\[A_{\max}=xy\]

\[A_{\max}=(\sqrt{2}r)(\sqrt{2}r)\]

\[\boxed{A_{\max}=2r^2}.\]
Since the rectangle has equal length and breadth, it is a square.

🎯 Exam Significance

Exam Significance

›

Always identify the geometric constraint before forming the function to be optimized.
For an inscribed rectangle, the diagonal equals the diameter of the circle.
Use Pythagoras' theorem to obtain $x^2+y^2=4r^2$.
Do not stop after finding $A'(x)=0$; verify that the stationary point gives a maximum.
The conclusion $x=y$ directly establishes that the maximum-area rectangle is a square.

Significance for Competitive Entrance Examinations

This result can also be understood through the identity

\[ (x-y)^2\geq0. \]

Therefore,

\[ x^2+y^2\geq2xy. \]

Since

\[ x^2+y^2=4r^2, \]

we obtain

\[ 4r^2\geq2xy \]

\[ xy\leq2r^2. \]

Equality holds precisely when

\[ x=y. \]

Hence,

\[ \boxed{A=xy\leq2r^2}, \]

with equality when $x=y=\sqrt{2}r$. Thus, the square gives the maximum possible area.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 7 points

The diagonal of an inscribed rectangle is the diameter of the circle.
The fundamental constraint is $x^2+y^2=4r^2$.
The area is $A=xy$.
Differentiation gives the critical condition $x^2=2r^2$.
This leads to $x=y=\sqrt{2}r$.
Equal sides mean that the rectangle is a square.
The maximum area is $\boxed{2r^2}$.

📘 Concept & Theory

Theory and Concept

›

This is a standard optimization problem. We have to maximize the volume of a right circular cylinder while its total surface area is fixed.

Let the radius of the base of the cylinder be $r$ and its height be $h$.

The total surface area of a closed right circular cylinder is

\[ S=2\pi r^2+2\pi rh. \]

Since the surface area is given and fixed, $S$ is a constant.

The volume of the cylinder is

\[ V=\pi r^2h. \]

Our objective is to find the relationship between $h$ and $r$ for which $V$ is maximum.

🗺️ Solution Roadmap

Step-by-step Plan

›

Let $r$ be the radius and $h$ be the height of the cylinder.
Use the fixed total surface area to express $h$ in terms of $r$.
Form the volume function in terms of $r$ alone.
Differentiate the volume and find the critical point.
Use the second derivative to verify that the volume is maximum.
Show that the resulting relation is $h=2r$, which means that the height equals the diameter of the base.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 26 steps

Let the radius of the circular base be $r$ and the height of the cylinder be $h$
Let the fixed total surface area be $S$
Therefore,
\[2\pi r^2+2\pi rh=S\]
Express the height in terms of the radius
From
\[2\pi r^2+2\pi rh=S\]
we get
\[2\pi rh=S-2\pi r^2\]
Dividing by $2\pi r$
\[h=\frac{S-2\pi r^2}{2\pi r}\]
Therefore,
\[h=\frac{S}{2\pi r}-r\]
Form the volume function
The volume of the cylinder is
\[V=\pi r^2h\]
Substituting
\[h=\frac{S}{2\pi r}-r\]
we obtain
\[V=\pi r^2\left(\frac{S}{2\pi r}-r\right)\]
Simplifying,
\[V=\frac{Sr}{2}-\pi r^3\]
Thus,
\[\boxed{V(r)=\frac{S}{2}r-\pi r^3}\]
Differentiate the volume
\[\frac{dV}{dr}=\frac{S}{2}-3\pi r^2\]
For a stationary value of the volume,
\[\frac{dV}{dr}=0\]
Hence,
\[\frac{S}{2}-3\pi r^2=0\]
Therefore,
\[\frac{S}{2}=3\pi r^2\]

\[S=6\pi r^2\]
Find the corresponding height
From the fixed surface-area equation,
\[S=2\pi r^2+2\pi rh\]
We have found that at the stationary point,
\[S=6\pi r^2\]
Therefore,
\[6\pi r^2=2\pi r^2+2\pi rh\]
Subtracting $2\pi r^2$ from both sides,
\[4\pi r^2=2\pi rh\]
Dividing by $2\pi r$,
\[h=2r\]
But the diameter of the base is
\[d=2r\]
Hence,
\[\boxed{h=d}\]
Thus, the height of the cylinder is equal to the diameter of its base.
Verify that the volume is maximum
We have
\[V(r)=\frac{S}{2}r-\pi r^3\]
Differentiating again,
\[\frac{d^2V}{dr^2}=-6\pi r\]
Since $r>0$,
\[-6\pi r < 0\]
Therefore,
\[\boxed{\frac{d^2V}{dr^2} < 0}\]
Hence, the stationary value of $V$ is a maximum.

💡 Answer

Final Answer

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\[ \boxed{\text{Height}=\text{Diameter of the base}}. \]

🎯 Exam Significance

Exam Significance

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Clearly identify whether the given surface means total surface area or curved surface area.
For a closed cylinder, use $S=2\pi r^2+2\pi rh$.
Express one variable in terms of the other before differentiating.
After finding a stationary point, use the second derivative to establish maximum volume.
State the final geometric relationship explicitly: $h=2r$.

Significance for Competitive Entrance Examinations

This problem illustrates an important optimization principle: a geometric quantity can be maximized by converting a constraint into a one-variable function and then applying differential calculus.

Here,

\[ V=\frac{S}{2}r-\pi r^3. \]

The competition-level shortcut is to recognize that

\[ V'(r)=\frac{S}{2}-3\pi r^2=0 \]

gives

\[ S=6\pi r^2. \]

Substituting this immediately into the surface-area constraint yields

\[ 6\pi r^2=2\pi r^2+2\pi rh \]

\[ \boxed{h=2r}. \]

The second derivative

\[ V''(r)=-6\pi r<0 \]

confirms that this stationary value corresponds to maximum volume. This type of constrained optimization is frequently useful in calculus-based questions involving cylinders, cones, spheres and cuboids.

🔑 Key Takeaways

Key Takeaways

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Key Takeaways · 6 points

Fixed total surface area:
\[ S=2\pi r^2+2\pi rh. \]
Volume:
\[ V=\pi r^2h. \]
After eliminating $h$:
\[ V=\frac{S}{2}r-\pi r^3. \]
The critical condition is:
\[ \frac{S}{2}-3\pi r^2=0. \]
The maximum occurs when:
\[ \boxed{h=2r}. \]
Since $2r$ is the diameter of the base:
\[ \boxed{\text{Height}=\text{Diameter of the base}}. \]

📘 Concept & Theory

Theory and Concept

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Let the radius of the cylindrical can be $r$ cm and its height be $h$ cm.

Since the can is closed, its total surface area is

\[S=2\pi r^2+2\pi rh\]

The volume is fixed at $100\text{ cm}^3$. Therefore,

\[\pi r^2h=100\]

We need to use this volume constraint to express $h$ in terms of $r$, substitute it into the surface-area formula, and then optimize the resulting one-variable function.

🗺️ Solution Roadmap

Step-by-step Plan

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Let the radius and height of the cylinder be $r$ and $h$.
Use the fixed-volume condition to express $h$ in terms of $r$.
Form the surface-area function $S(r)$.
Differentiate $S(r)$ and find the critical point.
Use the second derivative to verify that the critical point gives minimum surface area.
Find the corresponding height.
Explain why the literal request for maximum surface area has no finite solution.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 38 steps

Let the radius of the base be $r$ cm and the height be $h$ cm.
The given volume is
\[V=100\text{ cm}^3\]
For a right circular cylinder,
\[V=\pi r^2h\]
Therefore,
\[\pi r^2h=100\]
Express $h$ in terms of $r$
\[\pi r^2h=100\]
Hence,
\[h=\frac{100}{\pi r^2}\]
Form the surface-area function
Since the can is closed, its total surface area is
\[S=2\pi r^2+2\pi rh\]
Substituting
\[h=\frac{100}{\pi r^2}\]
we get
\[S=2\pi r^2+2\pi r\left(\frac{100}{\pi r^2}\right)\]
Simplifying,
\[S=2\pi r^2+\frac{200}{r}\]
Thus,
\[\boxed{S(r)=2\pi r^2+\frac{200}{r}}\quad r>0\]
Differentiate the surface area
\[S'(r)=\frac{d}{dr}\left(2\pi r^2+\frac{200}{r}\right)\]
Therefore,
\[S'(r)=4\pi r-\frac{200}{r^2}\]
For a stationary point,
\[S'(r)=0\]
Hence,
\[4\pi r-\frac{200}{r^2}=0\]
Multiplying by $r^2$,
\[4\pi r^3-200=0\]
Therefore,
\[4\pi r^3=200\]

