NCERT Class 12 Maths Ex 6.2 Solutions

📘 Concept & Theory

Theory and Concept

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A function $f(x)$ is said to be increasing on an interval if, for any two numbers $x_1$ and $x_2$ in that interval,

\[ x_1 < x_2 \Rightarrow f(x_1) < f(x_2). \]

In other words, whenever the value of $x$ increases, the corresponding value of $f(x)$ also increases. Graphically, an increasing function rises from left to right.

For a differentiable function, another important test for increasing behaviour is

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

If the derivative is positive throughout an interval, the function is increasing on that interval.

In this question, we will use the definition of an increasing function, which gives a direct proof.

🗺️ Solution Roadmap

Step-by-step Plan

›

Take any two real numbers $x_1$ and $x_2$.
Assume $x_1 < x_2$.
Multiply the inequality by the positive number $3$. The direction of the inequality remains unchanged.
Add $17$ to both sides.
Recognise the resulting expressions as $f(x_1)$ and $f(x_2)$.
Conclude from the definition that $f(x)$ is increasing on $\mathbb{R}$.

📊 Graph / Figure

Graph / Figure

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Graphical Representation

✏️ Solution

Complete Solution

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Step-by-step Solution · 7 steps

Given —
\[f(x)=3x+17\]
Let $x_1$ and $x_2$ be any two real numbers such that
\[x_1 < x_2\]
.
Since $3$ is a positive number, multiplying both sides of the inequality by $3$ does not change the direction of the inequality:
\[3x_1 < 3x_2\]
Adding $17$ to both sides gives
\[3x_1+17 < 3x_2+17\]
Now, from the definition of the function,
\[f(x_1)=3x_1+17\]
and
\[f(x_2)=3x_2+17\]
Therefore,
\[f(x_1) < f(x_2)\]
Hence,
\[x_1 < x_2\Rightarrow f(x_1) < f(x_2)\]
This satisfies the definition of an increasing function. Therefore,
\[\boxed{f(x)=3x+17\text{ is increasing on }\mathbb{R}.}\]

💡 Answer

Final Answer

›

The function $f(x)=3x+17$ is increasing on $\mathbb{R}$ because for any two real numbers $x_1$ and $x_2$ such that $x_1 < x_2$, we have

\[ f(x_1) < f(x_2). \]

This conclusion is drawn directly from the definition of an increasing function, and it is also supported by the fact that the derivative $f'(x)=3>0$ for all $x\in\mathbb{R}$.

🎯 Exam Significance

Exam Significance

›

This question tests the fundamental concept of increasing functions, which is an important part of the Applications of Derivatives chapter. In board examinations, such questions can be used to test whether a student understands the mathematical definition of increasing behaviour rather than merely applying a derivative formula.

A complete answer should explicitly state

\[ x_1 < x_2\Rightarrow f(x_1) < f(x_2), \]

because this establishes increasing behaviour directly from the definition.

The derivative method is also important because questions may ask students to determine intervals of increase and decrease for more complicated functions. The central criterion is

\[ f'(x)>0\Rightarrow f(x)\text{ is increasing}. \]

Significance for Competitive Entrance Examinations

For JEE Main, JEE Advanced and other entrance examinations, the concept of monotonicity is frequently used in problems involving derivatives, inequalities, number of roots, extrema and the behaviour of functions.

For a linear function

\[ f(x)=ax+b, \]

the sign of $a$ immediately determines its monotonic behaviour:

\[ a>0\Rightarrow f(x)\text{ is increasing}, \]

\[ a<0\Rightarrow f(x)\text{ is decreasing}, \]

and

\[ a=0\Rightarrow f(x)\text{ is constant}. \]

Therefore, recognising that the coefficient of $x$ is positive can provide a very quick solution in objective and multiple-choice questions.

🚫 Common Mistakes

Common Mistakes

›

When multiplying an inequality by a number, always check its sign. Here, $3>0$, so

\[ x_1 < x_2\Rightarrow3x_1 < 3x_2. \]

If the inequality were multiplied by a negative number, its direction would have to be reversed.

Another common mistake is to write

\[ 3x_1+17 < 3x_2+17 \]

without explaining why the inequality is preserved. In a step-by-step proof, it is better to explicitly state that multiplying by a positive number and adding the same number to both sides preserve the inequality.

🔑 Key Takeaways

Key Takeaways

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Key Takeaways · 8 points

An increasing function satisfies $x_1 < x_2\Rightarrow f(x_1) < f(x_2)$.
For $f(x)=3x+17$, the coefficient of $x$ is positive.
The derivative is $f'(x)=3$.
Since $f'(x)>0$ for every $x\in\mathbb{R}$, the function is strictly increasing on $\mathbb{R}$.
The graph of an increasing function rises from left to right.
For a linear function $ax+b$, $a>0$ implies increasing behaviour and $a<0$ implies decreasing behaviour.
The definition-based proof is especially useful for establishing the fundamental meaning of monotonicity.
The derivative test provides a faster method for more complex functions encountered in board and competitive examinations.

📘 Concept & Theory

Theory and Concept

›

A function $f(x)$ is said to be increasing on an interval if for any two numbers $x_1$ and $x_2$ in that interval,

\[ x_1 < x_2\Rightarrow f(x_1) < f(x_2). \]

For exponential functions, the important property used here is that

\[ e^u \]

is a strictly increasing function of $u$. Therefore,

\[ u_1 < u_2\Rightarrow e^{u_1} < e^{u_2}. \]

Since the exponent in this question is $2x$, and multiplication by the positive number $2$ preserves the order of real numbers, the value of $2x$ increases whenever $x$ increases. Consequently, $e^{2x}$ also increases.

The result can also be established directly using the derivative test:

\[ f'(x)>0\Rightarrow f(x)\text{ is increasing}. \]

🗺️ Solution Roadmap

Step-by-step Plan

›

Take any two real numbers $x_1$ and $x_2$.
Assume $x_1 < x_2$.
Multiply the inequality by $2>0$ to obtain $2x_1 < 2x_2$.
Use the fact that the exponential function $e^x$ is strictly increasing.
Therefore, $e^{2x_1} < e^{2x_2}$.
Recognise these expressions as $f(x_1)$ and $f(x_2)$.
Conclude that $f(x)=e^{2x}$ is increasing on $\mathbb{R}$.

📊 Graph / Figure

Graph / Figure

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Graphical Representation

✏️ Solution

Complete Solution

›

Step-by-step Solution · 6 steps

Given —
\[f(x)=e^{2x}\]
Let $x_1$ and $x_2$ be any two real numbers such that
\[x_1 < x_2\]
Since $2>0$, multiplying both sides of the inequality by $2$ does not change the direction of the inequality:
\[2x_1 < 2x_2\]
Now, the exponential function $e^x$ is strictly increasing. Hence, if
\[2x_1 < 2x_2,\]
then
\[e^{2x_1} < e^{2x_2}\]
But
\[f(x_1)=e^{2x_1}\]
and
\[f(x_2)=e^{2x_2}\]
Therefore,
\[f(x_1) < f(x_2)\]
Thus,
\[x_1 < x_2\Rightarrow f(x_1) < f(x_2)\]
This satisfies the definition of an increasing function.

💡 Answer

Final Answer

›

Let $x_1,x_2\in\mathbb{R}$ such that

\[x_1 < x_2\]

Since $2>0$,

\[2x_1 < 2x_2\]

As $e^x$ is a strictly increasing function,

\[e^{2x_1} < e^{2x_2}\]

Therefore,

\[f(x_1) < f(x_2)\]

Hence,

\[\boxed{f(x)=e^{2x}\text{ is increasing on }\mathbb{R}.}\]

🎯 Exam Significance

Exam Significance

›

This question tests the concept of monotonicity and the use of the definition of an increasing function. It is important to show the logical sequence clearly:

\[ x_1 < x_2\Rightarrow2x_1 < 2x_2\Rightarrow e^{2x_1} < e^{2x_2}. \]

The derivative approach is equally important for board examinations because it connects monotonicity directly with the Applications of Derivatives:

\[ f'(x)>0\Rightarrow f(x)\text{ is increasing}. \]

A complete solution should not simply state that $e^{2x}$ is increasing. The reason should be established either from the definition or by showing that its derivative is positive throughout the required domain.

Significance for Competitive Entrance Examinations

The derivative method is particularly useful in competitive examinations because it gives a fast way to establish monotonicity:

\[ f'(x)=2e^{2x}>0. \]

Since the exponential function is always positive, expressions of the form

\[ ke^{ax} \]

can often be analysed immediately. For example, when $k>0$ and $a>0$,

\[ \frac{d}{dx}\left(ke^{ax}\right)=kae^{ax}>0, \]

so the function is increasing on $\mathbb{R}$.

This observation is useful in objective questions involving monotonicity, inequalities, comparison of function values, number of roots and optimisation.

🚫 Common Mistakes

Common Mistakes

›

Do not write $e^{2x_1} < e^{2x_2}$ without first establishing $2x_1 < 2x_2$.
Remember that $e^x$ is strictly increasing because its derivative $e^x$ is always positive.
When differentiating $e^{2x}$, do not forget the derivative of the inner function $2x$.
The correct derivative is
\[ \frac{d}{dx}(e^{2x})=2e^{2x}, \]
not merely $e^{2x}$.
Do not conclude that a function is increasing merely because its values are positive. Positivity of $f(x)$ and positivity of $f'(x)$ are different concepts.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 7 points

An increasing function satisfies $x_1 < x_2\Rightarrow f(x_1) < f(x_2)$.
The exponential function $e^x$ is strictly increasing on $\mathbb{R}$.
Since $x_1 < x_2$ implies $2x_1 < 2x_2$, we have $e^{2x_1} < e^{2x_2}$.
The derivative of $e^{2x}$ is
\[ f'(x)=2e^{2x}. \]
Because $e^{2x}>0$ for every real $x$, $f'(x)>0$ for every real $x$.
Therefore, $f(x)=e^{2x}$ is strictly increasing on $\mathbb{R}$.
The graph rises continuously from left to right, confirming the increasing nature of the function.

Q3

LONG ANSWER4 marks

Show that the function given by \[f(x)=\sin x\] is:

Increasing in \[ \left(0,\frac{\pi}{2}\right) \]
Decreasing in \[ \left(\frac{\pi}{2},\pi\right) \]

📘 Concept & Theory

Theory and Concept

›

A function $f(x)$ is increasing on an interval if, for any two numbers $x_1$ and $x_2$ belonging to that interval,

\[x_1 < x_2\Rightarrow f(x_1) < f(x_2)\]

Similarly, a function is decreasing on an interval if

\[x_1 < x_2\Rightarrow f(x_1)>f(x_2)\]

For a differentiable function, the derivative provides a convenient test:

\[f'(x)>0\Rightarrow f(x)\text{ is increasing},\]

while

\[f'(x)<0\Rightarrow f(x)\text{ is decreasing}\]

For the given function,

\[ f(x)=\sin x, \]

its derivative is

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

The sign of $\cos x$ changes at

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

In the first interval,

\[ 0 we have

\[ \cos x>0. \]

Therefore, $\sin x$ is increasing there.

In the second interval,

\[ \frac{\pi}{2} we have

\[ \cos x<0. \]

Therefore, $\sin x$ is decreasing there.

🗺️ Solution Roadmap

Step-by-step Plan

›

Write the given function $f(x)=\sin x$.
Differentiate the function to obtain $f'(x)=\cos x$.
Determine the sign of $\cos x$ in $\left(0,\frac{\pi}{2}\right)$.
Use $f'(x)>0$ to establish that the function is increasing in the first interval.
Determine the sign of $\cos x$ in $\left(\frac{\pi}{2},\pi\right)$.
Use $f'(x)<0$ to establish that the function is decreasing in the second interval.
Conclude the required intervals of increase and decrease.

📊 Graph / Figure

Graph / Figure

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Graphical Representation

✏️ Solution

Complete Solution

›

Step-by-step Solution · 12 steps

Given —
\[f(x)=\sin x\]
Differentiating with respect to $x$,
\[f'(x)=\frac{d}{dx}(\sin x)\]
Therefore,
\[f'(x)=\cos x\]
(a) Showing that $f(x)=\sin x$ is increasing in $\left(0,\frac{\pi}{2}\right)$
Consider the interval
\[0 < x < \frac{\pi}{2}\]
In the first quadrant, the cosine function is positive. Hence,
\[\cos x>0\]
for every
\[x\in\left(0,\frac{\pi}{2}\right)\]
Since
\[f'(x)=\cos x,\]
we obtain
\[f'(x)>0\]
for every
\[x\in\left(0,\frac{\pi}{2}\right)\]
By the derivative test for monotonicity,
\[f'(x)>0\Rightarrow f(x)\text{ is increasing}\]
Therefore,
\[\boxed{f(x)=\sin x\text{ is increasing in }\left(0,\frac{\pi}{2}\right)}\]
(b) Showing that $f(x)=\sin x$ is decreasing in $\left(\frac{\pi}{2},\pi\right)$
Now consider the interval
\[\frac{\pi}{2}
In the second quadrant, the cosine function is negative. Hence,
\[\cos x < 0\]
for every
\[x\in\left(\frac{\pi}{2},\pi\right)\]
Since
\[f'(x)=\cos x,\]
we have
\[f'(x)<0\]
for every
\[x\in\left(\frac{\pi}{2},\pi\right)\]
By the derivative test for monotonicity,
\[f'(x)<0\Rightarrow f(x)\text{ is decreasing}\]
Therefore,
\[\boxed{f(x)=\sin x\text{ is decreasing in }\left(\frac{\pi}{2},\pi\right)}\]

💡 Answer

Final Answer

›

🎯 Exam Significance

Exam Significance

›

This question is important because it tests the application of derivatives to determine intervals of increase and decrease. The key step is to calculate

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

and then determine its sign on the given intervals.

A board examination solution should clearly mention the sign of $\cos x$ in each interval:

\[ \cos x>0\quad\text{for}\quad0 < x < \frac{\pi}{2}, \]

and

\[ \cos x < 0\quad\text{for}\quad\frac{\pi}{2} < x < \pi. \]

This directly leads to the required conclusion using the first derivative test.

Significance for Competitive Entrance Examinations

For JEE Main, JEE Advanced and other competitive examinations, the sign of a derivative is frequently used to determine monotonicity, local extrema and the shape of a graph.

This example also demonstrates an important connection:

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

Therefore, the intervals in which $\sin x$ increases or decreases are determined by the sign of $\cos x$.

The point

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

is particularly important because

\[ \cos\frac{\pi}{2}=0. \]

At this point, the derivative changes from positive to negative, corresponding to the maximum value

\[ \sin\frac{\pi}{2}=1. \]

Recognising such derivative sign changes can substantially reduce the time required to solve objective questions involving trigonometric functions.

