Ch 5  ·  Q–
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Chapter 5 Exercise 5.1 Solutions

Continuity and Differentiability

Step-by-Step Solutions to the NCERT Class 12 Mathematics Chapter 5 Exercise 5.1

Class 12 Mathematics Exercise 5.1 NCERT Solutions Continuity and Differentiability Class 12 Mathematics Chapter 5 CBSE Board Exam JEE Main CUET Continuity Continuous Functions Discontinuity Points of Discontinuity Left Hand Limit Right Hand Limit Continuity Criterion Absolute Value Functions
34 Questions
75–110 min Ideal time
Q1 Now at
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Q1
NUMERIC3 marks

Prove that the function

\[ f(x)=5x-3 \]

is continuous at \(x=0\), \(x=-3\), and \(x=5\).

📘 Concept & Theory
Concept/Theory
›

A function \(f(x)\) is said to be continuous at \(x=a\) if the following three conditions are satisfied:

  1. The function value \(f(a)\) exists.
  2. The left-hand limit and right-hand limit exist and are equal:
    \[ \lim_{x\to a^-}f(x)=\lim_{x\to a^+}f(x) \]
  3. The common limiting value is equal to the actual value of the function:
    \[ \lim_{x\to a}f(x)=f(a) \]

Therefore, the most useful test for continuity at \(x=a\) is

\[ \boxed{\lim_{x\to a^-}f(x)=\lim_{x\to a^+}f(x)=f(a)} \]

The given function \(f(x)=5x-3\) is a linear polynomial function. Every polynomial function is continuous for every real value of \(x\). However, since the question specifically asks us to prove continuity at three particular points, we will verify the continuity criterion separately at \(x=0\), \(x=-3\), and \(x=5\).

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Write the given function \(f(x)=5x-3\).

  2. For \(x=0\), calculate the left-hand limit, right-hand limit, and \(f(0)\).

  3. Verify that all three values are equal.

  4. Repeat the same procedure for \(x=-3\).

  5. Repeat the procedure for \(x=5\).

  6. Conclude that the function is continuous at all three specified points.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  35 steps
  1. Given — Function
    \[f(x)=5x-3\]
  2. Continuity at \(x=0\)
  3. To prove continuity at \(x=0\), we need to verify
    \[\lim_{x\to0^-}f(x)=\lim_{x\to0^+}f(x)=f(0)\]
  4. Find the left-hand limit
  5. Since
    \[ f(x)=5x-3, \]
  6. we have
    \[\lim_{x\to0^-}f(x)=\lim_{x\to0^-}(5x-3)\]
  7. Since \(5x-3\) is a polynomial expression, we can substitute \(x=0\) directly:
  8. \[\lim_{x\to0^-}(5x-3)=5(0)-3=-3\]
  9. Therefore,
    \[\boxed{\lim_{x\to0^-}f(x)=-3}\]
  10. Find the right-hand limit
  11. Similarly,
    \[\lim_{x\to0^+}f(x)=\lim_{x\to0^+}(5x-3)\]
  12. Substituting \(x=0\),
    \[\lim_{x\to0^+}(5x-3)=5(0)-3=-3\]
  13. Therefore,
    \[\boxed{\lim_{x\to0^+}f(x)=-3}\]
  14. Find the actual value \(f(0)\)
  15. From
    \[f(x)=5x-3,\]
    putting \(x=0\), we get
    \[f(0)=5(0)-3=-3\]
  16. Hence,
    \[\boxed{f(0)=-3}\]
  17. Apply the continuity criterion
  18. We have obtained
    \[\lim_{x\to0^-}f(x)=-3\]
    \[\lim_{x\to0^+}f(x)=-3\]
    and
    \[f(0)=-3\]
  19. Therefore,
    \[\lim_{x\to0^-}f(x)=\lim_{x\to0^+}f(x)=f(0)=-3\]
  20. Hence, \(f(x)=5x-3\) is continuous at \(x=0\).
  21. Continuity at \(x=-3\)
  22. To prove continuity at \(x=-3\), we need to verify
    \[\lim_{x\to-3^-}f(x)=\lim_{x\to-3^+}f(x)=f(-3)\]
  23. Find the left-hand limit
  24. We have
    \[\lim_{x\to-3^-}f(x)=\lim_{x\to-3^-}(5x-3)\]
  25. Substituting \(x=-3\),
    \[\begin{aligned}\lim_{x\to-3^-}(5x-3)&=5(-3)-3\\&=-15-3\\&=-18\end{aligned}\]
  26. Therefore,
    \[\boxed{\lim_{x\to-3^-}f(x)=-18}\]
  27. Find the right-hand limit
  28. Similarly,
    \[\lim_{x\to-3^+}f(x)=\lim_{x\to-3^+}(5x-3)\]
  29. Substituting \(x=-3\),
    \[\begin{aligned}\lim_{x\to-3^+}(5x-3)&=5(-3)-3\\&=-15-3\\&=-18\end{aligned}\]
  30. Therefore,
    \[\boxed{\lim_{x\to-3^+}f(x)=-18}\]
  31. Find the actual value \(f(-3)\)
  32. From
    \[f(x)=5x-3,\]
    putting \(x=-3\), we get
    \[f(-3)=5(-3)-3=-18\]
  33. Hence,
    \[\boxed{f(-3)=-18}\]
  34. Apply the continuity criterion
  35. We have obtained
    \[\lim_{x\to-3^-}f(x)=-18\]
    \[\lim_{x\to-3^+}f(x)=-18\]
    and
    \[f(-3)=-18\]
  36. Therefore,
    \[\lim_{x\to-3^-}f(x)=\lim_{x\to-3^+}f(x)=f(-3)=-18\]
  37. Hence, \(f(x)=5x-3\) is continuous at \(x=-3\).
  38. Continuity at \(x=5\)
  39. To prove continuity at \(x=5\), we need to verify
    \[\lim_{x\to5^-}f(x)=\lim_{x\to5^+}f(x)=f(5)\]
  40. Find the left-hand limit
  41. We have
    \[\lim_{x\to5^-}f(x)=\lim_{x\to5^-}(5x-3)\]
  42. Substituting \(x=5\),
    \[\begin{aligned}\lim_{x\to5^-}(5x-3)&=5(5)-3\\&=25-3\\&=22\end{aligned}\]
  43. Therefore,
    \[\boxed{\lim_{x\to5^-}f(x)=22}\]
  44. Find the right-hand limit
  45. Similarly,
    \[\lim_{x\to5^+}f(x)=\lim_{x\to5^+}(5x-3)\]
  46. Substituting \(x=5\),
    \[\begin{aligned}\lim_{x\to5^+}(5x-3)&=5(5)-3\\&=25-3\\&=22\end{aligned}\]
  47. Therefore,
    \[\boxed{\lim_{x\to5^+}f(x)=22}\]
  48. Find the actual value \(f(5)\)
  49. From
    \[f(x)=5x-3,\]
    putting \(x=5\), we get
    \[f(5)=5(5)-3=22\]
  50. Hence,
    \[\boxed{f(5)=22}\]
  51. Apply the continuity criterion
  52. We have obtained
    \[\lim_{x\to5^-}f(x)=22\]
    \[\lim_{x\to5^+}f(x)=22\]
    and
    \[f(5)=22\]
  53. Therefore,
    \[\lim_{x\to5^-}f(x)=\lim_{x\to5^+}f(x)=f(5)=22\]
  54. Hence, \(f(x)=5x-3\) is continuous at \(x=5\).
💡 Answer
Final Answer
›

Therefore, the function \(f(x)=5x-3\) is continuous at \(x=0\), \(x=-3\), and \(x=5\).

🎯 Exam Significance
Exam Significance
›

This problem introduces the fundamental operational test for continuity:

\[ \lim_{x\to a^-}f(x)=\lim_{x\to a^+}f(x)=f(a). \]
It is particularly important for Class 12 board examinations because questions involving continuity commonly require students to calculate the left-hand limit, right-hand limit, and function value separately before drawing the conclusion.

A complete solution should not merely state that a polynomial is continuous. When the question says “prove”, showing the continuity criterion explicitly demonstrates the required mathematical reasoning and reduces the possibility of losing marks for an incomplete justification.

Significance for Competitive Entrance Examinations

For JEE and other competitive examinations, the deeper takeaway is that continuity can often be recognised immediately from standard classes of functions. Since every polynomial function is continuous for every real number, the function

\[ f(x)=5x-3 \]
is continuous on \(\mathbb{R}\).

+

However, competitive problems frequently modify a polynomial or combine it with rational, modulus, trigonometric, logarithmic, or piecewise functions. In such cases, the fundamental criterion used in this question becomes essential for identifying points where continuity must be checked.

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  7 points

  1. A function \(f(x)\) is continuous at \(x=a\) when

    \[ \lim_{x\to a^-}f(x)=\lim_{x\to a^+}f(x)=f(a). \]

  2. For a polynomial function, direct substitution can be used to evaluate the limit at every real point.

  3. The function \(f(x)=5x-3\) is a polynomial and is therefore continuous for every real \(x\).

  4. At \(x=0\),

    \[ \lim_{x\to0^-}f(x)=\lim_{x\to0^+}f(x)=f(0)=-3. \]

  5. At \(x=-3\),

    \[ \lim_{x\to-3^-}f(x)=\lim_{x\to-3^+}f(x)=f(-3)=-18. \]

  6. At \(x=5\),

    \[ \lim_{x\to5^-}f(x)=\lim_{x\to5^+}f(x)=f(5)=22. \]

  7. For a “prove continuity” question, explicitly checking LHL, RHL, and \(f(a)\) provides a complete and logically structured solution.
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1 / 34  ·  3%
Q2 →
Q2
NUMERIC3 marks
Examine the continuity of the function \(f(x)=2x^2-1\) at \(x=3\).
📘 Concept & Theory
Concept/Theory
›

A function \(f(x)\) is continuous at \(x=a\) if its left-hand limit, right-hand limit, and actual value of the function at \(x=a\) are equal. Mathematically,

\[ \boxed{ \lim_{x\to a^-}f(x) = \lim_{x\to a^+}f(x) = f(a) } \]

In this question,

\[ a=3. \]
Therefore, we must verify

\[ \lim_{x\to3^-}f(x) = \lim_{x\to3^+}f(x) = f(3). \]

The given function

\[ f(x)=2x^2-1 \]
is a polynomial function. Every polynomial function is continuous for all real values of \(x\). Nevertheless, because the question asks us to examine continuity specifically at \(x=3\), we will verify the continuity criterion step by step.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Write the given function \(f(x)=2x^2-1\).

  2. Calculate the left-hand limit as \(x\to3^-\).

  3. Calculate the right-hand limit as \(x\to3^+\).

  4. Calculate the actual value \(f(3)\).

  5. Compare the three values.

  6. If all three are equal, conclude that the function is continuous at \(x=3\).

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  11 steps
  1. Given —
    \[f(x)=2x^2-1\]
  2. To examine continuity at \(x=3\), we check
    \[\lim_{x\to3^-}f(x),\qquad\lim_{x\to3^+}f(x),\quad f(3).\]
  3. Find the Left-Hand Limit
  4. The left-hand limit is
    \[\lim_{x\to3^-}f(x)=\lim_{x\to3^-}(2x^2-1)\]
  5. Since \(2x^2-1\) is a polynomial, we can substitute \(x=3\) directly:
    \[\lim_{x\to3^-}(2x^2-1)=2(3)^2-1\]
  6. Calculating,
    \[\lim_{x\to3^-}(2x^2-1)=2(9)-1=18-1=17\]
  7. Hence,
    \[\boxed{\lim_{x\to3^-}f(x)=17}\]
  8. Find the Right-Hand Limit
  9. The right-hand limit is
    \[\lim_{x\to3^+}f(x)=\lim_{x\to3^+}(2x^2-1)\]
  10. Since \(2x^2-1\) is a polynomial, we can substitute \(x=3\) directly:
    \[\lim_{x\to3^+}(2x^2-1)=2(3)^2-1\]
  11. Calculating,
    \[\lim_{x\to3^+}(2x^2-1)=2(9)-1=18-1=17\]
  12. Hence,
    \[\boxed{\lim_{x\to3^+}f(x)=17}\]
  13. Find the Actual Value
  14. The actual value is
    \[f(3)=2(3)^2-1=2(9)-1=18-1=17\]
  15. Hence,
    \[\boxed{f(3)=17}\]
  16. Apply the Continuity Criterion
  17. Since all three values are equal, the function is continuous at \(x=3\).
🎯 Exam Significance
Exam Significance
›

This question tests the basic definition of continuity and the correct use of one-sided limits. For a board examination, it is important to write the three required quantities separately:

\[ \text{LHL},\qquad \text{RHL},\qquad f(3). \]

The conclusion should then explicitly show

\[ \text{LHL}=\text{RHL}=f(3). \]

This makes the verification complete and clearly demonstrates why the function is continuous at the specified point.

Significance for Competitive Entrance Examinations

For competitive examinations, the important recognition is that \(2x^2-1\) is a polynomial. Therefore, it is continuous throughout \(\mathbb{R}\). This observation can save considerable time in objective questions.

At the same time, the formal criterion used here becomes especially important when a function is piecewise defined or contains expressions such as rational functions, modulus functions, or functions involving parameters. In those cases, continuity often has to be checked at specific boundary points.

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  8 points

  1. The continuity criterion at \(x=a\) is

    \[ \lim_{x\to a^-}f(x) = \lim_{x\to a^+}f(x) = f(a). \]

  2. For the given function,

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

  3. The left-hand limit at \(x=3\) is

    \[ \lim_{x\to3^-}f(x)=17. \]

  4. The right-hand limit at \(x=3\) is

    \[ \lim_{x\to3^+}f(x)=17. \]

  5. The actual value of the function is

    \[ f(3)=17. \]

  6. Thus,

    \[ \lim_{x\to3^-}f(x) = \lim_{x\to3^+}f(x) = f(3). \]

  7. Hence, \(f(x)=2x^2-1\) is continuous at \(x=3\).

  8. More generally, every polynomial function is continuous for every real value of \(x\).
← Q1
2 / 34  ·  6%
Q3 →
Q3
NUMERIC3 marks
Examine the following functions for continuity:
\( \begin{aligned} (i)\quad &f(x)=x-5,\\[4pt] (ii)\quad &f(x)=\frac{1}{x-5},\quad x\neq5,\\[4pt] (iii)\quad &f(x)=\frac{x^2-25}{x+5},\quad x\neq-5,\\[4pt] (iv)\quad &f(x)=|x-5|. \end{aligned} \)
📘 Concept & Theory
Concept/Theory
›

A function \(f(x)\) is continuous at \(x=a\) if and only if

\[ \boxed{ \lim_{x\to a^-}f(x) = \lim_{x\to a^+}f(x) = f(a) } \]

Thus, three things must be checked:

  1. The function must be defined at \(x=a\).
  2. The left-hand limit and right-hand limit must exist and be equal.
  3. The common limit must be equal to the actual value \(f(a)\).

Different types of functions require slightly different approaches. Polynomial functions are continuous everywhere, rational functions are continuous wherever their denominators are non-zero, and modulus functions are continuous everywhere.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. For the linear function in part (i), verify continuity at an arbitrary point \(x=a\).

  2. For the rational function in part (ii), identify the point excluded from its domain.

  3. For the rational expression in part (iii), factor and simplify it, while remembering that the original function is still undefined at \(x=-5\).

  4. For the modulus function in part (iv), write its piecewise form according to whether \(x-5\) is negative or non-negative, and check continuity at the critical point \(x=5\).

  5. State the complete interval/domain on which each function is continuous.

✏️ Solution
Solution (i): \(f(x)=x-5\)
›
Step-by-step Solution  ·  12 steps
  1. Given — f(x)=x-5
  2. Since this is a polynomial function of degree one, it is expected to be continuous for every real number. We verify this using the continuity criterion at an arbitrary point \(x=a\).
  3. Find the left-hand limit
  4. Consider
    \[\lim_{x\to a^-}f(x)\]
  5. Substituting \(f(x)=x-5\), we get
    \[\lim_{x\to a^-}f(x)=\lim_{x\to a^-}(x-5)\]
  6. Since \(x-5\) is a polynomial expression, direct substitution is valid:
    \[\lim_{x\to a^-}(x-5)=a-5.\]
  7. Therefore,
    \[\boxed{\lim_{x\to a^-}f(x)=a-5}\]
  8. Find the right-hand limit
  9. Similarly,
    \[\lim_{x\to a^+}f(x)=\lim_{x\to a^+}(x-5)\]
  10. Substituting \(x=a\),
    \[\lim_{x\to a^+}(x-5)=a-5\]
  11. Hence,
    \[\boxed{\lim_{x\to a^+}f(x)=a-5}\]
  12. Find the actual value \(f(a)\)
  13. From
    \[f(x)=x-5,\]
    putting \(x=a\), we obtain
    \[f(a)=a-5\]
  14. Thus,
    \[\boxed{f(a)=a-5}\]
  15. Apply the continuity criterion
  16. We have
    \[\lim_{x\to a^-}f(x)=a-5,\]
    \[\lim_{x\to a^+}f(x)=a-5,\]
    and
    \[f(a)=a-5\]
  17. Therefore,
    \[\boxed{\lim_{x\to a^-}f(x)=\lim_{x\to a^+}f(x)=f(a)=a-5}\]
  18. Therefore,
    \[\boxed{f(x)=x-5\text{ is continuous on }\mathbb{R}.}\]
✏️ Solution
Solution (ii): \(f(x)=\frac{1}{x-5}\quad x\neq5\)
›
Step-by-step Solution  ·  9 steps
  1. Given —
    \[f(x)=\frac{1}{x-5},\qquad x\neq5\]
  2. The first step with a rational function is to examine its domain. The denominator must not be zero.
  3. Determine the excluded point
  4. The denominator is
    \[x-5\]
  5. Setting it equal to zero,
    \[x-5=0\]
  6. Therefore,
    \[x=5\]
    Thus, the function is not defined at \(x=5\)
  7. In fact,
    \[f(5)=\frac{1}{5-5}=\frac{1}{0}\]
    which is undefined.
  8. Examine continuity at \(x=5\)
  9. A necessary condition for continuity at \(x=5\) is that \(f(5)\) must exist.
    But
    \[\boxed{f(5)\text{ is not defined}.}\]
  10. Therefore, the function cannot be continuous at \(x=5\).
  11. For completeness, the one-sided behaviour also confirms the problem. We have
    \[\lim_{x\to5^-}\frac{1}{x-5}=-\infty\]
    and
    \[\lim_{x\to5^+}\frac{1}{x-5}=+\infty\]
    Hence, the two-sided finite limit does not exist.
  12. Therefore,
    \[\boxed{f(x)=\frac{1}{x-5}\text{ is discontinuous at }x=5.}\]
  13. Continuity on its Domain
  14. A rational function is continuous wherever its denominator is non-zero. Hence,
    \[\boxed{f(x)=\frac{1}{x-5}\text{ is continuous on }(-\infty,5)\cup(5,\infty).}\]
✏️ Solution
Solution (iii): \(f(x)=\frac{x^2-25}{x+5},\quad x\neq-5\)
›
Step-by-step Solution  ·  16 steps
  1. Given —
    \[f(x)=\frac{x^2-25}{x+5},\qquad x\neq-5\]
  2. The important point in this part is \(x=-5\), because the original denominator becomes zero there.
  3. Factorise the numerator
  4. The numerator is a difference of two squares:
    \[x^2-25=x^2-5^2\]
  5. Using
    \[a^2-b^2=(a-b)(a+b),\]
  6. we obtain
    \[x^2-25=(x-5)(x+5)\]
  7. Therefore,
    \[f(x)=\frac{(x-5)(x+5)}{x+5},\qquad x\neq-5\]
  8. Simplify the function
  9. Since \(x\neq-5\), we have \(x+5\neq0\), so cancellation is valid:
    \[f(x)=x-5,\qquad x\neq-5\]
    Thus, the given function behaves exactly like \(x-5\) for every \(x\neq-5\), but the original function remains undefined at \(x=-5\).
  10. Calculate the left-hand limit at \(x=-5\)
  11. Using the simplified expression,
    \[\lim_{x\to-5^-}f(x)=\lim_{x\to-5^-}(x-5)\]
  12. Substituting \(x=-5\),
    \[\lim_{x\to-5^-}(x-5)=(-5)-5=-10\]
  13. Therefore,
    \[\boxed{\lim_{x\to-5^-}f(x)=-10}\]
  14. Calculate the right-hand limit at \(x=-5\)
  15. Similarly,
    \[\lim_{x\to-5^+}f(x)=\lim_{x\to-5^+}(x-5)\]
  16. Substituting \(x=-5\),
    \[\lim_{x\to-5^+}(x-5)=(-5)-5=-10\]
  17. Hence,
    \[\boxed{\lim_{x\to-5^+}f(x)=-10}\]
  18. Find the actual value \(f(-5)\)
  19. From the original definition,
    \[f(-5)=\frac{(-5)^2-25}{-5+5}\]
  20. Simplifying,
    \[f(-5)=\frac{25-25}{0}=\frac{0}{0},\]
    which is undefined.
  21. Therefore,
    \[\boxed{f(-5)\text{ is not defined}.}\]
  22. Apply the continuity criterion
  23. Although the left-hand and right-hand limits are equal,
    \[\lim_{x\to-5^-}f(x)=\lim_{x\to-5^+}f(x)=-10,\]
    the actual function value \(f(-5)\) does not exist.
  24. Hence,
    \[\lim_{x\to-5^-}f(x)=\lim_{x\to-5^+}f(x)\neq f(-5),\]
    because \(f(-5)\) is undefined.
  25. Therefore,
    \[ \boxed{ f(x)=\frac{x^2-25}{x+5} \text{ is discontinuous at }x=-5. } \]
✏️ Solution
Solution (iv): \(f(x)=|x-5|\)
›
Step-by-step Solution  ·  22 steps
  1. Given —
    \[f(x)=|x-5|\]
  2. The modulus function changes its algebraic form according to the sign of \(x-5\). Therefore, \(x=5\) is the critical point that needs special attention.
  3. Write the modulus function in piecewise form
  4. Recall that
    \[|u|=\begin{cases}-u,&u < 0,\\u,&u\geq0.\end{cases}\]
  5. Here,
    \[u=x-5.\]
  6. When
    \[x-5 < 0,\]
    we have
    \[x < 5,\]
    and
  7. hence
    \[|x-5|=-(x-5)=5-x\]
  8. When
    \[x-5\geq0,\]
    we have
    \[x\geq5,\]
  9. and hence
    \[|x-5|=x-5\]
  10. Therefore,
    \[\boxed{f(x)=\begin{cases}5-x,&x < 5,\\x-5,&x\geq5\end{cases}}\]
  11. Calculate the left-hand limit at \(x=5\)
  12. For \(x < 5\),
    \[f(x)=5-x\]
  13. Therefore,
    \[\lim_{x\to5^-}f(x)=\lim_{x\to5^-}(5-x)\]
  14. Substituting \(x=5\),
    \[\lim_{x\to5^-}(5-x)=5-5=0\]
  15. Hence,
    \[\boxed{\lim_{x\to5^-}f(x)=0}\]
  16. Calculate the right-hand limit at \(x=5\)
  17. For \(x > 5\),
    \[f(x)=x-5\]
  18. Therefore,
    \[\lim_{x\to5^+}f(x)=\lim_{x\to5^+}(x-5)\]
  19. Substituting \(x=5\)
    \[\lim_{x\to5^+}(x-5)=5-5=0\]
  20. Hence,
    \[\boxed{\lim_{x\to5^+}f(x)=0}\]
  21. Find the actual value \(f(5)\)
  22. From
    \[f(x)=|x-5|,\]
    putting \(x=5\),
  23. we get
    \[f(5)=|5-5|=0\]
  24. Therefore,
    \[\boxed{f(5)=0}\]
  25. Apply the continuity criterion
  26. We have
    \[\lim_{x\to5^-}f(x)=0\]
    \[\lim_{x\to5^+}f(x)=0\]
    and
    \[f(5)=0\]
  27. Hence,
    \[\boxed{\lim_{x\to5^-}f(x)=\lim_{x\to5^+}f(x)=f(5)=0}\]
  28. Therefore, \(f(x)=|x-5|\) is continuous at \(x=5\).
  29. In fact, the modulus function is continuous for every real number. Hence,
    \[\boxed{f(x)=|x-5|\text{ is continuous on }\mathbb{R}.}\]
🎯 Exam Significance
Exam Significance
›

This question covers three important classes of functions that frequently appear in Class 12 board examinations: polynomial functions, rational functions, and modulus functions. The key skill is not simply calculating limits but identifying the points at which continuity needs to be examined.