\[\pi r^3=50\]
Hence,
\[r^3=\frac{50}{\pi}\]
Therefore,
\[\boxed{r=\left(\frac{50}{\pi}\right)^{1/3}\text{ cm}}\]
Verify that the surface area is minimum
Differentiate $S'(r)$ again:
\[S''(r)=4\pi+\frac{400}{r^3}\]
Since $r>0$,
\[4\pi+\frac{400}{r^3}>0\]
Thus,
\[\boxed{S''(r)>0}\]
Hence, the stationary value of the surface area is a minimum
Find the corresponding height
We have
\[h=\frac{100}{\pi r^2}\]
From the critical-point condition,
\[4\pi r^3=200\]
Dividing by $2\pi r^2$,
\[2r=\frac{100}{\pi r^2}\]
But
\[\frac{100}{\pi r^2}=h\]
Therefore,
\[\boxed{h=2r}\]
Hence, the height is equal to the diameter of the base.
Substituting the value of $r$,
\[h=2\left(\frac{50}{\pi}\right)^{1/3}\text{ cm}\]
Thus,
\[\boxed{r=\left(\frac{50}{\pi}\right)^{1/3}\text{ cm},\quad h=2\left(\frac{50}{\pi}\right)^{1/3}\text{ cm}}\]
Approximate dimensions
Since
\[\left(\frac{50}{\pi}\right)^{1/3}\approx2.515,\]
we obtain approximately
\[r\approx2.515\text{ cm}\]
and
\[h\approx5.030\text{ cm}\]
Therefore, the approximate dimensions are
\[\boxed{r\approx2.515\text{ cm},\qquad h\approx5.030\text{ cm}}\]
Why the Literal "Maximum Surface Area" Does Not Exist
The question as written asks for the maximum surface area. However, from
\[S(r)=2\pi r^2+\frac{200}{r}\]
consider what happens as $r$ becomes arbitrarily large:
\[r\to\infty\]
Then
\[2\pi r^2\to\infty\]
and
\[\frac{200}{r}\to0\]
Consequently,
\[S(r)\to\infty\]
Therefore, there is no finite maximum surface area for a cylinder whose volume is fixed at $100\text{ cm}^3$.
The surface area has a finite minimum, and that minimum occurs when
\[\boxed{h=2r}\]

🎯 Exam Significance

Exam Significance

›

Read optimization questions carefully and identify whether the quantity is to be maximized or minimized.
For a closed cylinder, total surface area is $2\pi r^2+2\pi rh$.
For fixed volume $100\text{ cm}^3$, use $\pi r^2h=100$ as the constraint.
After finding the critical point, always use the second derivative to determine whether it represents a maximum or minimum.
If the textbook states "maximum surface area," mention the mathematical inconsistency rather than forcing a false maximum.

Significance for Competitive Entrance Examinations

The essential one-variable optimization function is

\[ S(r)=2\pi r^2+\frac{200}{r}. \]

Differentiating,

\[ S'(r)=4\pi r-\frac{200}{r^2}. \]

Setting $S'(r)=0$ gives

\[ 4\pi r^3=200. \]

Therefore,

\[ \pi r^3=50. \]

The fixed-volume condition then immediately gives

\[ h=2r. \]

Also,

\[ S''(r)=4\pi+\frac{400}{r^3}>0, \]

confirming that the stationary value is a minimum.

🔑 Key Takeaways

Key Takeaways

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Key Takeaways · 7 points

For a closed cylinder:
\[ S=2\pi r^2+2\pi rh. \]
Given volume:
\[ \pi r^2h=100. \]
Therefore:
\[ S(r)=2\pi r^2+\frac{200}{r}. \]
The critical radius satisfies:
\[ r^3=\frac{50}{\pi}. \]
The corresponding height is:
\[ h=2r. \]
The finite optimum is a minimum surface area, not a maximum.
For the question literally asking for maximum surface area, no finite maximum exists.

📘 Concept & Theory

Theory and Concept

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This is a constrained optimization problem. The total length of the wire is fixed at $28$ m, but it must be divided between a square and a circle. We need to determine the division that gives the minimum combined area.

Let $x$ m of wire be used to form the square. Then the remaining $28-x$ m is used to form the circle.

For the square, if its side is $a$, then

\[ 4a=x \]

\[ a=\frac{x}{4}. \]

Therefore, its area is

\[ A_s=\left(\frac{x}{4}\right)^2=\frac{x^2}{16}. \]

For the circle, if its radius is $r$, then its circumference is $28-x$:

\[ 2\pi r=28-x. \]

Hence,

\[ r=\frac{28-x}{2\pi}. \]

Its area is therefore

\[ A_c=\pi r^2 =\pi\left(\frac{28-x}{2\pi}\right)^2 =\frac{(28-x)^2}{4\pi}. \]

Thus, the total area can be expressed as a function of $x$, and calculus can be used to minimize it.

🗺️ Solution Roadmap

Step-by-step Plan

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Let $x$ m be used for the square.
Then $28-x$ m is used for the circle.
Express the area of the square in terms of $x$.
Express the area of the circle in terms of $x$.
Form the total area function.
Differentiate and find the critical point.
Use the second derivative to verify that the critical point gives minimum area.
Find the lengths of the two wire pieces.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 36 steps

Let $x$ m of wire be used to make the square.
Therefore, the remaining wire used to make the circle is
\[28-x\text{ m}\]
Since both figures require a positive length,
\[\boxed{0 < x < 28}\]
Find the area of the square
Let the side of the square be $a$ m.
Since the perimeter of the square is $x$ m,
\[4a=x\]
Therefore,
\[a=\frac{x}{4}\]
Hence, the area of the square is
\[A_s=a^2\]

\[A_s=\left(\frac{x}{4}\right)^2\]

\[\boxed{A_s=\frac{x^2}{16}}\]
Find the area of the circle
Let the radius of the circle be $r$ m.
The circumference of the circle is the length of the second piece of wire:
\[2\pi r=28-x\]
Therefore,
\[r=\frac{28-x}{2\pi}\]
The area of the circle is
\[A_c=\pi r^2\]
Substituting the value of $r$,
\[A_c=\pi\left(\frac{28-x}{2\pi}\right)^2\]

\[A_c=\frac{\pi(28-x)^2}{4\pi^2}\]
Hence,
\[\boxed{A_c=\frac{(28-x)^2}{4\pi}}\]
Form the total area function
The combined area is
\[A(x)=A_s+A_c\]
Therefore,
\[A(x)=\frac{x^2}{16}+\frac{(28-x)^2}{4\pi}\]
Thus,
\[\boxed{A(x)=\frac{x^2}{16}+\frac{(28-x)^2}{4\pi}}\]
Differentiate the total area
\[A'(x)=\frac{2x}{16}+\frac{2(28-x)(-1)}{4\pi}\]
Therefore,
\[A'(x)=\frac{x}{8}-\frac{28-x}{2\pi}\]
For a stationary point,
\[A'(x)=0\]
Hence,
\[\frac{x}{8}-\frac{28-x}{2\pi}=0\]
Therefore,
\[\frac{x}{8}=\frac{28-x}{2\pi}\]
Cross-multiplying,
\[2\pi x=8(28-x)\]

\[2\pi x=224-8x\]
Therefore,
\[2\pi x+8x=224\]

\[x(2\pi+8)=224\]
Hence,
\[x=\frac{224}{2\pi+8}\]
Dividing numerator and denominator by $2$,
\[\boxed{x=\frac{112}{\pi+4}}\]
Find the length of the second piece
The length used for the circle is
\[28-x\]
Substituting $x=\frac{112}{\pi+4}$,
\[28-x=28-\frac{112}{\pi+4}\]
Taking the common denominator,
\[28-x=\frac{28(\pi+4)-112}{\pi+4}\]

\[28-x=\frac{28\pi+112-112}{\pi+4}\]
Therefore,
\[\boxed{28-x=\frac{28\pi}{\pi+4}}\]
Verify that the combined area is minimum
Differentiate $A'(x)$ again:
\[A''(x)=\frac{1}{8}+\frac{1}{2\pi}\]
Since $\pi>0$,
\[A''(x)>0\]
Therefore, the stationary value gives a minimum of the combined area.
\[\boxed{A''(x)>0\Rightarrow A(x)\text{ is minimum}}.\]
Determine the two wire lengths
Therefore, the length of wire used for the square is
\[\boxed{\frac{112}{\pi+4}\text{ m}}\]
The length of wire used for the circle is
\[\boxed{\frac{28\pi}{\pi+4}\text{ m}}\]
Using $\pi\approx3.14159$,
\[\frac{112}{\pi+4}\approx15.68\text{ m},\]
and
\[\frac{28\pi}{\pi+4}\approx12.32\text{ m}\]
Hence, approximately $15.68$ m of wire should be used for the square and $12.32$ m for the circle.