🚫 Common Mistakes

Common Mistakes

›

Comparing only $\sin0$ and $\sin\frac{\pi}{2}$ does not prove that $\sin x$ is increasing throughout the interval.
Similarly, comparing only $\sin\frac{\pi}{2}$ and $\sin\pi$ does not by itself prove decreasing behaviour throughout the interval.
Do not confuse the signs of $\sin x$ and $\cos x$. On $\left(\frac{\pi}{2},\pi\right)$, $\sin x>0$ but $\cos x<0$.
The derivative must be examined on the entire interval, not just at one point.
At $x=\frac{\pi}{2}$, $f'(x)=0$, so the derivative test must be applied separately on the two intervals.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 7 points

For $f(x)=\sin x$, the derivative is
\[ f'(x)=\cos x. \]
Since $\cos x>0$ on
\[ \left(0,\frac{\pi}{2}\right), \]
$\sin x$ is increasing there.
Since $\cos x<0$ on
\[ \left(\frac{\pi}{2},\pi\right), \]
$\sin x$ is decreasing there.
The point
\[ x=\frac{\pi}{2} \]
is where the increasing behaviour changes to decreasing behaviour.
The maximum value of $\sin x$ on $[0,\pi]$ is
\[ 1 \]
at
\[ x=\frac{\pi}{2}. \]
Checking a few function values is not enough to prove monotonicity over an entire interval.
The sign of the first derivative provides a systematic method for determining intervals of increase and decrease.

📘 Concept & Theory

Theory and Concept

›

To determine the intervals in which a differentiable function is increasing or decreasing, we examine the sign of its first derivative.

If

\[ f'(x)>0 \]

throughout an interval, then $f(x)$ is increasing on that interval.

If

\[ f'(x)<0 \]

throughout an interval, then $f(x)$ is decreasing on that interval.

Points at which

\[ f'(x)=0 \]

are called critical points. These points divide the real number line into intervals on which the sign of $f'(x)$ can be investigated separately.

🗺️ Solution Roadmap

Step-by-step Plan

›

Write the given function.
Differentiate the function to obtain $f'(x)$.
Set $f'(x)=0$ to find the critical point.
Use the critical point to divide $\mathbb{R}$ into two intervals.
Determine the sign of $f'(x)$ on each interval.
Where $f'(x)>0$, the function is increasing.
Where $f'(x)<0$, the function is decreasing.

📊 Graph / Figure

Graph / Figure

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Graphical Representation

✏️ Solution

Complete Solution

›

Step-by-step Solution · 22 steps

Given —
\[f(x)=2x^2-3x\]
Differentiating with respect to $x$,
\[f'(x)=\frac{d}{dx}(2x^2-3x)\]
Using
\[\frac{d}{dx}(x^2)=2x\]
and
\[\frac{d}{dx}(x)=1,\]
we get
\[f'(x)=2(2x)-3(1)\]
Therefore,
\[f'(x)=4x-3\]
To Find — the critical point, put
\[f'(x)=0\]
Hence,
\[4x-3=0\]
Adding $3$ to both sides,
\[4x=3\]
Dividing both sides by $4$,
\[x=\frac{3}{4}\]
Thus, the critical point is
\[\boxed{x=\frac{3}{4}}\]
This critical point divides the real number line into the two intervals
\[\left(-\infty,\frac{3}{4}\right)\]
and
\[\left(\frac{3}{4},\infty\right)\]
(a) Interval in which the function is increasing
We have
\[f'(x)=4x-3\]
Consider
\[x\in\left(\frac{3}{4},\infty\right)\]
Therefore,
\[x>\frac{3}{4}\]
Multiplying both sides by $4$,
\[4x>3\]
Subtracting $3$ from both sides,
\[4x-3>0\]
Hence,
\[f'(x)>0\]
for every
\[x\in\left(\frac{3}{4},\infty\right)\]
Therefore, $f(x)$ is increasing on
\[\boxed{\left(\frac{3}{4},\infty\right)}\]
(b) Interval in which the function is decreasing
Now consider
\[x\in\left(-\infty,\frac{3}{4}\right)\]
Therefore,
\[x<\frac{3}{4}\]
Multiplying both sides by $4$,
\[4x<3\]
Subtracting $3$ from both sides,
\[4x-3<0\]
Hence,
\[f'(x)<0\]
for every
\[x\in\left(-\infty,\frac{3}{4}\right)\]
Therefore, $f(x)$ is decreasing on
\[\boxed{\left(-\infty,\frac{3}{4}\right)}\]
Sign Chart for $f'(x)$

The sign of the derivative can be summarised as follows:

Interval	Sign of $4x-3$	Sign of $f'(x)$	Behaviour of $f(x)$
$\left(-\infty,\frac{3}{4}\right)$	Negative	$f'(x) < 0$	Decreasing
$x=\frac{3}{4}$	Zero	$f'(x)=0$	Critical point
$\left(\frac{3}{4},\infty\right)$	Positive	$f'(x)>0$	Increasing

💡 Answer

Final Answer

›

🎯 Exam Significance

Exam Significance

›

This question is a standard application of the first derivative test. It tests whether a student can correctly differentiate a polynomial, identify critical points, construct intervals and determine the sign of the derivative.

A complete board-level solution should show all four essential stages:

Find $f'(x)$.
Solve $f'(x)=0$.
Determine the sign of $f'(x)$ on the resulting intervals.
State the intervals of increase and decrease clearly.

The arithmetic correction is particularly important because an incorrect derivative changes the critical point and consequently gives incorrect intervals.

Significance for Competitive Entrance Examinations

For JEE Main, JEE Advanced and other competitive examinations, the ability to determine monotonic intervals quickly is important in questions involving maxima, minima, inequalities, graphs and the number of roots.

Once

\[ f'(x)=4x-3 \]

is obtained, its sign can be determined immediately:

\[ 4x-3 < 0\quad\text{when}\quad x < \frac{3}{4}, \]

and

\[ 4x-3>0\quad\text{when}\quad x>\frac{3}{4}. \]

Hence, an objective question can often be solved without calculating function values or plotting the complete graph.

🚫 Common Mistakes

Common Mistakes

›

Do not differentiate $2x^2$ as $2x$. The correct derivative is $4x$.
Do not use $x=\frac{3}{2}$ as the critical point. The correct critical point is $x=\frac{3}{4}$.
Do not omit the critical point while constructing the intervals.
Remember that $f'(x)<0$ indicates decreasing behaviour, while $f'(x)>0$ indicates increasing behaviour.
The point where $f'(x)=0$ itself is not included in either open interval of increase or decrease in this answer.
Do not confuse the critical point $x=\frac{3}{4}$ with the minimum function value
\[ f\left(\frac{3}{4}\right)=-\frac{9}{8}. \]

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 7 points

For
\[ f(x)=2x^2-3x, \]
the first derivative is
\[ f'(x)=4x-3. \]
The critical point is obtained from
\[ 4x-3=0, \]
giving
\[ x=\frac{3}{4}. \]
The function is decreasing when
\[ x < \frac{3}{4}. \]
The function is increasing when
\[ x>\frac{3}{4}. \]
The function has a minimum at
\[ x=\frac{3}{4}. \]
The minimum value is
\[ -\frac{9}{8}. \]
The graph is an upward-opening parabola because the coefficient of $x^2$ is positive.

📘 Concept & Theory

Theory and Concept

›

For a differentiable function $f(x)$, the sign of the first derivative determines whether the function is increasing or decreasing.

If

\[ f'(x)>0 \]

throughout an interval, then $f(x)$ is increasing on that interval.

If

\[ f'(x) < 0 \]

throughout an interval, then $f(x)$ is decreasing on that interval.

The values of $x$ for which

\[ f'(x)=0 \]

are called critical points. These points divide the real number line into intervals. The sign of $f'(x)$ is then determined separately on each interval.

🗺️ Solution Roadmap

Step-by-step Plan

›

Write the given polynomial function.
Differentiate it to obtain $f'(x)$.
Set $f'(x)=0$ and solve the resulting quadratic equation.
Use the critical points to divide the real number line into three intervals.
Factorise $f'(x)$ to make its sign easy to determine.
Determine the sign of $f'(x)$ on each interval.
Use $f'(x)>0$ for increasing intervals and $f'(x)<0$ for decreasing intervals.

📊 Graph / Figure

Graph / Figure

›

Graphical Representation

✏️ Solution

Complete Solution

›

Step-by-step Solution · 34 steps

Given —
\[f(x)=2x^3-3x^2-36x+7\]
Differentiating with respect to $x$,
\[f'(x)=\frac{d}{dx}\left(2x^3-3x^2-36x+7\right)\]
Using the power rule,
\[\frac{d}{dx}(x^n)=nx^{n-1},\]
we obtain
\[f'(x)=2(3x^2)-3(2x)-36+0\]
Therefore,
\[f'(x)=6x^2-6x-36\]
Factor out $6$:
\[f'(x)=6(x^2-x-6)\]
Now factorise the quadratic:
\[x^2-x-6=x^2-3x+2x-6\]
Grouping the terms,
\[x^2-3x+2x-6=x(x-3)+2(x-3)\]
Taking the common factor $(x-3)$,
\[x(x-3)+2(x-3)=(x-3)(x+2)\]
Hence,
\[\boxed{f'(x)=6(x-3)(x+2)}\]
Finding the Critical Points
Critical points are obtained by putting
\[f'(x)=0.\]
Therefore,
\[6(x-3)(x+2)=0\]
Since $6\neq0$,
\[(x-3)(x+2)=0\]
Hence,
\[x-3=0\]
or
\[x+2=0\]
Therefore,
\[x=3\]
or
\[x=-2\]
Thus, the critical points are
\[ \boxed{x=-2,\;3}. \]
These two critical points divide the real number line into three intervals:
\[\left(-\infty,-2\right),\]

\[\left(-2,3\right),\]
and
\[\left(3,\infty\right).\]
Determining the Sign of $f'(x)$
We have
\[f'(x)=6(x-3)(x+2).\]
Since $6$ is positive, it does not affect the sign of the derivative. Therefore, we only need to determine the signs of
\[(x-3)(x+2)\]
Interval 1: $\left(-\infty,-2\right)$
For $x < -2,$
we have
\[x-3 < 0\]
and
\[x+2 < 0\]
Therefore,
\[(x-3)(x+2)=(-)(-)>0\]
Since $6>0$,
\[f'(x)=6(x-3)(x+2)>0\]
Hence, $f(x)$ is increasing on
\[\boxed{\left(-\infty,-2\right)}\]
Interval 2: $\left(-2,3\right)$
For
\[-2 < x < 3,\]
we have
\[x-3 < 0\]
but
\[x+2>0\]
Therefore,
\[(x-3)(x+2)=(-)(+) < 0\]
Hence,
\[f'(x) < 0\]
Therefore, $f(x)$ is decreasing on
\[\boxed{\left(-2,3\right)}\]
Interval 3: $\left(3,\infty\right)$
For
\[x>3,\]
we have
\[x-3>0\]
and
\[x+2>0.\]
Therefore,
\[(x-3)(x+2)=(+)(+)>0.\]
Hence,
\[f'(x)>0.\]
Therefore, $f(x)$ is increasing on
\[\boxed{\left(3,\infty\right)}.\]
Sign Chart

Interval	$x-3$	$x+2$	Sign of $f'(x)$	Behaviour
$\left(-\infty,-2\right)$	−	−	+	Increasing
$x=-2$	−	0	0	Critical point
$\left(-2,3\right)$	−	+	−	Decreasing
$x=3$	0	+	0	Critical point
$\left(3,\infty\right)$	+	+	+	Increasing

🎯 Exam Significance

Exam Significance

›

This problem is an important application of the first derivative test. It combines polynomial differentiation, factorisation, critical-point analysis and the determination of intervals of monotonicity.

In a board examination, students should show the factorisation

\[ f'(x)=6(x-3)(x+2) \]

because it makes the sign analysis transparent.

The final answer should clearly distinguish the increasing and decreasing intervals rather than merely listing the critical points.

Significance for Competitive Entrance Examinations

This question develops a standard technique used extensively in JEE Main, JEE Advanced and other competitive entrance examinations: factorising the derivative and constructing its sign pattern.

Once

\[ f'(x)=6(x-3)(x+2) \]

is obtained, the entire monotonicity behaviour can be determined from the two critical points

\[ x=-2,\qquad x=3. \]

The derivative sign pattern

\[ +,\;-,\;+ \]

immediately shows that the function has a local maximum at $x=-2$ and a local minimum at $x=3$. This technique is particularly useful in questions involving maximum and minimum values, graph behaviour, inequalities and the number of solutions of equations.

🚫 Common Mistakes

Common Mistakes

›

Do not omit the factor $6$ from the derivative without understanding that it is positive and therefore does not affect the sign.
Do not write the critical points as points of inflection. The values $x=-2$ and $x=3$ are critical points; they are turning points of the function, not points of inflection.
Remember that a point of inflection is associated with a change in concavity, not simply with $f'(x)=0$.
Do not write
\[ \left(\infty,-2\right) \]
for the first interval. The correct interval is
\[ \left(-\infty,-2\right). \]
Do not forget the interval $(-2,3)$, where the derivative is negative.
For a product such as
\[ (x-3)(x+2), \]
determine the signs of both factors before determining the sign of the product.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 8 points

The first derivative of
\[ f(x)=2x^3-3x^2-36x+7 \]
is
\[ f'(x)=6x^2-6x-36. \]
The factorised derivative is
\[ f'(x)=6(x-3)(x+2). \]
The critical points are
\[ x=-2,\quad x=3. \]
The function is increasing on
\[ \left(-\infty,-2\right)\cup\left(3,\infty\right). \]
The function is decreasing on
\[ \left(-2,3\right). \]
The derivative sign pattern is
\[ +\rightarrow-\rightarrow+. \]
Therefore, $x=-2$ gives a local maximum and $x=3$ gives a local minimum.
The local maximum value is $51$, while the local minimum value is $-74$.

Q6

LONG ANSWER4 marks

Find the intervals in which each of the following functions is strictly increasing or strictly decreasing:

\[ f(x)=x^2+2x-5 \]
\[ f(x)=10-6x-2x^2 \]
\[ f(x)=-2x^3-9x^2-12x+1 \]
\[ f(x)=6-9x-x^2 \]
\[ f(x)=(x+1)^3(x-3)^3 \]

📘 Concept & Theory

Theory and Concept

›

For a differentiable function $f(x)$, the sign of its first derivative determines its increasing or decreasing behaviour.

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.

The points at which

\[ f'(x)=0 \]

are critical points. These points divide the real number line into intervals on which the sign of $f'(x)$ is determined.

A critical point should not automatically be called a point of inflection. A point of inflection is associated with a change in concavity. In monotonicity questions, the values satisfying

\[ f'(x)=0 \]

should be called critical points unless a change in concavity has also been established.

🗺️ Solution Roadmap

Step-by-step Plan

›

Differentiate each given function.
Set $f'(x)=0$ to find the critical points.
Use the critical points to divide the real number line into suitable intervals.
Determine the sign of $f'(x)$ on each interval.
Use $f'(x)>0$ to identify strictly increasing intervals.
Use $f'(x)<0$ to identify strictly decreasing intervals.
For part (e), pay special attention to repeated factors because a squared factor does not change sign at its zero.