Part (iii) is particularly important. Cancelling \(x+5\) gives \(x-5\), but the original function is explicitly undefined at \(x=-5\). Cancellation simplifies the expression for \(x\neq-5\); it does not automatically redefine the original function at \(x=-5\). This distinction is essential for correctly identifying a removable discontinuity.

For board answers, always distinguish between the simplified expression and the original function's domain.

Significance for Competitive Entrance Examinations

This problem develops several high-value recognition skills for JEE and other entrance examinations. Polynomial functions can immediately be identified as continuous everywhere. Rational functions are continuous wherever their denominators are non-zero. Modulus functions are continuous everywhere, although their algebraic form changes at the point where the expression inside the modulus becomes zero.

Part (iii) is a standard model for questions involving a removable discontinuity. A common competitive-exam technique is to simplify the expression, calculate the limiting value at the excluded point, and then compare it with the defined value of the function.

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  9 points

  1. Every polynomial function is continuous on \(\mathbb{R}\).

  2. A rational function is continuous wherever its denominator is non-zero.

  3. For

    \[ f(x)=\frac{1}{x-5}, \]
    \(x=5\) is excluded from the domain, so the function is discontinuous there.

  4. For

    \[ f(x)=\frac{x^2-25}{x+5}, \]
    factorisation gives
    \[ f(x)=x-5,\qquad x\neq-5. \]
    The original function remains undefined at \(x=-5\), producing a removable discontinuity.

  5. For the third function,

    \[ \lim_{x\to-5}f(x)=-10, \]
    but \(f(-5)\) is undefined.

  6. The modulus function

    \[ f(x)=|x-5| \]
    is continuous at \(x=5\), because
    \[ \lim_{x\to5^-}f(x) = \lim_{x\to5^+}f(x) = f(5) = 0. \]

  7. When checking continuity, always inspect the domain first.

  8. For a piecewise or modulus function, identify the points where the algebraic definition changes and check continuity there.

  9. Equality of LHL and RHL alone is not sufficient for continuity. The common limit must also equal the actual function value \(f(a)\).
← Q2
3 / 34  ·  9%
Q4 →
Q4
NUMERIC3 marks
Prove that the function \(f(x)=x^n\) is continuous at \(x=n\), where \(n\) is a positive integer.
📘 Concept & Theory
Concept/Theory
›

A function \(f(x)\) is continuous at \(x=a\) if

\[\boxed{\lim_{x\to a^-}f(x)=\lim_{x\to a^+}f(x)=f(a)}\]

In this question, the point at which continuity is to be examined is

\[ x=n. \]
Therefore, we have to verify

\[ \lim_{x\to n^-}f(x) = \lim_{x\to n^+}f(x) = f(n). \]

Since \(n\) is a positive integer, \(x^n\) is a polynomial function. Every polynomial function is continuous for every real value of \(x\). Nevertheless, we will prove continuity at \(x=n\) directly using the definition of continuity.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Write the given function \(f(x)=x^n\).

  2. Calculate the left-hand limit as \(x\to n^-\).

  3. Calculate the right-hand limit as \(x\to n^+\).

  4. Calculate the actual value \(f(n)\).

  5. Compare the three quantities.

  6. If they are equal, conclude that \(f(x)\) is continuous at \(x=n\).

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  14 steps
  1. Given —
    \[f(x)=x^n\]
    where \(n\) is a positive integer.
  2. We have to examine the continuity of \(f(x)\) at
    \[x=n\]
  3. Thus, we need to verify
    \[\lim_{x\to n^-}f(x)=\lim_{x\to n^+}f(x)=f(n)\]
  4. Find the Left-Hand Limit
  5. The left-hand limit of \(f(x)\) at \(x=n\) is
    \[\lim_{x\to n^-}f(x)=\lim_{x\to n^-}x^n.\]
  6. Since \(n\) is a positive integer, \(x^n\) is a polynomial function. Therefore, direct substitution of \(x=n\) is valid:
    \[\lim_{x\to n^-}x^n=n^n.\]
  7. Hence,
    \[\boxed{\lim_{x\to n^-}f(x)=n^n}\]
  8. Find the Right-Hand Limit
  9. The right-hand limit of \(f(x)\) at \(x=n\) is
    \[\lim_{x\to n^+}f(x)=\lim_{x\to n^+}x^n\]
  10. Again, since \(x^n\) is a polynomial function, we can substitute \(x=n\) directly:
    \[\lim_{x\to n^+}x^n=n^n\]
  11. Therefore,
    \[\boxed{\lim_{x\to n^+}f(x)=n^n}\]
  12. Find the Actual Value \(f(n)\)
  13. From the given function,
    \[f(x)=x^n\]
  14. Putting \(x=n\), we obtain
    \[f(n)=n^n\]
  15. Hence,
    \[\boxed{f(n)=n^n}\]
  16. Apply the Continuity Criterion
  17. From the above calculations,
    \[\lim_{x\to n^-}f(x)=n^n,\]
    \[\lim_{x\to n^+}f(x)=n^n,\]
    and
    \[f(n)=n^n\]
  18. Therefore,
    \[\boxed{\lim_{x\to n^-}f(x)=\lim_{x\to n^+}f(x)=f(n)=n^n}\]
  19. Hence, all the conditions for continuity are satisfied. Therefore,
    \[\boxed{f(x)=x^n\text{ is continuous at }x=n.}\]
🎯 Exam Significance
Exam Significance
›

This question is a direct application of the definition of continuity and is useful for learning how to present a proof systematically. For a board examination, the safest approach is to calculate the left-hand limit, right-hand limit, and function value separately and then explicitly show their equality.

A complete conclusion should contain the statement

\[ \lim_{x\to n^-}f(x) = \lim_{x\to n^+}f(x) = f(n). \]

This establishes continuity rather than merely asserting it.

Significance for Competitive Entrance Examinations

For competitive examinations, the key recognition is that \(x^n\), with \(n\) a positive integer, is a polynomial and hence is continuous everywhere. Therefore, if an objective question asks about continuity of \(x^n\) at any real point, no lengthy limit calculation is normally required.

This recognition becomes particularly valuable when \(x^n\) appears as one component of a more complicated function. Polynomial continuity can then be combined with standard continuity results for sums, products, quotients, and compositions.

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  9 points

  1. The function given is

    \[ f(x)=x^n, \]
    where \(n\) is a positive integer.

  2. The point at which continuity is examined is \(x=n\).

  3. The left-hand limit is

    \[ \lim_{x\to n^-}f(x)=n^n. \]

  4. The right-hand limit is

    \[ \lim_{x\to n^+}f(x)=n^n. \]

  5. The actual value of the function is

    \[ f(n)=n^n. \]

  6. Therefore,

    \[ \lim_{x\to n^-}f(x) = \lim_{x\to n^+}f(x) = f(n) = n^n. \]

  7. Hence, \(f(x)=x^n\) is continuous at \(x=n\).

  8. More generally, every polynomial function is continuous at every real number.

  9. For competitive examinations, recognising \(x^n\) as a polynomial can provide a much faster solution than explicitly evaluating one-sided limits.
← Q3
4 / 34  ·  12%
Q5 →
Q5
NUMERIC3 marks
Is the function \(f\) defined by \[f(x)=\begin{cases}x, & \text{if }x\leq1,\\5, & \text{if }x>1\end{cases}\] continuous at \(x=0\), at \(x=1\), and at \(x=2\)?
📘 Concept & Theory
Concept/Theory
›

For a function \(f(x)\) to be continuous at \(x=a\), the following condition must hold:

\[\boxed{\lim_{x\to a^-}f(x)=\lim_{x\to a^+}f(x)=f(a)}\]

For a piecewise defined function, special attention must be given to the point where the definition changes. Here the definition changes at

\[x=1\]

Therefore:

  • At \(x=0\), both values immediately to the left and right of \(0\) satisfy \(x\leq1\), so the expression \(f(x)=x\) is used on both sides.
  • At \(x=1\), the expression changes from \(f(x)=x\) to \(f(x)=5\). This is the critical point and requires separate calculation of LHL, RHL, and \(f(1)\).
  • At \(x=2\), both values immediately to the left and right of \(2\) satisfy \(x>1\) in a neighbourhood of \(2\). Thus, \(f(x)=5\) on both sides.
🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Check continuity at \(x=0\) using the first branch \(f(x)=x\).

  2. Check continuity at \(x=1\), where the two branches meet.

  3. Check continuity at \(x=2\) using the second branch \(f(x)=5\).

  4. At each point, compare the left-hand limit, right-hand limit, and actual function value.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  38 steps
  1. Continuity at \(x=0\)
  2. Since \(0<1\), values of \(x\) sufficiently close to \(0\) satisfy \(x\leq1\). Therefore, near \(x=0\),
    \[f(x)=x\]
  3. Find the left-hand limit
  4. We have
    \[\lim_{x\to0^-}f(x)=\lim_{x\to0^-}x\]
  5. Hence,
    \[\lim_{x\to0^-}x=0\]
  6. Therefore,
    \[\boxed{\lim_{x\to0^-}f(x)=0}\]
  7. Find the right-hand limit
  8. Similarly, for \(x\) approaching \(0\) from the right, we still have \(x\leq1\) in a sufficiently small neighbourhood of \(0\). Therefore,
    \[\lim_{x\to0^+}f(x)=\lim_{x\to0^+}x=0\]
  9. Thus,
    \[\boxed{\lim_{x\to0^+}f(x)=0}\]
  10. Find the actual value \(f(0)\)
  11. Since \(0\leq1\), the first branch applies:
    \[f(0)=0\]
  12. Therefore,
    \[\boxed{f(0)=0}\]
  13. Apply the continuity criterion
  14. We have
    \[\lim_{x\to0^-}f(x)=\lim_{x\to0^+}f(x)=f(0)=0\]
  15. Hence,
    \[\boxed{f(x)\text{ is continuous at }x=0}\]
  16. Continuity at \(x=1\)
  17. The point \(x=1\) is particularly important because the definition of the function changes at this point:
    \[f(x)=x,\quad x\leq1\]
  18. whereas
    \[f(x)=5,\quad x>1\]
  19. Therefore, we must calculate the left-hand limit, right-hand limit, and \(f(1)\) separately.
  20. Find the left-hand limit
  21. As \(x\to1^-\), we have \(x<1\), so the first branch applies:
    \[f(x)=x\]
  22. Therefore,
    \[\lim_{x\to1^-}f(x)=\lim_{x\to1^-}x=1\]
  23. Hence,
    \[\boxed{\lim_{x\to1^-}f(x)=1}\]
  24. Find the right-hand limit
  25. As \(x\to1^+\), we have \(x>1\), so the second branch applies:
    \[f(x)=5\]
  26. Therefore,
    \[\lim_{x\to1^+}f(x)=\lim_{x\to1^+}5=5.\]
  27. Hence,
    \[\boxed{\lim_{x\to1^+}f(x)=5}\]
  28. Find the actual value \(f(1)\)
  29. At \(x=1\), the condition \(x\leq1\) is satisfied. Therefore, the first branch is used:
    \[f(1)=1\]
  30. Thus,
    \[\boxed{f(1)=1}\]
  31. Compare LHL, RHL, and \(f(1)\)
  32. We have
    \[\lim_{x\to1^-}f(x)=1\]
  33. while
    \[\lim_{x\to1^+}f(x)=5\]
  34. Therefore,
    \[\lim_{x\to1^-}f(x)\neq\lim_{x\to1^+}f(x)\]
  35. Since the left-hand and right-hand limits are unequal, the two-sided limit
    \[ \lim_{x\to1}f(x) \]
    does not exist.
  36. Consequently, the continuity condition cannot be satisfied, even though \(f(1)\) exists.
  37. Hence,
    \[\boxed{f(x)\text{ is discontinuous at }x=1.}\]
  38. Continuity at \(x=2\)
  39. Since \(2>1\), the second branch applies at \(x=2\). Moreover, all values of \(x\) sufficiently close to \(2\) are also greater than \(1\). Hence, in a neighbourhood of \(x=2\),
    \[f(x)=5\]
  40. Find the left-hand limit
  41. As \(x\to2^-\), values of \(x\) sufficiently close to \(2\) still satisfy \(x>1\). Therefore,
    \[\lim_{x\to2^-}f(x)=\lim_{x\to2^-}5=5\]
  42. Hence,
    \[\boxed{\lim_{x\to2^-}f(x)=5}\]
  43. Find the right-hand limit
  44. As \(x\to2^+\), we again have \(x>1\), so
    \[f(x)=5\]
  45. Therefore,
    \[\lim_{x\to2^+}f(x)=\lim_{x\to2^+}5=5\]
  46. Thus,
    \[\boxed{\lim_{x\to2^+}f(x)=5}\]
  47. Find the actual value \(f(2)\)
  48. Since \(2>1\), the second branch applies:
    \[f(2)=5\]
  49. Hence,
    \[\boxed{f(2)=5}\]
  50. Apply the continuity criterion
  51. We have
    \[\lim_{x\to2^-}f(x)=5\]
    \[\lim_{x\to2^+}f(x)=5\]
    and
    \[f(2)=5\]
  52. Therefore,
    \[\boxed{\lim_{x\to2^-}f(x)=\lim_{x\to2^+}f(x)=f(2)=5}\]
  53. Hence,
    \[\boxed{f(x)\text{ is continuous at }x=2.}\]
🎯 Exam Significance
Exam Significance
›

This is an important model problem for piecewise functions. The main examination skill is identifying which branch of the function must be used for the left-hand limit, right-hand limit, and actual function value.

The point \(x=1\) requires special attention because the definition changes there. Since the first branch includes \(x=1\), we have

\[ f(1)=1. \]
However, immediately to the right of \(1\), the second branch gives the value \(5\). This produces unequal one-sided limits:
\[ 1\neq5. \]

For board examinations, it is important not to conclude continuity merely because \(f(a)\) exists. Equality of the two one-sided limits is also necessary.

Significance for Competitive Entrance Examinations

This problem illustrates a common competitive-examination pattern: a piecewise function can be continuous throughout portions of its domain but discontinuous at the point where its definition changes.

The fastest strategy is to identify the boundary point between the pieces. Here that point is \(x=1\). At points away from the boundary, the function behaves like a single simple function in a neighbourhood, so continuity can often be recognised immediately.

At \(x=1\), the mismatch

\[ \text{LHL}=1,\qquad \text{RHL}=5 \]
immediately establishes a jump discontinuity.

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  8 points

  1. For a piecewise function, first identify the points where the definition changes.

  2. The continuity criterion is

    \[ \lim_{x\to a^-}f(x) = \lim_{x\to a^+}f(x) = f(a). \]

  3. At \(x=0\), the first branch \(f(x)=x\) applies on both sides:

    \[ \text{LHL}=\text{RHL}=f(0)=0. \]
    Hence the function is continuous at \(x=0\).

  4. At \(x=1\), the left branch gives

    \[ \text{LHL}=1, \]
    while the right branch gives
    \[ \text{RHL}=5. \]
    Hence the function is discontinuous at \(x=1\).

  5. The discontinuity at \(x=1\) is a jump discontinuity.

  6. At \(x=2\), the second branch \(f(x)=5\) applies on both sides:

    \[ \text{LHL}=\text{RHL}=f(2)=5. \]
    Hence the function is continuous at \(x=2\).

  7. Do not confuse the value \(f(a)\) with the right-hand limit. At \(x=1\), \(f(1)=1\) because the condition \(x\leq1\) includes \(x=1\).

  8. For piecewise continuity problems, always pay careful attention to whether an inequality is \(<\), \(\leq\), \(>\), or \(\geq\).
← Q4
5 / 34  ·  15%
Q6 →
Q6
NUMERIC3 marks
Find the points of discontinuity of the function \(f\), where \[ f(x)= \begin{cases} 2x+3, & \text{if }x\leq2,\\ 2x-3, & \text{if }x>2. \end{cases} \]
📘 Concept & Theory
Concept/Theory
›

A function is continuous at \(x=a\) if

\[ \boxed{ \lim_{x\to a^-}f(x) = \lim_{x\to a^+}f(x) = f(a) } \]

For a piecewise function, each individual branch may be continuous on its own interval. The only point that generally requires special examination is the boundary point where the definition changes.

Here the definition changes at

\[ x=2. \]
Both \(2x+3\) and \(2x-3\) are polynomial functions and are continuous everywhere on their respective domains. Therefore, the only possible point of discontinuity is \(x=2\).

We therefore need to calculate the left-hand limit, right-hand limit, and function value at \(x=2\).

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Identify the point where the definition of the function changes.

  2. Check continuity on each side of this point.

  3. Calculate the left-hand limit at \(x=2\).

  4. Calculate the right-hand limit at \(x=2\).

  5. Calculate \(f(2)\) using the branch whose condition includes \(x=2\).

  6. Compare LHL, RHL, and \(f(2)\).

  7. State the point of discontinuity and its type.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  15 steps
  1. Identify the Possible Point of Discontinuity
  2. The function is defined as
    \[f(x)=\begin{cases}2x+3, & x\leq2,\\2x-3, & x>2\end{cases}\]
  3. The definition changes at
    \[x=2\]
  4. For \(x<2\), the function is
    \[f(x)=2x+3,\]
    which is a polynomial and hence continuous.
  5. Therefore, there can be no discontinuity away from the joining point \(x=2\). We only need to examine continuity at \(x=2\). For \(x>2\), the function is
    \[f(x)=2x-3,\]
    which is also a polynomial and hence continuous.
  6. Find the Left-Hand Limit
  7. As \(x\to2^-\), we have \(x<2\), so the first branch applies:
    \[f(x)=2x+3\]
  8. Therefore,
    \[\lim_{x\to2^-}f(x)=\lim_{x\to2^-}(2x+3)=2(2)+3=7\]
  9. Hence,
    \[\boxed{\lim_{x\to2^-}f(x)=7}\]
  10. Find the Right-Hand Limit
  11. As \(x\to2^+\), we have \(x>2\), so the second branch applies:
    \[f(x)=2x-3\]
  12. Therefore,
    \[\lim_{x\to2^+}f(x)=\lim_{x\to2^+}(2x-3)=2(2)-3=1\]
  13. Hence,
    \[\boxed{\lim_{x\to2^+}f(x)=1}\]
  14. Find the Actual Value \(f(2)\)
  15. Since \(x=2\) satisfies the condition for the first branch, we use:
    \[f(2)=2(2)+3=7\]
  16. Hence,
    \[\boxed{f(2)=7}\]
  17. Apply the Continuity Criterion
  18. We have:
    \[\lim_{x\to2^-}f(x)=7, \quad \lim_{x\to2^+}f(x)=1, \quad f(2)=7\]
  19. Since the left-hand limit and the function value are equal, but the right-hand limit is different, the function is not continuous at \(x=2\).
  20. The discontinuity is a jump discontinuity.
🎯 Exam Significance
Exam Significance
›

This is a standard Class 12 problem on continuity of a piecewise function. The key examination skill is to correctly identify the branch applicable to each one-sided limit.

In particular, at \(x=2\), the condition \(x\leq2\) includes the point \(2\), so

\[ f(2)=2(2)+3=7. \]
The right-hand limit, however, must use the second branch because values immediately to the right of \(2\) satisfy \(x>2\).

Writing LHL, RHL, and \(f(2)\) separately makes the reasoning transparent and provides a complete board-examination solution.

Significance for Competitive Entrance Examinations

This problem illustrates a frequently tested pattern in which two individually continuous polynomial expressions are joined at a boundary point. The only possible discontinuity occurs at the joining point.

For objective questions, the calculation can be reduced to comparing the two branch values at the boundary:

\[ 2(2)+3=7 \]

and

\[ 2(2)-3=1. \]

Since these values differ, the function has a jump discontinuity at \(x=2\).

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  10 points

  1. For a piecewise function, first identify where its definition changes.

  2. Here the critical point is

    \[ x=2. \]

  3. For \(x<2\), use

    \[ f(x)=2x+3. \]

  4. For \(x>2\), use

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

  5. The left-hand limit is

    \[ \lim_{x\to2^-}f(x)=7. \]

  6. The right-hand limit is

    \[ \lim_{x\to2^+}f(x)=1. \]

  7. Since \(x=2\) satisfies \(x\leq2\),

    \[ f(2)=7. \]

  8. Because

    \[ \lim_{x\to2^-}f(x)\neq\lim_{x\to2^+}f(x), \]
    the function is discontinuous at \(x=2\).

  9. The discontinuity is a jump discontinuity.

  10. Both branches are polynomial functions and are therefore continuous away from the joining point.
← Q5
6 / 34  ·  18%
Q7 →
Q7
NUMERIC3 marks
Find the points of discontinuity of the function \(f\), where \[ f(x)= \begin{cases} |x|+3, & \text{if }x\leq-3,\\ -2x, & \text{if }-3 < x < 3,\\ 6x+2, & \text{if }x\geq3. \end{cases} \]
📘 Concept & Theory
Concept/Theory
›

A function \(f(x)\) is continuous at \(x=a\) if

\[\boxed{\lim_{x\to a^-}f(x)=\lim_{x\to a^+}f(x)=f(a)}\]

For a piecewise defined function, the individual branches are usually continuous on their respective intervals. Therefore, the main points that require examination are the boundary points where the definition of the function changes.

In the present function, the definition changes at

\[ x=-3 \]
and
\[ x=3. \]
Hence, these are the only points that need to be tested for discontinuity.

Notice that the first branch contains \(|x|\). Although modulus functions are continuous everywhere, we still use the correct branch at \(x=-3\). Since \(-3<0\),

\[ |-3|=3. \]

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Identify the boundary points \(x=-3\) and \(x=3\).