🎯 Exam Significance

Exam Significance

›

Convert the wire lengths into geometric constraints before differentiating.
For a square, perimeter $=4a$.
For a circle, circumference $=2\pi r$.
Express the combined area as a function of one variable.
Use the second derivative to confirm that the critical point gives a minimum.
Clearly state which piece of wire is used for the square and which is used for the circle.

Significance for Competitive Entrance Examinations

The optimization function is

\[ A(x)=\frac{x^2}{16}+\frac{(28-x)^2}{4\pi}. \]

The critical-point equation can be written directly as

\[\frac{x}{8}=\frac{28-x}{2\pi}\]

This gives

\[ 2\pi x=8(28-x), \]

and hence

\[\boxed{x=\frac{112}{\pi+4}}\]

A useful structural observation is that at the minimum,

\[\boxed{a=2r},\]

where $a$ is the side of the square and $r$ is the radius of the circle. This relation provides a useful check on the final result.

🔑 Key Takeaways

Key Takeaways

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Key Takeaways · 8 points

Total wire length:
\[ x+(28-x)=28. \]
Square area:
\[ A_s=\frac{x^2}{16}. \]
Circle area:
\[ A_c=\frac{(28-x)^2}{4\pi}. \]
Total area:
\[ A(x)=\frac{x^2}{16}+\frac{(28-x)^2}{4\pi}. \]
Minimum occurs at:
\[ \boxed{x=\frac{112}{\pi+4}}. \]
Wire for the circle:
\[ \boxed{\frac{28\pi}{\pi+4}\text{ m}}. \]
Approximately $15.68$ m is used for the square and $12.32$ m for the circle.
At the optimum, the square side and circle radius satisfy:
\[ \boxed{a=2r}. \]

📘 Concept & Theory

Theory and Concept

›

This is a three-dimensional optimization problem. We have a sphere of fixed radius $R$, and we need to determine the right circular cone inscribed in it whose volume is maximum.

Let the radius of the base of the cone be $r$ and its height be $h$. The centre of the sphere need not coincide with the centre of the cone's base. Therefore, we first establish the geometric relation between $r$, $h$, and $R$.

If the cone is inscribed in a sphere of radius $R$, the centre of the sphere lies on the axis of the cone. From the right triangle formed by the sphere's centre, the vertex of the cone, and a point on the circular base, we obtain

\[ r^2+(R-h)^2=R^2. \]

Hence,

\[ r^2=2Rh-h^2. \]

The volume of the cone is

\[ V=\frac{1}{3}\pi r^2h. \]

Substituting the relation between $r$ and $h$, the volume becomes a function of $h$ alone. We can then use differentiation to find its maximum value.

🗺️ Solution Roadmap

Step-by-step Plan

›

Let the sphere have radius $R$, and let the cone have height $h$ and base radius $r$.
Use the geometry of the inscribed cone to obtain $r^2=2Rh-h^2$.
Express the volume of the cone as a function of $h$.
Differentiate and find the critical point.
Use the second derivative to prove that the critical point gives maximum volume.
Calculate the maximum cone volume.
Compare it with the volume of the sphere and prove the required ratio.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 41 steps

Define the dimensions
Let a right circular cone be inscribed in a sphere of radius $R$. Let
\[h=\text{height of the cone}\]

\[r=\text{radius of the base of the cone}.\]
Since the cone is inscribed in the sphere, its vertex lies on the sphere and its circular base is also contained in the sphere.
Establish the relation between $r$, $h$, and $R$
Let $O$ be the centre of the sphere and $A$ be the vertex of the cone. Let $C$ be the centre of the circular base and $P$ be any point on the circumference of the base.
The axis of the cone passes through the centre $O$ of the sphere. Since $OA=R$ and $AC=h$, the distance from $O$ to the plane of the base is
\[OC=R-h\]
In right triangle $OCP$,
\[OP^2=OC^2+CP^2\]
Since $OP=R$ and $CP=r$,
\[R^2=(R-h)^2+r^2\]
Therefore,
\[r^2=R^2-(R-h)^2\]
Expanding,
\[r^2=R^2-\left(R^2-2Rh+h^2\right)\]
Hence,
\[\boxed{r^2=2Rh-h^2}\]
Form the volume function
The volume of a right circular cone is
\[V=\frac{1}{3}\pi r^2h\]
Substituting
\[r^2=2Rh-h^2\]
we get
\[V=\frac{1}{3}\pi(2Rh-h^2)h\]
Therefore,
\[V=\frac{\pi}{3}(2Rh^2-h^3)\]
Thus,
\[\boxed{V(h)=\frac{\pi}{3}(2Rh^2-h^3)}\]
Find the critical point
Differentiate with respect to $h$:
\[\frac{dV}{dh}=\frac{\pi}{3}(4Rh-3h^2)\]
Factorising,
\[\frac{dV}{dh}=\frac{\pi h}{3}(4R-3h)\]
For a stationary value,
\[\frac{dV}{dh}=0\]
Therefore,
\[\frac{\pi h}{3}(4R-3h)=0\]
Hence,
\[h=0\]
or
\[4R-3h=0\]
Thus,
\[h=\frac{4R}{3}\]
The value $h=0$ does not represent a cone of positive volume. Therefore, the relevant critical point is
\[\boxed{h=\frac{4R}{3}}\]
Verify that the volume is maximum
Differentiate again:
\[\frac{d^2V}{dh^2}=\frac{\pi}{3}(4R-6h)\]
At
\[h=\frac{4R}{3},\]
we get
\[\frac{d^2V}{dh^2}=\frac{\pi}{3}\left(4R-6\cdot\frac{4R}{3}\right)\]

\[=\frac{\pi}{3}(4R-8R)\]

\[=-\frac{4\pi R}{3}\]
Since $R>0$,
\[\frac{d^2V}{dh^2} < 0\]
Hence, the volume is maximum when
\[\boxed{h=\frac{4R}{3}}\]
Find the corresponding base radius
We have
\[r^2=2Rh-h^2\]
Substituting
\[h=\frac{4R}{3}\]
we get
\[r^2=2R\left(\frac{4R}{3}\right)-\left(\frac{4R}{3}\right)^2\]

\[r^2=\frac{8R^2}{3}-\frac{16R^2}{9}\]
Taking the common denominator,
\[r^2=\frac{24R^2-16R^2}{9}\]
Therefore,
\[r^2=\frac{8R^2}{9}\]
Hence,
\[\boxed{r=\frac{2\sqrt{2}}{3}R}\]
Calculate the maximum volume of the cone
The volume is
\[V_{\max}=\frac{1}{3}\pi r^2h\]
Substituting
\[r^2=\frac{8R^2}{9}\]
and
\[h=\frac{4R}{3}\]
we get
\[V_{\max}=\frac{1}{3}\pi\left(\frac{8R^2}{9}\right)\left(\frac{4R}{3}\right)\]
Therefore,
\[V_{\max}=\frac{32\pi R^3}{81}\]
Thus,
\[ \boxed{V_{\max}=\frac{32\pi R^3}{81}}. \]
Find the volume of the sphere
The volume of a sphere of radius $R$ is
\[V_{\text{sphere}}=\frac{4}{3}\pi R^3\]
Compare the two volumes
Therefore,
\[\frac{V_{\max}}{V_{\text{sphere}}}=\frac{\frac{32\pi R^3}{81}}{\frac{4\pi R^3}{3}}\]
Simplifying,
\[\frac{V_{\max}}{V_{\text{sphere}}}=\frac{32}{81}\times\frac{3}{4}\]

\[=\frac{96}{324}\]
Therefore,
\[\boxed{\frac{V_{\max}}{V_{\text{sphere}}}=\frac{8}{27}}\]
Hence,
\[\boxed{V_{\max}=\frac{8}{27}V_{\text{sphere}}}\]

🎯 Exam Significance

Exam Significance

›

First establish the geometric constraint before differentiating.
For the inscribed cone, use
\[ r^2+(R-h)^2=R^2. \]
Hence,
\[ r^2=2Rh-h^2. \]
Express the cone volume as a function of one variable.
Use the second derivative to verify the maximum.
Finally compare the maximum cone volume with the sphere's volume to obtain the required ratio.