✏️ Solution

(a) ${f(x)=x^2+2x-5}$

›

Step-by-step Solution · 16 steps

Given —
\[f(x)=x^2+2x-5\]
Find the derivative
Differentiating with respect to $x$,
\[f'(x)=2x+2\]
Find the critical point
Put
\[f'(x)=0\]
Therefore,
\[2x+2=0\]
Subtracting $2$ from both sides,
\[2x=-2\]
Dividing by $2$,
\[x=-1\]
Hence, the critical point is
\[\boxed{x=-1}\]
Determine the sign of $f'(x)$
The critical point $x=-1$ divides the real number line into
\[\left(-\infty,-1\right)\]
and
\[\left(-1,\infty\right)\]
For
\[x < -1,\]
we have
\[2x+2 < 0\]
Hence,
\[f'(x)<0\]
Therefore, $f(x)$ is strictly decreasing on
\[\boxed{\left(-\infty,-1\right)}\]
For
\[x>-1,\]
we have
\[2x+2>0\]
Hence,
\[f'(x)>0\]
Therefore, $f(x)$ is strictly increasing on
\[\boxed{\left(-1,\infty\right)}\]
$Graph of function $x^2+2x-5$$

✏️ Solution

(b) ${f(x)=10-6x-2x^2}$

›

Step-by-step Solution · 16 steps

Given —
\[f(x)=10-6x-2x^2\]
Find the derivative
Differentiating,
\[f'(x)=-6-4x\]
Find the critical point
Put
\[f'(x)=0\]
Therefore,
\[-6-4x=0\]
Adding $6$ to both sides,
\[-4x=6\]
Dividing by $-4$,
\[x=-\frac{6}{4}=-\frac{3}{2}\]
Thus, the critical point is
\[\boxed{x=-\frac{3}{2}}\]
Determine the sign of $f'(x)$
The critical point divides the real number line into
\[\left(-\infty,-\frac{3}{2}\right)\]
and
\[\left(-\frac{3}{2},\infty\right).\]
For
\[x < -\frac{3}{2},\]
we have
\[-4x>6,\]
and hence
\[-6-4x>0\]
Therefore,
\[f'(x)>0\]
Thus, $f(x)$ is strictly increasing on
\[\boxed{\left(-\infty,-\frac{3}{2}\right)}\]
For
\[x>-\frac{3}{2},\]
we have
\[-6-4x < 0\]
Therefore,
\[f'(x) < 0\]
Thus, $f(x)$ is strictly decreasing on
\[\boxed{\left(-\frac{3}{2},\infty\right)}\]
Graphical Representation
$Graph of function $10-6x-2x^2$$

✏️ Solution

(c) ${f(x)=-2x^3-9x^2-12x+1}$

›

Step-by-step Solution · 29 steps

Given —
\[f(x)=-2x^3-9x^2-12x+1\]
Find the derivative
Differentiating,
\[f'(x)=-6x^2-18x-12\]
Factor out $-6$:
\[f'(x)=-6(x^2+3x+2)\]
Factorising the quadratic,
\[x^2+3x+2=x^2+2x+x+2\]
Therefore,
\[x^2+2x+x+2=x(x+2)+1(x+2)\]
Taking $(x+2)$ common,
\[x(x+2)+1(x+2)=(x+2)(x+1)\]
Hence,
\[\boxed{f'(x)=-6(x+2)(x+1)}\]
Find the critical points
Put
\[f'(x)=0\]
Therefore,
\[-6(x+2)(x+1)=0\]
Hence,
\[x+2=0\]
or
\[x+1=0\]
Therefore,
\[\boxed{x=-2,\;-1}\]
Determine the sign of $f'(x)$
The critical points divide the real number line into
\[\left(-\infty,-2\right),\quad\left(-2,-1\right),\quad\left(-1,\infty\right)\]
Since
\[f'(x)=-6(x+2)(x+1),\]
the factor $-6$ is negative and must be included in the sign analysis.
For
\[x < -2,\]
both
\[x+2 < 0,\quad x+1 < 0\]
Therefore,
\[(x+2)(x+1)>0\]
Multiplication by $-6$ gives
\[f'(x) < 0\]
Hence $f(x)$ is strictly decreasing on
\[\boxed{\left(-\infty,-2\right)}\]
For
\[-2 < x < -1,\]
we have
\[x+2>0,\quad x+1 < 0\]
Hence,
\[(x+2)(x+1)<0\]
Since $-6$ is negative,
\[f'(x)>0\]
Therefore, $f(x)$ is strictly increasing on
\[\boxed{\left(-2,-1\right)}\]
For
\[x>-1,\]
both factors are positive:
\[x+2>0,\quad x+1>0\]
Hence,
\[(x+2)(x+1)>0\]
Multiplying by $-6$,
\[f'(x) < 0\]
Therefore, $f(x)$ is strictly decreasing on
\[\boxed{\left(-1,\infty\right)}\]
Sign Pattern

Interval	Sign of $f'(x)$	Behaviour
$\left(-\infty,-2\right)$	$-$	Decreasing
$\left(-2,-1\right)$	$+$	Increasing
$\left(-1,\infty\right)$	$-$	Decreasing

Graphical Representation
$Graph of function $-2x^3-9x^2-12x+1$$

✏️ Solution

(d) ${f(x)=6-9x-x^2}$

›

Step-by-step Solution · 16 steps

Given —
\[f(x)=6-9x-x^2\]
Find the derivative
Differentiating,
\[f'(x)=-9-2x\]
Find the critical point
Put
\[f'(x)=0\]
Therefore,
\[-9-2x=0\]
Adding $9$ to both sides,
\[-2x=9\]
Dividing by $-2$,
\[x=-\frac{9}{2}\]
Thus, the critical point is
\[\boxed{x=-\frac{9}{2}}\]
Determine the sign of $f'(x)$
For
\[x < -\frac{9}{2},\]
we have
\[-2x>9\]
Therefore,
\[-9-2x>0\]
Hence,
\[f'(x)>0\]
Thus, the function is strictly increasing on
\[\boxed{\left(-\infty,-\frac{9}{2}\right)}\]
For
\[x>-\frac{9}{2},\]
we have
\[-2x < 9,\]
so
\[-9-2x < 0\]
Hence,
\[f'(x) < 0\]
Therefore, the function is strictly decreasing on
\[\boxed{\left(-\frac{9}{2},\infty\right)}\]
Graphical Representation
$Graph of function $f(x)=(6-9x-x^2$$

✏️ Solution

(e) ${f'(x)=(x+1)^3(x-3)^3}$

›

Step-by-step Solution · 34 steps

Given —
\[f(x)=(x+1)^3(x-3)^3\]
Find the derivative
Since the function is a product of two functions, use the product rule:
\[\frac{d}{dx}[uv]=u'v+uv'\]
Let
\[u=(x+1)^3\]
and
\[v=(x-3)^3\]
Then,
\[u'=3(x+1)^2\]
and
\[v'=3(x-3)^2\]
Therefore,
\[f'(x)=3(x+1)^2(x-3)^3+3(x+1)^3(x-3)^2\]
Taking the common factor
\[3(x+1)^2(x-3)^2,\]
we get
\[f'(x)=3(x+1)^2(x-3)^2\left[(x-3)+(x+1)\right]\]
Simplifying the expression inside the bracket,
\[(x-3)+(x+1)=2x-2\]
Therefore,
\[f'(x)=3(x+1)^2(x-3)^2(2x-2)\]
Taking $2$ common from $2x-2$,
\[2x-2=2(x-1)\]
Hence,
\[\boxed{f'(x)=6(x+1)^2(x-3)^2(x-1)}\]
Find the critical points
Put
\[f'(x)=0\]
Therefore,
\[6(x+1)^2(x-3)^2(x-1)=0\]
Since $6\neq0$, we obtain
\[(x+1)^2(x-3)^2(x-1)=0\]
Therefore,
\[(x+1)^2=0,\]
or
\[(x-3)^2=0,\]
or
\[x-1=0\]
Hence,
\[x=-1,\quad x=3,\quad x=1\]
Thus, the critical points are
\[ \boxed{x=-1,\;1,\;3}. \]
Determine the sign of $f'(x)$
We have
\[f'(x)=6(x+1)^2(x-3)^2(x-1)\]
Notice that
\[(x+1)^2\geq0\]
and
\[(x-3)^2\geq0\]
Except at $x=-1$ and $x=3$, these squared factors are positive.
Therefore, away from the critical points, the sign of $f'(x)$ is determined entirely by
\[x-1\]
Interval 1: $\left(-\infty,-1\right)$
Here,
\[x-1 < 0\]
The squared factors are positive, so
\[f'(x) < 0\]
Hence, $f(x)$ is strictly decreasing on
\[\left(-\infty,-1\right)\]
Interval 2: $\left(-1,1\right)$
Here also,
\[x-1 < 0\]
Therefore,
\[f'(x)<0\]
Hence, $f(x)$ is strictly decreasing on
\[\left(-1,1\right)\]
Since the derivative does not change from negative to positive or from positive to negative at $x=-1$, the point $x=-1$ does not separate the overall decreasing interval.
Therefore, the two adjacent intervals can be combined:
\[ \boxed{f(x)\text{ is strictly decreasing on }\left(-\infty,1\right)}. \]
Interval 3: $\left(1,3\right)$
Here,
\[x-1 > 0\]
The squared factors are positive, so
\[f'(x)>0\]
Hence, $f(x)$ is strictly increasing on
\[\left(1,3\right)\]
Interval 4: $\left(3,\infty\right)$
Here,
\[x-1 > 0\]
Again,
\[f'(x)>0\]
Hence, $f(x)$ is strictly increasing on
\[\left(3,\infty\right)\]
Since the derivative remains positive on both sides of $x=3$, the point $x=3$ does not interrupt the overall increasing behaviour. Therefore, the intervals can be combined:
\[ \boxed{f(x)\text{ is strictly increasing on }\left(1,\infty\right)}. \]
Sign Chart for Part (e)

Interval	Sign of $f'(x)$	Behaviour
$\left(-\infty,-1\right)$	$-$	Decreasing
$\left(-1,1\right)$	$-$	Decreasing
$\left(1,3\right)$	$+$	Increasing
$\left(3,\infty\right)$	$+$	Increasing

Graphical Representation
$Graph of function $(x+1)^3(x-3)^3$$

💡 Answer

Final Answer

›

Part	Function	Strictly Increasing	Strictly Decreasing
(a)	$x^2+2x-5$	$\left(-1,\infty\right)$	$\left(-\infty,-1\right)$
(b)	$10-6x-2x^2$	$\left(-\infty,-\frac{3}{2}\right)$	$\left(-\frac{3}{2},\infty\right)$
(c)	$-2x^3-9x^2-12x+1$	$\left(-2,-1\right)$	$\left(-\infty,-2\right)\cup\left(-1,\infty\right)$
(d)	$6-9x-x^2$	$\left(-\infty,-\frac{9}{2}\right)$	$\left(-\frac{9}{2},\infty\right)$
(e)	$f'(x)=(x+1)^3(x-3)^3$	$\left(1,\infty\right)$	$\left(-\infty,1\right)$

🎯 Exam Significance

Exam Significance

›

This question is important because it covers several standard applications of the first derivative: finding critical points, constructing intervals, analysing the sign of the derivative and identifying increasing and decreasing behaviour.

In board examinations, students should avoid jumping directly to the final intervals. The expected logical sequence is

\[ f'(x)\rightarrow f'(x)=0\rightarrow\text{critical points}\rightarrow\text{sign of }f'(x) \rightarrow\text{monotonicity}. \]

Part (e) is particularly valuable because it tests whether the student understands that repeated roots of $f'(x)$ do not necessarily produce a change in the sign of the derivative.

Significance for Competitive Entrance Examinations

For JEE Main, JEE Advanced and other competitive entrance examinations, these problems strengthen the ability to analyse the derivative sign pattern rapidly.

Factorisation is especially useful. For example, in part (c),

\[ f'(x)=-6(x+2)(x+1), \]

immediately gives the critical points and allows a sign chart to be constructed.

Part (e) is a higher-value conceptual problem because

\[ f'(x)=6(x+1)^2(x-3)^2(x-1) \]

contains repeated factors. The squared factors never become negative, so only $x-1$ determines the sign change. This allows the final intervals to be combined:

\[ \text{Decreasing on }(-\infty,1), \qquad \text{Increasing on }(1,\infty). \]

🚫 Common Mistakes

Common Mistakes

›

Do not call every point where $f'(x)=0$ a point of inflection.
Always include the numerical sign of any constant factor in $f'(x)$. In part (c), the factor $-6$ reverses the sign.
Check the sign of the derivative rather than the sign of the original function.
Do not make an arithmetic sign error when solving $f'(x)=0$.
Repeated factors such as $(x+1)^2$ and $(x-3)^2$ do not change sign when crossing their zeros.
Do not automatically keep every critical point as a separate monotonicity interval. Adjacent intervals can be combined when the derivative has the same sign on both sides.
Strictly increasing or decreasing behaviour should be established from the derivative, not from a few selected function values.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 8 points

\[ f'(x)>0 \]
indicates strictly increasing behaviour.
\[ f'(x)<0 \]
indicates strictly decreasing behaviour.
Critical points are obtained by solving
\[ f'(x)=0. \]
The critical points divide the real number line into intervals for sign analysis.
A negative constant factor in the derivative must be included when determining its sign.
Repeated even-power factors do not change the sign of the derivative.
Not every critical point is a point of inflection.
In part (e), the derivative changes sign only at $x=1$, so the final monotonicity intervals are
\[ (-\infty,1) \]
and
\[ (1,\infty). \]

📘 Concept & Theory

Theory and Concept

›

To show that a differentiable function is increasing throughout an interval, we use the first derivative test. If

\[ \frac{dy}{dx}>0 \]

throughout the given domain, then $y$ is strictly increasing on that domain.

Here,

\[ y=\log(1+x)-\frac{2x}{2+x}. \]

The given restriction is

\[ x>-1. \]

This restriction is important because the logarithm is defined only when

\[ 1+x>0, \]

which gives

\[ x>-1. \]

We differentiate both terms carefully and simplify the resulting expression. The crucial result will be

\[ \frac{dy}{dx}=\frac{x^2}{(1+x)(2+x)^2}. \]

Since $x^2\geq0$, $1+x>0$, and $(2+x)^2>0$ throughout the domain $x>-1$, the derivative is non-negative. At $x=0$, the derivative is zero, but the function is still strictly increasing throughout the domain. This can be established by observing that the derivative is positive on either side of $x=0$, with only an isolated zero at $x=0$.