  2. At \(x=-3\), calculate the left-hand limit using \(|x|+3\), the right-hand limit using \(-2x\), and \(f(-3)\) using the first branch.

  3. Compare these three quantities.

  4. At \(x=3\), calculate the left-hand limit using \(-2x\), the right-hand limit using \(6x+2\), and \(f(3)\) using the third branch.

  5. Compare these three quantities and identify the discontinuity.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  43 steps
  1. Identify the Points of Consideration
  2. The function is
    \[f(x)=\begin{cases}|x|+3, & x\leq-3,\\-2x, & -3 < x < 3,\\6x+2, & x\geq3\end{cases}\]
  3. The formula changes at
    \[\boxed{x=-3\quad\text{and}\quad x=3}\]
  4. These are therefore the only possible points of discontinuity.
  5. Examine Continuity at \(x=-3\)
  6. We need to verify
    \[\lim_{x\to-3^-}f(x)=\lim_{x\to-3^+}f(x)=f(-3)\]
  7. Find the Left-Hand Limit at \(x=-3\)
  8. As \(x\to-3^-\), the values of \(x\) are less than \(-3\). Therefore, the first branch applies:
    \[f(x)=|x|+3\]
  9. Hence,
    \[\lim_{x\to-3^-}f(x)=\lim_{x\to-3^-}(|x|+3)\]
  10. Since \(x\) is negative near \(-3\),
    \[|x|=-x\]
  11. Therefore,
    \[\lim_{x\to-3^-}(|x|+3)=\lim_{x\to-3^-}(-x+3)\]
  12. Substituting \(x=-3\),
    \[-(-3)+3=3+3=6\]
  13. Thus,
    \[\boxed{\lim_{x\to-3^-}f(x)=6}\]
  14. Find the Right-Hand Limit at \(x=-3\)
  15. As \(x\to-3^+\), the values of \(x\) satisfy
    \[-3 < x < 3\]
  16. Therefore, the second branch applies:
    \[f(x)=-2x\]
  17. Hence,
    \[\lim_{x\to-3^+}f(x)=\lim_{x\to-3^+}(-2x)\]
  18. Substituting \(x=-3\),
    \[-2(-3)=6\]
  19. Thus,
    \[\boxed{\lim_{x\to-3^+}f(x)=6}\]
  20. Find \(f(-3)\)
  21. Using the first branch, since \(x=-3\),
    \[f(-3)=|-3|+3=3+3=6\]
  22. Therefore,
    \[\boxed{f(-3)=6}\]
  23. Find the Actual Value \(f(-3)\)
  24. At \(x=-3\), the condition
    \[x\leq-3\]
    is satisfied. Therefore, the first branch is used:
    \[f(-3)=|-3|+3\]
  25. Since
    \[|-3|=3\]
  26. we obtain
    \[f(-3)=3+3=6\]
  27. Hence,
    \[\boxed{f(-3)=6}\]
  28. Apply the Continuity Criterion at \(x=-3\)
  29. We have
    \[\lim_{x\to-3^-}f(x)=6\]
    \[\lim_{x\to-3^+}f(x)=6\]
    and
    \[f(-3)=6\]
  30. Therefore,
    \[\boxed{\lim_{x\to-3^-}f(x)=\lim_{x\to-3^+}f(x)=f(-3)=6}\]
  31. Hence, the function is continuous at \(x=-3\)
  32. Examine Continuity at \(x=3\)
  33. We now check the other boundary point \(x=3\). We need to verify
    \[\lim_{x\to3^-}f(x)=\lim_{x\to3^+}f(x)=f(3)\]
  34. Find the Left-Hand Limit at \(x=3\)
  35. As \(x\to3^-\), the values of \(x\) satisfy
    \[-3 < x < 3.\]
  36. Therefore, the second branch applies:
    \[f(x)=-2x\]
  37. Hence,
    \[\lim_{x\to3^-}f(x)=\lim_{x\to3^-}(-2x).\]
  38. Substituting \(x=3\),
    \[-2(3)=-6\]
  39. Therefore,
    \[\boxed{\lim_{x\to3^-}f(x)=-6}.\]
  40. Find the Right-Hand Limit at \(x=3\)
  41. As \(x\to3^+\), the values of \(x\) are greater than \(3\). Therefore, the third branch applies:
    \[f(x)=6x+2\]
  42. Hence,
    \[\lim_{x\to3^+}f(x)=\lim_{x\to3^+}(6x+2).\]
  43. Substituting \(x=3\),
    \[6(3)+2=18+2=20\]
  44. Therefore,
    \[\boxed{\lim_{x\to3^+}f(x)=20}\]
  45. Find the Actual Value \(f(3)\)
  46. At \(x=3\), the condition
    \[x\geq3\]
    is satisfied. Therefore, the third branch is used:
    \[\begin{aligned}f(3)&=6(3)+2\\&=18+2\\&=20\end{aligned}\]
  47. Thus,
    \[\boxed{f(3)=20}\]
  48. Apply the Continuity Criterion at \(x=3\)
  49. We have
    \[\lim_{x\to3^-}f(x)=-6\]
  50. whereas
    \[\lim_{x\to3^+}f(x)=20\]
  51. Therefore,
    \[\lim_{x\to3^-}f(x)\neq\lim_{x\to3^+}f(x)\]
  52. Hence, the two-sided limit
    \[\lim_{x\to3}f(x)\]
    does not exist.
  53. Although
    \[f(3)=20,\]
    the equality required for continuity is not satisfied because the left-hand and right-hand limits are unequal.
  54. Therefore,
    \[\boxed{f(x)\text{ is discontinuous at }x=3.}\]
  55. Since the two one-sided limits are finite but unequal, the discontinuity at \(x=3\) is a jump discontinuity.
🎯 Exam Significance
Exam Significance
›

This problem is an important application of continuity for piecewise functions. It tests whether the student can identify the correct branch for each one-sided limit and for the actual function value.

At \(x=-3\), the first branch applies to the function value because the condition is \(x\leq-3\). The right-hand limit uses the middle branch because values immediately greater than \(-3\) satisfy \(-3

At \(x=3\), the left-hand limit uses the middle branch, whereas both the right-hand limit and \(f(3)\) use the third branch because the condition there is \(x\geq3\).

Writing these distinctions explicitly is important in a board examination because an incorrect choice of branch can produce an incorrect conclusion even when the subsequent arithmetic is correct.

Significance for Competitive Entrance Examinations

For competitive examinations, this problem demonstrates a fast and reliable method for piecewise continuity questions: identify the boundary points first, then compare the two branch values at each boundary.

At \(x=-3\), the two limiting values agree:

\[ |-3|+3=6 \]

and

\[ -2(-3)=6. \]

Hence there is no discontinuity at \(-3\).

At \(x=3\), the corresponding values are

\[ -2(3)=-6 \]

and

\[ 6(3)+2=20. \]

Since these are unequal, \(x=3\) is immediately identified as a jump discontinuity.

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  9 points

  1. For a piecewise function, the points where the formula changes are the primary points to test for discontinuity.

  2. Here the points of consideration are

    \[ x=-3\quad\text{and}\quad x=3. \]

  3. At \(x=-3\),

    \[ \lim_{x\to-3^-}f(x)=6, \]
    \[ \lim_{x\to-3^+}f(x)=6, \]
    and
    \[ f(-3)=6. \]
    Hence the function is continuous at \(x=-3\).

  4. At \(x=3\),

    \[ \lim_{x\to3^-}f(x)=-6, \]
    \[ \lim_{x\to3^+}f(x)=20, \]
    and
    \[ f(3)=20. \]

  5. Since

    \[ -6\neq20, \]
    the function is discontinuous at \(x=3\).

  6. The discontinuity at \(x=3\) is a jump discontinuity.

  7. The modulus expression \(|x|\) is continuous, so it does not itself introduce a discontinuity.

  8. Always use the correct branch when calculating LHL, RHL, and \(f(a)\).

  9. At an endpoint included by a condition such as \(x\leq-3\) or \(x\geq3\), that branch determines the actual function value.
← Q6
7 / 34  ·  21%
Q8 →
Q8
NUMERIC3 marks
Find the points of discontinuity of the function \(f\), where \[ f(x)= \begin{cases} \dfrac{|x|}{x}, & \text{if }x\neq0,\\[6pt] 0, & \text{if }x=0. \end{cases} \]
📘 Concept & Theory
Concept/Theory
›

A function \(f(x)\) is continuous at \(x=a\) if

\[\boxed{\lim_{x\to a^-}f(x)=\lim_{x\to a^+}f(x)=f(a)}\]

In this problem, the expression

\[ \frac{|x|}{x} \]
is defined only for \(x\neq0\), while a separate value \(f(0)=0\) is assigned at \(x=0\). This makes \(x=0\) the natural point at which continuity must be examined.

The important property of the modulus function is

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

Consequently,

\[\frac{|x|}{x}=\begin{cases}-1,&x<0,\\1,&x>0.\end{cases}\]

Thus, the function approaches different values from the two sides of \(x=0\). This is the key observation behind the discontinuity.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Identify \(x=0\) as the point requiring examination.

  2. For \(x<0\), replace \(|x|\) by \(-x\) and calculate the left-hand limit.

  3. For \(x>0\), replace \(|x|\) by \(x\) and calculate the right-hand limit.

  4. Calculate the actual function value \(f(0)\).

  5. Compare the left-hand limit, right-hand limit, and \(f(0)\).

  6. Conclude whether the function is continuous or discontinuous at \(x=0\).

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  24 steps
  1. Identify the Point of Consideration
  2. The function is defined separately at \(x=0\):
    \[f(0)=0\]
  3. For \(x\neq0\)
    \[f(x)=\frac{|x|}{x}\]
  4. Therefore, the only point at which continuity needs to be examined is
    \[\boxed{x=0}\]
  5. Find the Left-Hand Limit at \(x=0\)
  6. When
    \[x<0,\]
  7. we have
    \[|x|=-x\]
  8. Therefore, for \(x<0\),
    \[f(x)=\frac{|x|}{x}=\frac{-x}{x}.\]
  9. Since \(x\neq0\),
    \[\frac{-x}{x}=-1\]
  10. Hence,
    \[\lim_{x\to0^-}f(x)=\lim_{x\to0^-}\frac{|x|}{x}=-1\]
  11. Therefore,
    \[\boxed{\lim_{x\to0^-}f(x)=-1}\]
  12. Find the Right-Hand Limit at \(x=0\)
  13. When
    \[x>0,\]
  14. we have
    \[|x|=x\]
  15. Therefore, for \(x>0\),
    \[f(x)=\frac{|x|}{x}=\frac{x}{x}\]
  16. Since \(x\neq0\),
    \[\frac{x}{x}=1\]
  17. Hence,
    \[\lim_{x\to0^+}f(x)=\lim_{x\to0^+}\frac{|x|}{x}=1\]
  18. Therefore,
    \[\boxed{\lim_{x\to0^+}f(x)=1}\]
  19. Find the Actual Value \(f(0)\)
  20. The function is explicitly defined at \(x=0\) by the second branch:
    \[f(x)=0,\qquad x=0\]
  21. Therefore,
    \[\boxed{f(0)=0}\]
  22. Apply the Continuity Criterion
  23. We have obtained
    \[\lim_{x\to0^-}f(x)=-1\]
    \[\lim_{x\to0^+}f(x)=1\]
    and
    \[f(0)=0\]
  24. Since
    \[-1\neq1\]
  25. we have
    \[\boxed{\lim_{x\to0^-}f(x)\neq\lim_{x\to0^+}f(x)}\]
  26. Therefore, the two-sided limit
    \[\lim_{x\to0}f(x)\]
    does not exist.
  27. Hence the continuity condition
    \[\lim_{x\to0}f(x)=f(0)\]
    cannot be satisfied.
  28. Therefore,
    \[\boxed{f(x)\text{ is discontinuous at }x=0.}\]
  29. Since the left-hand and right-hand limits are finite but unequal, the discontinuity is a jump discontinuity.
🎯 Exam Significance
Exam Significance
›

This problem is an important application of the modulus function and one-sided limits. The crucial step is to use

\[ |x|=-x\quad\text{for }x<0 \]
and
\[ |x|=x\quad\text{for }x>0. \]

At \(x=0\), the function has been assigned the value \(0\), but that does not make it continuous. Continuity depends on the behaviour of the function as \(x\) approaches the point from both sides.

For a complete board-examination answer, students should explicitly calculate LHL, RHL, and \(f(0)\), followed by the comparison.

Significance for Competitive Entrance Examinations

This is a standard example of a function involving the sign of \(x\). For \(x\neq0\),

\[ \frac{|x|}{x} = \begin{cases} -1,&x<0,\\ 1,&x>0. \end{cases} \]

Therefore, the function behaves like a step function around the origin. The two one-sided limits immediately reveal a jump discontinuity:

\[ \text{LHL}=-1,\qquad\text{RHL}=1. \]

This pattern is particularly useful in JEE and other entrance examinations, where recognising standard modulus and sign-function behaviour can considerably shorten calculations.

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  10 points

  1. The only point requiring examination is

    \[ x=0. \]

  2. For \(x<0\),

    \[ |x|=-x, \]
    so
    \[ \frac{|x|}{x}=-1. \]

  3. For \(x>0\),

    \[ |x|=x, \]
    so
    \[ \frac{|x|}{x}=1. \]

  4. The left-hand limit is

    \[ \lim_{x\to0^-}f(x)=-1. \]

  5. The right-hand limit is

    \[ \lim_{x\to0^+}f(x)=1. \]

  6. The actual value is

    \[ f(0)=0. \]

  7. Since

    \[ \lim_{x\to0^-}f(x)\neq\lim_{x\to0^+}f(x), \]
    the two-sided limit does not exist.

  8. Therefore, the function is discontinuous at \(x=0\).

  9. The discontinuity is a jump discontinuity.

  10. Assigning a value to a function at a point does not necessarily make the function continuous there.
← Q7
8 / 34  ·  24%
Q9 →
Q9
NUMERIC3 marks
Find the points of discontinuity where \(f\) is defined by \[ f(x)= \begin{cases} \dfrac{x}{|x|}, & x<0,\\[6pt] -1, & x\geq 0 \end{cases} \]
📘 Concept & Theory
Concept/Theory
›

A function \(f(x)\) is continuous at \(x=a\) if and only if all three of the following conditions are satisfied:

\[\lim_{x\to a^-}f(x)=\lim_{x\to a^+}f(x)=f(a)\]

For a piecewise-defined function, discontinuity can occur at the point where the definition changes. Here, the two branches meet at \(x=0\), so \(x=0\) is the only point that requires special examination.

Also, remember that

\[|x|=\begin{cases}-x,&x<0,\\x,&x\geq0.\end{cases}\]
🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Identify the point where the definition of \(f(x)\) changes.

  2. Evaluate the left-hand limit at that point.

  3. Evaluate the right-hand limit at that point.

  4. Find the actual value of the function at the point.

  5. Compare the two one-sided limits with the function value.

  6. State the points of discontinuity.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  16 steps
  1. Identify the critical point
  2. The definition of \(f(x)\) changes at \(x=0\):
    \[ f(x)= \begin{cases} \dfrac{x}{|x|},&x<0,\\[6pt] -1,&x\geq0 \end{cases} \]
  3. Therefore, \(x=0\) is the only point at which continuity needs to be checked.
  4. Evaluate the left-hand limit at \(x=0\)
  5. For \(x<0\), we have
    \[|x|=-x\]
  6. Hence, for \(x<0\),
    \[f(x)=\frac{x}{|x|}=\frac{x}{-x}=-1\]
  7. Therefore, as \(x\) approaches \(0\) from the left, the function remains equal to \(-1\):
    \[\boxed{\lim_{x\to0^-}f(x)=-1}\]
  8. Evaluate the right-hand limit at \(x=0\)
  9. For \(x>0\), the second branch applies because \(x\geq0\):
    \[f(x)=-1\]
  10. Hence,
    \[\boxed{\lim_{x\to0^+}f(x)=-1}\]
  11. Find \(f(0)\)
  12. Since the second branch is applicable for \(x\geq0\),
    \[f(0)=-1\]
  13. Test continuity at \(x=0\)
  14. We have obtained
    \[\lim_{x\to0^-}f(x)=-1\]
    \[\lim_{x\to0^+}f(x)=-1\]
    and
    \[f(0)=-1\]
  15. Thus,
    \[\lim_{x\to0^-}f(x)=\lim_{x\to0^+}f(x)=f(0)=-1\]
  16. Therefore, \(f\) is continuous at \(x=0\).
  17. Check the remaining points
  18. For \(x<0\)
    \[f(x)=\frac{x}{|x|}=-1\]
  19. which is a constant function and hence continuous on \((-\infty,0)\).
  20. For \(x\geq0\)
    \[f(x)=-1\]
  21. which is also a constant function and hence continuous on \([0,\infty)\).
  22. Since the function is continuous on both intervals and is also continuous at their joining point \(x=0\), there are no points of discontinuity.
📊 Graph / Figure
Graph / Figure
›
x y 0 −1 f(x) = −1 f(x) = −1 x < 0 x ≥ 0
Visual Interpretation
🎯 Exam Significance
Exam Significance
›
  • Always identify the point where the definition of a piecewise function changes.
  • At the joining point, calculate the LHL, RHL, and the actual value \(f(a)\) separately.
  • Do not assume that the presence of \(|x|\) automatically causes discontinuity.
  • For \(x<0\), use \(|x|=-x\). This immediately simplifies \(\dfrac{x}{|x|}\) to \(-1\).
  • A complete continuity test should explicitly establish \(\mathrm{LHL}=\mathrm{RHL}=f(a)\).
Significance for Competitive Entrance Examinations

This is a useful recognition-based problem. The expression

\[ \frac{x}{|x|} \]
represents the sign of \(x\) wherever \(x\neq0\):
\[\frac{x}{|x|}=\begin{cases}-1,&x<0,\\1,&x>0.\end{cases}\]
However, the present function deliberately assigns the value \(-1\) for \(x\geq0\). Therefore, both branches reduce to the same constant value \(-1\), making the function continuous everywhere.

The key competitive-exam observation is:

\[ \boxed{ \frac{x}{|x|}=-1\quad\text{for }x<0 } \]

Since the second branch is also \(-1\), there is no jump at the joining point \(x=0\).

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  6 points

  1. The only possible critical point is \(x=0\).

  2. For \(x<0\), \(|x|=-x\), so \(\dfrac{x}{|x|}=-1\).

  3. For \(x\geq0\), \(f(x)=-1\).

  4. Thus the entire function is effectively the constant function \(f(x)=-1\).

  5. At \(x=0\), \(\mathrm{LHL}=\mathrm{RHL}=f(0)=-1\).

  6. Hence, there is no discontinuity anywhere on \(\mathbb{R}\).
← Q8
9 / 34  ·  26%
Q10 →
Q10
NUMERIC3 marks
Find the points of discontinuity where \(f\) is defined by \[ f(x)= \begin{cases} x+1, & x\geq1,\\[6pt] x^2+1, & x<1 \end{cases} \]
📘 Concept & Theory
Concept/Theory
›

A function \(f(x)\) is continuous at \(x=a\) if and only if

\[ \lim_{x\to a^-}f(x) = \lim_{x\to a^+}f(x) = f(a). \]

For a piecewise-defined function, the points where the defining formula changes are the primary points that must be checked for continuity. In this question, the definition changes at \(x=1\). Therefore, \(x=1\) is the only possible point of discontinuity.

An important point is that the branch \(x\geq1\) includes \(x=1\). Therefore,

\[ f(1)=1+1=2. \]
However, for the left-hand limit, values approaching \(1\) from the left satisfy \(x<1\), so the branch \(x^2+1\) must be used.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Identify the point where the definition of \(f(x)\) changes.

  2. Calculate the left-hand limit at \(x=1\).

  3. Calculate the right-hand limit at \(x=1\).

  4. Find the actual value \(f(1)\).

  5. Compare the two one-sided limits with \(f(1)\).

  6. State whether \(x=1\) is a point of discontinuity.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  16 steps
  1. Identify the critical point
  2. The function is defined using two different expressions, with the definition changing at \(x=1\):
    \[ f(x)= \begin{cases} x+1, & x\geq1,\\[6pt] x^2+1, & x<1 \end{cases} \]
  3. Both \(x+1\) and \(x^2+1\) are polynomials and are continuous everywhere in their respective domains. Hence, the only point that needs to be checked is
    \[\boxed{x=1}\]
  4. Find the left-hand limit at \(x=1\)
  5. For \(x\to1^-\), we approach \(1\) through values less than \(1\). Therefore, we must use the branch
    \[f(x)=x^2+1\]
  6. Thus,
    \[ \begin{aligned} \lim_{x\to1^-}f(x) &=\lim_{x\to1^-}(x^2+1)\\ &=1^2+1\\ &=2 \end{aligned} \]
  7. Therefore,
    \[\boxed{\lim_{x\to1^-}f(x)=2}\]
  8. Find the right-hand limit at \(x=1\)
  9. For \(x\to1^+\), we approach \(1\) through values greater than \(1\). Therefore, we use the branch
    \[f(x)=x+1\]
  10. Hence,
    \[ \begin{aligned} \lim_{x\to1^+}f(x) &=\lim_{x\to1^+}(x+1)\\ &=1+1\\ &=2 \end{aligned} \]
  11. Therefore,
    \[\boxed{\lim_{x\to1^+}f(x)=2}\]
  12. Find \(f(1)\)
  13. Since the first branch applies for \(x\geq1\), it includes \(x=1\). Therefore,
    \[ \begin{aligned} f(1) &=1+1\\ &=2 \end{aligned} \]
  14. Test continuity at \(x=1\)
  15. We have
    \[\lim_{x\to1^-}f(x)=2,\]
    \[\lim_{x\to1^+}f(x)=2,\]
    and
    \[f(1)=2\]
  16. Therefore,
    \[\boxed{\lim_{x\to1^-}f(x)=\lim_{x\to1^+}f(x)=f(1)=2}\]
  17. Hence, \(f\) is continuous at \(x=1\)
  18. Check all other points
  19. On \(x<1\), the function is \(x^2+1\), a polynomial, so it is continuous throughout \((-\infty,1)\).
  20. On \(x\geq1\), the function is \(x+1\), also a polynomial, so it is continuous throughout \([1,\infty)\).
  21. Since the function is also continuous at the joining point \(x=1\), there is no point of discontinuity.
  22. Continuity Check at \(x=1\)
  23. Quantity Calculation Value
    Left-hand limit \(\displaystyle\lim_{x\to1^-}(x^2+1)\) \(2\)
    Right-hand limit \(\displaystyle\lim_{x\to1^+}(x+1)\) \(2\)
    Function value \(f(1)=1+1\) \(2\)
  24. Since all three values are equal, the function is continuous at \(x=1\).
🎯 Exam Significance
Exam Significance
›
  • The most important step in a piecewise continuity problem is selecting the correct branch for each one-sided limit.
  • For \(x\to1^-\), use the expression valid for \(x<1\), namely \(x^2+1\).
  • For \(x\to1^+\), use the expression valid for \(x>1\), namely \(x+1\).
  • For \(f(1)\), use the branch whose condition actually contains \(x=1\), namely \(x\geq1\).
  • Writing LHL, RHL, and \(f(1)\) separately makes the continuity argument complete and earns full credit.
Significance for Competitive Entrance Examinations

This problem tests a fundamental pattern in piecewise continuity. The expressions on either side of the joining point need not be identical. They only need to produce the same limiting value at the joining point for continuity.