Significance for Competitive Entrance Examinations

The key optimization function is

\[ V(h)=\frac{\pi}{3}(2Rh^2-h^3). \]

Its derivative factors conveniently:

\[ V'(h)=\frac{\pi h}{3}(4R-3h). \]

Therefore, the non-trivial stationary point is immediately

\[ \boxed{h=\frac{4R}{3}}. \]

A useful algebraic check is that the corresponding radius satisfies

\[ r^2=2R\left(\frac{4R}{3}\right) - \left(\frac{4R}{3}\right)^2 = \frac{8R^2}{9}. \]

Thus,

\[ r=\frac{2\sqrt{2}}{3}R. \]

Substitution directly gives

\[ V_{\max}=\frac{32\pi R^3}{81}, \]

and comparison with

\[ \frac{4}{3}\pi R^3 \]

yields the required ratio $\frac{8}{27}$.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 7 points

For an inscribed cone:
\[ r^2=2Rh-h^2. \]
The volume function is:
\[ V(h)=\frac{\pi}{3}(2Rh^2-h^3). \]
The maximum occurs at:
\[ \boxed{h=\frac{4R}{3}}. \]
The corresponding base radius is:
\[ \boxed{r=\frac{2\sqrt{2}}{3}R}. \]
The maximum cone volume is:
\[ \boxed{\frac{32\pi R^3}{81}}. \]
The volume of the sphere is:
\[ \frac{4}{3}\pi R^3. \]
Therefore:
\[ \boxed{V_{\max}=\frac{8}{27}V_{\text{sphere}}}. \]

📘 Concept & Theory

Theory and Concept

›

This is a constrained optimization problem. The volume of the cone is fixed, while its curved surface area is to be minimized.

Let the radius of the base of the cone be $r$, its altitude be $h$, and its slant height be $l$.

For a right circular cone,

\[ l^2=r^2+h^2. \]

Its curved surface area is

\[ S=\pi rl. \]

Its volume is

\[ V=\frac{1}{3}\pi r^2h. \]

Since the volume is given and fixed, we can express $h$ in terms of $r$, substitute this relation into the curved surface area, and minimize the resulting function.

🗺️ Solution Roadmap

Step-by-step Plan

›

Let $r$ be the base radius and $h$ the altitude of the cone.
Use the fixed-volume condition to express $h$ in terms of $r$.
Use $l^2=r^2+h^2$ to express the slant height in terms of $r$.
Form the curved surface-area function.
Differentiate and find the stationary point.
Verify that the stationary point gives minimum curved surface area.
Show that the resulting relation is $h=\sqrt{2}r$.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 31 steps

Let the radius of the base be $r$ cm and the altitude be $h$ cm. Let the slant height be $l$ cm.
Since the cone is right circular, the radius, altitude and slant height form a right triangle. Hence,
\[l^2=r^2+h^2\]
Use the given volume condition
The volume of the cone is
\[V=\frac{1}{3}\pi r^2h\]
Since the volume is given and fixed, let its value be $V_0$. Then
\[\frac{1}{3}\pi r^2h=V_0\]
Multiplying by $3$,
\[\pi r^2h=3V_0\]
Therefore,
\[h=\frac{3V_0}{\pi r^2}\]
For convenience, put
\[k=\frac{3V_0}{\pi}\]
Then
\[\boxed{h=\frac{k}{r^2}}\]
where $k$ is a positive constant.
Express the slant height in terms of $r$
From
\[l^2=r^2+h^2\]
and
\[h=\frac{k}{r^2}\]
we obtain
\[l^2=r^2+\frac{k^2}{r^4}\]
Taking the positive square root,
\[l=\sqrt{r^2+\frac{k^2}{r^4}}\]
Form the curved surface-area function
The curved surface area of the cone is
\[S=\pi rl\]
Substituting the expression for $l$,
\[S=\pi r\sqrt{r^2+\frac{k^2}{r^4}}\]
Since $r>0$,
\[S=\pi\sqrt{r^4+\frac{k^2}{r^2}}\]
Since $\pi>0$ and the square-root function is increasing for positive arguments, minimizing $S$ is equivalent to minimizing
\[F(r)=r^4+\frac{k^2}{r^2}\]
This form makes the differentiation considerably simpler.
Differentiate the function to be minimized
F(r)=r^4+k^2r^{-2}
Differentiating
\[F'(r)=4r^3-2k^2r^{-3}\]
Hence,
\[F'(r)=4r^3-\frac{2k^2}{r^3}\]
For a stationary point,
\[F'(r)=0\]
Therefore,
\[4r^3-\frac{2k^2}{r^3}=0\]
Multiplying by $r^3$,
\[4r^6-2k^2=0\]
Hence,
\[4r^6=2k^2\]

\[2r^6=k^2\]
Since $k>0$,
\[k=\sqrt{2}\,r^3\]
Obtain the relationship between $h$ and $r$
Recall that
\[h=\frac{k}{r^2}\]
Substituting
\[k=\sqrt{2}\,r^3\]
we get
\[h=\frac{\sqrt{2}\,r^3}{r^2}\]
Therefore,
\[\boxed{h=\sqrt{2}\,r}\]
Thus, the altitude of the cone is $\sqrt{2}$ times the radius of its base.
Verify that the curved surface area is minimum
Differentiate $F'(r)$ again:
\[F''(r)=12r^2+\frac{6k^2}{r^4}\]
Since $r>0$ and $k>0$,
\[12r^2+\frac{6k^2}{r^4}>0\]
Therefore,
\[\boxed{F''(r)>0}\]
Hence, the stationary point gives a minimum value of $F(r)$, and therefore also a minimum value of the curved surface area $S$.

💡 Answer

Final Answer

›

For a right circular cone of fixed volume, the curved surface area is minimum when

\[ \boxed{h=\sqrt{2}\,r}. \]

Hence, the altitude of the cone is $\sqrt{2}$ times the radius of its base.

🎯 Exam Significance

Exam Significance

›

Distinguish between curved surface area and total surface area.
For a cone, curved surface area is $\pi rl$, not $\pi r(r+l)$.
Use the right-triangle relation $l^2=r^2+h^2$.
Use the fixed-volume condition to eliminate one variable.
Verify the minimum using the second derivative.
The required result is
\[ \boxed{h=\sqrt{2}r}. \]

Significance for Competitive Entrance Examinations

A particularly efficient approach is to minimize the square of the curved surface area because

\[ S=\pi rl. \]

Therefore,

\[ S^2=\pi^2r^2l^2. \]

Since

\[ l^2=r^2+h^2 \]

and the fixed-volume condition gives

\[ h=\frac{k}{r^2}, \]

we obtain

\[S^2=\pi^2\left(r^4+\frac{k^2}{r^2}\right)\]

Since $\pi^2$ is a positive constant, it is sufficient to minimize

\[ r^4+\frac{k^2}{r^2}. \]

Its derivative gives

\[ 4r^3-\frac{2k^2}{r^3}=0, \]

which leads directly to

\[ k=\sqrt{2}r^3. \]

Since $h=k/r^2$,

\[ \boxed{h=\sqrt{2}r}. \]

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 6 points

Curved surface area:
\[ S=\pi rl. \]
Slant-height relation:
\[ l^2=r^2+h^2. \]
For fixed volume:
\[ h=\frac{k}{r^2}, \qquad k=\frac{3V_0}{\pi}. \]
The minimum occurs when:
\[ \boxed{h=\sqrt{2}r}. \]
At the optimum:
\[ \boxed{l=\sqrt{3}r}. \]
Therefore:
\[ \boxed{h:r:l=\sqrt{2}:1:\sqrt{3}}. \]

📘 Concept & Theory

Theory and Concept

›

This is a maximization problem involving a right circular cone whose slant height is fixed. We need to determine the semi-vertical angle for which the volume of the cone is maximum.

Let the radius of the base be $r$, the height be $h$, and the fixed slant height be $l$.

Let $\theta$ be the semi-vertical angle of the cone. In the right triangle formed by the radius, height and slant height,

\[\sin\theta=\frac{r}{l},\quad\cos\theta=\frac{h}{l},\quad\tan\theta=\frac{r}{h}\]

Since $l$ is fixed, we can express the volume in terms of $r$ and then maximize it using differentiation.

🗺️ Solution Roadmap

Step-by-step Plan

›

Let $r$, $h$, and $l$ denote the radius, height, and slant height of the cone.
Use the fixed-slant-height relation $r^2+h^2=l^2$.
Express the volume in terms of one variable.
Differentiate and find the critical point.
Verify that the critical point gives maximum volume.
Use $\tan\theta=r/h$ to find the semi-vertical angle.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 35 steps