🗺️ Solution Roadmap

Step-by-step Plan

›

Write the given function and its domain.
Differentiate $\log(1+x)$.
Differentiate $\frac{2x}{2+x}$ using the quotient rule.
Combine the two derivative terms over a common denominator.
Carefully simplify the numerator to obtain $x^2$.
Analyse the signs of all factors in the denominator for $x>-1$.
Show that the derivative is non-negative throughout the domain and positive except at $x=0$.
Conclude that the function is strictly increasing throughout its domain.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 27 steps

Given —
\[y=\log(1+x)-\frac{2x}{2+x},\quad x>-1\]
Differentiate the logarithmic term
Differentiating
\[\log(1+x)\]
with respect to $x$,
\[\frac{d}{dx}\log(1+x)=\frac{1}{1+x}\]
Differentiate the rational term
We need to differentiate
\[\frac{2x}{2+x}\]
Using the quotient rule,
\[\frac{d}{dx}\left(\frac{u}{v}\right)=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}\]
Here,
\[u=2x,\quad v=2+x\]
Therefore,
\[\frac{du}{dx}=2\]
and
\[\frac{dv}{dx}=1\]
Hence,
\[\frac{d}{dx}\left(\frac{2x}{2+x}\right)=\frac{(2+x)(2)-(2x)(1)}{(2+x)^2}\]
Expanding the numerator,
\[\frac{d}{dx}\left(\frac{2x}{2+x}\right)=\frac{4+2x-2x}{(2+x)^2}\]
Therefore,
\[\boxed{\frac{d}{dx}\left(\frac{2x}{2+x}\right)=\frac{4}{(2+x)^2}}\]
Find $\frac{dy}{dx}$
Since
\[y=\log(1+x)-\frac{2x}{2+x},\]
we have
\[\frac{dy}{dx}=\frac{1}{1+x}-\frac{4}{(2+x)^2}\]
Taking the common denominator,
\[\frac{dy}{dx}=\frac{(2+x)^2-4(1+x)}{(1+x)(2+x)^2}\]
Expand the numerator:
\[(2+x)^2=x^2+4x+4\]
Therefore,
\[\frac{dy}{dx}=\frac{x^2+4x+4-4-4x}{(1+x)(2+x)^2}\]
Combining like terms,
\[4x-4x=0\]
and
\[4-4=0\]
Hence,
\[\boxed{\frac{dy}{dx}=\frac{x^2}{(1+x)(2+x)^2}}\]
Determine the Sign of $\frac{dy}{dx}$
The given domain is
\[x>-1\]
Therefore,
\[1+x>0\]
Also, if $x>-1$, then
\[x+2>1>0\]
Hence,
\[(x+2)^2>0\]
Finally,
\[x^2\geq0\]
for every real number $x$.
Consequently,
\[\frac{x^2}{(1+x)(2+x)^2}\geq0\]
for every
\[x>-1\]
More specifically,
\[\frac{dy}{dx}>0\]
for
\[-1 < x < 0\]
and for
\[x>0,\]
while
\[\frac{dy}{dx}=0\]
only at
\[x=0\]
Thus, the derivative never becomes negative anywhere in the domain.
Why the Function Is Still Strictly Increasing at $x=0$
It is important to understand that having
\[ f'(0)=0 \]
does not mean that the function ceases to be increasing.
A zero derivative at an isolated point does not prevent a function from being strictly increasing. For example, the function
\[ f(x)=x^3 \]
has
\[ f'(0)=0, \]
but $x^3$ is strictly increasing on $\mathbb{R}$.
Here, $\frac{dy}{dx}$ is positive on both sides of $x=0$. Hence, the function continues to increase as $x$ passes through $0$.

Therefore,
\[ \boxed{ y=\log(1+x)-\frac{2x}{2+x} } \]
is strictly increasing throughout its domain
\[ \boxed{x>-1}. \]
Graphical Representation
Sign Analysis at a Glance

Quantity	For $x>-1$
$x^2$	$\geq0$
$1+x$	$>0$
$(2+x)^2$	$>0$
$\frac{x^2}{(1+x)(2+x)^2}$	$\geq0$
At $x=0$	$f'(0)=0$
For $x\neq0$	$f'(x)>0$

🎯 Exam Significance

Exam Significance

›

This question is important because it combines logarithmic differentiation, the quotient rule, algebraic simplification and monotonicity.

The most important examination step is simplifying the derivative correctly:

\[ \frac{dy}{dx} = \frac{x^2}{(1+x)(2+x)^2}. \]

Students should also justify why the denominator is positive under the given condition $x>-1$. Simply stating that the derivative is positive without analysing its factors leaves an important logical step incomplete.

Another important board-level concept is that

\[ f'(x)=0 \]

at an isolated point does not necessarily mean that the function stops being strictly increasing.

Significance for Competitive Entrance Examinations

This problem is useful for competitive examinations because it demonstrates how a complicated-looking derivative can often be reduced to a simple sign expression.

Once we obtain

\[ f'(x)=\frac{x^2}{(1+x)(2+x)^2}, \]

the sign is almost immediate:

\[ x^2\geq0,\qquad 1+x>0,\qquad (2+x)^2>0. \]

Therefore, no detailed interval testing is required. This type of factor-based sign analysis is valuable in JEE Main, JEE Advanced and other entrance examinations involving monotonicity, inequalities and function behaviour.

🚫 Common Mistakes

Common Mistakes

›

Do not simplify
\[ (x+2)^2-4(x+1) \]
incorrectly as $x^2-2x$. The correct result is $x^2$.
Do not forget the negative sign before the derivative of the second term.
Do not ignore the domain condition $x>-1$.
Do not assume that $f'(x)=0$ at $x=0$ means the function is not increasing.
Do not say that $x=-2$ belongs to the domain. It does not, and it is already excluded by $x>-1$.
Do not confuse $f(x)>0$ with $f'(x)>0$. It is the sign of the derivative that determines monotonicity.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 8 points

The domain of the function is
\[ (-1,\infty). \]
The derivative of the logarithmic term is
\[ \frac{1}{1+x}. \]
\[ \frac{d}{dx}\left(\frac{2x}{2+x}\right) = \frac{4}{(2+x)^2}. \]
The simplified derivative is
\[ \frac{dy}{dx} = \frac{x^2}{(1+x)(2+x)^2}. \]
For $x>-1$,
\[ 1+x>0 \]
and
\[ (2+x)^2>0. \]
\[ x^2\geq0 \]
for every real $x$, with equality only at $x=0$.
The derivative is positive everywhere in the domain except at the isolated point $x=0$, where it is zero.
The function is therefore strictly increasing throughout its domain.

📘 Concept & Theory

Theory and Concept

›

To determine where a differentiable function is increasing, we examine the sign of its first derivative.

If

\[ \frac{dy}{dx}>0 \]

throughout an interval, then the function is strictly increasing on that interval.

The values of $x$ for which

\[ \frac{dy}{dx}=0 \]

are critical points. These points divide the real number line into intervals on which the sign of the derivative can be studied.

Here, the function is a square of a quadratic expression:

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

The chain rule is therefore required when differentiating.

🗺️ Solution Roadmap

Step-by-step Plan

›

Rewrite the function as $y=(x^2-2x)^2$.
Differentiate using the chain rule.
Factorise the derivative completely.
Find the critical points by putting $\frac{dy}{dx}=0$.
Divide the real number line using the critical points $0,1,2$.
Determine the sign of the derivative on each interval.
Identify the intervals where $\frac{dy}{dx}>0$.

📊 Graph / Figure

Graph / Figure

›

Graph of function $y=\left[x(x-2)\right]^2$ — Graphical Representation

✏️ Solution

Complete Solution

›

Step-by-step Solution · 36 steps

Given —
\[y=\left[x(x-2)\right]^2\]
First simplify the expression inside the square:
\[x(x-2)=x^2-2x\]
Therefore,
\[y=(x^2-2x)^2\]
Differentiate using the Chain Rule
Let
\[u=x^2-2x\]
Then
\[y=u^2\]
Differentiating $y=u^2$ with respect to $u$,
\[\frac{dy}{du}=2u\]
Also,
\[\frac{du}{dx}=\frac{d}{dx}(x^2-2x)=2x-2\]
By the chain rule,
\[\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}\]
Hence,
\[\frac{dy}{dx}=2(x^2-2x)(2x-2)\]
Factorise each expression:
\[x^2-2x=x(x-2)\]
and
\[2x-2=2(x-1)\]
Therefore,
\[\frac{dy}{dx}=2[x(x-2)]\cdot2(x-1)\]
Hence,
\[\boxed{\frac{dy}{dx}=4x(x-1)(x-2)}\]
Find the Critical Points
For the critical points, put
\[\frac{dy}{dx}=0\]
Therefore,
\[4x(x-1)(x-2)=0\]
Since $4\neq0$,
\[x(x-1)(x-2)=0\]
Hence,
\[x=0,\]
or
\[x-1=0,\]
or
\[x-2=0.\]
Therefore,
\[ \boxed{x=0,\;1,\;2}. \]
These three critical points divide the real number line into four intervals:
\[(-\infty,0),\]

\[(0,1),\]

\[(1,2),\]
and
\[(2,\infty)\]
Determine the Sign of the Derivative
We have
\[\frac{dy}{dx}=4x(x-1)(x-2)\]
Since $4>0$, the sign of the derivative is determined by
\[x(x-1)(x-2)\]
Interval 1: ${(-\infty,0)}$
For
\[x < 0,\]
all three factors are negative:
\[x < 0,\]

\[x-1 < 0,\]

\[x-2 < 0.\]
Therefore,
\[(-)(-)(-)=-.\]
Hence,
\[\frac{dy}{dx} < 0\]
Therefore, the function is decreasing on
\[\boxed{(-\infty,0)}\]
Interval 2: ${(0,1)}$
For
\[0 < x < 1,\]
we have
\[x>0,\]

\[x-1 < 0,\]
and
\[x-2 < 0.\]
Therefore,
\[(+)(-)(-)=+.\]
Hence,
\[\frac{dy}{dx}>0.\]
Therefore, the function is increasing on
\[\boxed{(0,1)}\]
Interval 3: ${(1,2)}$
For
\[1 < x < 2,\]
we have
\[x>0,\]

\[x-1>0,\]
and
\[x-2 < 0.\]
Therefore,
\[(+)(+)(-)=-.\]
Hence,
\[\frac{dy}{dx} < 0.\]
Therefore, the function is decreasing on
\[\boxed{(1,2)}\]
Interval 4: ${(2,\infty)}$
For
\[x>2,\]
all three factors are positive:
\[x>0,\]

\[x-1>0,\]

\[x-2>0.\]
Therefore,
\[(+)(+)(+)=+.\]
Hence,
\[ \frac{dy}{dx}>0. \]
Therefore, the function is increasing on
\[ \boxed{(2,\infty)}. \]
Sign Chart

Interval	$x$	$x-1$	$x-2$	Sign of $\frac{dy}{dx}$	Behaviour
$(-\infty,0)$	−	−	−	−	Decreasing
$(0,1)$	+	−	−	+	Increasing
$(1,2)$	+	+	−	−	Decreasing
$(2,\infty)$	+	+	+	+	Increasing

🎯 Exam Significance

Exam Significance

›

This question tests the use of the chain rule, factorisation of the derivative and the first derivative test for monotonicity.

The key board-level sequence is

\[ y\rightarrow\frac{dy}{dx}\rightarrow\frac{dy}{dx}=0 \rightarrow\text{critical points}\rightarrow\text{sign analysis}. \]

A complete solution should not stop after finding

\[ x=0,1,2. \]

These are only critical points. The increasing intervals must be obtained by analysing the sign of $\frac{dy}{dx}$.

Significance for Competitive Entrance Examinations

This problem is useful for competitive examinations because the derivative factorises into three linear factors:

\[ \frac{dy}{dx}=4x(x-1)(x-2). \]

Once this form is obtained, the sign pattern can be determined rapidly:

\[ -,\; +,\; -,\; +. \]

This immediately gives the intervals of increase and decrease and also identifies the local maxima and minima. Such sign-chart techniques are frequently useful in JEE Main, JEE Advanced and other entrance examinations.

🚫 Common Mistakes

Common Mistakes

›

Do not forget the chain rule when differentiating $[x(x-2)]^2$.
The derivative is
\[ 4x(x-1)(x-2), \]
not merely $2x(x-2)$.
Do not call $x=0,1,2$ points of inflection merely because the derivative is zero there.
Finding critical points alone does not determine the intervals of increase.
Always analyse the sign of the derivative between consecutive critical points.
Remember that the factor $4$ is positive and therefore does not change the sign of the derivative.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 8 points

The given function can be written as
\[ y=(x^2-2x)^2. \]
Using the chain rule,
\[ \frac{dy}{dx}=4x(x-1)(x-2). \]
The critical points are
\[ x=0,\;1,\;2. \]
The derivative is negative on
\[ (-\infty,0) \]
and
\[ (1,2). \]
The derivative is positive on
\[ (0,1) \]
and
\[ (2,\infty). \]
Therefore, the function is increasing on
\[ (0,1)\cup(2,\infty). \]
The function is decreasing on
\[ (-\infty,0)\cup(1,2). \]
The points $x=0$ and $x=2$ are local minima, while $x=1$ is a local maximum.

📘 Concept & Theory

Theory and Concept

›

To prove that a differentiable function is increasing on an interval, we examine the sign of its first derivative.

If

\[ \frac{dy}{d\theta}>0 \]

throughout the interval, then $y$ is strictly increasing there.

The given function contains a quotient involving $\sin\theta$ and $\cos\theta$, so the quotient rule is required for differentiation.

🗺️ Solution Roadmap

Step-by-step Plan

›

Differentiate the quotient using the quotient rule.
Remember to differentiate the term $-\theta$, which contributes $-1$.
Simplify the numerator using
\[ \sin^2\theta+\cos^2\theta=1. \]
Determine the sign of the derivative on
\[ \left[0,\frac{\pi}{2}\right]. \]
Conclude that $y$ is increasing on the given interval.

📊 Graph / Figure

Graph / Figure

›

Graph of function $y=\frac{4\sin\theta}{2+\cos\theta}-\theta$ — Graphical Representation

✏️ Solution

Complete Solution

›

Step-by-step Solution · 21 steps

Given —
\[y=\frac{4\sin\theta}{2+\cos\theta}-\theta\]
Differentiate the function
For the first term, let
\[u=4\sin\theta\]
and
\[v=2+\cos\theta\]
Then
\[u'=4\cos\theta\]
and
\[v'=-\sin\theta\]
By the quotient rule,
\[\frac{d}{d\theta}\left(\frac{u}{v}\right)=\frac{vu'-uv'}{v^2}\]
Therefore,
\[\frac{dy}{d\theta}=\frac{(2+\cos\theta)(4\cos\theta)-(4\sin\theta)(-\sin\theta)}{(2+\cos\theta)^2}-1\]
Simplifying the numerator,
\[\frac{dy}{d\theta}=\frac{8\cos\theta+4\cos^2\theta+4\sin^2\theta}{(2+\cos\theta)^2}-1\]
Using
\[\sin^2\theta+\cos^2\theta=1,\]
we get
\[4\cos^2\theta+4\sin^2\theta=4\]
Hence,
\[\frac{dy}{d\theta}=\frac{8\cos\theta+4}{(2+\cos\theta)^2}-1\]
Combine the terms correctly
Write $1$ with the same denominator:
\[\frac{dy}{d\theta}=\frac{8\cos\theta+4-(2+\cos\theta)^2}{(2+\cos\theta)^2}\]
Expand the square:
\[(2+\cos\theta)^2=4+4\cos\theta+\cos^2\theta\]
Therefore,
\[\frac{dy}{d\theta}=\frac{8\cos\theta+4-4-4\cos\theta-\cos^2\theta}{(2+\cos\theta)^2}\]
Simplifying,
\[\frac{dy}{d\theta}=\frac{4\cos\theta-\cos^2\theta}{(2+\cos\theta)^2}\]
Factorise the numerator:
\[\boxed{\frac{dy}{d\theta}=\frac{\cos\theta(4-\cos\theta)}{(2+\cos\theta)^2}}\]
Determine the Sign of $\frac{dy}{d\theta}$
We are given
\[0\leq\theta\leq\frac{\pi}{2}\]
On this interval,
\[0\leq\cos\theta\leq1\]
Therefore,
\[\cos\theta\geq0\]
Also,
\[4-\cos\theta\geq4-1=3>0\]
Finally,
\[2+\cos\theta\geq2>0,\]
so
\[(2+\cos\theta)^2>0\]
Hence,
\[\frac{\cos\theta(4-\cos\theta)}{(2+\cos\theta)^2}\geq0\]
More precisely,
\[ \frac{dy}{d\theta}>0 \quad\text{for}\quad 0\leq\theta<\frac{\pi}{2}, \]
while
\[ \frac{dy}{d\theta}=0 \quad\text{at}\quad \theta=\frac{\pi}{2}. \]
Thus $y$ is increasing throughout
\[\left[0,\frac{\pi}{2}\right]\]

🎯 Exam Significance

Exam Significance

›

This problem tests three important skills:

Application of the quotient rule.
Correct differentiation of a linear term such as $-\theta$.
Using the sign of the first derivative to establish monotonicity.