Here,

\[ \lim_{x\to1^-}(x^2+1)=2 \]

and

\[ \lim_{x\to1^+}(x+1)=2. \]

Therefore, even though the two branches are different functions, they join continuously at \(x=1\).

A useful time-saving observation is that both branches are polynomials. Consequently, no point other than the joining point \(x=1\) can cause a discontinuity.

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  8 points

  1. The only possible point of discontinuity is \(x=1\).

  2. For the left-hand limit, use \(x^2+1\).

  3. For the right-hand limit, use \(x+1\).

  4. \(\displaystyle\lim_{x\to1^-}f(x)=2\).

  5. \(\displaystyle\lim_{x\to1^+}f(x)=2\).

  6. \(f(1)=2\), because the condition \(x\geq1\) includes \(x=1\).

  7. Since \(\mathrm{LHL}=\mathrm{RHL}=f(1)\), the function is continuous at \(x=1\).

  8. Both branches are polynomials, so the function is continuous everywhere.
← Q9
10 / 34  ·  29%
Q11 →
Q11
NUMERIC3 marks
Find the points of discontinuity where \(f\) is defined by \[ f(x)= \begin{cases} x^3-3, & x\leq2,\\[6pt] x^2+1, & x>2. \end{cases} \]
📘 Concept & Theory
Concept/Theory
›

A function \(f(x)\) is continuous at \(x=a\) if and only if

\[ \lim_{x\to a^-}f(x) = \lim_{x\to a^+}f(x) = f(a). \]

In a piecewise-defined function, the points where the defining formula changes are the primary points that must be checked for continuity. Here, the definition changes at \(x=2\). Therefore, \(x=2\) is the only possible point of discontinuity.

Notice that the condition \(x\leq2\) includes \(x=2\). Therefore, the actual value \(f(2)\) must be calculated from the first branch:

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

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Identify the point where the definition of \(f(x)\) changes.

  2. Calculate the left-hand limit at \(x=2\).

  3. Calculate the right-hand limit at \(x=2\).

  4. Find the actual value \(f(2)\).

  5. Compare the two one-sided limits with \(f(2)\).

  6. State whether \(x=2\) is a point of discontinuity.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  15 steps
  1. Identify the critical point
  2. The function is defined as
    \[ f(x)= \begin{cases} x^3-3, & x\leq2,\\[6pt] x^2+1, & x>2. \end{cases} \]
  3. The definition changes at \(x=2\). Hence, we only need to test continuity at
    \[\boxed{x=2}\]
  4. Both \(x^3-3\) and \(x^2+1\) are polynomials, so each branch is continuous throughout its respective interval. Thus, no other point can produce a discontinuity.
  5. Find the left-hand limit at \(x=2\)
  6. For \(x\to2^-\), we approach \(2\) through values less than \(2\). Therefore, we use the branch
    \[f(x)=x^3-3\]
    \[ \begin{aligned} \lim_{x\to2^-}f(x) &=\lim_{x\to2^-}(x^3-3)\\ &=2^3-3\\ &=8-3\\ &=5 \end{aligned} \]
  7. Therefore,
    \[\boxed{\lim_{x\to2^-}f(x)=5}\]
  8. Find the right-hand limit at \(x=2\)
  9. For \(x\to2^+\), we approach \(2\) through values greater than \(2\). Therefore, we use the second branch
    \[ f(x)=x^2+1 \]
    \[ \begin{aligned} \lim_{x\to2^+}f(x) &=\lim_{x\to2^+}(x^2+1)\\ &=2^2+1\\ &=4+1\\ &=5 \end{aligned} \]
  10. Therefore,
    \[\boxed{\lim_{x\to2^+}f(x)=5}\]
  11. Find \(f(2)\)
  12. Since the first branch is defined for \(x\leq2\), it includes \(x=2\). Hence,
    \[ \begin{aligned} f(2) &=2^3-3\\ &=8-3\\ &=5 \end{aligned} \]
    \[ \boxed{f(2)=5} \]
  13. Test continuity at \(x=2\)
  14. We have
    \[\lim_{x\to2^-}f(x)=5,\]
    \[\lim_{x\to2^+}f(x)=5,\]
    and
    \[f(2)=5.\]
  15. Therefore,
    \[\boxed{\lim_{x\to2^-}f(x)=\lim_{x\to2^+}f(x)=f(2)=5}\]
  16. Hence, \(f\) is continuous at \(x=2\).
  17. Check the remaining points
  18. For \(x<2\), the function is \(x^3-3\), which is a polynomial and therefore continuous.
  19. For \(x>2\), the function is \(x^2+1\), which is also a polynomial and therefore continuous.
  20. Since the function is continuous at the only joining point \(x=2\), it is continuous for every real number.
  21. Continuity Check at \(x=2\)
  22. Quantity Calculation Value
    Left-hand limit \(\displaystyle\lim_{x\to2^-}(x^3-3)\) \(5\)
    Right-hand limit \(\displaystyle\lim_{x\to2^+}(x^2+1)\) \(5\)
    Function value \(f(2)=2^3-3\) \(5\)
🎯 Exam Significance
Exam Significance
›
  • The point where the formula changes, \(x=2\), must always be checked carefully.
  • For \(x\to2^-\), use the branch valid for \(x<2\), namely \(x^3-3\).
  • For \(x\to2^+\), use the branch valid for \(x>2\), namely \(x^2+1\).
  • Since \(x\leq2\) includes \(2\), the actual value \(f(2)\) must be obtained from \(x^3-3\).
  • A complete answer should explicitly show
    \[ \mathrm{LHL}=\mathrm{RHL}=f(2). \]
Significance for Competitive Entrance Examinations

This problem illustrates an important principle for piecewise functions: the two formulas do not have to be the same at the joining point. They only need to produce the same limiting value, and that common value must equal the actual function value.

Here, the two different expressions satisfy

\[ 2^3-3=5 \]

and

\[ 2^2+1=5. \]

Therefore, the branches join continuously at \(x=2\).

A useful competitive-exam shortcut is to observe that both branches are polynomials. Hence, the only possible discontinuity is at the boundary \(x=2\), reducing the problem to a three-way comparison:

\[\boxed{\lim_{x\to2^-}f(x),\quad \lim_{x\to2^+}f(x),\quad f(2)}\]
🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  7 points

  1. The only possible point of discontinuity is \(x=2\).

  2. \(\displaystyle\lim_{x\to2^-}f(x)=5\).

  3. \(\displaystyle\lim_{x\to2^+}f(x)=5\).

  4. \(f(2)=5\), because \(x\leq2\) includes \(x=2\).

  5. Therefore,

    \[ \mathrm{LHL}=\mathrm{RHL}=f(2)=5. \]

  6. Both branches are polynomials and are continuous on their respective intervals.

  7. Hence, \(f\) is continuous on the entire real line.
← Q10
11 / 34  ·  32%
Q12 →
Q12
NUMERIC3 marks
Find the points of discontinuity where \(f\) is defined by \[ f(x)= \begin{cases} x^{10}-1, & x\leq1,\\[6pt] x^2, & x>0 \end{cases} \]
📘 Concept & Theory
Concept/Theory
›

A function \(f(x)\) is continuous at \(x=a\) if and only if

\[\lim_{x\to a^-}f(x)=\lim_{x\to a^+}f(x)=f(a)\]

For a piecewise-defined function, the point where the defining formula changes is the first point that must be checked.

In the intended question, the definition changes at \(x=1\):

\[f(x)=\begin{cases}x^{10}-1, & x\leq1,\\x^2, & x>1\end{cases}\]

Both \(x^{10}-1\) and \(x^2\) are polynomials and are therefore continuous everywhere. Hence, the only possible point of discontinuity is \(x=1\).

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Identify the joining point \(x=1\).

  2. Evaluate the left-hand limit using the branch \(x^{10}-1\).

  3. Evaluate the right-hand limit using the branch \(x^2\).

  4. Find the actual value \(f(1)\).

  5. Compare LHL, RHL, and \(f(1)\).

  6. State the point of discontinuity.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  17 steps
  1. Identify the critical point
  2. For the intended definition,
    \[ f(x)= \begin{cases} x^{10}-1, & x\leq1,\\[6pt] x^2, & x>1 \end{cases} \]
  3. The defining expression changes at \(x=1\). Therefore, we examine continuity at
    \[\boxed{x=1}\]
  4. Find the left-hand limit at \(x=1\)
  5. As \(x\to1^-\), we approach \(1\) through values less than \(1\). Therefore, we use the branch
    \[f(x)=x^{10}-1\]
  6. Hence,
    \[ \begin{aligned} \lim_{x\to1^-}f(x) &=\lim_{x\to1^-}(x^{10}-1)\\ &=1^{10}-1\\ &=1-1\\ &=0 \end{aligned} \]
  7. Therefore,
    \[\boxed{\lim_{x\to1^-}f(x)=0}\]
  8. Find the function value \(f(1)\)
  9. Since the first branch is defined for \(x\leq1\), it includes \(x=1\). Therefore,
    \[ \begin{aligned} f(1) &=1^{10}-1\\ &=1-1\\ &=0 \end{aligned} \]
    \[ \boxed{f(1)=0} \]
  10. Find the right-hand limit at \(x=1\)
  11. As \(x\to1^+\), we approach \(1\) through values greater than \(1\). Therefore, we use the second branch
    \[f(x)=x^2\]
  12. Thus,
    \[ \begin{aligned} \lim_{x\to1^+}f(x) &=\lim_{x\to1^+}x^2\\ &=1^2\\ &=1 \end{aligned} \]
  13. Therefore,
    \[\boxed{\lim_{x\to1^+}f(x)=1}\]
  14. Test continuity at \(x=1\)
  15. We have
    \[\lim_{x\to1^-}f(x)=0,\]
    \[\lim_{x\to1^+}f(x)=1,\]
    and
    \[f(1)=0.\]
  16. Since
    \[\lim_{x\to1^-}f(x)\neq\lim_{x\to1^+}f(x),\]
  17. the two-sided limit at \(x=1\) does not exist. Consequently, the function cannot be continuous at \(x=1\).

    More specifically, because the left-hand and right-hand limits are finite but unequal, \(x=1\) is a jump discontinuity.
  18. Check the remaining points
  19. For \(x<1\), the function is \(x^{10}-1\), which is a polynomial and hence continuous.
  20. For \(x>1\), the function is \(x^2\), which is also a polynomial and hence continuous.
  21. Therefore, the only point of discontinuity is \(x=1\).
  22. Continuity Check at \(x=1\)
  23. Quantity Calculation Value
    Left-hand limit \(\displaystyle\lim_{x\to1^-}(x^{10}-1)\) \(0\)
    Right-hand limit \(\displaystyle\lim_{x\to1^+}x^2\) \(1\)
    Function value \(f(1)=1^{10}-1\) \(0\)
  24. Since
    \[0\neq1\]
    the function is discontinuous at \(x=1\).
🎯 Exam Significance
Exam Significance
›
  • Always identify the branch applicable to the direction from which \(x\) approaches the critical point.
  • For \(x\to1^-\), use \(x^{10}-1\).
  • For \(x\to1^+\), use \(x^2\).
  • Since \(x\leq1\) includes \(1\), calculate \(f(1)\) from the first branch.
  • The correct continuity test is
    \[ \mathrm{LHL}=\mathrm{RHL}=f(1). \]
    Here, the LHL and RHL are unequal, so continuity fails.
Significance for Competitive Entrance Examinations

This problem is a standard test of branch selection in piecewise functions. The quickest approach is to recognize that both branches are polynomials and hence continuous individually. Therefore, only the joining point \(x=1\) needs to be checked.

The decisive calculation is

\[ (1^{10}-1)=0 \]

whereas

\[ 1^2=1. \]

Since the two one-sided limits differ, the function has a jump discontinuity at \(x=1\).

A useful competitive-exam pattern is:

\[ \boxed{ \mathrm{LHL}\neq\mathrm{RHL} \quad\Longrightarrow\quad \text{discontinuous} } \]
🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  8 points

  1. The intended joining point is \(x=1\).

  2. \(\displaystyle\lim_{x\to1^-}f(x)=0\).

  3. \(\displaystyle f(1)=0\).

  4. \(\displaystyle\lim_{x\to1^+}f(x)=1\).

  5. Since \(0\neq1\), the function is discontinuous at \(x=1\).

  6. The discontinuity is a jump discontinuity.

  7. Both polynomial branches are continuous away from \(x=1\).

  8. Therefore, \(x=1\) is the only point of discontinuity of the intended function.
← Q11
12 / 34  ·  35%
Q13 →
Q13
NUMERIC3 marks
Is the function defined by \[ f(x)= \begin{cases} x+5, & x\leq1,\\[6pt] x-5, & x>1 \end{cases} \] a continuous function?
📘 Concept & Theory
Concept/Theory
›

A function \(f(x)\) is continuous at \(x=a\) if and only if

\[\lim_{x\to a^-}f(x)=\lim_{x\to a^+}f(x)=f(a)\]

For a piecewise-defined function, the point where the defining formula changes is the primary point that must be checked. Here, the definition changes at \(x=1\). Therefore, we only need to examine continuity at \(x=1\).

Notice carefully that \(x\leq1\) includes \(x=1\), whereas \(x>1\) does not. Hence, \(f(1)\) must be calculated using the first branch.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Identify the joining point \(x=1\).

  2. Calculate the left-hand limit at \(x=1\).

  3. Calculate the right-hand limit at \(x=1\).

  4. Calculate the actual function value \(f(1)\).

  5. Compare LHL, RHL, and \(f(1)\).

  6. Conclude whether the function is continuous.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  14 steps
  1. Identify the critical point
  2. Given —
    \[ f(x)= \begin{cases} x+5, & x\leq1,\\[6pt] x-5, & x>1 \end{cases} \]
  3. The formula changes at \(x=1\). Thus, \(x=1\) is the only point that needs to be checked.
    \[\boxed{x=1}\]
  4. Both \(x+5\) and \(x-5\) are polynomials and hence are continuous on their respective intervals.
  5. Find the left-hand limit at \(x=1\)
  6. As \(x\to1^-\), we approach \(1\) through values less than \(1\). Therefore, we use the first branch:
    \[f(x)=x+5\]
  7. Hence,
    \[ \begin{aligned} \lim_{x\to1^-}f(x) &=\lim_{x\to1^-}(x+5)\\ &=1+5\\ &=6 \end{aligned} \]
  8. Therefore,
    \[\boxed{\lim_{x\to1^-}f(x)=6}\]
  9. Find \(f(1)\)
  10. Since the first branch applies for \(x\leq1\), it includes \(x=1\). Therefore,
    \[ \begin{aligned} f(1) &=1+5\\ &=6 \end{aligned} \]
    \[ \boxed{f(1)=6} \]
  11. Find the right-hand limit at \(x=1\)
  12. As \(x\to1^+\), we approach \(1\) through values greater than \(1\). Therefore, we use the second branch:
    \[f(x)=x-5\]
  13. Hence,
    \[ \begin{aligned} \lim_{x\to1^+}f(x) &=\lim_{x\to1^+}(x-5)\\ &=1-5\\ &=-4 \end{aligned} \]
  14. Therefore,
    \[\boxed{\lim_{x\to1^+}f(x)=-4}\]
  15. Test continuity at \(x=1\)
  16. We have
    \[\lim_{x\to1^-}f(x)=6,\]
    \[\lim_{x\to1^+}f(x)=-4,\]
    and
    \[f(1)=6\]
  17. Since
    \[ \lim_{x\to1^-}f(x) \neq \lim_{x\to1^+}f(x), \]
  18. the two-sided limit at \(x=1\) does not exist. Hence, \(f\) is not continuous at \(x=1\).
  19. Since both one-sided limits are finite but unequal, the function has a jump discontinuity at \(x=1\).
    \[\boxed{\text{\(f\) is discontinuous at }x=1.}\]
🎯 Exam Significance
Exam Significance
›
  • The point where the definition changes, \(x=1\), is the critical point.
  • For the left-hand limit, use \(x+5\), because \(x<1\).
  • For the right-hand limit, use \(x-5\), because \(x>1\).
  • For \(f(1)\), use \(x+5\), because the condition \(x\leq1\) includes \(x=1\).
  • Clearly writing LHL, RHL, and \(f(1)\) separately makes the continuity test complete.
  • Since LHL and RHL are unequal, the function is discontinuous at \(x=1\).
Significance for Competitive Entrance Examinations

This is a standard piecewise-function question that tests rapid identification of a jump discontinuity. Since both branches are linear functions, they are individually continuous. Therefore, only the joining point \(x=1\) needs to be tested.

The fastest decisive calculation is

\[ (1+5)=6 \]

and

\[ (1-5)=-4. \]

Since these values are different, the one-sided limits cannot be equal:

\[ \boxed{\mathrm{LHL}\neq\mathrm{RHL}\Rightarrow\text{discontinuous}}. \]

This recognition can save considerable time in objective questions, particularly when the question asks only for the nature or location of discontinuity.

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  7 points

  1. The only possible point of discontinuity is \(x=1\).

  2. \(\displaystyle\lim_{x\to1^-}f(x)=6\).

  3. \(\displaystyle\lim_{x\to1^+}f(x)=-4\).

  4. \(f(1)=6\).

  5. Since \(6\neq-4\), the function is discontinuous at \(x=1\).

  6. The discontinuity is a jump discontinuity.

  7. Both branches are linear and therefore continuous away from \(x=1\).
← Q12
13 / 34  ·  38%
Q14 →
Q14
NUMERIC3 marks
Discuss the continuity of the function \(f\), where \(f\) is defined by \[ f(x)= \begin{cases} 3, & 0\leq x\leq1,\\[6pt] 4, & 1 < x < 3,\\[6pt] 5, & 3\leq x\leq10 \end{cases} \]
📘 Concept & Theory
Concept/Theory
›

A constant function is continuous at every point in its interval of definition. Therefore, each individual branch

\[ f(x)=3,\qquad f(x)=4,\qquad f(x)=5 \]
is continuous within its respective interval.

Hence, discontinuity can occur only at the points where the definition changes. Here, the defining formula changes at

\[ x=1\quad\text{and}\quad x=3. \]

At each such point, we use the continuity criterion:

\[ \lim_{x\to a^-}f(x) = \lim_{x\to a^+}f(x) = f(a). \]
🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Observe that all three branches are constant functions.

  2. Conclude that the function is continuous within each open portion of its intervals.

  3. Check continuity at the joining point \(x=1\).

  4. Check continuity at the joining point \(x=3\).

  5. State the intervals on which \(f\) is continuous and the points of discontinuity.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  15 steps
  1. Examine the three branches
  2. The function is
    \[ f(x)= \begin{cases} 3, & 0\leq x\leq1,\\ 4, & 1 < x < 3,\\ 5, & 3\leq x\leq10 \end{cases} \]
  3. Each branch is a constant function. Therefore, \(f\) is continuous wherever the same constant branch applies.

    The only points requiring a continuity check are the points where the value changes:
    \[\boxed{x=1\quad\text{and}\quad x=3}\]
  4. Check continuity at \(x=1\)
  5. Left-hand limit
  6. For \(x\to1^-\), values of \(x\) are less than \(1\), so we use \(f(x)=3\).
    \[ \begin{aligned} \lim_{x\to1^-}f(x) &=\lim_{x\to1^-}3\\ &=3 \end{aligned} \]
    \[ \boxed{\lim_{x\to1^-}f(x)=3} \]
  7. Right-hand limit
  8. For \(x\to1^+\), values of \(x\) are greater than \(1\) but less than \(3\), so we use \(f(x)=4\).
    \[ \begin{aligned} \lim_{x\to1^+}f(x) &=\lim_{x\to1^+}4\\ &=4. \end{aligned} \]
    \[ \boxed{\lim_{x\to1^+}f(x)=4} \]
  9. Function value
  10. Since \(0\leq x\leq1\) includes \(x=1\), the first branch applies:
    \[\boxed{f(1)=3}\]
  11. Continuity test at \(x=1\)
  12. We have
    \[\lim_{x\to1^-}f(x)=3\neq4=\lim_{x\to1^+}f(x)\]
  13. Therefore, the two-sided limit does not exist and \(f\) is discontinuous at \(x=1\).
    \[\boxed{\text{\(f\) is discontinuous at }x=1.}\]
    Since the one-sided limits are finite but unequal, this is a jump discontinuity.
  14. Check continuity at \(x=3\)
  15. Left-hand limit
  16. For \(x\to3^-\), values of \(x\) are less than \(3\) but greater than \(1\), so we use \(f(x)=4\).
    \[ \begin{aligned} \lim_{x\to3^-}f(x) &=\lim_{x\to3^-}4\\ &=4. \end{aligned} \]
    \[ \boxed{\lim_{x\to3^-}f(x)=4} \]
  17. Right-hand limit
  18. For \(x\to3^+\), values of \(x\) are greater than \(3\), so we use \(f(x)=5\).
    \[ \begin{aligned} \lim_{x\to3^+}f(x) &=\lim_{x\to3^+}5\\ &=5. \end{aligned} \]
    \[ \boxed{\lim_{x\to3^+}f(x)=5} \]
  19. Function value
  20. Since the third branch is defined for \(3\leq x\leq10\), it includes \(x=3\). Therefore,
    \[\boxed{f(3)=5}\]
  21. Continuity test at \(x=3\)
  22. We have
    \[ \lim_{x\to3^-}f(x)=4 \neq 5=\lim_{x\to3^+}f(x). \]
  23. Therefore, \(f\) is discontinuous at \(x=3\).
  24. This is also a jump discontinuity.
  25. Continuity on the remaining intervals
  26. The function is constant on each of the following intervals:
    \[f(x)=3\quad\text{for }0\leq x\leq1\]
    \[f(x)=4\quad\text{for }1
    \[f(x)=5\quad\text{for }3\leq x\leq10\]
  27. Therefore, \(f\) is continuous at every point of these intervals except at the joining points \(x=1\) and \(x=3\), where the function changes value.
  28. Continuity Summary
  29. Point LHL RHL Function Value Conclusion
    \(x=1\) \(3\) \(4\) \(3\) Discontinuous
    \(x=3\) \(4\) \(5\) \(5\) Discontinuous
🎯 Exam Significance
Exam Significance
›
  • A constant function is continuous throughout its interval of definition.
  • Therefore, only the points where the constant value changes need to be checked.
  • At \(x=1\), the left-hand value is \(3\), while the right-hand value is \(4\).
  • At \(x=3\), the left-hand value is \(4\), while the right-hand value is \(5\).
  • Since the one-sided limits are unequal at both points, both are discontinuities.
  • Writing LHL, RHL, and \(f(a)\) separately gives a complete and logically sound continuity test.
Significance for Competitive Entrance Examinations

This is a useful recognition-based problem. Because all three branches are constants, there is no need to perform lengthy limit calculations. The only possible discontinuities are the boundaries where the constant value changes.