Define the dimensions
Let the radius of the base of the cone be $r$, its height be $h$, and its slant height be $l$.
The slant height $l$ is given and fixed.
Since the cone is right circular, the radius, height and slant height form a right-angled triangle. Therefore, by Pythagoras' theorem,
\[l^2=r^2+h^2\]
Hence,
\[h^2=l^2-r^2\]
Since $h>0$,
\[h=\sqrt{l^2-r^2}\]
Form the volume function
The volume of a right circular cone is
\[V=\frac{1}{3}\pi r^2h\]
Substituting
\[h=\sqrt{l^2-r^2}\]
we get
\[V(r)=\frac{\pi}{3}r^2\sqrt{l^2-r^2}\]
Thus,
\[\boxed{V(r)=\frac{\pi}{3}r^2\sqrt{l^2-r^2}}\]
Differentiate the volume
Differentiate using the product rule:
\[V'(r)=\frac{\pi}{3}\frac{d}{dr}\left[r^2(l^2-r^2)^{1/2}\right]\]
Therefore,
\[V'(r)=\frac{\pi}{3}\left[2r\sqrt{l^2-r^2}+r^2\left(\frac{1}{2}\right)(l^2-r^2)^{-1/2}(-2r)\right]\]
Hence,
\[V'(r)=\frac{\pi}{3}\left[2r\sqrt{l^2-r^2}-\frac{r^3}{\sqrt{l^2-r^2}}\right]\]
Taking the common denominator,
\[V'(r)=\frac{\pi}{3}\left[\frac{2r(l^2-r^2)-r^3}{\sqrt{l^2-r^2}}\right]\]
Simplifying the numerator,
\[2r(l^2-r^2)-r^3=2rl^2-2r^3-r^3\]

\[=2rl^2-3r^3\]

\[=r(2l^2-3r^2)\]
Therefore,
\[\boxed{V'(r)=\frac{\pi r(2l^2-3r^2)}{3\sqrt{l^2-r^2}}}\]
Find the critical point
For a stationary value of the volume,
\[V'(r)=0\]
Since $r>0$ and $l>r$, the denominator is positive. Hence,
\[2l^2-3r^2=0\]
Therefore,
\[3r^2=2l^2\]
\[r^2=\frac{2l^2}{3}\]
Hence,
\[\boxed{r=l\sqrt{\frac{2}{3}}}\]
Find the corresponding height
From
\[r^2+h^2=l^2\]
we have
\[h^2=l^2-r^2\]
Substituting
\[r^2=\frac{2l^2}{3}\]
gives
\[h^2=l^2-\frac{2l^2}{3}\]
Therefore,
\[ h^2=\frac{l^2}{3}\]
Since $h>0$,
\[\boxed{h=\frac{l}{\sqrt{3}}}\]
Verify that the volume is maximum
For $0 < r < l$, the denominator in
\[V'(r)=\frac{\pi r(2l^2-3r^2)}{3\sqrt{l^2-r^2}}\]
is positive. Also, $r>0$. Therefore, the sign of $V'(r)$ is determined by $2l^2-3r^2$.
For
\[0 < r < l\sqrt{\frac{2}{3}},\]
we have
\[2l^2-3r^2>0,\]
so
\[V'(r)>0.\]
Hence, the volume is increasing.
For
\[l\sqrt{\frac{2}{3}} we have
\[2l^2-3r^2<0,\]
so
\[V'(r) < 0\]
Hence, the volume is decreasing.
Therefore, the volume changes from increasing to decreasing at
\[r=l\sqrt{\frac{2}{3}}\]
Hence, this critical point gives the maximum volume.
Determine the semi-vertical angle
Let $\theta$ be the semi-vertical angle of the cone. From the right triangle formed by the radius, height and slant height,
\[\tan\theta=\frac{r}{h}\]
Using
\[r=l\sqrt{\frac{2}{3}}\]
and
\[h=\frac{l}{\sqrt{3}}\]
we obtain
\[\tan\theta=\frac{l\sqrt{\frac{2}{3}}}{\frac{l}{\sqrt{3}}}\]
Cancelling $l$,
\[\tan\theta=\sqrt{\frac{2}{3}}\sqrt{3}\]
Therefore,
\[\tan\theta=\sqrt{2}\]
Hence,
\[\boxed{\theta=\tan^{-1}\sqrt{2}}\]

💡 Answer

Final Answer

›

\[ \boxed{r=l\sqrt{\frac{2}{3}}}, \quad \boxed{h=\frac{l}{\sqrt{3}}}, \]

and consequently,

\[ \boxed{\frac{r}{h}=\sqrt{2}}. \]

🎯 Exam Significance

Exam Significance

›

Use $r^2+h^2=l^2$ because the radius, height and slant height form a right triangle.
Since $l$ is fixed, express $h$ in terms of $r$.
Form the volume function before differentiating.
Verify that the critical point corresponds to maximum volume.
For the semi-vertical angle, remember:
\[ \tan\theta=\frac{r}{h}. \]
The required result is:
\[ \boxed{\theta=\tan^{-1}\sqrt{2}}. \]

Significance for Competitive Entrance Examinations

A compact route is to maximize $V^2$ instead of $V$. Since

\[ V=\frac{\pi}{3}r^2h \]

and

\[ h^2=l^2-r^2, \]

we have

\[ V^2=\frac{\pi^2}{9}r^4(l^2-r^2). \]

Since $\frac{\pi^2}{9}$ is constant, maximize

\[ F(r)=r^4(l^2-r^2). \]

Differentiating,

\[ F'(r) = 4r^3(l^2-r^2)-2r^5. \]

\[ F'(r) = 2r^3(2l^2-3r^2). \]

Thus, the non-zero critical point satisfies

\[ 2l^2-3r^2=0, \]

giving

\[ r^2=\frac{2l^2}{3}. \]

Hence,

\[ h^2=l^2-r^2=\frac{l^2}{3}. \]

Therefore,

\[ \tan\theta=\frac{r}{h} = \frac{\sqrt{2}l/\sqrt{3}}{l/\sqrt{3}} = \sqrt{2}. \]

Thus,

\[ \boxed{\theta=\tan^{-1}\sqrt{2}}. \]

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 6 points

Fixed slant height:
\[ r^2+h^2=l^2. \]
Volume:
\[ V=\frac{1}{3}\pi r^2h. \]
Maximum volume occurs when:
\[ \boxed{r^2=\frac{2l^2}{3}}. \]
The corresponding height is:
\[ \boxed{h=\frac{l}{\sqrt{3}}}. \]
Therefore:
\[ \boxed{\frac{r}{h}=\sqrt{2}}. \]
Hence, the semi-vertical angle is:
\[ \boxed{\theta=\tan^{-1}\sqrt{2}}. \]

📘 Concept & Theory

Theory and Concept

›

This is a constrained optimization problem. The total surface area of the right circular cone is fixed, and we have to determine the semi-vertical angle for which its volume is maximum.

Let the radius of the base be $r$, the height be $h$, and the slant height be $l$. Let $\theta$ be the semi-vertical angle of the cone.

For a right circular cone,

\[ l^2=r^2+h^2. \]

If $\theta$ is the semi-vertical angle, then from the right triangle formed by $r$, $h$, and $l$,

\[ \sin\theta=\frac{r}{l}, \qquad \cos\theta=\frac{h}{l}. \]

Hence,

\[ r=l\sin\theta, \qquad h=l\cos\theta. \]

Since the problem refers to the surface area of a closed cone including its base, the total surface area is

\[ S=\pi r^2+\pi rl \]

\[ S=\pi r(r+l). \]

🗺️ Solution Roadmap

Step-by-step Plan

›

Let $\theta$ be the semi-vertical angle of the cone.
Express the slant height $l$ in terms of $r$ and $\theta$.
Use the fixed total surface-area condition to express $r$ in terms of $\theta$.
Express the volume as a function of $\theta$.
Differentiate and find the critical value of $\theta$.
Verify that this critical value gives maximum volume.
Show that $\sin\theta=\frac{1}{3}$.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 44 steps