A complete board-level solution should explicitly establish the signs of

\[ \cos\theta,\qquad 4-\cos\theta,\qquad (2+\cos\theta)^2 \]

on the given interval rather than simply stating that the derivative is positive.

Significance for Competitive Entrance Examinations

The problem illustrates an efficient trigonometric simplification after differentiation. The identity

\[ \sin^2\theta+\cos^2\theta=1 \]

reduces the derivative to the factored form

\[ \frac{\cos\theta(4-\cos\theta)} {(2+\cos\theta)^2}. \]

Once this form is obtained, the sign can be determined immediately from the known range

\[ 0\leq\cos\theta\leq1. \]

This type of sign analysis is useful in JEE Main, JEE Advanced and other entrance examinations.

🚫 Common Mistakes

Common Mistakes

›

Forgetting the $-1$ obtained from differentiating $-\theta$.
Using the quotient rule incorrectly by reversing the order of $vu'-uv'$.
Forgetting that
\[ \frac{d}{d\theta}(\cos\theta)=-\sin\theta. \]
Writing $\frac{dy}{dt}$ instead of $\frac{dy}{d\theta}$.
Assuming $\frac{dy}{d\theta}>0$ at every point of the closed interval. At
\[ \theta=\frac{\pi}{2}, \]
the derivative is actually zero.
Ignoring the given interval. The range
\[ 0\leq\cos\theta\leq1 \]
is essential for establishing the sign.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 6 points

The given function is
\[ y=\frac{4\sin\theta}{2+\cos\theta}-\theta. \]
Its correct derivative is
\[ \boxed{ \frac{dy}{d\theta} = \frac{\cos\theta(4-\cos\theta)} {(2+\cos\theta)^2} }. \]
For
\[ 0\leq\theta\leq\frac{\pi}{2}, \]
we have
\[ 0\leq\cos\theta\leq1. \]
Thus,
\[ \cos\theta\geq0, \qquad 4-\cos\theta>0, \qquad (2+\cos\theta)^2>0. \]
Therefore,
\[ \frac{dy}{d\theta}\geq0 \]
throughout the interval, with equality only at
\[ \theta=\frac{\pi}{2}. \]
Hence the function is increasing on
\[ \boxed{\left[0,\frac{\pi}{2}\right]}. \]

📘 Concept & Theory

Theory and Concept

›

To determine whether a differentiable function is increasing on an interval, we examine the sign of its first derivative.

If

\[ f'(x)>0 \]

throughout an interval, then $f(x)$ is strictly increasing on that interval.

Here, the logarithmic function is

\[ y=\log x. \]

Its domain is

\[ x>0, \]

which is exactly the interval

\[ (0,\infty). \]

🗺️ Solution Roadmap

Step-by-step Plan

›

Consider the logarithmic function $y=\log x$.
Differentiate it with respect to $x$.
Determine the sign of $\frac{dy}{dx}$ for $x>0$.
Use the first derivative test to conclude that the function is increasing.

📊 Graph / Figure

Graph / Figure

›

Graph of function $y=\log x$ — Graphical Representation

✏️ Solution

Complete Solution

›

Step-by-step Solution · 9 steps

Consider the logarithmic function
\[y=\log x\]
For the logarithmic function to be defined,
\[x>0\]
Therefore, its domain is
\[(0,\infty)\]
Differentiate
Differentiating with respect to $x$,
\[\frac{dy}{dx}=\frac{1}{x}\]
Determine the Sign of the Derivative
On the given interval,
\[x\in(0,\infty),\]
so
\[x>0\]
Therefore,
\[\frac{1}{x}>0\]
Hence,
\[\boxed{\frac{dy}{dx}>0\quad\text{for all }x\in(0,\infty)}\]
Apply the First Derivative Test
Since the derivative is strictly positive throughout the entire domain,
\[\frac{dy}{dx}>0\quad\text{for all }x>0,\]
the logarithmic function is strictly increasing on $(0,\infty)$.
Therefore,
\[\boxed{\log x\text{ is an increasing function on }(0,\infty).}\]

🎯 Exam Significance

Exam Significance

›

This is a fundamental application of derivatives to monotonicity. The expected approach is short but must contain the essential logical steps:

\[ y=\log x \]

followed by

\[ \frac{dy}{dx}=\frac1x>0 \quad\text{for }x>0. \]

Therefore, the function is increasing on its entire domain. It is important to state explicitly why $\frac1x>0$: the given interval ensures that $x$ is positive.

Significance for Competitive Entrance Examinations

The result

\[ \frac{d}{dx}\log x=\frac1x>0 \]

is frequently used when solving questions involving inequalities, inverse functions, logarithmic equations, extrema and monotonicity.

It also immediately establishes the standard property:

\[ 0 This monotonicity is particularly useful when applying logarithms to inequalities.

🚫 Common Mistakes

Common Mistakes

›

Forgetting that the domain of $\log x$ is $x>0$.
Writing $\frac{d}{dx}(\log x)=x$; the correct derivative is
\[ \frac1x. \]
Simply stating that $\frac1x>0$ without connecting it to the condition $x>0$.
Claiming that the logarithmic function is increasing for negative $x$. The real logarithmic function is not defined there.
Confusing an increasing function with a function whose derivative is increasing. Here, the function is increasing, while its derivative
\[ \frac1x \]
is actually decreasing on $(0,\infty)$.

🔑 Key Takeaways

Key Takeaways

›

📘 Concept & Theory

Theory and Concept

›

To determine whether a differentiable function is increasing on an interval, we examine the sign of its first derivative.

If

\[ f'(x)>0 \]

throughout an interval, then $f(x)$ is strictly increasing on that interval.

Here, the logarithmic function is

\[ y=\log x. \]

Its domain is

\[ x>0, \]

which is exactly the interval

\[ (0,\infty). \]

🗺️ Solution Roadmap

Step-by-step Plan

›

Consider the logarithmic function $y=\log x$.
Differentiate it with respect to $x$.
Determine the sign of $\frac{dy}{dx}$ for $x>0$.
Use the first derivative test to conclude that the function is increasing.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 9 steps

Consider the logarithmic function
Let
\[y=\log x\]
For the logarithmic function to be defined,
\[x>0\]
Therefore, its domain is
\[(0,\infty)\]
Differentiate
Differentiating with respect to $x$,
\[\frac{dy}{dx}=\frac{1}{x}\]
Determine the Sign of the Derivative
On the given interval,
\[x\in(0,\infty),\]
so
\[x>0\]
Therefore,
\[\frac{1}{x}>0\]
Hence,
\[\boxed{\frac{dy}{dx}>0\quad\text{for all }x\in(0,\infty)}\]
Apply the First Derivative Test
Since the derivative is strictly positive throughout the entire domain,
\[ \frac{dy}{dx}>0 \quad\text{for all }x>0, \]
the logarithmic function is strictly increasing on $(0,\infty)$.
Therefore,
\[ \boxed{\log x\text{ is an increasing function on }(0,\infty).} \]

🎯 Exam Significance

Exam Significance

›

This is a fundamental application of derivatives to monotonicity. The expected approach is short but must contain the essential logical steps:

\[ y=\log x \]

followed by

\[ \frac{dy}{dx}=\frac1x>0 \quad\text{for }x>0. \]

Therefore, the function is increasing on its entire domain. It is important to state explicitly why $\frac1x>0$: the given interval ensures that $x$ is positive.

Significance for Competitive Entrance Examinations

The result

\[ \frac{d}{dx}\log x=\frac1x>0 \]

is frequently used when solving questions involving inequalities, inverse functions, logarithmic equations, extrema and monotonicity.

It also immediately establishes the standard property:

\[ 0 < x_1 < x_2 \quad\Longrightarrow\quad \log x_1 < \log x_2. \]

This monotonicity is particularly useful when applying logarithms to inequalities.

🚫 Common Mistakes

Common Mistakes

›

Forgetting that the domain of $\log x$ is $x>0$.
Writing $\frac{d}{dx}(\log x)=x$; the correct derivative is
\[ \frac1x. \]
Simply stating that $\frac1x>0$ without connecting it to the condition $x>0$.
Claiming that the logarithmic function is increasing for negative $x$. The real logarithmic function is not defined there.
Confusing an increasing function with a function whose derivative is increasing. Here, the function is increasing, while its derivative
\[ \frac1x \]
is actually decreasing on $(0,\infty)$.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 6 points

The logarithmic function is
\[ y=\log x. \]
Its domain is
\[ (0,\infty). \]
Its derivative is
\[ \frac{dy}{dx}=\frac1x. \]
Since $x>0$,
\[ \frac1x>0. \]
Therefore, $\log x$ is strictly increasing throughout
\[ (0,\infty). \]
Consequently,
\[ 0 < x_1 < x_2 \quad\Longrightarrow\quad \log x_1 < \log x_2. \]

📘 Concept & Theory

Theory and Concept

›

A function $f(x)$ is strictly increasing on an interval if

\[ x_1 < x_2\quad\Longrightarrow\quad f(x_1) < f(x_2). \]

A function is strictly decreasing if

\[ x_1 < x_2\quad\Longrightarrow\quad f(x_1)>f(x_2). \]

For a differentiable function, the first derivative helps determine monotonicity:

\[ f'(x)>0 \]

indicates increasing behaviour, while

\[ f'(x) < 0 \]

indicates decreasing behaviour.

Therefore, we need to examine the sign of $f'(x)$ throughout the interval $(-1,1)$.

🗺️ Solution Roadmap

Step-by-step Plan

›

Differentiate $f(x)$.
Find the critical point by solving $f'(x)=0$.
Divide $(-1,1)$ into intervals using the critical point.
Determine the sign of $f'(x)$ on each interval.
Show that the function is decreasing on one part and increasing on another.
Conclude that it cannot be strictly increasing or strictly decreasing throughout $(-1,1)$.

📊 Graph / Figure

Graph / Figure

›

Graph of function $f(x)=x^2-x+1$ — Graphical Representation

✏️ Solution

Complete Solution

›

Step-by-step Solution · 25 steps

Given —
\[f(x)=x^2-x+1\]
Find the First Derivative
Differentiating with respect to $x$,
\[ f'(x)=2x-1. \]
Thus,
\[ \boxed{f'(x)=2x-1}. \]
Find the Critical Point
Set
\[f'(x)=0\]
Therefore,
\[2x-1=0\]
Hence,
\[2x=1\]
and
\[\boxed{x=\frac12}\]
The value
\[x=\frac12\]
is a critical point.
It should not be called a point of inflection merely because $f'(x)=0$. A point of inflection requires a change in concavity. Here, the relevant concept is a critical point.
Determine the Sign of $f'(x)$
The critical point
\[x=\frac12\]
divides the interval $(-1,1)$ into
\[\left(-1,\frac12\right)\]
and
\[\left(\frac12,1\right)\]
Interval 1: ${\left(-1,\frac12\right)}$
For
\[-1 < x < \frac12,\]
we have
\[2x < 1\]
Therefore,
\[2x-1 < 0\]
Hence,
\[f'(x) < 0\]
Therefore, $f(x)$ is strictly decreasing on
\[\boxed{\left(-1,\frac12\right)}\]
Interval 2: ${\left(\frac12,1\right)}$
For
\[\frac12 < x < 1,\]
we have
\[2x>1\]
Therefore,
\[2x-1>0\]
Hence,
\[f'(x)>0\]
Therefore, $f(x)$ is strictly increasing on
\[\boxed{\left(\frac12,1\right)}\]
Monotonicity Table

Interval	Sign of $f'(x)=2x-1$	Behaviour of $f(x)$
$\left(-1,\frac12\right)$	$f'(x)<0$	Strictly decreasing
$x=\frac12$	$f'(x)=0$	Critical point
$\left(\frac12,1\right)$	$f'(x)>0$	Strictly increasing

Why Is the Function Neither Increasing Nor Decreasing on $(-1,1)$?
We have established that
\[ f'(x) < 0 \]
on
\[ \left(-1,\frac12\right), \]
so the function decreases there.
On the other hand,
\[ f'(x)>0 \]
on
\[ \left(\frac12,1\right), \]
so the function increases there.
Therefore, the function changes its monotonic behaviour at
\[ x=\frac12. \]
Consequently, it cannot be strictly increasing throughout the entire interval $(-1,1)$, because it is decreasing on
\[ \left(-1,\frac12\right). \]
Similarly, it cannot be strictly decreasing throughout $(-1,1)$, because it is increasing on
\[ \left(\frac12,1\right). \]
Hence,
\[ \boxed{ f(x)=x^2-x+1 \text{ is neither strictly increasing nor strictly decreasing on }(-1,1). } \]

🎯 Exam Significance

Exam Significance

›

This question is an important application of the first derivative test. A complete answer should not merely find

\[ x=\frac12. \]

It must establish the sign of $f'(x)$ on both sides of this critical point.

The essential reasoning is

\[ f'(x)<0 \quad\text{on}\quad \left(-1,\frac12\right) \]

and

\[ f'(x)>0 \quad\text{on}\quad \left(\frac12,1\right). \]

Therefore, the function decreases first and increases later, proving that it is neither strictly increasing nor strictly decreasing on the complete interval.

Significance for Competitive Entrance Examinations

This problem develops the ability to identify monotonicity intervals from the sign of a derivative. It also reinforces the distinction between a critical point and a point of inflection.

For a quadratic function, the critical point can often be identified rapidly. Here,

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

gives

\[ x=\frac12. \]

The sign change

\[ -\rightarrow+ \]

immediately identifies the point as a local minimum.

🚫 Common Mistakes

Common Mistakes

›

Writing $f(x)=0$ instead of $f'(x)=0$ when finding the critical point.
Calling $x=\frac12$ a point of inflection without checking concavity.
Checking only the value of the function at $x=\frac12$ instead of analysing $f'(x)$.
Forgetting to divide the interval $(-1,1)$ at the critical point.
Concluding that a function is increasing or decreasing on the whole interval merely because its derivative is zero at one point.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 6 points

Given:
\[ f(x)=x^2-x+1. \]
First derivative:
\[ f'(x)=2x-1. \]
Critical point:
\[ f'(x)=0 \Rightarrow x=\frac12. \]
On
\[ \left(-1,\frac12\right), \]

\[ f'(x)<0, \]
so $f$ is strictly decreasing.
On
\[ \left(\frac12,1\right), \]

\[ f'(x)>0, \]
so $f$ is strictly increasing.
Therefore, $f$ is neither strictly increasing nor strictly decreasing on
\[ (-1,1). \]

Q12

LONG ANSWER4 marks

Which of the following functions are decreasing on \[ \left(0,\frac{\pi}{2}\right)? \]

\[ f(x)=\cos x \]
\[ f(x)=\cos 2x \]
\[ f(x)=\cos 3x \]
\[ f(x)=\tan x \]

📘 Concept & Theory

Theory and Concept

›

A differentiable function $f(x)$ is decreasing on an interval if

\[ f'(x)<0 \]

throughout that interval.

Therefore, we differentiate each option and determine the sign of its derivative for

\[ 0 < x < \frac{\pi}{2} \]

🗺️ Solution Roadmap

Step-by-step Plan

›

Differentiate each function.
Use the appropriate trigonometric range on the given interval.
Determine whether the derivative remains negative throughout the interval.
Select the functions for which $f'(x)<0$ everywhere on the interval.