The key observations are

\[ 3\neq4 \]

at \(x=1\), and

\[ 4\neq5 \]

at \(x=3\). Thus, both points are jump discontinuities.

A useful shortcut is:

\[ \boxed{ \text{If the left and right constant values differ at a boundary, there is a jump discontinuity.} } \]
🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  7 points

  1. The function consists of three constant pieces.

  2. Constant functions are continuous within their intervals.

  3. The only critical points are \(x=1\) and \(x=3\).

  4. At \(x=1\):

    \[ \mathrm{LHL}=3,\qquad \mathrm{RHL}=4,\qquad f(1)=3. \]

  5. At \(x=3\):

    \[ \mathrm{LHL}=4,\qquad \mathrm{RHL}=5,\qquad f(3)=5. \]

  6. The function has jump discontinuities at both \(x=1\) and \(x=3\).

  7. There are no other points of discontinuity.
← Q13
14 / 34  ·  41%
Q15 →
Q15
NUMERIC3 marks
Discuss the continuity of the function \(f\), where \(f\) is defined by \[ f(x)= \begin{cases} 2x, & x<0,\\[6pt] 0, & 0\leq x\leq1,\\[6pt] 4x, & x>1 \end{cases} \]
📘 Concept & Theory
Concept/Theory
›

A function \(f(x)\) is continuous at \(x=a\) if and only if

\[\lim_{x\to a^-}f(x)=\lim_{x\to a^+}f(x)=f(a).\]

In a piecewise-defined function, the points where the defining formula changes are the points that require special attention. Here, the formula changes at

\[ x=0\quad\text{and}\quad x=1. \]
Therefore, these two points must be checked separately.

Away from these points, each branch is either a linear function or a constant function, all of which are continuous on their respective intervals.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Identify the joining points \(x=0\) and \(x=1\).

  2. Check continuity at \(x=0\) using LHL, RHL, and \(f(0)\).

  3. Check continuity at \(x=1\) using LHL, RHL, and \(f(1)\).

  4. Use the continuity of linear and constant functions to conclude continuity elsewhere.

  5. State the intervals of continuity and the points of discontinuity.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  23 steps
  1. Identify the critical points
  2. Given —
    \[ f(x)= \begin{cases} 2x, & x<0,\\ 0, & 0\leq x\leq1,\\ 4x, & x>1 \end{cases} \]
  3. The defining expression changes at
    \[\boxed{x=0\quad\text{and}\quad x=1}\]
  4. Thus, only \(x=0\) and \(x=1\) need to be checked for possible discontinuities.
  5. Check continuity at \(x=0\)
  6. Left-hand limit at \(x=0\)
  7. As \(x\to0^-\), we approach \(0\) through values less than \(0\). Therefore, we use the first branch:
    \[f(x)=2x\]
  8. Hence,
    \[ \begin{aligned} \lim_{x\to0^-}f(x) &=\lim_{x\to0^-}2x\\ &=2(0)\\ &=0 \end{aligned} \]
    \[\boxed{\lim_{x\to0^-}f(x)=0}\]
  9. Right-hand limit at \(x=0\)
  10. As \(x\to0^+\), we approach \(0\) through positive values. Since \(0 < x\leq1\), the second branch applies:
    \[f(x)=0\]
  11. Therefore,
    \[\boxed{\lim_{x\to0^+}f(x)=0}\]
  12. Function value at \(x=0\)
  13. Since \(0\leq x\leq1\) includes \(x=0\),
    \[\boxed{f(0)=0}\]
  14. Continuity test at \(x=0\)
  15. We have
    \[\lim_{x\to0^-}f(x)=\lim_{x\to0^+}f(x)=f(0)=0\]
  16. Hence, \(f\) is continuous at \(x=0\).
  17. Check continuity at \(x=1\)
  18. Left-hand limit at \(x=1\)
  19. As \(x\to1^-\), we approach \(1\) through values less than \(1\). Since \(0\leq x\leq1\), the second branch applies:
    \[f(x)=0\]
  20. Therefore,
    \[\boxed{\lim_{x\to1^-}f(x)=0}\]
  21. Right-hand limit at \(x=1\)
  22. As \(x\to1^+\), we approach \(1\) through values greater than \(1\). Therefore, the third branch applies:
    \[f(x)=4x\]
  23. Hence,
    \[ \begin{aligned} \lim_{x\to1^+}f(x) &=\lim_{x\to1^+}4x\\ &=4(1)\\ &=4 \end{aligned} \]
    \[ \boxed{\lim_{x\to1^+}f(x)=4} \]
  24. Function value at \(x=1\)
  25. Since the middle branch is defined for \(0\leq x\leq1\), it includes \(x=1\). Therefore,
    \[\boxed{f(1)=0}\]
  26. Continuity test at \(x=1\)
  27. We have
    \[\lim_{x\to1^-}f(x)=0\]
  28. but
    \[\lim_{x\to1^+}f(x)=4\]
  29. Thus,
    \[ \lim_{x\to1^-}f(x) \neq \lim_{x\to1^+}f(x). \]
  30. Hence, the two-sided limit at \(x=1\) does not exist, and \(f\) is discontinuous at \(x=1\).
    \[\boxed{\text{\(f\) is discontinuous at }x=1.}\]
  31. Since the one-sided limits are finite but unequal, the discontinuity at \(x=1\) is a jump discontinuity.
  32. Check continuity elsewhere
  33. For \(x<0\),
    \[ f(x)=2x, \]
    which is a linear function and therefore continuous.
  34. For \(0 < x < 1\),
    \[ f(x)=0, \]
    which is a constant function and therefore continuous.
  35. For \(x>1\),
    \[ f(x)=4x, \]
    which is a linear function and therefore continuous.
  36. Therefore, there are no other points of discontinuity.
🎯 Exam Significance
Exam Significance
›
  • In a piecewise function, first identify all points where the definition changes.
  • Here, the critical points are \(x=0\) and \(x=1\).
  • At \(x=0\), use \(2x\) for the left-hand limit and \(0\) for the right-hand limit.
  • At \(x=1\), use \(0\) for the left-hand limit and \(4x\) for the right-hand limit.
  • Remember that \(0\leq x\leq1\) includes both \(x=0\) and \(x=1\), so
    \[ f(0)=f(1)=0. \]
  • Clearly showing LHL, RHL, and the function value is the safest way to establish continuity or discontinuity in a board examination.
Significance for Competitive Entrance Examinations

This problem is particularly useful for recognizing jump discontinuities quickly. Since each branch is either linear or constant, all possible discontinuities occur only at the transition points \(0\) and \(1\).

At \(x=0\),

\[ 2(0)=0, \]

and the middle branch also has value \(0\). Hence, there is no jump.

At \(x=1\), however,

\[ 0\neq4(1)=4. \]

Therefore, \(x=1\) is immediately identified as a jump discontinuity.

The time-saving principle is:

\[ \boxed{ \text{For piecewise linear/constant functions, check only the transition points.} } \]
🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  6 points

  1. The possible discontinuity points are \(x=0\) and \(x=1\).

  2. At \(x=0\):

    \[ \mathrm{LHL}=\mathrm{RHL}=f(0)=0. \]
    Hence, \(f\) is continuous at \(0\).

  3. At \(x=1\):

    \[ \mathrm{LHL}=0,\qquad \mathrm{RHL}=4,\qquad f(1)=0. \]

  4. Since \(\mathrm{LHL}\neq\mathrm{RHL}\), \(f\) is discontinuous at \(x=1\).

  5. The discontinuity at \(x=1\) is a jump discontinuity.

  6. There are no other points of discontinuity.
← Q14
15 / 34  ·  44%
Q16 →
Q16
NUMERIC3 marks
Discuss the continuity of the function \(f\), where \(f\) is defined by \[ f(x)= \begin{cases} -2, & x\leq -1,\\[6pt] 2x, & -1 < x \leq1,\\[6pt] 2, & x>1. \end{cases} \]
📘 Concept & Theory
Concept/Theory
›

A function \(f(x)\) is continuous at \(x=a\) if and only if

\[ \lim_{x\to a^-}f(x) = \lim_{x\to a^+}f(x) = f(a). \]

For a piecewise-defined function, the points where the defining formula changes are the critical points that must be checked. Here, the definition changes at

\[ \boxed{x=-1\quad\text{and}\quad x=1}. \]

Each individual branch is a constant or linear function, so it is continuous within its respective interval. Therefore, only \(x=-1\) and \(x=1\) need to be examined.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Identify the transition points \(x=-1\) and \(x=1\).

  2. Check continuity at \(x=-1\) using LHL, RHL, and \(f(-1)\).

  3. Check continuity at \(x=1\) using LHL, RHL, and \(f(1)\).

  4. Use the continuity of constant and linear functions to establish continuity elsewhere.

  5. State the complete interval of continuity and the points of discontinuity.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  20 steps
  1. Identify the critical points
  2. We have
    \[ f(x)= \begin{cases} -2, & x\leq -1,\\ 2x, & -1 < x\leq1,\\ 2, & x>1. \end{cases} \]
  3. The formula changes at \(x=-1\) and \(x=1\). Hence, these are the only possible points of discontinuity.
  4. Left-hand limit
  5. As \(x\to-1^-\), we approach \(-1\) through values less than \(-1\). Therefore, we use the first branch:
    \[f(x)=-2\]
    \[\boxed{\lim_{x\to-1^-}f(x)=-2}\]
  6. Right-hand limit
  7. As \(x\to-1^+\), we approach \(-1\) through values greater than \(-1\). Therefore, the second branch applies:
    \[f(x)=2x.\]
  8. Hence,
    \[ \begin{aligned} \lim_{x\to-1^+}f(x) &=\lim_{x\to-1^+}2x\\ &=2(-1)\\ &=-2 \end{aligned} \]
    \[ \boxed{\lim_{x\to-1^+}f(x)=-2} \]
  9. Function value at \(x=-1\)
  10. Since the first branch is defined for \(x\leq-1\), it includes \(x=-1\). Therefore,
    \[\boxed{f(-1)=-2}\]
  11. >Continuity test at \(x=-1\)
  12. Thus,
    \[\boxed{\lim_{x\to-1^-}f(x)=\lim_{x\to-1^+}f(x)=f(-1)=-2}.\]
  13. Hence, \(f\) is continuous at \(x=-1\).
  14. Check continuity at \(x=1\)
  15. Left-hand limit
  16. As \(x\to1^-\), we approach \(1\) through values less than \(1\). Since \(-1 < x\leq1\), the second branch applies:
    \[ f(x)=2x. \]
  17. Hence,
    \[ \begin{aligned} \lim_{x\to1^-}f(x) &=\lim_{x\to1^-}2x\\ &=2(1)\\ &=2 \end{aligned} \]
    \[ \boxed{\lim_{x\to1^-}f(x)=2} \]
  18. Right-hand limit
  19. As \(x\to1^+\), we approach \(1\) through values greater than \(1\). Therefore, the third branch applies:
    \[ f(x)=2. \]
  20. Thus,
    \[\boxed{\lim_{x\to1^+}f(x)=2}\]
  21. Function value at \(x=1\)
  22. Since the second branch is defined for \(-1 < x\leq1\), it includes \(x=1\). Therefore,
  23. \[ \begin{aligned} f(1) &=2(1)\\ &=2 \end{aligned} \]
    \[ \boxed{f(1)=2} \]
  24. Therefore,
    \[\boxed{\lim_{x\to1^-}f(x)=\lim_{x\to1^+}f(x)=f(1)=2}.\]
  25. Hence, \(f\) is continuous at \(x=1\).
  26. Check continuity elsewhere
  27. For \(x < -1\),
    \[ f(x)=-2, \]
    which is a constant function and therefore continuous.
  28. For \(-1 < x < 1\),
    \[ f(x)=2x, \]
    which is a linear function and therefore continuous.
  29. For \(x>1\),
    \[ f(x)=2, \]
    which is a constant function and therefore continuous.
  30. Since the function is also continuous at both transition points, it is continuous everywhere.
🎯 Exam Significance
Exam Significance
›
  • First identify every point where the definition of a piecewise function changes.
  • Here, those points are \(x=-1\) and \(x=1\).
  • At \(x=-1\), the left branch gives \(-2\), while the right branch gives \(2(-1)=-2\).
  • At \(x=1\), the left branch gives \(2(1)=2\), while the right branch is the constant \(2\).
  • In both cases, LHL, RHL, and the function value are equal.
  • Showing these three quantities explicitly is the clearest way to establish continuity in a board examination.
Significance for Competitive Entrance Examinations

This is a useful recognition-based continuity problem. Since the branches are only linear or constant functions, they are individually continuous. Consequently, only the transition points need to be tested.

At \(x=-1\),

\[ -2=2(-1), \]

so the two branches join without a jump.

At \(x=1\),

\[ 2(1)=2, \]

so the middle and right branches also join without a jump.

Thus, the fastest observation is

\[ \boxed{ \text{Both joining values match, so there is no discontinuity.} } \]
🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  7 points

  1. The possible points of discontinuity are \(x=-1\) and \(x=1\).

  2. At \(x=-1\):

    \[ \mathrm{LHL}=\mathrm{RHL}=f(-1)=-2. \]

  3. At \(x=1\):

    \[ \mathrm{LHL}=\mathrm{RHL}=f(1)=2. \]

  4. Therefore, the function is continuous at both transition points.

  5. All three branches are individually continuous.

  6. Hence, \(f\) is continuous on the entire real line.

  7. There are no points of discontinuity.
← Q15
16 / 34  ·  47%
Q17 →
Q17
NUMERIC3 marks
Find the relationship between \(a\) and \(b\) so that the function \[ f(x)= \begin{cases} ax+1, & x\leq3,\\[6pt] bx+3, & x>3 \end{cases} \] is continuous at \(x=3\).
📘 Concept & Theory
Concept/Theory
›

A function \(f(x)\) is continuous at \(x=a\) if and only if

\[ \lim_{x\to a^-}f(x) = \lim_{x\to a^+}f(x) = f(a). \]

Since the given function changes its definition at \(x=3\), continuity at \(x=3\) requires the two branches to meet at the same value.

The first branch is applicable at \(x=3\) because its condition is \(x\leq3\). Therefore,

\[ f(3)=3a+1. \]
The right-hand limit must be calculated using the second branch \(bx+3\).

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Identify the joining point \(x=3\).

  2. Calculate the left-hand limit using \(ax+1\).

  3. Calculate the right-hand limit using \(bx+3\).

  4. Calculate \(f(3)\) from the branch containing \(x=3\).

  5. Apply the continuity condition and solve for the relationship between \(a\) and \(b\).

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  12 steps
  1. Identify the critical point
  2. The function changes its definition at \(\boxed{x=3}\). Therefore, continuity needs to be checked at \(x=3\).
  3. Find the left-hand limit at \(x=3\)
  4. As \(x\to3^-\), we approach \(3\) through values less than \(3\). Hence, the first branch
    \[ \begin{aligned} f(x) &= ax + 1 \\ &\implies \lim_{x \to 3^-} f(x) = \lim_{x \to 3^-} (ax + 1) = 3a + 1 \end{aligned} \]
  5. Thus,
    \[\boxed{\lim_{x\to3^-}f(x)=3a+1}\]
  6. Find the right-hand limit at \(x=3\)
  7. As \(x\to3^+\), we approach \(3\) through values greater than \(3\). Hence, the second branch
    \[f(x)=bx+3\]
    \[\begin{aligned} \lim_{x\to3^+}f(x) &=\lim_{x\to3^+}(bx+3)\\ &=3b+3. \end{aligned} \]
  8. Thus,
    \[\boxed{\lim_{x\to3^+}f(x)=3b+3}\]
  9. Find \(f(3)\)
  10. Since \(x=3\) satisfies the condition \(x\leq3\), we use the first branch:
    \[ f(3)=3a+1. \]
    Therefore,
    \[\boxed{f(3)=3a+1}\]
  11. Apply the continuity condition
  12. For \(f(x)\) to be continuous at \(x=3\)
  13. \[ \lim_{x\to3^-}f(x) = \lim_{x\to3^+}f(x) = f(3). \]
  14. Substituting the values obtained above,
    \[3a+1=3b+3\]
  15. Rearranging,
    \[ \begin{aligned} 3a+1&=3b+3\\ 3a-3b&=3-1\\ 3a-3b&=2\\ 3(a-b)&=2 \end{aligned} \]
  16. Dividing both sides by \(3\),
    \[\boxed{a-b=\frac{2}{3}}\]
  17. Equivalently,
    \[\boxed{a=b+\frac{2}{3}}\]
🎯 Exam Significance
Exam Significance
›
  • In a piecewise function, first locate the point where the definition changes. Here it is \(x=3\).
  • For \(x\to3^-\), use \(ax+1\).
  • For \(x\to3^+\), use \(bx+3\).
  • Since \(x\leq3\) includes \(3\), calculate \(f(3)\) using \(ax+1\).
  • For continuity, equate the left-hand limit and right-hand limit:
    \[ 3a+1=3b+3. \]
  • Simplifying gives the required relationship:
    \[ a-b=\frac23. \]
Significance for Competitive Entrance Examinations

This is a standard parameter-based continuity problem. The key observation is that both branches are linear and therefore continuous individually. Hence, only the joining point \(x=3\) needs to be considered.

For such questions, continuity can be tested directly by equating the values obtained from the two branches at the joining point:

\[ \boxed{3a+1=3b+3}. \]

This immediately yields

\[ \boxed{a-b=\frac23}. \]

This shortcut is especially useful in objective questions where only the relationship between the parameters is required.

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  7 points

  1. The critical point is \(x=3\).

  2. \(\displaystyle\lim_{x\to3^-}f(x)=3a+1\).

  3. \(\displaystyle f(3)=3a+1\).

  4. \(\displaystyle\lim_{x\to3^+}f(x)=3b+3\).

  5. Continuity requires

    \[ 3a+1=3b+3. \]

  6. Therefore,

    \[ \boxed{a-b=\frac23}. \]

  7. Equivalently,

    \[ \boxed{a=b+\frac23}. \]
← Q16
17 / 34  ·  50%
Q18 →
Q18
NUMERIC3 marks
For what value of \(\lambda\) is the function defined by \[ f(x)= \begin{cases} \lambda(x^2-2x), & x\le 0,\\ 4x+1, & x>0 \end{cases} \] continuous at \(x=0\)? What about continuity at \(x=1\)?
📘 Concept & Theory
Concept/Theory
›

A function \(f(x)\) is continuous at \(x=a\) if and only if

\[ \boxed{\lim_{x\to a^-}f(x)=\lim_{x\to a^+}f(x)=f(a)} \]

For a piecewise-defined function, the branch used for each one-sided limit must be selected according to the direction of approach:

  • For \(x\to a^-\), use the branch valid for \(x
  • For \(x\to a^+\), use the branch valid for \(x>a\).
  • For \(f(a)\), use the branch whose condition includes \(x=a\).

Here, the only point where the definition changes is \(x=0\). Therefore, \(x=0\) is the only possible point of discontinuity. At \(x=1\), the function is simply \(4x+1\) in a neighbourhood of \(1\).

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Check the continuity condition at \(x=0\).

  2. Find the left-hand limit, right-hand limit and \(f(0)\).

  3. Determine whether any value of \(\lambda\) can make these three quantities equal.

  4. Check continuity at \(x=1\), where the second branch applies on both sides.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  17 steps
  1. Continuity at \(x=0\)
  2. Left-hand limit at \(x=0\)
  3. For \(x\to0^-\), we have \(x\le0\), so the first branch is applicable:
    \[f(x)=\lambda(x^2-2x)\]
  4. Therefore,
    \[ \begin{aligned} \lim_{x\to0^-}f(x) &=\lim_{x\to0^-}\lambda(x^2-2x)\\ &=\lambda\left(0^2-2(0)\right)\\ &=0 \end{aligned} \]
    \[ \boxed{\lim_{x\to0^-}f(x)=0} \]
  5. Right-hand limit at \(x=0\)
  6. For \(x\to0^+\), we have \(x>0\), so the second branch is applicable:
    \[f(x)=4x+1\]
  7. Hence,
    \[ \begin{aligned} \lim_{x\to0^+}f(x) &=\lim_{x\to0^+}(4x+1)\\ &=4(0)+1\\ &=1. \end{aligned} \]
    \[ \boxed{\lim_{x\to0^+}f(x)=1} \]
  8. Function value at \(x=0\)
  9. Since \(x=0\) satisfies the condition \(x\le0\), the first branch must be used:
    \[ \begin{aligned} f(0) &=\lambda(0^2-2\cdot0)\\ &=0 \end{aligned} \]
    \[ \boxed{f(0)=0} \]
  10. Apply the continuity criterion
  11. For continuity at \(x=0\), we require
    \[\lim_{x\to0^-}f(x)=\lim_{x\to0^+}f(x)=f(0)\]
  12. But we have
    \[\lim_{x\to0^-}f(x)=0,\qquad\lim_{x\to0^+}f(x)=1,\qquad f(0)=0.\]
  13. Thus,
    \[0\ne1\]
  14. Hence the left-hand and right-hand limits are unequal. Therefore, the function cannot be continuous at \(x=0\), regardless of the value of \(\lambda\).
    \[\boxed{\text{There is no value of }\lambda\text{ for which }f(x)\text{ is continuous at }x=0.}\]
  15. Continuity at \(x=1\)
  16. Notice that \(1>0\). Therefore, in a neighbourhood of \(x=1\), the function is given by the second branch:
    \[f(x)=4x+1\]
  17. The first branch containing \(\lambda\) is irrelevant at \(x=1\).
  18. Left-hand limit at \(x=1\)
  19. \[ \begin{aligned} \lim_{x\to1^-}f(x) &=\lim_{x\to1^-}(4x+1)\\ &=4(1)+1\\ &=5 \end{aligned} \]
    \[ \boxed{\lim_{x\to1^-}f(x)=5} \]
  20. Right-hand limit at \(x=1\)
  21. \[ \begin{aligned} \lim_{x\to1^+}f(x) &=\lim_{x\to1^+}(4x+1)\\ &=4(1)+1\\ &=5 \end{aligned} \]
    \[ \boxed{\lim_{x\to1^+}f(x)=5} \]
  22. Function value at \(x=1\)
  23. \[ f(1)=4(1)+1=5. \]
    \[ \boxed{f(1)=5} \]
  24. Continuity at \(x=1\)
  25. Therefore,
    \[\lim_{x\to1^-}f(x)=\lim_{x\to1^+}f(x)=f(1)=5.\]
  26. Hence,
    \[\boxed{\text{\(f(x)\) is continuous at \(x=1\) for every value of \(\lambda\).}}\]
  27. \[ \boxed{\text{No value of }\lambda\text{ makes }f\text{ continuous at }x=0.} \]
    \[ \boxed{f\text{ is continuous at }x=1\text{ for every value of }\lambda.} \]
🎯 Exam Significance
Exam Significance
›

This question tests the most important skill in piecewise continuity problems: selecting the correct branch for each limit. At \(x=0\), the value \(f(0)\) also comes from the first branch because its condition is \(x\le0\). At \(x=1\), the second branch \(4x+1\) applies on both sides.