Define the semi-vertical angle
Let $\theta$ be the semi-vertical angle of the cone.
From the right triangle formed by the radius $r$, height $h$, and slant height $l$,
\[\sin\theta=\frac{r}{l}\]
Therefore,
\[r=l\sin\theta\]
Hence,
\[l=\frac{r}{\sin\theta}.\]
Also,
\[\cos\theta=\frac{h}{l},\]
so
\[h=l\cos\theta\]
Substituting $l=\frac{r}{\sin\theta}$,
\[h=\frac{r}{\sin\theta}\cos\theta\]
Therefore,
\[\boxed{h=r\cot\theta}\]
Use the fixed surface-area condition
Let the given total surface area of the cone be $S$.
The total surface area of a closed right circular cone is
\[S=\pi r^2+\pi rl\]
Therefore,
\[S=\pi r(r+l)\]
Since
\[l=\frac{r}{\sin\theta}\]
we obtain
\[S=\pi r\left(r+\frac{r}{\sin\theta}\right)\]
Hence,
\[S=\pi r^2\left(1+\frac{1}{\sin\theta}\right)\]
Taking the common denominator,
\[S=\pi r^2\frac{1+\sin\theta}{\sin\theta}\]
Therefore,
\[r^2=\frac{S\sin\theta}{\pi(1+\sin\theta)}\]
Thus,
\[\boxed{r^2=\frac{S\sin\theta}{\pi(1+\sin\theta)}}\]
Express the volume in terms of $\theta$
The volume of the cone is
\[V=\frac{1}{3}\pi r^2h\]
Since
\[h=r\cot\theta\]
we get
\[V=\frac{1}{3}\pi r^3\cot\theta\]
Therefore,
\[V=\frac{\pi}{3}r^3\frac{\cos\theta}{\sin\theta}\]
From the surface-area constraint,
\[r^2=\frac{S\sin\theta}{\pi(1+\sin\theta)}\]
Hence,
\[r^3=\left[\frac{S\sin\theta}{\pi(1+\sin\theta)}\right]^{3/2}.\]
Substituting this into the volume formula,
\[V=\frac{\pi}{3}\left[\frac{S\sin\theta}{\pi(1+\sin\theta)}\right]^{3/2}\frac{\cos\theta}{\sin\theta}.\]
The quantities involving only $S$ and $\pi$ are constants. Therefore, maximizing $V$ is equivalent to maximizing
\[f(\theta)=\frac{\cos\theta\sqrt{\sin\theta}}{(1+\sin\theta)^{3/2}}.\]
Simplify the optimization using logarithmic differentiation
Since all quantities are positive for $0<\theta<\frac{\pi}{2}$, we can maximize $f(\theta)$ by maximizing $\ln f(\theta)$.
\[\ln f=\ln(\cos\theta)+\frac{1}{2}\ln(\sin\theta)-\frac{3}{2}\ln(1+\sin\theta).\]
Differentiating with respect to $\theta$,
\[\frac{1}{f}\frac{df}{d\theta}=-\tan\theta+\frac{1}{2}\cot\theta-\frac{3}{2}\frac{\cos\theta}{1+\sin\theta}.\]
For a stationary value,
\[-\tan\theta+\frac{1}{2}\cot\theta-\frac{3}{2}\frac{\cos\theta}{1+\sin\theta}=0.\]
Multiplying throughout by $2$,
\[-2\tan\theta+\cot\theta-\frac{3\cos\theta}{1+\sin\theta}=0.\]
Put $s=\sin\theta$
Let
\[s=\sin\theta.\]
Since $0<\theta<\frac{\pi}{2}$, we have $0 < s < 1$.
Using
\[\tan\theta=\frac{s}{\cos\theta},\qquad\cot\theta=\frac{\cos\theta}{s},\]
the stationary-point equation becomes
\[-\frac{2s}{\cos\theta}+\frac{\cos\theta}{s}-\frac{3\cos\theta}{1+s}=0.\]
Multiplying by the positive quantity $s\cos\theta(1+s)$,
\[-2s^2(1+s)+\cos^2\theta(1+s)-3s\cos^2\theta=0.\]
Since
\[\cos^2\theta=1-\sin^2\theta=1-s^2,\]
we obtain
\[-2s^2(1+s)+(1-s^2)(1+s)-3s(1-s^2)=0\]
Expanding,
\[-2s^2-2s^3+1+s-s^2-s^3-3s+3s^3=0.\]
Combining like terms,
\[1-2s-3s^2=0\]
Therefore,
\[3s^2+2s-1=0\]
Factorising,
\[(3s-1)(s+1)=0.\]
Hence,
\[s=\frac{1}{3}\]
or
\[s=-1.\]
But $0 < s < 1$, so $s=-1$ is inadmissible.
Therefore,
\[\boxed{\sin\theta=\frac{1}{3}}\]
Determine the semi-vertical angle
Since
\[\sin\theta=\frac{1}{3}\]
we obtain
\[\boxed{\theta=\sin^{-1}\left(\frac{1}{3}\right)}.\]
Verify that the volume is maximum
Consider
\[f(\theta)=\frac{\cos\theta\sqrt{\sin\theta}}{(1+\sin\theta)^{3/2}}.\]
For $0<\theta<\frac{\pi}{2}$, the volume tends to zero as $\theta\to0^+$, because $\sqrt{\sin\theta}\to0$, and it also tends to zero as $\theta\to\frac{\pi}{2}^-$, because $\cos\theta\to0$.
The only admissible stationary point is
\[ \sin\theta=\frac{1}{3}. \]
Therefore, the function increases before this point and decreases after this point, making it the maximum-volume configuration.

💡 Answer

Final Answer

›

The semi-vertical angle of a right circular cone having a given total surface area and maximum volume is

\[ \boxed{ \theta=\sin^{-1}\left(\frac{1}{3}\right) }. \]

🎯 Exam Significance

Exam Significance

›

For a closed right circular cone, total surface area is
\[ S=\pi r(r+l). \]
The semi-vertical angle satisfies
\[ \sin\theta=\frac{r}{l}, \qquad \tan\theta=\frac{r}{h}. \]
Use the fixed surface-area condition to eliminate $r$.
Logarithmic differentiation is especially convenient because the resulting function contains products and powers.
The admissible solution must satisfy $0<\theta<\frac{\pi}{2}$.
The required result is
\[ \boxed{\theta=\sin^{-1}\left(\frac{1}{3}\right)}. \]

Significance for Competitive Entrance Examinations

A useful compact form of the volume, after eliminating $r$ using the fixed total surface area, is

\[ V \propto \frac{\cos\theta\sqrt{\sin\theta}} {(1+\sin\theta)^{3/2}}. \]

Letting $s=\sin\theta$, the logarithmic derivative condition reduces to

\[ 1-2s-3s^2=0. \]

Thus,

\[ (3s-1)(s+1)=0. \]

Since \(0

\[ s=\frac{1}{3}. \]

Therefore,

\[ \boxed{\theta=\sin^{-1}\left(\frac{1}{3}\right)}. \]

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 5 points

Total surface area of a closed cone:
\[ S=\pi r(r+l). \]
For semi-vertical angle $\theta$:
\[ r=l\sin\theta, \qquad h=l\cos\theta. \]
Under the fixed-surface-area constraint, maximizing volume leads to
\[ 1-2\sin\theta-3\sin^2\theta=0. \]
The admissible solution is
\[ \boxed{\sin\theta=\frac{1}{3}}. \]
Hence, the semi-vertical angle is
\[ \boxed{\theta=\sin^{-1}\left(\frac{1}{3}\right)}. \]

📘 Concept & Theory

Theory and Concept

›

To find the point on a curve that is nearest to a given point, we minimize the distance between the two points.

Let $P(x,y)$ be a point on the curve $x^2=2y$, and let $Q=(0,5)$. The distance between $P$ and $Q$ is

\[ d=\sqrt{(x-0)^2+(y-5)^2}. \]

Since the square-root function is increasing, minimizing $d$ is equivalent to minimizing $d^2$. This avoids unnecessary square roots during differentiation.

🗺️ Solution Roadmap

Step-by-step Plan

›

Use the equation of the curve to express $y$ in terms of $x$.
Form the square of the distance from $(0,5)$.
Differentiate the distance-squared function.
Find the critical points.
Verify which point gives the minimum distance.
Compare the result with the given options.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 30 steps

Express $y$ in terms of $x$
The point lies on
\[x^2=2y\]
Therefore,
\[\boxed{y=\frac{x^2}{2}}\]
Form the distance function
Let $P(x,y)$ be a point on the parabola. Its distance from $Q(0,5)$ is
\[d=\sqrt{x^2+(y-5)^2}\]
Instead of minimizing $d$, minimize $d^2$:
\[d^2=x^2+(y-5)^2\]
Substituting
\[y=\frac{x^2}{2},\]
we get
\[d^2=x^2+\left(\frac{x^2}{2}-5\right)^2\]
Let
\[D(x)=x^2+\left(\frac{x^2}{2}-5\right)^2\]
Simplify the function
\[D(x)=x^2+\frac{(x^2-10)^2}{4}\]
Expanding,
\[D(x)=x^2+\frac{x^4-20x^2+100}{4}\]
Therefore,
\[D(x)=\frac{x^4}{4}-5x^2+25+x^2\]

\[D(x)=\frac{x^4}{4}-4x^2+25.\]
Hence,
\[\boxed{D(x)=\frac{x^4}{4}-4x^2+25}.\]
Differentiate
\[D'(x)=\frac{d}{dx}\left(\frac{x^4}{4}-4x^2+25\right)\]
Therefore,
\[D'(x)=x^3-8x\]
Factorising,
\[D'(x)=x(x^2-8)\]
For a stationary point,
\[D'(x)=0\]
Hence,
\[x(x^2-8)=0\]
Therefore,
\[x=0\]
and
\[x^2=8\]
Thus,
\[x=\pm2\sqrt{2}\]
Find the corresponding $y$-coordinates
Since
\[y=\frac{x^2}{2},\]
for $x=\pm2\sqrt{2}$,
\[y=\frac{8}{2}=4\]
Hence, the two symmetric points are
\[\boxed{(2\sqrt{2},4)}\]
and
\[\boxed{(-2\sqrt{2},4)}.\]
For $x=0$,
\[y=0\]
Thus, the other stationary point is
\[(0,0)\]
Determine which stationary point gives the minimum distance
The second derivative is
\[D''(x)=3x^2-8\]
At $x=0$,
\[D''(0)=-8<0\]
Therefore, $(0,0)$ gives a local maximum of the squared distance, not a minimum.
At $x=\pm2\sqrt{2}$,
\[D''(\pm2\sqrt{2})=3(8)-8\]

\[=24-8\]

\[=16>0\]
Therefore, both
\[(2\sqrt{2},4)\]
and
\[(-2\sqrt{2},4)\]
give minimum distance.
Verify the minimum distance
At either of the points $(\pm2\sqrt{2},4)$,
\[d^2=(2\sqrt{2})^2+(4-5)^2\]

\[=8+1\]

\[=9.\]
Therefore,
\[\boxed{d_{\min}=3}.\]
At $(0,0)$,
\[d^2=0^2+(0-5)^2=25,\]
so its distance is $5$, which is greater than $3$.