✏️ Solution

Option (A): ${f(x)=\cos x}$

›

Step-by-step Solution · 8 steps

Given —
\[f(x)=\cos x\]
Differentiating,
\[ f'(x)=-\sin x. \]
For
\[ 0 < x < \frac{\pi}{2}, \]
we have
\[ 0 < \sin x < 1. \]
Therefore,
\[ -\sin x < 0. \]
Hence,
\[ f'(x)<0 \quad\text{for all}\quad x\in\left(0,\frac{\pi}{2}\right). \]
Therefore,
\[ \boxed{\cos x\text{ is decreasing on }\left(0,\frac{\pi}{2}\right).} \]
Option (A) is correct.
$Graph of function $f(x)=x^2-x+1$$

✏️ Solution

Option (B): ${f(x)=\cos 2x}$

›

Step-by-step Solution · 10 steps

Given —
\[f(x)=\cos 2x\]
Using the chain rule,
\[ f'(x)=-\sin(2x)\cdot2. \]
Hence,
\[ \boxed{f'(x)=-2\sin 2x}. \]
Now,
\[ 0 < x < \frac{\pi}{2}. \]
Multiplying the entire inequality by $2$,
\[ 0 < 2 x < \pi. \]
Since
\[ 0 < 2x < \pi, \]
we have
\[ \sin 2x>0. \]
Therefore,
\[ -2\sin2x < 0. \]
Thus,
\[ f'(x)<0 \quad\text{for all}\quad x\in\left(0,\frac{\pi}{2}\right). \]
Hence,
\[ \boxed{\cos2x\text{ is decreasing on }\left(0,\frac{\pi}{2}\right).} \]
Option (B) is correct.

✏️ Solution

Option (C): ${f(x)=\cos3x}$

›

Step-by-step Solution · 15 steps

Given —
\[f(x)=\cos3x\]
Differentiating using the chain rule,
\[ f'(x)=-3\sin3x. \]
Thus,
\[ \boxed{f'(x)=-3\sin3x}. \]
For
\[ 0 < x < \frac{\pi}{2}, \]
multiplying by $3$ gives
\[ 0 < 3 x < \frac{3\pi}{2}. \]
This interval crosses $\pi$. Therefore, $\sin3x$ does not have the same sign throughout the given interval.
For ${0 < x < \frac{\pi}{3}}$
We have
\[ 0 < 3 x < \pi. \]
Hence,
\[ \sin3x>0. \]
Therefore,
\[ f'(x)=-3\sin3x<0. \]
So $f(x)$ is decreasing on
\[ \boxed{\left(0,\frac{\pi}{3}\right)}. \]
For ${\frac{\pi}{3} < x < \frac{\pi}{2}}$
We have
\[ \pi < 3x < \frac{3\pi}{2}. \]
In this interval,
\[ \sin3x < 0. \]
Therefore,
\[ f'(x)=-3\sin3x>0. \]
Hence $f(x)$ is increasing on
\[ \boxed{\left(\frac{\pi}{3},\frac{\pi}{2}\right)}. \]
Thus, $\cos3x$ is decreasing on one part of the interval and increasing on another part. It is therefore not decreasing throughout
\[ \left(0,\frac{\pi}{2}\right). \]
Option (C) is incorrect.

✏️ Solution

Option (D): ${f(x)=\tan x}$

›

Step-by-step Solution · 6 steps

Given —
\[f(x)=\tan x\]
Differentiating,
\[ \boxed{f'(x)=\sec^2x}. \]
For
\[ 0 < x < \frac{\pi}{2}, \]
$\cos x>0$, so
\[ \sec^2x>0. \]
Therefore,
\[ f'(x)>0 \]
Hence,
\[ \boxed{\tan x\text{ is increasing on }\left(0,\frac{\pi}{2}\right).} \]
Therefore, Option (D) is incorrect.
Comparison of All Options

Option	Function	Derivative	Sign on $\left(0,\frac{\pi}{2}\right)$	Behaviour
(A)	$\cos x$	$-\sin x$	Negative	Decreasing
(B)	$\cos2x$	$-2\sin2x$	Negative	Decreasing
(C)	$\cos3x$	$-3\sin3x$	Changes sign	Neither throughout
(D)	$\tan x$	$\sec^2x$	Positive	Increasing

💡 Answer

Final Answer

›

The functions that are decreasing on

\[ \left(0,\frac{\pi}{2}\right) \]

are

\[ \boxed{(A)\ \cos x\quad\text{and}\quad(B)\ \cos2x}. \]

Therefore, the correct answer is

\[ \boxed{\text{A and B}}. \]

🎯 Exam Significance

Exam Significance

›

This question tests the direct application of derivatives to determine monotonicity. The important step is to differentiate each function correctly and then use the appropriate trigonometric range.

In particular, the chain rule must be applied to

\[ \cos2x \]

and

\[ \cos3x. \]

Significance for Competitive Entrance Examinations

This is a useful multiple-choice pattern because the derivative can quickly eliminate incorrect options. The most important observation is that the argument of the trigonometric function also changes the sign analysis.

For example,

\[ 0 < x < \frac{\pi}{2} \]

does not imply that

\[ 3x<\pi. \]

Instead,

\[ 0 < 3 x < \frac{3\pi}{2}, \]

so $\sin3x$ changes sign.

🚫 Common Mistakes

Common Mistakes

›

Forgetting the chain-rule multiplier in $\cos2x$ and $\cos3x$.
Writing
\[ \frac{d}{dx}(\cos2x)=-2\sin x \]
instead of
\[ -2\sin2x. \]
Assuming that $3x$ remains in the first quadrant simply because $x$ is in the first quadrant.
Forgetting that $\sin3x$ changes sign at $x=\frac{\pi}{3}$.
Writing
\[ \frac{d}{dx}(\tan x)=\sec x+\tan x. \]
The correct derivative is
\[ \sec^2x. \]

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 4 points

\[ \cos x \]
is decreasing on
\[ \left(0,\frac{\pi}{2}\right). \]
\[ \cos2x \]
is also decreasing on
\[ \left(0,\frac{\pi}{2}\right). \]
\[ \cos3x \]
is decreasing on
\[ \left(0,\frac{\pi}{3}\right) \]
and increasing on
\[ \left(\frac{\pi}{3},\frac{\pi}{2}\right). \]
\[ \tan x \]
is increasing on
\[ \left(0,\frac{\pi}{2}\right). \]

Q13

LONG ANSWER4 marks

On which of the following intervals is the function \[ f(x)=x^{100}+\sin x-1 \] decreasing?

\[ (0,1) \]
\[ \left(\frac{\pi}{2},\pi\right) \]
\[ \left(0,\frac{\pi}{2}\right) \]
\[ \text{None of these} \]

📘 Concept & Theory

Theory and Concept

›

A differentiable function $f(x)$ is decreasing on an interval if

\[ f'(x)<0 \]

throughout that interval.

Therefore, we first calculate the derivative and then determine whether it can become negative on any of the given intervals.

🗺️ Solution Roadmap

Step-by-step Plan

›

Differentiate the given function.
Analyse the sign of $f'(x)$ on each option.
For interval (B), remember that although $\cos x$ is negative, the term $100x^{99}$ is sufficiently large and positive.
Determine whether $f'(x)$ is negative anywhere on the listed intervals.
Select the appropriate option.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 28 steps

Given —
\[f(x)=x^{100}+\sin x-1\]
Find the First Derivative
Differentiating with respect to $x$,
\[ f'(x)=100x^{99}+\cos x. \]
Hence,
\[ \boxed{f'(x)=100x^{99}+\cos x}. \]
Option (A): ${(0,1)}$
For
\[ 0 < x < 1, \]
we have
\[ x^{99}>0. \]
Therefore,
\[ 100x^{99}>0. \]
Also, since
\[ 0 < x < 1 < \frac{\pi}{2}, \]
we have
\[ \cos x>0. \]
Therefore,
\[ f'(x)=100x^{99}+\cos x>0. \]
Hence, $f(x)$ is increasing on
\[ (0,1). \]
Therefore, option (A) is not correct.
Option (B): ${\left(\frac{\pi}{2},\pi\right)}$
Here,
\[ \frac{\pi}{2} < x < \pi. \]
Thus,
\[ \cos x < 0 . \]
At first glance, one might think that the negative value of $\cos x$ could make $f'(x)$ negative. However, the term $100x^{99}$ is much larger and remains positive.
Since
\[ x>\frac{\pi}{2}>1, \]
we have
\[ x^{99}>1. \]
Hence,
\[ 100x^{99}>100. \]
For every real $x$,
\[ -1\leq\cos x\leq1. \]
Therefore,
\[ \cos x\geq-1. \]
Consequently,
\[ f'(x) = 100x^{99}+\cos x > 100-1 = 99. \]
Thus,
\[ \boxed{f'(x)>0} \]
throughout
\[ \left(\frac{\pi}{2},\pi\right). \]
Hence the function is increasing, not decreasing, on this interval.
Therefore, option (B) is not correct.
Option (C): ${\left(0,\frac{\pi}{2}\right)}$
For
\[ 0 < x < \frac{\pi}{2}, \]
we have
\[ x^{99}>0 \]
and
\[ \cos x>0. \]
Therefore,
\[ f'(x)=100x^{99}+\cos x>0. \]
Hence, $f(x)$ is increasing on
\[ \left(0,\frac{\pi}{2}\right). \]
Therefore, option (C) is not correct.
Conclusion
In all the three given intervals,
\[ f'(x)>0. \]
Therefore, the function is increasing on each of them.

Consequently, there is no interval among (A), (B), and (C) on which the function is decreasing.

Hence,
\[ \boxed{\text{Option (D) is correct.}} \]
Compact Sign Analysis

Option	Interval	Sign of $100x^{99}$	Sign/Bound for $\cos x$	Sign of $f'(x)$	Behaviour
(A)	$(0,1)$	Positive	$\cos x>0$	Positive	Increasing
(B)	$\left(\frac{\pi}{2},\pi\right)$	$>100$	$\cos x\geq-1$	$>99$	Increasing
(C)	$\left(0,\frac{\pi}{2}\right)$	Positive	$\cos x>0$	Positive	Increasing

💡 Answer

Final Answer

›

\[\boxed{\text{(D) None of these}}\]

🎯 Exam Significance

Exam Significance

›

This question reinforces the use of the first derivative for determining monotonicity. It also tests careful handling of trigonometric signs on different intervals.

A complete solution should not incorrectly assume that $\cos x$ is positive throughout

\[ \left(\frac{\pi}{2},\pi\right). \]

Instead, use the bound

\[ -1\leq\cos x\leq1 \]

to establish the sign of the complete derivative.

Significance for Competitive Entrance Examinations

This is a useful multiple-choice problem because the expression

\[ 100x^{99}+\cos x \]

can be handled quickly using simple inequalities.

For

\[ x>\frac{\pi}{2}>1, \]

we have

\[ 100x^{99}>100, \]

while

\[ \cos x\geq-1. \]

Hence,

\[ 100x^{99}+\cos x>99>0. \]

This type of bounding argument is often faster and more reliable than trying to determine the exact value of the derivative.

🚫 Common Mistakes

Common Mistakes

›

Assuming $\cos x>0$ on $\left(\frac{\pi}{2},\pi\right)$. In fact, $\cos x<0$ there.
Ignoring the large positive term $100x^{99}$.
Concluding that $f'(x)$ is negative merely because $\cos x$ is negative.
Forgetting that
\[ -1\leq\cos x\leq1. \]
Confusing $x^{99}$ with $x^{100}$ when differentiating.
Assuming that a negative trigonometric term necessarily makes the entire derivative negative.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 6 points

\[ f'(x)=100x^{99}+\cos x. \]
On
\[ (0,1), \]
both terms are positive, so
\[ f'(x)>0. \]
On
\[ \left(0,\frac{\pi}{2}\right), \]
both terms are positive, so
\[ f'(x)>0. \]
On
\[ \left(\frac{\pi}{2},\pi\right), \]
$\cos x<0$, but
\[ 100x^{99}>100 \]
and
\[ \cos x\geq-1. \]
Hence,
\[ f'(x)>99>0. \]
Therefore, the function is increasing on every interval listed in (A), (B), and (C).
Hence, it is decreasing on none of them.

📘 Concept & Theory

Theory and Concept

›

For a differentiable function to be increasing on a closed interval $[1,2]$, its derivative must satisfy

\[ f'(x)\geq0 \]

throughout the interval.

For strict increase, it is sufficient that

\[ f'(x)>0 \]

throughout the interval. In this problem, the derivative is linear, so we can determine the required condition on $a$ by finding its minimum value on $[1,2]$.

🗺️ Solution Roadmap

Step-by-step Plan

›

Differentiate $f(x)$.
Set the derivative condition for increasing behaviour.
Find where $2x+a$ attains its minimum on $[1,2]$.
Apply the condition at that endpoint.
Obtain the required range of $a$.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 15 steps

Given —
\[f(x)=x^2+ax+1\]
Find the First Derivative
Differentiating with respect to $x$,
\[ f'(x)=2x+a. \]
Thus,
\[ \boxed{f'(x)=2x+a}. \]
Apply the Increasing Function Condition
For $f(x)$ to be increasing on
\[ [1,2], \]
we require
\[ f'(x)\geq0 \]
for every
\[ x\in[1,2]. \]
Therefore,
\[ 2x+a\geq0. \]
Rearranging,
\[ a\geq-2x. \]
Find the Most Restrictive Condition
Since
\[ x\in[1,2], \]
the expression
\[ -2x \]
decreases as $x$ increases.
Therefore, its greatest value occurs at
\[ x=1. \]
At $x=1$,
\[ a\geq-2(1). \]
Hence,
\[ \boxed{a\geq-2}. \]
The supplied solution concludes
\[ a>-2. \]
This is unnecessarily restrictive.
Because the interval includes the endpoint $x=1$, the derivative is allowed to be zero at that single endpoint while the function remains strictly increasing on the entire interval.
For
\[ a=-2, \]
the function becomes
\[ f(x)=x^2-2x+1=(x-1)^2. \]
If
\[ 1\leq x_1 < x_2\leq2, \]
then
\[ 0\leq x_1-1 < x_2-1, \]
and therefore
\[ (x_1-1)^2 < (x_2-1)^2. \]
Hence $f(x)$ is strictly increasing on $[1,2]$.
Therefore, the equality case $a=-2$ must be included.

💡 Answer

Final Answer

›

\[ \boxed{a\geq-2}. \]

🎯 Exam Significance

Exam Significance

›

Parameter-based monotonicity questions are common in competitive examinations. A fast method is to identify the minimum of the derivative on the given interval.

Here,

\[ f'(x)=2x+a \]

has positive slope with respect to $x$, so its minimum occurs at the left endpoint $x=1$:

\[ \min_{[1,2]}f'(x)=f'(1)=a+2. \]

Therefore,

\[ a+2\geq0 \]

immediately gives

\[ \boxed{a\geq-2}. \]

🚫 Common Mistakes

Common Mistakes

›

Checking $f'(x)>0$ at only an arbitrary point instead of considering the entire interval.
Writing both $a>-2$ and $a>-4$ without identifying which condition is restrictive.
Forgetting that $x\in[1,2]$, so the smallest value of $2x+a$ occurs at $x=1$.
Excluding $a=-2$ unnecessarily.
Confusing the condition for increasing behaviour on a closed interval with requiring $f'(x)>0$ at every endpoint.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 6 points

\[ f(x)=x^2+ax+1. \]
\[ f'(x)=2x+a. \]
On $[1,2]$, the minimum value of $f'(x)$ occurs at $x=1$.
Therefore,
\[ f'(1)=2+a\geq0. \]
Hence,
\[ \boxed{a\geq-2}. \]
The boundary value $a=-2$ is included because
\[ f(x)=(x-1)^2 \]
is strictly increasing on $[1,2]$, despite $f'(1)=0$.