A complete board-examination answer should explicitly write the LHL, RHL and function value before applying the continuity criterion.

Significance for Competitive Entrance Examinations

The key observation is that the parameter \(\lambda\) cannot affect the right-hand limit at \(0\). In fact, it also cannot affect the left-hand limit because

\[ \lambda(0^2-2\cdot0)=0 \]

for every real \(\lambda\). Thus, before performing lengthy algebra, comparing the one-sided limits immediately establishes that continuity at \(0\) is impossible.

At \(x=1\), the parameter-containing branch is completely irrelevant because \(1\) lies strictly inside the region \(x>0\).

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  7 points

  1. For \(x\to0^-\), use \(\lambda(x^2-2x)\).

  2. For \(x\to0^+\), use \(4x+1\).

  3. \(f(0)=0\) because the first branch includes \(x=0\).

  4. Since \(0\ne1\), continuity at \(0\) is impossible for every \(\lambda\).

  5. At \(x=1\), the function is \(4x+1\) on both sides.

  6. Therefore, \(f\) is continuous at \(1\) for every \(\lambda\).

  7. Final result: no value of \(\lambda\) gives continuity at \(0\); continuity at \(1\) holds for all \(\lambda\).
← Q17
18 / 34  ·  53%
Q19 →
Q19
NUMERIC3 marks
Show that the function defined by \(g(x)=x-[x]\) is discontinuous at all integral points, where \([x]\) denotes the greatest integer less than or equal to \(x\).
📘 Concept & Theory
Concept/Theory
›

The greatest integer function \([x]\) is defined as the greatest integer less than or equal to \(x\). Thus, if \(n\) is an integer, then

\[ [x]=n \quad \text{for } n\le x < n+1. \]

Consider the function

\[ g(x)=x-[x]. \]

This function gives the fractional part of \(x\). For example,

\[ g(2.7)=2.7-[2.7]=2.7-2=0.7. \]

At every integer \(n\), the greatest integer function changes its value abruptly. We shall show that this produces a discontinuity in \(g(x)\).

A function is continuous at \(x=a\) if

\[ \boxed{\lim_{x\to a^-}g(x)=\lim_{x\to a^+}g(x)=g(a)}. \]
🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Let \(n\) be any arbitrary integer.

  2. Find the left-hand limit of \(g(x)\) as \(x\to n^-\).

  3. Find the right-hand limit of \(g(x)\) as \(x\to n^+\).

  4. Find the actual value \(g(n)\).

  5. Compare the three quantities and establish discontinuity at every integer.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  15 steps
  1. Proof — Let \(n\in\mathbb Z\) be any integer. We examine the continuity of \(g(x)\) at \(x=n\).
  3. When \(x\to n^-\), we hav
    \[n-1 < x < n\]
  4. Therefore, the greatest integer less than or equal to \(x\) is
    \[[x]=n-1\]
  5. Hence,
    \[ \begin{aligned} \lim_{x\to n^-}g(x) &=\lim_{x\to n^-}\left(x-[x]\right)\\ &=\lim_{x\to n^-}\left(x-(n-1)\right)\\ &=n-(n-1)\\ &=1 \end{aligned} \]
    \[ \boxed{\lim_{x\to n^-}g(x)=1} \]
  6. Right-hand limit at \(x=n\)
  7. When \(x\to n^+\), we have
    \[n < x < n+1\]
  8. Therefore,
    \[[x]=n\]
  9. Thus,
    \[ \begin{aligned} \lim_{x\to n^+}g(x) &=\lim_{x\to n^+}\left(x-[x]\right)\\ &=\lim_{x\to n^+}(x-n)\\ &=n-n\\ &=0 \end{aligned} \]
    \[ \boxed{\lim_{x\to n^+}g(x)=0} \]
  10. Function value at \(x=n\)
  11. Since \(n\) is an integer,
    \[[n]=n\]
  12. Therefore,
    \[ \begin{aligned} g(n) &=n-[n]\\ &=n-n\\ &=0 \end{aligned} \]
    \[ \boxed{g(n)=0} \]
  13. Apply the continuity criterion
  14. We have obtained
    \[\lim_{x\to n^-}g(x)=1,\]
    \[\lim_{x\to n^+}g(x)=0,\]
    and
    \[g(n)=0.\]
  15. Since
    \[\boxed{\lim_{x\to n^-}g(x)\ne\lim_{x\to n^+}g(x)},\]
  16. the two-sided limit at \(x=n\) does not exist. Hence, \(g(x)\) is not continuous at \(x=n\).
  17. But \(n\) was an arbitrary integer. Therefore, the function is discontinuous at every integral point.
    \[\boxed{\text{\(g(x)=x-[x]\) is discontinuous at every integer }n\in\mathbb Z.}\]
  18. Nature of the Discontinuity
  19. At every integer \(n\),
    \[ \lim_{x\to n^-}g(x)=1 \quad\text{and}\quad \lim_{x\to n^+}g(x)=0. \]
  20. The two one-sided limits exist and are finite, but they are unequal. Therefore, the discontinuity at every integer is a jump discontinuity.
    \[\boxed{\text{Every integral point is a point of jump discontinuity.}}\]
🎯 Exam Significance
Exam Significance
›

This is a standard application of the greatest integer function and one-sided limits. In a board examination, the most important step is to write explicitly that for an integer \(n\),

\[ [x]=n-1\quad\text{when }x\to n^-, \]

whereas

\[ [x]=n\quad\text{when }x\to n^+. \]

Failing to distinguish these two cases is the most common source of error in this question.

Significance for Competitive Entrance Examinations

The expression

\[ x-[x] \]

represents the fractional part of \(x\). It always lies in the interval

\[ 0\le x-[x]<1. \]

At every integer, the fractional part resets from values approaching \(1\) from the left to \(0\) from the right. This immediately signals a jump discontinuity at every integer.

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  8 points

  1. For an integer \(n\), \([n]=n\).

  2. As \(x\to n^-\), \([x]=n-1\).

  3. As \(x\to n^+\), \([x]=n\).

  4. Hence,

    \[ \lim_{x\to n^-}g(x)=1. \]

  5. Also,

    \[ \lim_{x\to n^+}g(x)=0. \]

  6. Since the one-sided limits are unequal, \(g(x)\) is discontinuous at \(n\).

  7. Because \(n\) is arbitrary, \(g(x)\) is discontinuous at all integral points.

  8. The discontinuity at every integer is a jump discontinuity.
← Q18
19 / 34  ·  56%
Q20 →
Q20
NUMERIC3 marks
Is the function defined by \(f(x)=x^2-\sin x+5\;\) continuous at \(x=\pi\)?
📘 Concept & Theory
Concept/Theory
›

A function \(f(x)\) is continuous at \(x=a\) if and only if

\[ \boxed{ \lim_{x\to a^-}f(x) = \lim_{x\to a^+}f(x) = f(a) } \]

The given function

\[ f(x)=x^2-\sin x+5 \]

is a combination of a polynomial function and a trigonometric function. Both \(x^2\) and \(\sin x\) are continuous for every real \(x\). Therefore, their sum/difference, along with the constant \(5\), is also continuous for every real \(x\).

Nevertheless, since the question specifically asks about \(x=\pi\), we verify continuity directly using the one-sided limits and the function value.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Find the left-hand limit as \(x\to\pi^-\).

  2. Find the right-hand limit as \(x\to\pi^+\).

  3. Evaluate \(f(\pi)\).

  4. Compare the three quantities using the continuity criterion.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  9 steps
  1. Left-Hand Limit at \(x=\pi\)
  2. \[ \begin{aligned} \lim_{x\to\pi^-}f(x) &=\lim_{x\to\pi^-}(x^2-\sin x+5)\\ &=\pi^2-\sin\pi+5 \end{aligned} \]
  3. Since
    \[\sin\pi=0\]
  4. we get
    \[\boxed{\lim_{x\to\pi^-}f(x)=\pi^2+5}\]
  5. Right-Hand Limit at \(x=\pi\)
  6. \[ \begin{aligned} \lim_{x\to\pi^+}f(x) &=\lim_{x\to\pi^+}(x^2-\sin x+5)\\ &=\pi^2-\sin\pi+5\\ &=\pi^2+5 \end{aligned} \]
  7. Therefore,
    \[\boxed{\lim_{x\to\pi^+}f(x)=\pi^2+5}\]
  8. Function Value at \(x=\pi\)
  9. \[\begin{aligned}f(\pi)&=\pi^2-\sin\pi+5\\&=\pi^2-0+5\\&=\pi^2+5\end{aligned}\]
  10. Apply the Continuity Criterion
  11. We have
    \[\lim_{x\to\pi^-}f(x)=\pi^2+5,\]
    \[\lim_{x\to\pi^+}f(x)=\pi^2+5,\]
    and
    \[f(\pi)=\pi^2+5\]
  12. Hence,
    \[\boxed{\lim_{x\to\pi^-}f(x)=\lim_{x\to\pi^+}f(x)=f(\pi)=\pi^2+5}\]
  13. Therefore, the function is continuous at \(x=\pi\).
    \[\boxed{\therefore\ f(x)=x^2-\sin x+5\text{ is continuous at }x=\pi.}\]
🎯 Exam Significance
Exam Significance
›

This is a straightforward application of the three-part continuity test. For full marks, clearly show:

\[ \text{LHL},\qquad \text{RHL},\qquad f(\pi), \]

and then conclude that all three are equal.

A useful shortcut is that polynomial functions and trigonometric functions such as \(\sin x\) are continuous everywhere. Hence, any polynomial expression involving \(\sin x\) is continuous throughout \(\mathbb R\).

Significance for Competitive Entrance Examinations

For objective questions, the continuity can be recognized immediately because \(x^2\) and \(\sin x\) are continuous everywhere. Their linear combination

\[ x^2-\sin x+5 \]

is therefore continuous for every \(x\in\mathbb R\).

Thus, \(x=\pi\) is not a special point of discontinuity; it is simply the point chosen for verification.

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  4 points

  1. \(x^2\) is continuous for every real \(x\).

  2. \(\sin x\) is continuous for every real \(x\).

  3. Therefore, \(x^2-\sin x+5\) is continuous on \(\mathbb R\).

  4. At \(x=\pi\),

    \[ \lim_{x\to\pi^-}f(x) = \lim_{x\to\pi^+}f(x) = f(\pi) = \pi^2+5. \]
← Q19
20 / 34  ·  59%
Q21 →
Q21
NUMERIC3 marks
Discuss the continuity of the following functions:
  1. \(f(x)=\sin x+\cos x\)
  2. \(f(x)=\sin x-\cos x\)
  3. \(f(x)=\sin x\cdot\cos x\)
📘 Concept & Theory
Concept/Theory
›

The basic continuity results for elementary functions are:

  • \(\sin x\) is continuous for every \(x\in\mathbb R\).
  • \(\cos x\) is continuous for every \(x\in\mathbb R\).
  • The sum or difference of two continuous functions is continuous.
  • The product of two continuous functions is continuous.

Therefore, each of the three given functions is continuous for every real value of \(x\). We can also verify this directly using the limit laws.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. >Use the fact that \(\sin x\) and \(\cos x\) are continuous everywhere.

  2. Apply the sum, difference, and product rules for continuous functions.

  3. State the interval of continuity for each function.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  15 steps
  1. (a) \(f(x)=\sin x+\cos x\)
  2. Let \(x=a\), where \(a\in\mathbb R\). Then
  3. \[ \begin{aligned} \lim_{x\to a}f(x) &=\lim_{x\to a}(\sin x+\cos x)\\ &=\lim_{x\to a}\sin x+\lim_{x\to a}\cos x\\ &=\sin a+\cos a \end{aligned} \]
  4. But
    \[f(a)=\sin a+\cos a\]
  5. Hence,
    \[\boxed{\lim_{x\to a}f(x)=f(a)}\]
  6. for every \(a\in\mathbb R\).
  7. Therefore,
    \[\boxed{f(x)=\sin x+\cos x\text{ is continuous on }\mathbb R.}\]
  8. (b) \(f(x)=\sin x-\cos x\)
  9. Let \(x=a\), where \(a\in\mathbb R\). Then
    \[ \begin{aligned} \lim_{x\to a}f(x) &=\lim_{x\to a}(\sin x-\cos x)\\ &=\lim_{x\to a}\sin x-\lim_{x\to a}\cos x\\ &=\sin a-\cos a \end{aligned} \]
  10. Also,
    \[f(a)=\sin a-\cos a\]
  11. Therefore,
    \[\boxed{\lim_{x\to a}f(x)=f(a)}\]
    for every \(a\in\mathbb R\).
  12. Hence,
    \[\boxed{f(x)=\sin x-\cos x\text{ is continuous on }\mathbb R.}\]
  13. (c) \(f(x)=\sin x\cdot\cos x\)
  14. Let \(x=a\), where \(a\in\mathbb R\). Using the product law of limits,
    \[ \begin{aligned} \lim_{x\to a}f(x) &=\lim_{x\to a}(\sin x\cdot\cos x)\\ &=\left(\lim_{x\to a}\sin x\right) \left(\lim_{x\to a}\cos x\right)\\ &=\sin a\cos a \end{aligned} \]
  15. Also,
    \[f(a)=\sin a\cos a\]
  16. Thus,
    \[\boxed{\lim_{x\to a}f(x)=f(a)}\]
  17. for every \(a\in\mathbb R\).
  18. Therefore,
    \[\boxed{f(x)=\sin x\cos x\text{ is continuous on }\mathbb R.}\]
🎯 Exam Significance
Exam Significance
›

This question tests the standard algebra of continuous functions. Once it is known that \(\sin x\) and \(\cos x\) are continuous on \(\mathbb R\), the results follow directly:

  • Sum of continuous functions → continuous.
  • Difference of continuous functions → continuous.
  • Product of continuous functions → continuous.

For a short board-examination answer, these standard results can be cited directly. For a detailed answer, the limit verification given above provides a complete justification.

Significance for Competitive Entrance Examinations

For objective questions, recognizing the closure properties of continuous functions saves substantial time. Since both \(\sin x\) and \(\cos x\) are continuous everywhere, any finite algebraic combination involving only their sums, differences, and products is also continuous everywhere.

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  6 points

  1. \(\sin x\) and \(\cos x\) are continuous for all real \(x\).

  2. The sum of continuous functions is continuous.

  3. The difference of continuous functions is continuous.

  4. The product of continuous functions is continuous.

  5. All three functions in the question are continuous on the entire real line.

  6. There are no points of discontinuity for any of the three functions.
← Q20
21 / 34  ·  62%
Q22 →
Q22
NUMERIC3 marks
Discuss the continuity of the cosine, cosecant, secant and cotangent functions.
📘 Concept & Theory
Concept/Theory
›

The continuity of reciprocal trigonometric functions depends on whether their denominators are zero.

The basic trigonometric functions \(\sin x\) and \(\cos x\) are continuous for every \(x\in\mathbb R\). For a quotient

\[ \frac{u(x)}{v(x)}, \]
continuity is guaranteed at points where \(u(x)\) and \(v(x)\) are continuous and \(v(x)\ne0\).

We use the following identities:

\[ \csc x=\frac{1}{\sin x},\qquad \sec x=\frac{1}{\cos x},\qquad \cot x=\frac{\cos x}{\sin x}. \]
🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Identify the domain of each trigonometric function.

  2. Use the continuity of \(\sin x\) and \(\cos x\).

  3. Exclude points where a denominator becomes zero.

  4. State the intervals on which each function is continuous.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  27 steps
  1. (a) Continuity of the Cosine Function
  2. Consider
    \[f(x)=\cos x\]
  3. For any \(a\in\mathbb R\),
    \[\lim_{x\to a}\cos x=\cos a\]
  4. Since
    \[f(a)=\cos a\]
  5. we have
    \[\boxed{\lim_{x\to a}f(x)=f(a)}\]
  6. Therefore, \(\cos x\) is continuous at every real number.
    \[\boxed{\cos x\text{ is continuous on }\mathbb R.}\]
  7. (b) Continuity of the Cosecant Function
  8. Consider
    \[f(x)=\operatorname{cosec} x=\frac{1}{\sin x}\]
  9. Since \(\sin x\) is continuous everywhere, \(\operatorname{cosec} x\) is continuous wherever \(\sin x\ne0\).
  10. Now,
    \[\sin x=0\]
  11. when
    \[x=n\pi,\qquad n\in\mathbb Z\]
  12. At these points, \(\csc x\) is not defined.
  13. Hence, \(\operatorname{cosec} x\) is continuous on each interval between consecutive multiples of \(\pi\):
    \[ \boxed{ (n\pi,(n+1)\pi),\quad n\in\mathbb Z. } \]
  14. Equivalently, its complete continuity domain is
    \[ \boxed{ \mathbb R\setminus\{n\pi:n\in\mathbb Z\}. } \]
  15. Thus, \(\operatorname{cosec} x\) is discontinuous at every integral multiple of \(\pi\).
  16. (c) Continuity of the Secant Function
  17. Consider
    \[f(x)=\sec x=\frac{1}{\cos x}\]
  18. Since \(\cos x\) is continuous everywhere, \(\sec x\) is continuous wherever \(\cos x\ne0\).
  19. Now,
    \[\cos x=0\]
  20. when
    \[x=\frac{(2n+1)\pi}{2},\quad n\in\mathbb Z.\]
    At these points, \(\sec x\) is not defined.
  21. Therefore, \(\sec x\) is continuous on every interval that does not contain a zero of \(\cos x\). Its continuity domain is
    \[ \boxed{ \mathbb R\setminus \left\{ \frac{(2n+1)\pi}{2}:n\in\mathbb Z \right\}. } \]
  22. Thus, \(\sec x\) is discontinuous at every odd multiple of \(\frac{\pi}{2}\).
  23. (d) Continuity of the Cotangent Function
  24. Consider
    \[f(x)=\cot x=\frac{\cos x}{\sin x}\]
  25. Both \(\cos x\) and \(\sin x\) are continuous everywhere. Therefore, their quotient is continuous wherever the denominator is non-zero.
  26. We require
    \[\sin x\ne0\]
  27. But
    \[ \sin x=0 \quad\text{when}\quad x=n\pi,\qquad n\in\mathbb Z. \]
  28. Hence, \(\cot x\) is continuous wherever \(x\ne n\pi\).
    \[ \boxed{ \cot x\text{ is continuous on } \mathbb R\setminus\{n\pi:n\in\mathbb Z\}. } \]
  29. Equivalently, it is continuous on each interval
  30. \[ \boxed{ (n\pi,(n+1)\pi),\qquad n\in\mathbb Z. } \]
  31. Summary Table
  32. Function Definition Points of Discontinuity Continuity
    \(\cos x\) \(\cos x\) None Continuous on \(\mathbb R\)
    \(\operatorname{cosec} x\) \(\dfrac{1}{\sin x}\) \(x=n\pi\) Continuous where \(x\ne n\pi\)
    \(\sec x\) \(\dfrac{1}{\cos x}\) \(x=\dfrac{(2n+1)\pi}{2}\) Continuous where \(\cos x\ne0\)
    \(\cot x\) \(\dfrac{\cos x}{\sin x}\) \(x=n\pi\) Continuous where \(x\ne n\pi\)

    Here, \(n\in\mathbb Z\).

🎯 Exam Significance
Exam Significance
›

The key rule to remember is:

\[ \boxed{ \frac{f(x)}{g(x)} \text{ is continuous wherever }f,g\text{ are continuous and }g(x)\ne0. } \]

Since \(\sin x\) and \(\cos x\) are continuous everywhere, the continuity of \(\csc x\), \(\sec x\), and \(\cot x\) is determined entirely by the zeros of their denominators.

Significance for Competitive Entrance Examinations

For quick problem solving, memorize the discontinuity sets:

\[ \boxed{ \begin{aligned} \csc x,\cot x &: \quad x=n\pi,\\[4pt] \sec x &: \quad x=\frac{(2n+1)\pi}{2}. \end{aligned} } \]

All three functions are continuous at every point of their respective domains. At the excluded points, the functions are undefined and hence cannot be continuous there.

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  5 points

  1. \(\boxed{\cos x}\) is continuous for every \(x\in\mathbb R\).

  2. \(\boxed{\ csc x}\) is continuous wherever \(\sin x\ne0\), i.e. \(x\ne n\pi\).

  3. \(\boxed{\sec x}\) is continuous wherever \(\cos x\ne0\), i.e. \(x\ne\dfrac{(2n+1)\pi}{2}\).

  4. \(\boxed{\cot x}\) is continuous wherever \(\sin x\ne0\), i.e. \(x\ne n\pi\).

  5. For reciprocal functions, always check where the denominator is zero.
← Q21
22 / 34  ·  65%
Q23 →
Q23
NUMERIC3 marks
Find all points of discontinuity of \(f\), where \[ f(x)= \begin{cases} \dfrac{\sin x}{x}, & x<0,\\[6pt] x+1, & x\ge0 \end{cases} \]
📘 Concept & Theory
Concept/Theory
›

For a piecewise-defined function, discontinuities can occur at:

  • points where one of the expressions is undefined, or
  • points where the definition changes from one branch to another.

Here, the first branch \(\dfrac{\sin x}{x}\) is undefined at \(x=0\), while the second branch includes \(x=0\). Therefore, \(x=0\) is the only point that requires special examination.

The function is continuous at \(x=a\) if

\[\boxed{\lim_{x\to a^-}f(x)=\lim_{x\to a^+}f(x)=f(a)}\]
🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Identify the possible point of discontinuity.

  2. Evaluate the left-hand limit at \(x=0\).

  3. Evaluate the right-hand limit at \(x=0\).

  4. Find \(f(0)\).

  5. Compare the three quantities.

  6. State the points of discontinuity.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  14 steps
  1. Identify the Possible Point of Discontinuity
  2. For \(x<0\),
    \[f(x)=\frac{\sin x}{x}\]
  3. This function is continuous for every \(x<0\), since \(x\ne0\) there.
  4. For \(x\ge0\),
    \[f(x)=x+1\]
    which is a polynomial function and is continuous everywhere.
  5. Thus, the only point that needs to be checked is the joining point
    \[\boxed{x=0}\]
  6. Left-Hand Limit at \(x=0\)
  7. As \(x\to0^-\), we use the first branch because \(x<0\):
    \[ \begin{aligned} \lim_{x\to0^-}f(x) &=\lim_{x\to0^-}\frac{\sin x}{x}\\ &=1 \end{aligned} \]
  8. Here we have used the standard limit
    \[\boxed{\lim_{x\to0}\frac{\sin x}{x}=1}\]
  9. Therefore,
    \[\boxed{\lim_{x\to0^-}f(x)=1}\]
  10. Right-Hand Limit at \(x=0\)
  11. As \(x\to0^+\), we use the second branch because \(x>0\):
    \[ \begin{aligned} \lim_{x\to0^+}f(x) &=\lim_{x\to0^+}(x+1)\\ &=0+1\\ &=1 \end{aligned} \]
    \[ \boxed{\lim_{x\to0^+}f(x)=1}. \]
  12. Function Value at \(x=0\)
  13. Since the condition \(x\ge0\) includes \(x=0\), the second branch must be used to calculate \(f(0)\):
    \[ \begin{aligned} f(0) &=0+1\\ &=1 \end{aligned} \]
    \[ \boxed{f(0)=1}. \]
  14. Apply the Continuity Criterion
  15. We have
    \[\lim_{x\to0^-}f(x)=1,\]
    \[\lim_{x\to0^+}f(x)=1,\]
    and
    \[f(0)=1\]
  16. Hence,
    \[\boxed{\lim_{x\to0^-}f(x)=\lim_{x\to0^+}f(x)=f(0)=1}\]
  17. Therefore, \(f(x)\) is continuous at \(x=0\).
  18. Since both branches are continuous everywhere in their respective domains, there are no other possible points of discontinuity.
  19. Conclusion
  20. \[ \boxed{\text{The function is continuous for every }x\in\mathbb R.} \]
    \[ \boxed{\text{There are no points of discontinuity.}} \]
🎯 Exam Significance
Exam Significance
›

This question is a standard test of continuity of a piecewise function. The most important point is to use the correct branch for each quantity:

  • LHL at \(0\) → use \(\dfrac{\sin x}{x}\).
  • RHL at \(0\) → use \(x+1\).
  • \(f(0)\) → use \(x+1\), because the condition is \(x\ge0\).