🎯 Exam Significance

Exam Significance

›

For nearest-point problems, minimize the square of the distance instead of the distance itself.
Use the curve equation to reduce the problem to one variable.
Do not assume that every stationary point gives a minimum.
Use the second derivative or a sign analysis to classify the stationary point.
Check symmetry of the curve because symmetric points can have the same minimum distance.

Significance for Competitive Entrance Examinations

A particularly quick approach is to put

\[ t=x^2. \]

Since $t\geq0$,

\[ D=x^2+\left(\frac{x^2}{2}-5\right)^2 \]

\[ D=t+\left(\frac{t}{2}-5\right)^2. \]

Expanding,

\[ D=\frac{t^2}{4}-4t+25. \]

Differentiating with respect to $t$,

\[ \frac{dD}{dt}=\frac{t}{2}-4. \]

Hence,

\[ \frac{t}{2}-4=0 \]

\[ t=8. \]

Since $t=x^2$,

\[ x^2=8 \]

\[ x=\pm2\sqrt2. \]

Then

\[ y=\frac{x^2}{2}=4. \]

Therefore,

\[ \boxed{(\pm2\sqrt2,4)} \]

are the two nearest points.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 7 points

Curve:
\[ x^2=2y \quad\Rightarrow\quad y=\frac{x^2}{2}. \]
Squared distance from $(0,5)$:
\[ D(x)=\frac{x^4}{4}-4x^2+25. \]
Stationary points:
\[ x=0,\quad x=\pm2\sqrt2. \]
$(0,0)$ gives a local maximum of the squared distance.
The minimum occurs at:
\[ \boxed{(\pm2\sqrt2,4)}. \]
Minimum distance:
\[ \boxed{3}. \]
Among the given options, the answer is:
\[ \boxed{\text{(A)}\ (2\sqrt2,4)}. \]

📘 Concept & Theory

Theory and Concept

›

We need to find the minimum value of the rational function

\[ f(x)=\frac{x^2-x+1}{x^2+x+1}, \qquad x\in\mathbb{R}. \]

First, observe that the denominator is always positive:

\[ x^2+x+1 = \left(x+\frac12\right)^2+\frac34>0. \]

Hence, the function is defined for every real $x$.

We can find its minimum either by differentiation or by converting the inequality $f(x)\geq k$ into a quadratic inequality. Since this is an application of derivatives, we use differentiation.

🗺️ Solution Roadmap

Step-by-step Plan

›

Define the given expression as $f(x)$.
Verify that its denominator never vanishes.
Differentiate using the quotient rule.
Find the critical points.
Determine which critical point gives the minimum value.
Evaluate the function at that point and identify the correct option.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 22 steps

Define the function
\[f(x)=\frac{x^2-x+1}{x^2+x+1}\]
Check the denominator
The denominator is
\[x^2+x+1\]
Completing the square,
\[x^2+x+1=\left(x+\frac12\right)^2+\frac34\]
Since
\[\left(x+\frac12\right)^2\geq0,\]
we have
\[x^2+x+1\geq\frac34>0\]
Therefore, $f(x)$ is defined for all real $x$.
Using the quotient rule,
\[f'(x)=\frac{(2x-1)(x^2+x+1)-(x^2-x+1)(2x+1)}{(x^2+x+1)^2}\]
Expand the first product:
\[(2x-1)(x^2+x+1)=2x^3+2x^2+2x-x^2-x-1\]

\[=2x^3+x^2+x-1\]
Expand the second product:
\[(x^2-x+1)(2x+1)=2x^3+x^2-2x^2-x+2x+1\]

\[=2x^3-x^2+x+1\]
Therefore, the numerator of $f'(x)$ is
\[(2x^3+x^2+x-1)-(2x^3-x^2+x+1)\]

\[=2x^2-2\]

\[=2(x^2-1)\]
Hence,
\[\boxed{f'(x)= \frac{2(x^2-1)}{(x^2+x+1)^2}}\]
Find the critical points
Since the denominator is always positive, $f'(x)=0$ when
\[x^2-1=0\]
Therefore,
\[x^2=1\]

\[\boxed{x=\pm1}\]
Classify the critical points
Since the denominator of $f'(x)$ is positive, the sign of $f'(x)$ depends only on $x^2-1$.
For $x < -1$,
\[x^2-1>0\]
so
\[f'(x)>0\]
For $-1 < x < 1$,
\[x^2-1<0\]
so
\[f'(x) < 0\]
For $x>1$,
\[x^2-1>0\]
so
\[f'(x)>0\]
Thus, $f(x)$ changes from increasing to decreasing at $x=-1$, so $x=-1$ gives a local maximum.
At $x=1$, $f(x)$ changes from decreasing to increasing. Therefore, $x=1$ gives a minimum.
At $x=1$,
\[f(1)=\frac{1-1+1}{1+1+1}\]
Therefore,
\[f(1)=\frac{1}{3}\]
Hence,
\[\boxed{f_{\min}=\frac13}\]

💡 Answer

Final Answer

›

Therefore, the correct option is

\[\boxed{\text{(D)}\ \frac13}\]

🎯 Exam Significance

Exam Significance

›

Always check whether the denominator can become zero.
For a rational function, the quotient rule must be applied carefully.
Finding $f'(x)=0$ only gives candidate critical points; their nature must be determined.
The sign-change method gives a clear proof that $x=1$ gives the minimum.
The inequality method provides an excellent independent verification of the answer.

Significance for Competitive Entrance Examinations

The fastest approach is often to compare the expression directly with a suitable constant. Let

\[ f(x)=\frac{x^2-x+1}{x^2+x+1}. \]

Testing the option $\frac13$,

\[ f(x)-\frac13 = \frac{3(x^2-x+1)-(x^2+x+1)} {3(x^2+x+1)}. \]

Simplifying,

\[ f(x)-\frac13 = \frac{2(x-1)^2} {3(x^2+x+1)}. \]

Both the numerator and denominator are non-negative and the denominator is strictly positive. Hence,

\[ f(x)-\frac13\geq0. \]

Therefore,

\[ \boxed{f(x)\geq\frac13}. \]

Equality occurs at $x=1$, so $\frac13$ is the exact minimum.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 7 points

Given function:
\[ f(x)=\frac{x^2-x+1}{x^2+x+1}. \]
The denominator is always positive:
\[ x^2+x+1= \left(x+\frac12\right)^2+\frac34>0. \]
Derivative:
\[ f'(x)= \frac{2(x^2-1)} {(x^2+x+1)^2}. \]
Critical points:
\[ x=\pm1. \]
$x=-1$ gives a local maximum, while $x=1$ gives the minimum.
Minimum value:
\[ \boxed{\frac13}. \]
Correct option:
\[ \boxed{\text{(D)}\ \frac13}. \]

Q29

LONG ANSWER4 marks

Find the maximum value of \[ \left[x(x-1)+1\right]^{\frac13}, \quad 0\leq x\leq1. \] The options are:

$\left(\dfrac13\right)^{1/3}$
$\dfrac12$
$1$
$0$

📘 Concept & Theory

Theory and Concept

›

The given function contains a cube root. Since the cube-root function $t^{1/3}$ is strictly increasing for all real $t$, maximizing the given expression is equivalent to maximizing its quantity inside the cube root.