📘 Concept & Theory

Theory and Concept

›

To prove that a differentiable function is increasing on an interval, we examine the sign of its first derivative.

If

\[ f'(x)>0 \]

throughout an interval, then $f(x)$ is strictly increasing on that interval.

The important condition in this question is that $I$ is disjoint from

\[ [-1,1]. \]

Therefore, every $x\in I$ satisfies either

\[ x>1 \]

or

\[ x<-1. \]

🗺️ Solution Roadmap

Step-by-step Plan

›

Differentiate the given function.
Factorise the derivative.
Use the condition that $I\cap[-1,1]=\varnothing$.
Show that $x^2>1$ for every $x\in I$.
Conclude that $f'(x)>0$ throughout $I$.
Hence prove that $f$ is increasing on $I$.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 19 steps

Given —
\[f(x)=x+\frac{1}{x}\]
Note that $x\neq0$, which is automatically satisfied because $I$ is disjoint from $[-1,1]$.
Differentiating,
\[ f'(x)=1-\frac{1}{x^2}. \]
Taking the terms over a common denominator,
\[f'(x)=\frac{x^2-1}{x^2}\]
Factorising the numerator,
\[\boxed{f'(x)=\frac{(x-1)(x+1)}{x^2}}\]
Use the Given Condition on $I$
We are told that $I$ is disjoint from
\[[-1,1]\]
Thus, no point of $I$ lies between $-1$ and $1$, and $I$ must lie entirely in one of the two regions
\[(-\infty,-1)\]
or
\[(1,\infty)\]
Therefore, for every $x\in I$,
\[|x|>1\]
Squaring both sides,
\[x^2>1\]
Hence,
\[x^2-1>0\]
Also,
\[x^2>0\]
Therefore,
\[\frac{x^2-1}{x^2}>0\]
Consequently,
\[ \boxed{f'(x)>0\quad\text{for every }x\in I}. \]
Apply the Monotonicity Test
Since
\[f'(x)>0\]
throughout $I$, the function $f(x)$ is strictly increasing on $I$.
Hence,
\[\boxed{f(x)=x+\frac1x\text{ is increasing on every interval }I\text{ disjoint from }[-1,1]}\]
Why Both Regions Are Included
The condition
\[ I\cap[-1,1]=\varnothing \]
does not mean that $x>1$ only.
There are two possible regions:
\[x>1\]
or
\[x < -1\]
In both cases,
\[x^2>1,\]
so
\[f'(x)=1-\frac1{x^2}>0\]
Thus the same derivative argument works on either side of the interval $[-1,1]$.
Sign Analysis

Region	$x^2$	$x^2-1$	$f'(x)$	Behaviour
$(-\infty,-1)$	$>1$	$>0$	$>0$	Increasing
$[-1,1]$	$\leq1$	$\leq0$	$\leq0$	Not part of $I$
$(1,\infty)$	$>1$	$>0$	$>0$	Increasing

💡 Answer

Final Answer

›

\[ \boxed{ f(x)=x+\frac1x \text{ is strictly increasing on every interval }I \text{ disjoint from }[-1,1]. } \]

🎯 Exam Significance

Exam Significance

›

This problem tests the use of the first derivative test together with an interval condition. A complete solution should explicitly connect

\[ I\cap[-1,1]=\varnothing \]

with

\[ |x|>1. \]

Once this is established, the derivative becomes immediately positive:

\[ f'(x)=\frac{x^2-1}{x^2}>0. \]

Significance for Competitive Entrance Examinations

The key competitive-examination technique is to factorise the derivative:

\[ f'(x)=\frac{(x-1)(x+1)}{x^2}. \]

The critical values are

\[ x=-1,\quad x=1, \]

while $x=0$ is excluded from the domain. The derivative is positive outside the interval $[-1,1]$.

Thus, the result can be obtained quickly by recognizing the condition

\[ |x|>1. \]

🚫 Common Mistakes

Common Mistakes

›

Writing $x>\pm1$, which is ambiguous. Write $x<-1$ or $x>1$.
Considering only $x>1$ and forgetting the region $x<-1$.
Forgetting that $x=0$ is not in the domain of $f(x)$.
Multiplying an inequality by $x^2$ without noting that $x^2>0$ for $x\neq0$.
Confusing the interval $[-1,1]$ with the domain of the function.
Assuming that the function is increasing on all of $\mathbb R\setminus\{0\}$. It is not increasing on $(-1,1)$.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 6 points

\[ f(x)=x+\frac1x. \]
\[ f'(x)=1-\frac1{x^2} = \frac{x^2-1}{x^2}. \]
$f'(x)>0$ when
\[ x^2>1. \]
Therefore,
\[ f'(x)>0 \quad\text{when}\quad x < -1\text{ or }x>1. \]
Any interval disjoint from $[-1,1]$ lies entirely in one of these regions.
Hence, $f$ is increasing on every interval $I$ disjoint from $[-1,1]$.

📘 Concept & Theory

Theory and Concept

›

To determine whether a differentiable function is increasing or decreasing on an interval, we study the sign of its first derivative.

The derivative test states:

If $f'(x)>0$ throughout an interval, then $f(x)$ is strictly increasing there.
If $f'(x) < 0$ throughout an interval, then $f(x)$ is strictly decreasing there.

Here,

\[ f(x)=\log(\sin x). \]

Since $\log(\sin x)$ is defined only when

\[ \sin x>0, \]

the function is defined on

\[ (0,\pi). \]

🗺️ Solution Roadmap

Step-by-step Plan

›

Identify the domain of the function.
Differentiate $f(x)=\log(\sin x)$ using the chain rule.
Express the derivative as $\cot x$.
Determine the sign of $\cot x$ on each given interval.
Apply the first derivative test to establish increasing and decreasing behaviour.

📊 Graph / Figure

Graph / Figure

›

Graph of function $f(x)=\log(\sin x)$ — Graphical Representation

✏️ Solution

Complete Solution

›

Step-by-step Solution · 14 steps

Given —
\[f(x)=\log(\sin x)\]
For the logarithm to be defined,
\[ \sin x>0. \]
In the interval under consideration, this gives
\[ 0 < x < \pi. \]
Hence the relevant domain is
\[ \boxed{(0,\pi)}. \]
Differentiate the Function
Using the chain rule,
\[\frac{d}{dx}\log(\sin x)=\frac{1}{\sin x}\cdot\frac{d}{dx}(\sin x)\]
Since
\[\frac{d}{dx}(\sin x)=\cos x,\]
we get
\[f'(x)=\frac{\cos x}{\sin x}\]
Therefore,
\[\boxed{f'(x)=\cot x}\]
Analyse the Interval $\left(0,\frac{\pi}{2}\right)$
For
\[0 < x < \frac{\pi}{2},\]
both sine and cosine are positive:
\[\sin x>0,\quad \cos x>0\]
Therefore,
\[ \cot x=\frac{\cos x}{\sin x}>0. \]
Hence,
\[f'(x)>0\quad\text{for all}\quad x\in\left(0,\frac{\pi}{2}\right)\]
By the first derivative test,
\[\boxed{f(x)=\log(\sin x)\text{ is strictly increasing on }\left(0,\frac{\pi}{2}\right)}\]
Analyse the Interval $\left(\frac{\pi}{2},\pi\right)$
For
\[\frac{\pi}{2} < x < \pi,\]
we have
\[\sin x>0\]
but
\[\cos x < 0\]
Therefore,
\[\cot x=\frac{\cos x}{\sin x} < 0\]
Hence,
\[f'(x) < 0\quad\text{for all}\quad x\in\left(\frac{\pi}{2},\pi\right)\]
By the first derivative test,
\[\boxed{f(x)=\log(\sin x)\text{ is strictly decreasing on }\left(\frac{\pi}{2},\pi\right)}\]
Derivative Sign Table

Interval	$\sin x$	$\cos x$	$\cot x=f'(x)$	Behaviour of $f$
$\left(0,\frac{\pi}{2}\right)$	Positive	Positive	Positive	Increasing
$\frac{\pi}{2}$	Positive	0	0	Stationary point
$\left(\frac{\pi}{2},\pi\right)$	Positive	Negative	Negative	Decreasing

🎯 Exam Significance

Exam Significance

›

This is a standard application of the first derivative test. A complete board-level solution should show the derivative and explicitly justify its sign on each interval.

The essential chain of reasoning is

\[ f(x)=\log(\sin x) \]

\[ \Downarrow \]

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

\[ \Downarrow \]

\[ \cot x>0 \text{ on } \left(0,\frac{\pi}{2}\right), \qquad \cot x<0 \text{ on } \left(\frac{\pi}{2},\pi\right). \]

Therefore,

\[ \boxed{ f\text{ is increasing on }\left(0,\frac{\pi}{2}\right) } \]

and

\[ \boxed{ f\text{ is decreasing on }\left(\frac{\pi}{2},\pi\right). } \]

Significance for Competitive Entrance Examinations

For competitive examinations, the quickest approach is to recognize that

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

The signs of $\sin x$ and $\cos x$ immediately determine the sign of $\cot x$.

In the first quadrant,

\[ \sin x>0,\qquad \cos x>0 \]

so

\[ \cot x>0. \]

In the second quadrant,

\[ \sin x>0,\qquad \cos x<0 \]

so

\[ \cot x<0. \]

This quadrant-based sign analysis is often faster than constructing a detailed numerical sign table.

🚫 Common Mistakes

Common Mistakes

›

Forgetting the chain rule while differentiating $\log(\sin x)$.
Writing $f'(x)=1/\sin x$ and omitting the derivative of $\sin x$.
Writing $f'(x)=\tan x$ instead of $\cot x$.
Incorrectly assuming that $\cos x$ is positive throughout $(0,\pi)$.
Forgetting that $\cos x<0$ in the second quadrant.
Ignoring the domain condition $\sin x>0$.
Using endpoint values alone to prove monotonicity instead of examining the derivative.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 5 points

The function is
\[ f(x)=\log(\sin x). \]
Its derivative is
\[ \boxed{f'(x)=\cot x}. \]
On
\[ \left(0,\frac{\pi}{2}\right), \]

\[ \cot x>0, \]
so $f$ is increasing.
On
\[ \left(\frac{\pi}{2},\pi\right), \]

\[ \cot x<0, \]
so $f$ is decreasing.
At
\[ x=\frac{\pi}{2}, \]

\[ f'\left(\frac{\pi}{2}\right)=0 \]
and
\[ f\left(\frac{\pi}{2}\right)=0, \]
giving the maximum value of the function on $(0,\pi)$.

📘 Concept & Theory

Theory and Concept

›

To determine whether a function is increasing or decreasing on an interval, we examine the sign of its first derivative.

For

\[ f(x)=\log|\cos x|, \]

we must remember that

\[ \cos x\neq0 \]

because the logarithm of zero is undefined.

A useful result is

\[ \frac{d}{dx}\log|u(x)| = \frac{u'(x)}{u(x)}, \]

provided $u(x)\neq0$.

Taking

\[ u(x)=\cos x, \]

we get

\[ f'(x) = \frac{-\sin x}{\cos x} = -\tan x. \]

🗺️ Solution Roadmap

Step-by-step Plan

›

Find the derivative of $\log|\cos x|$.
Use the signs of $\sin x$, $\cos x$, and $\tan x$ in the given intervals.
Show that $f'(x)<0$ on $\left(0,\frac{\pi}{2}\right)$.
Show that $f'(x)>0$ on $\left(\frac{3\pi}{2},2\pi\right)$.
Apply the first derivative test.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 16 steps

Given —
\[f(x)=\log|\cos x|\]
The function is defined whenever
\[|\cos x|>0,\]
that is,
\[\cos x\neq0\]
Find the First Derivative
Using
\[\frac{d}{dx}\log|u|=\frac{u'}{u},\]
with
\[u=\cos x,\]
we obtain
\[f'(x)=\frac{-\sin x}{\cos x}\]
Therefore,
\[\boxed{f'(x)=-\tan x}\]
Increasing or Decreasing on $\left(0,\frac{\pi}{2}\right)$
Consider
\[0 < x < \frac{\pi}{2}\]
This is the first quadrant. Therefore,
\[\sin x>0\]
and
\[\cos x>0\]
Hence,
\[\tan x=\frac{\sin x}{\cos x}>0.\]
Since
\[f'(x)=-\tan x,\]
we have
\[f'(x) < 0\]
Therefore, by the first derivative test,
\[\boxed{f(x)=\log|\cos x|\text{ is strictly decreasing on }\left(0,\frac{\pi}{2}\right)}\]
Increasing or Decreasing on $\left(\frac{3\pi}{2},2\pi\right)$
Now consider
\[\frac{3\pi}{2} < x < 2 \pi\]
This is the fourth quadrant. In this quadrant,
\[\sin x < 0\]
while
\[\cos x>0\]
Therefore,
\[\tan x=\frac{\sin x}{\cos x} < 0.\]
Hence,
\[f'(x)=-\tan x>0.\]
Therefore, by the first derivative test,
\[\boxed{f(x)=\log|\cos x|\text{ is strictly increasing on }\left(\frac{3\pi}{2},2\pi\right)}\]
Sign Analysis

Interval	Quadrant	$\sin x$	$\cos x$	$\tan x$	$f'(x)=-\tan x$	Behaviour
$\left(0,\frac{\pi}{2}\right)$	I	+	+	+	−	Decreasing
$\left(\frac{3\pi}{2},2\pi\right)$	IV	−	+	−	+	Increasing

🎯 Exam Significance

Exam Significance

›

This question tests the application of the first derivative test to a function involving an absolute value. The most efficient method is to use

\[ \frac{d}{dx}\log|u|=\frac{u'}{u}. \]

Therefore,

\[ f'(x) = \frac{-\sin x}{\cos x} = -\tan x. \]

The quadrant signs then immediately determine the monotonicity.

Significance for Competitive Entrance Examinations

The important shortcut is

\[ \boxed{f'(x)=-\tan x}. \]

Instead of differentiating the absolute-value expression separately in each interval, recognize the general derivative formula for $\log|u(x)|$.

Then use the signs of $\tan x$:

\[ \tan x>0 \quad\text{in Quadrants I and III}, \]

while

\[ \tan x<0 \quad\text{in Quadrants II and IV}. \]

Because $f'(x)=-\tan x$, the monotonicity is reversed:

\[ \tan x>0\Rightarrow f'(x)<0, \]

and

\[ \tan x<0\Rightarrow f'(x)>0. \]

🚫 Common Mistakes

Common Mistakes

›

Incorrectly stating that $\cos x<0$ on $\left(\frac{3\pi}{2},2\pi\right)$.
Forgetting that $\left(\frac{3\pi}{2},2\pi\right)$ lies in the fourth quadrant.
Writing $f'(x)=\tan x$ instead of $f'(x)=-\tan x$.
Ignoring the absolute value in $\log|\cos x|$.
Forgetting that $\cos x=0$ makes the logarithm undefined.
Using endpoint values instead of the derivative sign to establish monotonicity.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 4 points

For
\[ f(x)=\log|\cos x|, \]
the derivative is
\[ \boxed{f'(x)=-\tan x}. \]
On
\[ \left(0,\frac{\pi}{2}\right), \]

\[ \tan x>0 \Rightarrow f'(x) < 0, \]
so $f$ is decreasing.
On
\[ \left(\frac{3\pi}{2},2\pi\right), \]

\[ \tan x<0 \Rightarrow f'(x)>0, \]
so $f$ is increasing.
The interval
\[ \left(\frac{3\pi}{2},2\pi\right) \]
is in the fourth quadrant, where
\[ \sin x < 0,\quad \cos x>0. \]

📘 Concept & Theory

Theory and Concept

›

To determine whether a differentiable function is increasing on an interval, we examine the sign of its first derivative.