The standard limit

\[ \lim_{x\to0}\frac{\sin x}{x}=1 \]

is essential for completing the verification.

Significance for Competitive Entrance Examinations

The quickest approach is to notice that \(x=0\) is the only joining point. Then immediately use

\[ \lim_{x\to0^-}\frac{\sin x}{x}=1 \]

and

\[ \lim_{x\to0^+}(x+1)=1. \]

Since \(f(0)=1\), continuity follows immediately. There is therefore no need to investigate arbitrary \(x\ne0\).

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  6 points

  1. The only possible point of discontinuity is \(x=0\).

  2. \(\displaystyle \lim_{x\to0^-}\frac{\sin x}{x}=1\).

  3. \(\displaystyle \lim_{x\to0^+}(x+1)=1\).

  4. \(f(0)=1\) because \(0\) belongs to the second branch.

  5. All three quantities are equal.

  6. Therefore, the function is continuous at \(x=0\) and everywhere else.
← Q22
23 / 34  ·  68%
Q24 →
Q24
NUMERIC3 marks
Determine whether the function \[ f(x)= \begin{cases} x^2\sin\dfrac{1}{x}, & x\ne0,\\[6pt] 0, & x=0 \end{cases} \] is a continuous function.
📘 Concept & Theory
Concept/Theory
›

For a function \(f(x)\) to be continuous at \(x=a\), it must satisfy

\[\boxed{\lim_{x\to a}f(x)=f(a)}\]

Equivalently, for a piecewise function, we may verify

\[\boxed{\lim_{x\to a^-}f(x)=\lim_{x\to a^+}f(x)=f(a)}.\]

For the given function, the expression \(x^2\sin(1/x)\) is continuous for every \(x\ne0\). Therefore, the only point requiring investigation is \(x=0\).

The difficulty is that \(\sin(1/x)\) oscillates infinitely rapidly as \(x\to0\), so \(\lim_{x\to0}\sin(1/x)\) does not exist. However, the factor \(x^2\) forces the entire product to approach zero. This can be established using the squeeze theorem.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Observe that \(x=0\) is the only point requiring special attention.

  2. Bound \(\sin(1/x)\) using \(-1\le\sin(1/x)\le1\).

  3. Multiply by \(x^2\) and apply the squeeze theorem.

  4. Compare the resulting limit with \(f(0)\).

  5. Conclude continuity at \(0\) and hence on the entire real line.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  16 steps
  1. Continuity for \(x\ne0\)
  2. For \(x\ne0\),
    \[f(x)=x^2\sin\frac1x\]
  3. The functions \(x^2\), \(1/x\), and \(\sin x\) are continuous wherever they are defined. Since \(x\ne0\), \(1/x\) is defined, and therefore \(x^2\sin(1/x)\) is continuous for every \(x\ne0\).
  4. Thus, only \(x=0\) needs to be checked.
  5. Evaluate \(\displaystyle\lim_{x\to0}x^2\sin\dfrac1x\)
  6. We know that for every real number \(t\),
    \[-1\le\sin t\le1\]
  7. Putting \(t=\dfrac1x\), we obtain
    \[-1\le\sin\frac1x\le1\]
  8. Since \(x^2\ge0\), multiplying throughout by \(x^2\) gives
    \[-x^2\le x^2\sin\frac1x\le x^2.\]
  9. Now, as \(x\to0\),
    \[\lim_{x\to0}(-x^2)=0\]
    and
    \[\lim_{x\to0}x^2=0\]
  10. Therefore, by the Squeeze Theorem,
    \[\boxed{\lim_{x\to0}x^2\sin\frac1x=0}\]
  11. Find \(f(0)\)
  12. From the definition of the function,
    \[\boxed{f(0)=0}\]
  13. Apply the Continuity Criterion
  14. We have shown that
    \[\lim_{x\to0}f(x)=\lim_{x\to0}x^2\sin\frac1x=0\]
  15. Also,
    \[f(0)=0\]
  16. Hence,
    \[\boxed{\lim_{x\to0}f(x)=f(0)=0}\]
  17. Therefore, \(f(x)\) is continuous at \(x=0\).
  18. Since it is already continuous for every \(x\ne0\), we conclude that the function is continuous on the entire real line.
    \[\boxed{\text{\(f(x)\) is continuous for every }x\in\mathbb R.}\]
  19. One-Sided Limit Verification
  20. If continuity is checked using one-sided limits, the same squeeze argument gives
    \[ \boxed{ \lim_{x\to0^-}f(x)=0 } \]
    and
    \[ \boxed{ \lim_{x\to0^+}f(x)=0 } \]
  21. Since \(f(0)=0\),
    \[\boxed{\lim_{x\to0^-}f(x)=\lim_{x\to0^+}f(x)=f(0)=0}\]
🎯 Exam Significance
Exam Significance
›

This is an important example where the expression contains \(\sin(1/x)\), which is not itself convergent as \(x\to0\). Students should not incorrectly conclude that the given function is discontinuous merely because \(\sin(1/x)\) oscillates.

The factor \(x^2\) dominates the bounded oscillation because

\[ \left|x^2\sin\frac1x\right| \le x^2\to0. \]

This is the most efficient way to present the argument in an examination.

Significance for Competitive Entrance Examinations

The key observation is the boundedness

\[ \left|\sin\frac1x\right|\le1. \]

Therefore,

\[ \left|x^2\sin\frac1x\right| \le x^2. \]

Since \(x^2\to0\), the given limit is immediately \(0\). This absolute-value form of the Squeeze Theorem is particularly useful for rapidly solving such questions.

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  7 points

  1. The only potentially problematic point is \(x=0\).

  2. \(\sin(1/x)\) is bounded between \(-1\) and \(1\).

  3. Hence,

    \[ -x^2\le x^2\sin\frac1x\le x^2. \]

  4. Both bounding functions tend to \(0\) as \(x\to0\).

  5. Therefore,

    \[ \lim_{x\to0}x^2\sin\frac1x=0. \]

  6. Since \(f(0)=0\), the function is continuous at \(0\).

  7. The function is continuous for every \(x\ne0\) as well.
← Q23
24 / 34  ·  71%
Q25 →
Q25
NUMERIC3 marks
Examine the continuity of \(f\), where \[ f(x)= \begin{cases} \sin x-\cos x, & x\ne0,\\[4pt] -1, & x=0 \end{cases} \]
📘 Concept & Theory
Concept/Theory
›

A function \(f(x)\) is continuous at \(x=a\) if and only if

\[\boxed{\lim_{x\to a^-}f(x)=\lim_{x\to a^+}f(x)=f(a)}\]

Here, the formula changes only at \(x=0\). For every \(x\ne0\), the function is \(\sin x-\cos x\), which is continuous because both \(\sin x\) and \(\cos x\) are continuous everywhere.

Thus, \(x=0\) is the only point that needs to be examined.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Evaluate the left-hand limit at \(x=0\).

  2. Evaluate the right-hand limit at \(x=0\).

  3. Find the actual value \(f(0)\).

  4. Compare the three quantities.

  5. Conclude the continuity of the function.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  8 steps
  1. Left-Hand Limit at \(x=0\)
  2. As \(x\to0^-\), we have \(x\ne0\), so the first branch applies:
    \[ \begin{aligned} \lim_{x\to0^-}f(x) &=\lim_{x\to0^-}(\sin x-\cos x)\\ &=\sin0-\cos0\\ &=0-1\\ &=-1 \end{aligned} \]
    \[ \boxed{\lim_{x\to0^-}f(x)=-1} \]
  3. Right-Hand Limit at \(x=0\)
  4. As \(x\to0^+\), again \(x\ne0\), so the first branch applies:
  5. \[ \begin{aligned} \lim_{x\to0^+}f(x) &=\lim_{x\to0^+}(\sin x-\cos x)\\ &=\sin0-\cos0\\ &=0-1\\ &=-1 \end{aligned} \]
    \[ \boxed{\lim_{x\to0^+}f(x)=-1} \]
  6. Function Value at \(x=0\)
  7. The second branch explicitly defines the value at \(x=0\):
    \[\boxed{f(0)=-1}\]
  8. Apply the Continuity Criterion
  9. We have
    \[\lim_{x\to0^-}f(x)=-1,\]
    \[\lim_{x\to0^+}f(x)=-1,\]
    and
    \[f(0)=-1\]
  10. Therefore,
    \[\boxed{\lim_{x\to0^-}f(x)=\lim_{x\to0^+}f(x)=f(0)=-1}\]
  11. Hence, \(f(x)\) is continuous at \(x=0\).
  12. Moreover, for \(x\ne0\), \(f(x)=\sin x-\cos x\), which is continuous everywhere in its domain. Therefore, the entire function is continuous on \(\mathbb R\).
    \[\boxed{\text{\(f(x)\) is continuous for every }x\in\mathbb R.}\]
🎯 Exam Significance
Exam Significance
›

This question illustrates an important point about piecewise functions: the value assigned separately at a point does not automatically make the function discontinuous. What matters is whether that assigned value agrees with the limiting value.

Here, the separately assigned value \(-1\) exactly matches the limit of \(\sin x-\cos x\) at \(0\).

Significance for Competitive Entrance Examinations

For a quick solution, observe that \(\sin x\) and \(\cos x\) are continuous everywhere. Hence,

\[\lim_{x\to0}(\sin x-\cos x)=\sin0-\cos0=-1\]

Since \(f(0)=-1\), continuity follows immediately.

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  7 points

  1. The only point requiring special examination is \(x=0\).

  2. \(\displaystyle \lim_{x\to0^-}f(x)=-1\).

  3. \(\displaystyle \lim_{x\to0^+}f(x)=-1\).

  4. \(f(0)=-1\).

  5. Therefore, the three quantities are equal.

  6. The function is continuous at \(0\).

  7. Since \(\sin x-\cos x\) is continuous for \(x\ne0\), the function is continuous on all of \(\mathbb R\).
← Q24
25 / 34  ·  74%
Q26 →
Q26
NUMERIC3 marks
Find the value of \(k\) so that the function \(f\) is continuous at \(x=\dfrac{\pi}{2}\), where \[ f(x)= \begin{cases} \dfrac{k\cos x}{\pi-2x}, & x\ne\dfrac{\pi}{2},\\[8pt] 3, & x=\dfrac{\pi}{2} \end{cases} \]
📘 Concept & Theory
Concept/Theory
›

For a function to be continuous at \(x=a\), we require

\[ \boxed{ \lim_{x\to a}f(x)=f(a) }. \]

Here, the function is defined separately at \(x=\dfrac{\pi}{2}\), so we must ensure that the limiting value of the first branch is equal to the assigned value \(3\).

At \(x=\dfrac{\pi}{2}\), both the numerator and denominator of the first branch become zero:

\[ \cos\frac{\pi}{2}=0, \qquad \pi-2\left(\frac{\pi}{2}\right)=0. \]

Thus, the limit is of the indeterminate form \(\dfrac00\), and we need to simplify it using a standard trigonometric limit.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Evaluate \(\displaystyle\lim_{x\to\pi/2}f(x)\) using the first branch.

  2. Rewrite the expression so that the standard limit \(\displaystyle\lim_{t\to0}\frac{\sin t}{t}=1\) can be used.

  3. Equate the limiting value to \(f\left(\dfrac{\pi}{2}\right)=3\).

  4. Solve for \(k\).

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  15 steps
  1. Find the Required Limit
  2. For \(x\ne\dfrac{\pi}{2}\),
    \[ f(x)=\frac{k\cos x}{\pi-2x}\]
  3. Therefore,
    \[\lim_{x\to\pi/2}f(x)=\lim_{x\to\pi/2}\frac{k\cos x}{\pi-2x}.\]
  4. Rewrite the denominator:
    \[\pi-2x=-2\left(x-\frac{\pi}{2}\right).\]
  5. Also,
    \[\cos x=\cos\left(\frac{\pi}{2}+\left(x-\frac{\pi}{2}\right)\right)\]
  6. Using the identity
    \[\cos\left(\frac{\pi}{2}+t\right)=-\sin t,\]
  7. we get
    \[\cos x=-\sin\left(x-\frac{\pi}{2}\right)\]
  8. Hence,
    \[ \begin{aligned} \lim_{x\to\pi/2}f(x) &= \lim_{x\to\pi/2} \frac{-k\sin\left(x-\frac{\pi}{2}\right)} {-2\left(x-\frac{\pi}{2}\right)}\\[6pt] &= \frac{k}{2} \lim_{x\to\pi/2} \frac{\sin\left(x-\frac{\pi}{2}\right)} {x-\frac{\pi}{2}}. \end{aligned} \]
  9. Apply the Standard Limit
  10. Let
    \[t=x-\frac{\pi}{2}\]
  11. As \(x\to\dfrac{\pi}{2}\), we have \(t\to0\). Therefore, using
    \[\boxed{\lim_{t\to0}\frac{\sin t}{t}=1}\]
  12. we obtain
    \[ \begin{aligned} \lim_{x\to\pi/2}f(x) &= \frac{k}{2}\cdot1\\ &=\frac{k}{2}. \end{aligned} \]
    \[ \boxed{ \lim_{x\to\pi/2}f(x)=\frac{k}{2} } \]
  13. Find the Function Value
  14. From the second part of the definition,
    \[\boxed{f\left(\frac{\pi}{2}\right)=3}\]
  15. Apply the Continuity Condition
  16. For continuity at \(x=\dfrac{\pi}{2}\),
    \[\lim_{x\to\pi/2}f(x)=f\left(\frac{\pi}{2}\right).\]
  17. Therefore,
    \[\frac{k}{2}=3\]
  18. Multiplying both sides by \(2\),
    \[k=6\]
  19. Therefore, the function is continuous at \(x=\dfrac{\pi}{2}\) when \(k=6\)
🎯 Exam Significance
Exam Significance
›

This question is a standard application of the fundamental trigonometric limit

\[ \lim_{x\to0}\frac{\sin x}{x}=1. \]

Whenever a limit approaches \(\dfrac00\) near \(x=\dfrac{\pi}{2}\), expressions involving \(\cos x\) can often be converted into \(\sin\left(x-\dfrac{\pi}{2}\right)\).

The final continuity condition must always be written explicitly:

\[ \lim_{x\to\pi/2}f(x) = f\left(\frac{\pi}{2}\right). \]
Significance for Competitive Entrance Examinations

The fastest route is to recognize that

\[ \frac{\cos x}{\pi-2x} = \frac{-\sin\left(x-\frac{\pi}{2}\right)} {-2\left(x-\frac{\pi}{2}\right)} = \frac12 \frac{\sin\left(x-\frac{\pi}{2}\right)} {x-\frac{\pi}{2}}. \]

Therefore, the limit is immediately \(\dfrac12\), and the required condition becomes

\[ \frac{k}{2}=3. \]

Thus, \(k=6\).

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  7 points

  1. The point of interest is \(x=\dfrac{\pi}{2}\).

  2. Direct substitution gives the indeterminate form \(\dfrac00\).

  3. Use

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

  4. Use the standard limit

    \[ \lim_{t\to0}\frac{\sin t}{t}=1. \]

  5. The limiting value is

    \[ \frac{k}{2}. \]

  6. The defined value is \(3\).

  7. Continuity requires \(\dfrac{k}{2}=3\).
← Q25
26 / 34  ·  76%
Q27 →
Q27
NUMERIC3 marks
Find the value of \(k\) so that the function \(f\) is continuous at \(x=2\), where \[ f(x)= \begin{cases} kx^2, & x\le2,\\[4pt] 3, & x>2 \end{cases} \]
📘 Concept & Theory
Concept/Theory
›

For a function \(f(x)\) to be continuous at \(x=a\), the left-hand limit, right-hand limit and function value must be equal:

\[\boxed{\lim_{x\to a^-}f(x)=\lim_{x\to a^+}f(x)=f(a)}\]

Since the definition of the given function changes at \(x=2\), this is the point at which continuity must be checked.

Notice that \(x=2\) belongs to the first branch because its condition is \(x\le2\). Therefore, \(f(2)\) must be calculated from \(kx^2\), not from the second branch.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Find the left-hand limit using \(kx^2\).

  2. Find the right-hand limit using the constant branch \(3\).

  3. Calculate \(f(2)\) from the branch containing \(x=2\).

  4. Equate the quantities and solve for \(k\).

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  8 steps
  1. Left-Hand Limit at \(x=2\)
  2. For \(x\to2^-\), we have \(x<2\), so the first branch applies:
    \[ \begin{aligned} \lim_{x\to2^-}f(x) &=\lim_{x\to2^-}kx^2\\ &=k(2)^2\\ &=4k \end{aligned} \]
    \[ \boxed{\lim_{x\to2^-}f(x)=4k} \]
  3. Right-Hand Limit at \(x=2\)
  4. For \(x\to2^+\), we have \(x>2\), so the second branch applies:
    \[ \begin{aligned} \lim_{x\to2^+}f(x) &=\lim_{x\to2^+}3\\ &=3 \end{aligned} \]
    \[ \boxed{\lim_{x\to2^+}f(x)=3} \]
  5. Function Value at \(x=2\)
  6. Since \(2\le2\), the first branch applies:
    \[ \begin{aligned} f(2) &=k(2)^2\\ &=4k \end{aligned} \]
    \[ \boxed{f(2)=4k} \]
  7. Apply the Continuity Criterion
  8. \[\lim_{x\to2^-}f(x)=\lim_{x\to2^+}f(x)=f(2)\]
  9. Therefore,
    \[4k=3=4k\]
  10. The essential equation is
    \[4k=3\]
  11. Dividing both sides by \(4\),
    \[\boxed{k=\frac34}\]
  12. Therefore, the function is continuous at \(x=2\) when \(k=\dfrac34\).
🎯 Exam Significance
Exam Significance
›

In piecewise continuity questions, always identify which branch contains the point of interest. Here, because \(x=2\) satisfies \(x\le2\), the function value is obtained from \(kx^2\).

A complete board answer should show all three quantities:

\[ \text{LHL},\qquad \text{RHL},\qquad f(2). \]
Significance for Competitive Entrance Examinations

For a quick solution, continuity at the joining point requires the two branch values to match:

\[ k(2)^2=3. \]

Thus,

\[ 4k=3 \quad\Longrightarrow\quad k=\frac34. \]

The fact that \(x=2\) belongs to the first branch is important for determining \(f(2)\), although the same equation results from matching the two one-sided limits.

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  6 points

  1. The only point requiring examination is \(x=2\).

  2. \(\displaystyle \lim_{x\to2^-}f(x)=4k\).

  3. \(\displaystyle \lim_{x\to2^+}f(x)=3\).

  4. Since \(2\le2\), \(\displaystyle f(2)=4k\).

  5. Continuity requires \(4k=3\).

  6. Therefore, \(\displaystyle k=\frac34\).
← Q26
27 / 34  ·  79%
Q28 →
Q28
NUMERIC3 marks
Find the value of \(k\) so that the function \(f\) is continuous at \(x=\pi\), where

\[ f(x)= \begin{cases} kx+1, & x\le\pi,\\[4pt] \cos x, & x>\pi \end{cases} \]
📘 Concept & Theory
Concept/Theory
›

For a function \(f(x)\) to be continuous at \(x=a\), we require

\[\boxed{\lim_{x\to a^-}f(x)=\lim_{x\to a^+}f(x)=f(a)}.\]

Here, the definition changes at \(x=\pi\). Since the first branch has the condition \(x\le\pi\), the point \(x=\pi\) belongs to the first branch.

Therefore,

\[ f(\pi)=k\pi+1. \]

We must make the limiting value from the second branch agree with this value.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Find the left-hand limit using \(kx+1\).

  2. Find the right-hand limit using \(\cos x\).

  3. Evaluate \(f(\pi)\).

  4. Apply the continuity condition and solve for \(k\).

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  7 steps
  1. Left-Hand Limit at \(x=\pi\)
  2. For \(x\to\pi^-\), we have \(x<\pi\), so the first branch applies:
    \[ \begin{aligned} \lim_{x\to\pi^-}f(x) &=\lim_{x\to\pi^-}(kx+1)\\ &=k\pi+1 \end{aligned} \]
    \[ \boxed{\lim_{x\to\pi^-}f(x)=k\pi+1} \]
  3. Right-Hand Limit at \(x=\pi\)
  4. For \(x\to\pi^+\), we have \(x>\pi\), so the second branch applies:
    \[ \begin{aligned} \lim_{x\to\pi^+}f(x) &=\lim_{x\to\pi^+}\cos x\\ &=\cos\pi\\ &=-1 \end{aligned} \]
    \[ \boxed{\lim_{x\to\pi^+}f(x)=-1} \]
  5. Function Value at \(x=\pi\)
  6. Since the first branch includes \(x=\pi\),
    \[ \begin{aligned} f(\pi) &=k\pi+1. \end{aligned} \]
    \[ \boxed{f(\pi)=k\pi+1} \]
  7. Apply the Continuity Criterion
  8. For continuity at \(x=\pi\),
    \[ \lim_{x\to\pi^-}f(x) = \lim_{x\to\pi^+}f(x) = f(\pi). \]
  9. Thus,
    \[k\pi+1=-1\]
  10. Subtracting \(1\) from both sides,
    \[k\pi=-2\]
  11. Dividing by \(\pi\),
    \[\boxed{k=-\frac{2}{\pi}}\]
🎯 Exam Significance
Exam Significance
›

This question reinforces the importance of identifying the branch that contains the point of continuity. Since the condition is \(x\le\pi\), the value \(f(\pi)\) comes from \(kx+1\).

For full marks, write the LHL, RHL and \(f(\pi)\), then equate them.