Let

\[ f(x)=\left[x(x-1)+1\right]^{\frac13}. \]

Simplifying the expression inside the cube root,

\[ x(x-1)+1=x^2-x+1. \]

Thus,

\[ f(x)=(x^2-x+1)^{\frac13}. \]

🗺️ Solution Roadmap

Step-by-step Plan

›

Simplify the expression inside the cube root.
Differentiate the resulting function.
Find the critical point in the interval $[0,1]$.
Compare the function values at the critical point and both endpoints.
Identify the absolute maximum.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 15 steps

Define the function
\[ f(x)=(x^2-x+1)^{\frac13}, \quad 0\leq x\leq1. \]
Differentiate
Using the chain rule,
\[f'(x)=\frac13(x^2-x+1)^{-\frac23}(2x-1)\]
Therefore,
\[\boxed{f'(x)=\frac{2x-1}{3(x^2-x+1)^{2/3}}}\]
Find the critical point
We first check whether the denominator can become zero.
\[x^2-x+1=\left(x-\frac12\right)^2+\frac34>0\]
Hence, the denominator of $f'(x)$ is always positive. Therefore,
\[f'(x)=0\]
when
\[2x-1=0\]
Hence,
\[\boxed{x=\frac12}\]
Determine whether the critical point gives a maximum
Since the denominator of $f'(x)$ is positive, the sign of $f'(x)$ depends on $2x-1$.
For $0\leq x<\frac12$,
\[2x-1 < 0\]
so
\[f'(x) < 0\]
Thus, $f(x)$ is decreasing on $\left[0,\frac12\right]$
For $\frac12 < x\leq1$,
\[2x-1>0\]
so
\[f'(x)>0\]
Thus, $f(x)$ is increasing on $\left[\frac12,1\right]$. Therefore, $x=\frac12$ gives an absolute minimum, not a maximum. Consequently, for the maximum we must compare the endpoint values $x=0$ and $x=1$.
Evaluate the endpoint values
At $x=0$,
\[f(0)=(0^2-0+1)^{1/3}=1\]
At $x=1$,
\[f(1)=(1^2-1+1)^{1/3}=1\]
At the critical point $x=\frac12$,
\[f\left(\frac12\right)=\left(\frac14-\frac12+1\right)^{1/3}\]

\[=\left(\frac34\right)^{1/3}\]
Since
\[\left(\frac34\right)^{1/3} < 1\]
the maximum value is $1$

💡 Answer

Final Answer

›

\[\boxed{\text{Maximum value}=1}\]

Therefore, the correct option is

\[\boxed{\text{(C)}\ 1}\]

🎯 Exam Significance

Exam Significance

›

The interval $[0,1]$ is closed, so an absolute maximum and minimum must be checked among critical points and endpoints.
A critical point need not give a maximum; here $x=\frac12$ gives the minimum.
The cube-root function is strictly increasing, so the quantity inside the cube root can be optimized directly.
Always compare endpoint values when solving an absolute maximum/minimum problem on a closed interval.

Significance for Competitive Entrance Examinations

The fastest observation is

\[ 0\leq x\leq1 \quad\Longrightarrow\quad x(x-1)\leq0. \]

Hence,

\[ x(x-1)+1\leq1. \]

Taking cube roots,

\[ \left[x(x-1)+1\right]^{1/3}\leq1. \]

Equality is achieved at $x=0$ or $x=1$. Therefore, the answer can be obtained immediately without differentiation:

\[ \boxed{1}. \]

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 6 points

Given function:
\[ f(x)=(x^2-x+1)^{1/3}. \]
Critical point:
\[ x=\frac12. \]
$x=\frac12$ gives the minimum:
\[ f\left(\frac12\right)=\left(\frac34\right)^{1/3}. \]
Endpoint values:
\[ f(0)=f(1)=1. \]
Therefore, the maximum value is
\[ \boxed{1}. \]
Correct option:
\[ \boxed{\text{(C)}\ 1}. \]

Function	Critical Point	Second Derivative Test	Result
\(f(x)=(2x-1)^2+3\)	\(x=\frac12\)	\(f''(x)=8>0\)	Minimum \(=3\)
\(f(x)=9x^2+12x+3\)	\(x=-\frac23\)	\(f''(x)=18>0\)	Minimum \(=-1\)
\(f(x)=-(x-1)^2+10\)	\(x=1\)	\(f''(x)=-2<0\)	Maximum \(=10\)
\(g(x)=x^3+1\)	\(x=0\)	\(g''(0)=0\)	No maximum or minimum

Point	Function value
\(x=-2\)	\(-8\)
\(x=0\)	\(0\)
\(x=2\)	\(8\)

Point	Function value
\(x=0\)	\(1\)
\(\displaystyle x=\frac{\pi}{4}\)	\(\sqrt2\)
\(x=\pi\)	\(-1\)

Point	Function value
\(x=-2\)	\(-10\)
\(x=4\)	\(8\)
\(\displaystyle x=\frac92\)	\(\displaystyle \frac{63}{8}=7.875\)

Part	Function	Absolute Maximum	Point of Maximum	Absolute Minimum	Point of Minimum
(i)	\(\displaystyle x^3,\;[-2,2]\)	\(8\)	\(x=2\)	\(-8\)	\(x=-2\)
(ii)	\(\displaystyle \sin x+\cos x,\;[0,\pi]\)	\(\sqrt2\)	\(\displaystyle x=\frac{\pi}{4}\)	\(-1\)	\(x=\pi\)
(iii)	\(\displaystyle 4x-\frac12x^2,\;\left[-2,\frac92\right]\)	\(8\)	\(x=4\)	\(-10\)	\(x=-2\)
(iv)	\(\displaystyle (x-1)^2+3,\;[-3,1]\)	\(19\)	\(x=-3\)	\(3\)	\(x=1\)

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Part	Function	Maximum	Minimum
(i)	\(\|x+2\|-1\)	Does not exist	\(-1\) at \(x=-2\)
(ii)	\(-\|x+1\|+3\)	\(3\) at \(x=-1\)	Does not exist
(iii)	\(\sin(2x)+5\)	\(6\)	\(4\)
(iv)	\(\|\sin(4x)+3\|\)	\(4\)	\(2\)
(v)	\(x+1,\;x\in(-1,1)\)	Does not exist	Does not exist

Part	Function	Local Maximum	Local Minimum
(i)	\(x^2\)	None	\(0\) at \(x=0\)
(ii)	\(x^3-3x\)	\(2\) at \(x=-1\)	\(-2\) at \(x=1\)
(iii)	\(\sin x+\cos x,\;0	\(\sqrt2\) at \(x=\frac{\pi}{4}\)	None
(iv)	\(\sin x-\cos x,\;0	\(\sqrt2\) at \(x=\frac{3\pi}{4}\)	\(-\sqrt2\) at \(x=\frac{7\pi}{4}\)
(v)	\(x^3-6x^2+9x+15\)	\(19\) at \(x=1\)	\(15\) at \(x=3\)
(vi)	\(\frac{x}{2}+\frac{2}{x},\;x>0\)	None	\(2\) at \(x=2\)
(vii)	\(\frac{1}{x^2+2}\)	\(\frac12\) at \(x=0\)	None
(viii)	\(x\sqrt{1-x},\;0 < x < 1\)	\(\frac{2}{3\sqrt3}=\frac{2\sqrt3}{9}\) at \(x=\frac23\)	None

Function	Domain	Derivative	Nature	Maximum	Minimum
\(\displaystyle e^x\)	\(\mathbb R\)	\(\displaystyle e^x>0\)	Strictly increasing	None	None
\(\displaystyle \log x\)	\((0,\infty)\)	\(\displaystyle \frac1x>0\)	Strictly increasing	None	None
\(\displaystyle x^3+x^2+x+1\)	\(\mathbb R\)	\(\displaystyle 3x^2+2x+1>0\)	Strictly increasing	None	None

Point	Function value
\(x=0\)	\(0\)
\(x=\frac{\pi}{4}\)	\(1\)
\(x=\frac{3\pi}{4}\)	\(-1\)
\(x=\frac{5\pi}{4}\)	\(1\)
\(x=\frac{7\pi}{4}\)	\(-1\)
\(x=2\pi\)	\(0\)

Interval	Maximum Value	Point Where Maximum Occurs
\([1,3]\)	\(\boxed{89}\)	\(\boxed{x=3}\)
\([-3,1]\)	\(\boxed{139}\)	\(\boxed{x=-2}\)

Point	Function Value
\(x=0\)	\(0\)
\(x=\frac{\pi}{3}\)	\(\frac{\pi}{3}+\frac{\sqrt3}{2}\)
\(x=\frac{2\pi}{3}\)	\(\frac{2\pi}{3}-\frac{\sqrt3}{2}\)
\(x=\frac{4\pi}{3}\)	\(\frac{4\pi}{3}+\frac{\sqrt3}{2}\)
\(x=\frac{5\pi}{3}\)	\(\frac{5\pi}{3}-\frac{\sqrt3}{2}\)
\(x=2\pi\)	\(2\pi\)

Application of Derivatives

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Frequently Asked Questions

Q 01 What is NCERT Class 12 Maths Exercise 6.3 about?

Q 02 What concepts are covered in Exercise 6.3?

Q 03 How do I solve maxima and minima problems in Exercise 6.3?

Q 04 Why are endpoints checked in maxima and minima questions?

Q 05 What is the difference between local and absolute maximum?

Q 06 What is the role of the second derivative in Exercise 6.3?

Q 07 Is NCERT Exercise 6.3 important for CBSE Board exams?

Q 08 Is Exercise 6.3 useful for JEE Main and JEE Advanced?

Q 09 What is the quickest method for some maxima-minima MCQs?

Q 10 What should students remember while solving Exercise 6.3?

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