If

\[ f'(x)\geq0 \]

throughout an interval and the derivative is positive except possibly at isolated points, the function can still be strictly increasing. Therefore, it is important not to conclude that the function is not strictly increasing merely because $f'(x)=0$ at one point.

In this problem, the derivative will be a perfect square:

\[ f'(x)=3(x-1)^2. \]

It is zero at $x=1$ but positive everywhere else. We will verify that this still gives a strictly increasing function on $\mathbb{R}$.

🗺️ Solution Roadmap

Step-by-step Plan

›

Differentiate the given polynomial.
Factorise the derivative as a perfect square.
Determine its sign on $\mathbb{R}$.
Handle carefully the point $x=1$, where the derivative is zero.
Conclude that the function is strictly increasing on $\mathbb{R}$.

📊 Graph / Figure

Graph / Figure

›

Graph of function $f(x)=x^3-3x^2+3x-100$ — Graphical Representation

✏️ Solution

Complete Solution

›

Step-by-step Solution · 14 steps

Given —
\[f(x)=x^3-3x^2+3x-100\]
Find the First Derivative
Differentiating term by term,
\[f'(x)=3x^2-6x+3\]
Taking $3$ common,
\[f'(x)=3(x^2-2x+1)\]
Since
\[x^2-2x+1=(x-1)^2,\]
we obtain
\[\boxed{f'(x)=3(x-1)^2}\]
Determine the Sign of $f'(x)$
For every real number $x$,
\[(x-1)^2\geq0\]
Since
\[3>0,\]
it follows that
\[f'(x)=3(x-1)^2\geq0\]
for every
\[x\in\mathbb{R}.\]
More specifically,
\[f'(x)>0\quad\text{for }x\neq1,\]
while
\[f'(1)=0\]
Thus, the derivative is non-negative throughout $\mathbb{R}$, and it is positive everywhere except at the single point $x=1$.
Important Point About $x=1$
\[ \boxed{(x-1)^2\geq0} \]
for all $x\in\mathbb{R}$, with equality only when
\[ x=1. \]
Consequently,
\[ \boxed{f'(x)\geq0\quad\text{for all }x\in\mathbb{R}}. \]
Prove Strict Increase
To establish strict increase rigorously, take any two real numbers
\[x_1 < x_2\]
By the Mean Value Theorem, there exists some
\[ c\in(x_1,x_2) \]
such that
\[ f(x_2)-f(x_1)=f'(c)(x_2-x_1). \]
Now,
\[ x_2-x_1>0. \]
Also,
\[ f'(c)=3(c-1)^2\geq0. \]
However, if $c=1$, the derivative is zero. A more direct algebraic verification makes the strict increase completely clear.

🎯 Exam Significance

Exam Significance

›

This problem tests an important point about monotonicity: a derivative does not have to be strictly positive at every single point for the function to be strictly increasing.

Here,

\[ f'(x)=3(x-1)^2\geq0. \]

The derivative vanishes only at $x=1$, while remaining positive on both sides.

Therefore, the correct conclusion is that the function is increasing on all of $\mathbb{R}$.

Significance for Competitive Entrance Examinations

The expression can be recognized immediately as

\[ x^3-3x^2+3x-100=(x-1)^3-99. \]

Since $x^3$ is strictly increasing on $\mathbb{R}$, translating its graph horizontally and vertically does not change its increasing nature.

Alternatively,

\[ f'(x)=3(x-1)^2\geq0. \]

The point

\[ x=1 \]

is a stationary point, but not a turning point because the derivative remains positive on both sides.

🚫 Common Mistakes

Common Mistakes

›

Writing $(x-1)^2>0$ for every real $x$. The correct statement is $(x-1)^2\geq0$.
Forgetting that $f'(1)=0$.
Concluding that the function is not increasing because the derivative is zero at $x=1$.
Calling $x=1$ a local maximum or local minimum. The derivative does not change sign there.
Confusing a stationary point with a point of inflection or a turning point.
Ignoring the possibility of strict increase even when the derivative is zero at an isolated point.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 6 points

\[ f(x)=x^3-3x^2+3x-100. \]
Its derivative is
\[ \boxed{f'(x)=3(x-1)^2}. \]
Since
\[ (x-1)^2\geq0, \]
we have
\[ f'(x)\geq0 \]
for all $x\in\mathbb{R}$.
The only stationary point is
\[ x=1. \]
The derivative is positive on both sides of $x=1$, so the function continues increasing through the stationary point.
The function can also be written as
\[ f(x)=(x-1)^3-99, \]
which makes its strictly increasing nature evident.

📘 Concept & Theory

Theory and Concept

›

To determine the interval in which a differentiable function is increasing, we find its first derivative and determine where the derivative is positive.

If

\[ f'(x)>0, \]

then $f(x)$ is increasing on that interval.

Here,

\[ y=x^2e^{-x}. \]

Since this is a product of two functions, we use the product rule.

🗺️ Solution Roadmap

Step-by-step Plan

›

Differentiate $y=x^2e^{-x}$ using the product rule.
Factorise the derivative.
Determine the sign of the derivative.
Identify the interval where $y'>0$.
Compare the result with the given options.

📊 Graph / Figure

Graph / Figure

›

Graph of function $y=x^2e^{-x}$ — Graphical Representation

✏️ Solution

Complete Solution

›

Step-by-step Solution · 12 steps

Given — Function
\[y=x^2e^{-x}\]
Differentiate Using the Product Rule
Recall the product rule:
\[ \frac{d}{dx}(uv)=u'v+uv'. \]
Let
\[ u=x^2,\quad v=e^{-x}. \]
Therefore,
\[ u'=2x \]
and, using the chain rule,
\[ v'=-e^{-x}. \]
Hence,
\[\frac{dy}{dx}=(2x)e^{-x}+x^2(-e^{-x})\]
Therefore,
\[\frac{dy}{dx}=2xe^{-x}-x^2e^{-x}\]
Taking $xe^{-x}$ common,
\[\boxed{\frac{dy}{dx}=xe^{-x}(2-x)}\]
Determine the Sign of the Derivative
We know that
\[e^{-x}>0\]
for every real number $x$.
Therefore, the sign of
\[\frac{dy}{dx}=xe^{-x}(2-x)\]
depends only on
\[x(2-x)\]
The critical values are obtained from
\[x(2-x)=0\]
Thus,
\[x=0\]
or
\[x=2\]
Sign Analysis

Interval	$x$	$2-x$	$e^{-x}$	$y'=xe^{-x}(2-x)$	Behaviour
$(-\infty,0)$	−	+	+	−	Decreasing
$(0,2)$	+	+	+	+	Increasing
$(2,\infty)$	+	−	+	−	Decreasing

Hence,
\[ \boxed{ y=x^2e^{-x}\text{ is increasing on }(0,2). } \]
Checking the Options
- (A) $(-\infty,\infty)$: Incorrect. The function is not increasing for all real $x$.
- (B) $(-2,0)$: Incorrect. The derivative is negative on this interval.
- (C) $(2,\infty)$: Incorrect. The derivative is negative on this interval.
- (D) $(0,2)$: Correct.

💡 Answer

Final Answer

›

Therefore, the correct option is

\[ \boxed{\text{(D) }\left(0,2\right)}. \]

🎯 Exam Significance

Exam Significance

›

This question tests the product rule, chain rule, and the first derivative test. The key step is to recognize that the exponential factor

\[ e^{-x} \]

is always positive.

Therefore, the monotonicity is determined entirely by

\[ x(2-x). \]

Significance for Competitive Entrance Examinations

For a multiple-choice question, the fastest route is:

\[ y=x^2e^{-x} \]

\[ \Rightarrow y'=xe^{-x}(2-x). \]

Since

\[ e^{-x}>0, \]

we only need

\[ x(2-x)>0. \]

This inequality holds when

\[ \boxed{0 < x < 2}. \]

🚫 Common Mistakes

Common Mistakes

›

Forgetting the product rule.
Writing $\frac{d}{dx}(e^{-x})=e^{-x}$ and forgetting the factor $-1$.
Assuming $e^{-x}$ can be negative. It is always positive.
Ignoring the critical points $x=0$ and $x=2$.
Checking only one interval instead of determining the sign throughout the domain.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 5 points

\[ y=x^2e^{-x}. \]
Its derivative is
\[ \boxed{y'=xe^{-x}(2-x)}. \]
Since
\[ e^{-x}>0 \]
for every real $x$, the sign depends on $x(2-x)$.
The derivative is positive when
\[ 0 < x < 2. \]
Therefore, the function is increasing on
\[ \boxed{(0,2)}. \]

Interval	Sign of \(4x-3\)	Sign of \(f'(x)\)	Behaviour of \(f(x)\)
\(\left(-\infty,\frac{3}{4}\right)\)	Negative	\(f'(x) < 0\)	Decreasing
\(x=\frac{3}{4}\)	Zero	\(f'(x)=0\)	Critical point
\(\left(\frac{3}{4},\infty\right)\)	Positive	\(f'(x)>0\)	Increasing

Interval	Sign of \(f'(x)\)	Behaviour
\(\left(-\infty,-2\right)\)	\(-\)	Decreasing
\(\left(-2,-1\right)\)	\(+\)	Increasing
\(\left(-1,\infty\right)\)	\(-\)	Decreasing

Interval	Sign of \(f'(x)\)	Behaviour
\(\left(-\infty,-1\right)\)	\(-\)	Decreasing
\(\left(-1,1\right)\)	\(-\)	Decreasing
\(\left(1,3\right)\)	\(+\)	Increasing
\(\left(3,\infty\right)\)	\(+\)	Increasing

Part	Function	Strictly Increasing	Strictly Decreasing
(a)	\(x^2+2x-5\)	\(\left(-1,\infty\right)\)	\(\left(-\infty,-1\right)\)
(b)	\(10-6x-2x^2\)	\(\left(-\infty,-\frac{3}{2}\right)\)	\(\left(-\frac{3}{2},\infty\right)\)
(c)	\(-2x^3-9x^2-12x+1\)	\(\left(-2,-1\right)\)	\(\left(-\infty,-2\right)\cup\left(-1,\infty\right)\)
(d)	\(6-9x-x^2\)	\(\left(-\infty,-\frac{9}{2}\right)\)	\(\left(-\frac{9}{2},\infty\right)\)
(e)	\(f'(x)=(x+1)^3(x-3)^3\)	\(\left(1,\infty\right)\)	\(\left(-\infty,1\right)\)

Quantity	For \(x>-1\)
\(x^2\)	\(\geq0\)
\(1+x\)	\(>0\)
\((2+x)^2\)	\(>0\)
\(\frac{x^2}{(1+x)(2+x)^2}\)	\(\geq0\)
At \(x=0\)	\(f'(0)=0\)
For \(x\neq0\)	\(f'(x)>0\)

Application of Derivatives

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Chapter Complete!

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Interval	\(x-3\)	\(x+2\)	Sign of \(f'(x)\)	Behaviour
\(\left(-\infty,-2\right)\)	−	−	+	Increasing
\(x=-2\)	−	0	0	Critical point
\(\left(-2,3\right)\)	−	+	−	Decreasing
\(x=3\)	0	+	0	Critical point
\(\left(3,\infty\right)\)	+	+	+	Increasing

Interval	\(x\)	\(x-1\)	\(x-2\)	Sign of \(\frac{dy}{dx}\)	Behaviour
\((-\infty,0)\)	−	−	−	−	Decreasing
\((0,1)\)	+	−	−	+	Increasing
\((1,2)\)	+	+	−	−	Decreasing
\((2,\infty)\)	+	+	+	+	Increasing

Interval	Sign of \(f'(x)=2x-1\)	Behaviour of \(f(x)\)
\(\left(-1,\frac12\right)\)	\(f'(x)<0\)	Strictly decreasing
\(x=\frac12\)	\(f'(x)=0\)	Critical point
\(\left(\frac12,1\right)\)	\(f'(x)>0\)	Strictly increasing

Option	Function	Derivative	Sign on \(\left(0,\frac{\pi}{2}\right)\)	Behaviour
(A)	\(\cos x\)	\(-\sin x\)	Negative	Decreasing
(B)	\(\cos2x\)	\(-2\sin2x\)	Negative	Decreasing
(C)	\(\cos3x\)	\(-3\sin3x\)	Changes sign	Neither throughout
(D)	\(\tan x\)	\(\sec^2x\)	Positive	Increasing

Option	Interval	Sign of \(100x^{99}\)	Sign/Bound for \(\cos x\)	Sign of \(f'(x)\)	Behaviour
(A)	\((0,1)\)	Positive	\(\cos x>0\)	Positive	Increasing
(B)	\(\left(\frac{\pi}{2},\pi\right)\)	\(>100\)	\(\cos x\geq-1\)	\(>99\)	Increasing
(C)	\(\left(0,\frac{\pi}{2}\right)\)	Positive	\(\cos x>0\)	Positive	Increasing

Region	\(x^2\)	\(x^2-1\)	\(f'(x)\)	Behaviour
\((-\infty,-1)\)	\(>1\)	\(>0\)	\(>0\)	Increasing
\([-1,1]\)	\(\leq1\)	\(\leq0\)	\(\leq0\)	Not part of \(I\)
\((1,\infty)\)	\(>1\)	\(>0\)	\(>0\)	Increasing

Interval	Quadrant	\(\sin x\)	\(\cos x\)	\(\tan x\)	\(f'(x)=-\tan x\)	Behaviour
\(\left(0,\frac{\pi}{2}\right)\)	I	+	+	+	−	Decreasing
\(\left(\frac{3\pi}{2},2\pi\right)\)	IV	−	+	−	+	Increasing

Interval	\(x\)	\(2-x\)	\(e^{-x}\)	\(y'=xe^{-x}(2-x)\)	Behaviour
\((-\infty,0)\)	−	+	+	−	Decreasing
\((0,2)\)	+	+	+	+	Increasing
\((2,\infty)\)	+	−	+	−	Decreasing

Application of Derivatives

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

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Frequently Asked Questions

Q 01 What is covered in NCERT Class 12 Maths Exercise 6.2?

Q 02 How do you determine whether a function is increasing?

Q 03 How do you determine whether a function is decreasing?

Q 04 What is the first derivative test for increasing and decreasing functions?

Q 05 Why are critical points important in Exercise 6.2?

Q 06 What is the increasing interval for y = x²e?? in Exercise 6.2?

Q 07 How is log(sin x) analysed for monotonicity?

Q 08 How do you differentiate log|cos x|?

Q 09 Is a function still increasing if its derivative is zero at one point?

Q 10 Why are NCERT Exercise 6.2 solutions useful for JEE preparation?

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