Significance for Competitive Entrance Examinations

The problem can be solved quickly by equating the values of the two branches at the joining point:

\[ k\pi+1=\cos\pi=-1. \]

Hence,

\[ k\pi=-2 \quad\Longrightarrow\quad \boxed{k=-\frac2\pi}. \]
🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  6 points

  1. The only point requiring examination is \(x=\pi\).

  2. \(\displaystyle \lim_{x\to\pi^-}f(x)=k\pi+1\).

  3. \(\displaystyle \lim_{x\to\pi^+}f(x)=\cos\pi=-1\).

  4. Since \(\pi\le\pi\), \(\displaystyle f(\pi)=k\pi+1\).

  5. Continuity requires \(k\pi+1=-1\).

  6. Therefore,

    \[ \boxed{k=-\frac2\pi}. \]
← Q27
28 / 34  ·  82%
Q29 →
Q29
NUMERIC3 marks
Find the value of \(k\) so that the function \(f\) is continuous at \(x=5\), where \[ f(x)= \begin{cases} kx+1, & x\le5,\\[4pt] 3x-5, & x>5 \end{cases} \]
📘 Concept & Theory
Concept/Theory
›

For a function \(f(x)\) to be continuous at \(x=a\), we require

\[\boxed{\lim_{x\to a^-}f(x)=\lim_{x\to a^+}f(x)=f(a)}.\]

Here, th definition of \(f(x)\) changes at \(x=5\), so \(x=5\) is the point that must be checked.

Since the first branch has the condition \(x\le5\), the point \(x=5\) belongs to the first branch. Therefore,

\[(5)=5k+1\]
🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Evaluate the left-hand limit at \(x=5\).

  2. Evaluate the right-hand limit at \(x=5\).

  3. Evaluate \(f(5)\) using the branch containing \(x=5\).

  4. Apply the continuity condition and solve for \(k\).

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  6 steps
  1. Left-Hand Limit at \(x=5\)
  2. As \(x\to5^-\), we have \(x<5\), so the first branch applies:
    \[ \begin{aligned} \lim_{x\to5^-}f(x) &=\lim_{x\to5^-}(kx+1)\\ &=5k+1 \end{aligned} \]
    \[ \boxed{\lim_{x\to5^-}f(x)=5k+1} \]
  3. Right-Hand Limit at \(x=5\)
  4. As \(x\to5^+\), we have \(x>5\), so the second branch applies:
    \[ \begin{aligned} \lim_{x\to5^+}f(x) &=\lim_{x\to5^+}(3x-5)\\ &=3(5)-5\\ &=15-5\\ &=10 \end{aligned} \]
    \[ \boxed{\lim_{x\to5^+}f(x)=10} \]
  5. Function Value at \(x=5\)
  6. Since \(5\le5\), the first branch applies:
    \[ \begin{aligned} f(5) &=5k+1 \end{aligned} \]
    \[ \boxed{f(5)=5k+1} \]
  7. Apply the Continuity Criterion
  8. For continuity at \(x=5\), we require
    \[\lim_{x\to5^-}f(x)=\lim_{x\to5^+}f(x)=f(5)\]
  9. Therefore,
    \[\begin{aligned}5k+1&=10\\5k&=10-1\\k&=\dfrac95\end{aligned}\]
  10. \[\boxed{k=\frac95}\]
🎯 Exam Significance
Exam Significance
›

This question tests the standard method for finding an unknown parameter in a piecewise function. At the joining point, calculate LHL, RHL and the function value, then equate them.

Since \(x=5\) belongs to the first branch, it is essential to use \(kx+1\) for \(f(5)\).

Significance for Competitive Entrance Examinations

The problem can be solved rapidly by matching the two branch values at the joining point:

\[ 5k+1=3(5)-5=10. \]

Thus,

\[ 5k=9 \quad\Longrightarrow\quad \boxed{k=\frac95}. \]
🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  6 points

  1. The only point requiring examination is \(x=5\).

  2. \(\displaystyle \lim_{x\to5^-}f(x)=5k+1\).

  3. \(\displaystyle \lim_{x\to5^+}f(x)=10\).

  4. \(\displaystyle f(5)=5k+1\).

  5. Continuity requires \(5k+1=10\).

  6. Therefore,

    \[ \boxed{k=\frac95}. \]
← Q28
29 / 34  ·  85%
Q30 →
Q30
NUMERIC3 marks
Find the values of \(a\) and \(b\) such that the function defined by \[ f(x)= \begin{cases} 5, & x\le2,\\[4pt] ax+b, & 2 < x < 10,\\[4pt] 21, & x\ge10 \end{cases} \] is continuous.
📘 Concept & Theory
Concept/Theory
›

The function has three branches. The first and third branches are constant functions, so they are continuous on their respective intervals. The middle branch \(ax+b\) is a linear function and is also continuous wherever it is defined.

Therefore, the only possible points of discontinuity are the joining points:

\[ x=2\quad\text{and}\quad x=10. \]

For continuity at any point \(x=c\), we require

\[\boxed{\lim_{x\to c^-}f(x)=\lim_{x\to c^+}f(x)=f(c)}\]
🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Use continuity at \(x=2\) to obtain one equation in \(a\) and \(b\).

  2. Use continuity at \(x=10\) to obtain a second equation.

  3. Solve the resulting simultaneous equations.

  4. Verify the values of \(a\) and \(b\).

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  15 steps
  1. Continuity at \(x=2\)
  2. Left-Hand Limit
  3. For \(x\to2^-\), we have \(x < 2\), so the first branch applies:
    \[\boxed{\lim_{x\to2^-}f(x)=5}\]
  4. Right-Hand Limit
  5. For \(x\to2^+\), we have \( 2 < x < 10\), so the middle branch applies:
    \[ \begin{aligned} \lim_{x\to2^+}f(x) &=\lim_{x\to2^+}(ax+b)\\ &=2a+b. \end{aligned} \]
    \[ \boxed{ \lim_{x\to2^+}f(x)=2a+b } \]
  6. Function Value at \(x=2\)
  7. Since \(2\le2\), the first branch applies:
    \[\boxed{f(2)=5}\]
  8. Continuity Condition at \(x=2\)
  9. For continuity,
    \[\lim_{x\to2^-}f(x)=\lim_{x\to2^+}f(x)\]
  10. Hence,
    \[5=2a+b\]
  11. Therefore,
    \[\boxed{2a+b=5}\tag{1}\]
  12. Continuity at \(x=10\)
  13. Left-Hand Limit
  14. For \(x\to10^-\), we have \(2 < x < 10\), so the middle branch applies:
    \[ \begin{aligned} \lim_{x\to10^-}f(x) &=\lim_{x\to10^-}(ax+b)\\ &=10a+b \end{aligned} \]
    \[ \boxed{ \lim_{x\to10^-}f(x)=10a+b } \]
  15. Right-Hand Limit
  16. For \(x\to10^+\), we have \(x>10\), so the third branch applies:
    \[\boxed{\lim_{x\to10^+}f(x)=21}\]
  17. Function Value at \(x=10\)
  18. Since \(10\ge10\), the third branch applies:
    \[\boxed{f(10)=21}\]
  19. Continuity Condition at \(x=10\)
  20. For continuity,
    \[\lim_{x\to10^-}f(x)=\lim_{x\to10^+}f(x)\]
  21. Therefore,
    \[10a+b=21\]
  22. Hence,
    \[\boxed{10a+b=21}\tag{2}\]
  23. Solve the Simultaneous Equations
  24. From equations (1) and (2),
    \[2a+b=5\]
    \[10a+b=21\]
  25. Subtracting the first equation from the second,
    \[\begin{aligned}(10a+b)-(2a+b)&=21-5=\\8a&=16\\a&=2\end{aligned}\]
  26. Substitute \(a=2\) into equation (1):
    \[\begin{aligned}2(2)+b&=5\\4+b&=5\\b&=1\end{aligned}\]
🎯 Exam Significance
Exam Significance
›

This is an important type of question in which two unknown constants require two continuity conditions. The key is to identify every point where the definition of the function changes.

Here, the two joining points \(2\) and \(10\) provide the two equations:

\[ \boxed{2a+b=5} \]
\[ \boxed{10a+b=21}. \]

Solving these equations gives the required values of \(a\) and \(b\).

Significance for Competitive Entrance Examinations

For a quick solution, continuity means that the middle linear branch must meet the neighbouring constant branches without a jump.

At \(x=2\):

\[ 2a+b=5. \]

At \(x=10\):

\[ 10a+b=21. \]

Subtracting gives

\[ 8a=16\Rightarrow a=2, \]

and hence

\[ b=1. \]
🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  6 points

  1. Possible discontinuities occur at the joining points \(x=2\) and \(x=10\).

  2. Continuity at \(x=2\) gives

    \[ 2a+b=5. \]

  3. Continuity at \(x=10\) gives

    \[ 10a+b=21. \]

  4. Subtracting the equations gives

    \[ a=2. \]

  5. Substitution gives

    \[ b=1. \]

  6. For these values, the middle branch is \(2x+1\), which joins the constant branches continuously.
← Q29
30 / 34  ·  88%
Q31 →
Q31
NUMERIC3 marks
Show that the function defined by \[ f(x)=\cos(x^2) \] is a continuous function.
📘 Concept & Theory
Concept/Theory
›
A function \(f\) is continuous at a point \(x=a\) if
\[ \lim_{x\to a}f(x)=f(a). \]
Equivalently,
\[ \lim_{x\to a^-}f(x) = \lim_{x\to a^+}f(x) = f(a). \]

We use the following standard results:

  • The polynomial function \(x^2\) is continuous for every \(x\in\mathbb R\).
  • The trigonometric function \(\cos x\) is continuous for every \(x\in\mathbb R\).
  • The composition of two continuous functions is continuous wherever the composition is defined.
🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Take an arbitrary point \(x=a\in\mathbb R\).

  2. Evaluate \(f(a)\).

  3. Evaluate \(\displaystyle\lim_{x\to a}f(x)\).

  4. Compare the limit with \(f(a)\).

  5. Conclude continuity for every real \(a\).

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  6 steps
  1. Consider an arbitrary point
  2. Let \(a\) be any real number. Then
    \[ f(a)=\cos(a^2) \]
  3. Evaluate the limit
  4. Consider
    \[ \lim_{x\to a}f(x) = \lim_{x\to a}\cos(x^2). \]
  5. Since \(x^2\) is continuous at \(x=a\),
    \[ \lim_{x\to a}x^2=a^2. \]
  6. Also, \(\cos x\) is continuous at \(x=a^2\). Therefore,
    \[ \lim_{x\to a}\cos(x^2) = \cos\left(\lim_{x\to a}x^2\right). \]
    Hence,
    \[ \lim_{x\to a}\cos(x^2) = \cos(a^2). \]
  7. Compare with the function value
  8. We have
    \[ \lim_{x\to a}f(x)=\cos(a^2) \]
    and
    \[ f(a)=\cos(a^2). \]
    Therefore,
    \[ \boxed{\lim_{x\to a}f(x)=f(a)}. \]
  9. Since \(a\) was an arbitrary real number, the function is continuous at every \(a\in\mathbb R\).
🎯 Exam Significance
Exam Significance
›
  • Remember that \(x^2\) is a polynomial and hence continuous everywhere.
  • Remember that \(\cos x\) is continuous for every real \(x\).
  • The composition of continuous functions is continuous.
  • For a short-answer question, the composition argument provides a concise and rigorous proof.
Significance for Competitive Entrance Examinations

This question tests recognition of standard continuity results rather than lengthy limit calculations. A key pattern is:

\[ \text{continuous function}\circ\text{continuous function} \quad\Longrightarrow\quad \text{continuous function}. \]
Thus expressions such as
\[ \sin(x^3),\qquad \cos(\sqrt{x^2+1}),\qquad e^{\sin x} \]
can often be classified immediately by identifying the continuity of their component functions and their domains.

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  4 points

  1. \(x^2\) is continuous on \(\mathbb R\).

  2. \(\cos x\) is continuous on \(\mathbb R\).

  3. The composition \(\cos(x^2)\) is therefore continuous on \(\mathbb R\).

  4. For every \(a\in\mathbb R\),

    \[ \lim_{x\to a}\cos(x^2)=\cos(a^2)=f(a). \]
← Q30
31 / 34  ·  91%
Q32 →
Q32
NUMERIC3 marks
Show that the function defined by \[ f(x)=|\cos x| \] is a continuous function.
📘 Concept & Theory
Concept/Theory
›

A function \(f\) is continuous at \(x=a\) if

\[ \lim_{x\to a}f(x)=f(a). \]

We use the following standard results:

  • \(\cos x\) is continuous for every \(x\in\mathbb R\).
  • The absolute-value function \(g(x)=|x|\) is continuous for every \(x\in\mathbb R\).
  • The composition of two continuous functions is continuous wherever it is defined.
🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Take an arbitrary point \(x=a\in\mathbb R\).

  2. Find \(f(a)\).

  3. Evaluate \(\displaystyle\lim_{x\to a}f(x)\).

  4. Use the continuity of \(\cos x\) and \(|x|\).

  5. Verify that the limit equals the function value.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  7 steps
  1. Function value at \(x=a\)
  2. Let \(a\in\mathbb R\). Then
    \[ f(a)=|\cos a|. \]
  3. Evaluate the limit
  4. Consider
    \[ \lim_{x\to a}f(x) = \lim_{x\to a}|\cos x| \]
  5. Since \(\cos x\) is continuous at \(x=a\),
    \[ \lim_{x\to a}\cos x=\cos a. \]
  6. Since the absolute-value function is continuous,
    \[ \lim_{x\to a}|\cos x| = \left|\lim_{x\to a}\cos x\right|. \]
    Therefore,
    \[ \lim_{x\to a}|\cos x| = |\cos a|. \]
  7. Compare the limit with the function value
  8. We have
    \[ \lim_{x\to a}f(x)=|\cos a| \]
    and
    \[ f(a)=|\cos a|. \]
  9. Hence,
    \[\boxed{\lim_{x\to a}f(x)=f(a)}\]
  10. Thus, \(f(x)=|\cos x|\) is continuous at the arbitrary point \(a\). Since \(a\) was arbitrary, \(f\) is continuous for every \(x\in\mathbb R\).
🎯 Exam Significance
Exam Significance
›
  • Remember that \(\cos x\) is continuous on \(\mathbb R\).
  • Remember that \(|x|\) is continuous on \(\mathbb R\).
  • The composition of continuous functions is continuous.
  • Therefore, \(|\cos x|\) is continuous everywhere, including at points where \(\cos x=0\).
Significance for Competitive Entrance Examinations

The key skill tested here is recognizing continuity under standard operations and composition. In particular,

\[ g(x)\text{ continuous}\quad\Longrightarrow\quad |g(x)|\text{ continuous}. \]
Thus, once \(g(x)\) is known to be continuous on its domain, expressions such as
\[ |\sin x|,\qquad |\cos x|,\qquad |\sin x+\cos x| \]
are immediately recognized as continuous wherever the inner function is continuous.
🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  4 points

  1. \(\cos x\) is continuous on \(\mathbb R\).

  2. \(|x|\) is continuous on \(\mathbb R\).

  3. The composition \( |\cos x| \) is therefore continuous on \(\mathbb R\).

  4. For every \(a\in\mathbb R\),

    \[ \lim_{x\to a}|\cos x| = |\cos a| = f(a). \]
← Q31
32 / 34  ·  94%
Q33 →
Q33
NUMERIC3 marks
Examine whether the function \[ f(x)=\sin|x| \] is a continuous function.
📘 Concept & Theory
Concept/Theory
›

A function \(f\) is continuous at \(x=a\) if

\[ \lim_{x\to a}f(x)=f(a). \]

We use the following standard results:

  • The absolute-value function \(g(x)=|x|\) is continuous for every \(x\in\mathbb R\).
  • The sine function \(h(x)=\sin x\) is continuous for every \(x\in\mathbb R\).
  • The composition of two continuous functions is continuous.
🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Write \(f(x)=\sin|x|\) as a composition of two functions.

  2. Establish the continuity of the inner function \(g(x)=|x|\).

  3. Establish the continuity of the outer function \(h(x)=\sin x\).

  4. Apply the composition theorem for continuous functions.

  5. For a direct verification, check \(\lim_{x\to a}f(x)=f(a)\).

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  14 steps
  1. Method 1: Using the Composition Theorem
  2. Let
    \[ g(x)=|x| \]
    and
    \[ h(x)=\sin x. \]
    Then
    \[ f(x)=\sin|x|=h(g(x)). \]
  3. The function \(g(x)=|x|\) is continuous on \(\mathbb R\), and the function \(h(x)=\sin x\) is also continuous on \(\mathbb R\).
  4. Therefore, their composition
    \[ h(g(x))=\sin|x| \]
    is continuous on \(\mathbb R\).
  5. Method 2: Direct Verification
  6. Let \(a\in\mathbb R\) be arbitrary. Then
    \[ f(a)=\sin|a| \]
  7. Now,
    \[ \begin{aligned} \lim_{x\to a}f(x) &=\lim_{x\to a}\sin|x|\\ &=\sin\left(\lim_{x\to a}|x|\right) \end{aligned} \]
  8. Since \(|x|\) is continuous at \(a\),
    \[ \lim_{x\to a}|x|=|a|. \]
    Hence,
    \[ \lim_{x\to a}\sin|x| = \sin|a|. \]
  9. But
    \[ f(a)=\sin|a| \]
    Therefore,
    \[ \boxed{\lim_{x\to a}f(x)=f(a)}. \]
  10. Since \(a\) was arbitrary, \(f(x)=\sin|x|\) is continuous at every real number.
  11. Special Check at \(x=0\)
  12. Although the absolute-value function changes its algebraic form at \(x=0\), it remains continuous there. We can verify the continuity of \(f\) at this point directly.
  13. The function value is
    \[ f(0)=\sin|0|=0. \]
  14. The left-hand limit is
    \[ \begin{aligned} \lim_{x\to0^-}\sin|x| &=\lim_{x\to0^-}\sin(-x)\\ &=\sin 0\\ &=0 \end{aligned} \]
  15. The right-hand limit is
    \[ \begin{aligned} \lim_{x\to0^+}\sin|x| &=\lim_{x\to0^+}\sin x\\ &=\sin0\\ &=0. \end{aligned} \]
  16. Thus,
    \[ \lim_{x\to0^-}f(x) = \lim_{x\to0^+}f(x) = f(0)=0. \]
  17. Hence \(f\) is continuous at \(x=0\).
🎯 Exam Significance
Exam Significance
›
  • Know that \(|x|\) is continuous on \(\mathbb R\).
  • Know that \(\sin x\) is continuous on \(\mathbb R\).
  • The composition of continuous functions is continuous.
  • For a direct proof, show
    \[ \lim_{x\to a}\sin|x|=\sin|a|=f(a). \]
Significance for Competitive Entrance Examinations

This problem reinforces an important recognition rule: if \(g(x)\) is continuous and \(F\) is continuous, then \(F(g(x))\) is continuous. Thus,

\[ |\sin x|,\qquad \sin|x|,\qquad \cos|x| \]
are all continuous on \(\mathbb R\).

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  4 points

  1. \(|x|\) is continuous for every \(x\in\mathbb R\).

  2. \(\sin x\) is continuous for every \(x\in\mathbb R\).

  3. Therefore, \(\sin|x|\) is continuous on \(\mathbb R\).

  4. In particular, it is also continuous at the potentially critical point \(x=0\).
← Q32
33 / 34  ·  97%
Q34 →
Q34
NUMERIC3 marks
Find all the points of discontinuity of the function \(f(x)=|x|-|x+1|\).
📘 Concept & Theory
Concept/Theory
›

A function \(f\) is continuous at \(x=a\) if

\[ \lim_{x\to a^-}f(x) = \lim_{x\to a^+}f(x) = f(a). \]

The absolute-value expressions in

\[ f(x)=|x|-|x+1| \]
can change their algebraic form when their arguments become zero:
\[ x=0 \]
and
\[ x+1=0\quad\Rightarrow\quad x=-1. \]

Thus, \(x=-1\) and \(x=0\) are the only possible points where the algebraic form changes. We examine these points.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Identify the critical points \(x=-1\) and \(x=0\).

  2. Write the function in piecewise form on the relevant intervals.

  3. Check continuity at \(x=-1\).

  4. Check continuity at \(x=0\).

  5. Conclude all points of discontinuity.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  16 steps
  1. Write the Function in Piecewise Form
  2. We consider the intervals determined by \(-1\) and \(0\)
  3. For \(x < -1\) Since \(x<0\),
    \[ |x|=-x. \]
    Also, \(x+1<0\), so
    \[ |x+1|=-(x+1)=-x-1. \]
    Therefore,
    \[ f(x)=(-x)-(-x-1)=1. \]
  4. For \(-1\le x<0\) Here \(x<0\), so
    \[ |x|=-x. \]
    But \(x+1\ge0\), so
    \[ |x+1|=x+1. \]
    Hence,
    \[ f(x)=(-x)-(x+1)=-2x-1. \]
  5. For \(x\ge0\) Here \(x\ge0\) and \(x+1>0\). Thus,
    \[ |x|=x,\qquad |x+1|=x+1. \]
    Therefore,
    \[ f(x)=x-(x+1)=-1. \]
  6. Hence,
    \[ f(x)= \begin{cases} 1, & x<-1,\\[4pt] -2x-1, & -1\le x<0,\\[4pt] -1, & x\ge0. \end{cases} \]
  7. Check Continuity at \(x=-1\)
  8. Since \(-1\) belongs to the middle branch,
    \[ f(-1)=-2(-1)-1=1. \]
  9. The left-hand limit uses the branch \(x<-1\):
    \[ \lim_{x\to-1^-}f(x) = \lim_{x\to-1^-}1 = 1. \]
  10. The right-hand limit uses the branch \(-1\le x<0\):
    \[ \begin{aligned} \lim_{x\to-1^+}f(x) &=\lim_{x\to-1^+}(-2x-1)\\ &=-2(-1)-1\\ &=1 \end{aligned} \]
  11. Therefore,
    \[ \lim_{x\to-1^-}f(x) = \lim_{x\to-1^+}f(x) = f(-1)=1. \]
  12. Hence, \(f\) is continuous at \(x=-1\).
  13. Check Continuity at \(x=0\)
  14. Since \(0\) belongs to the last branch,
    \[ f(0)=-1. \]
  15. The left-hand limit uses the branch \(-1\le x<0\):
    \[ \begin{aligned} \lim_{x\to0^-}f(x) &=\lim_{x\to0^-}(-2x-1)\\ &=-1 \end{aligned} \]
  16. The right-hand limit uses the branch \(x\ge0\):
    \[ \lim_{x\to0^+}f(x) = \lim_{x\to0^+}(-1) = -1 \]
  17. Thus,
    \[ \lim_{x\to0^-}f(x) = \lim_{x\to0^+}f(x) = f(0)=-1. \]
  18. Hence, \(f\) is continuous at \(x=0\).
  19. Check Other Points
  20. On each of the intervals
    \[ (-\infty,-1),\qquad (-1,0),\qquad (0,\infty), \]
    the function is either constant or linear. Therefore, it is continuous throughout each of these intervals.
  21. Since the only possible transition points, \(x=-1\) and \(x=0\), have also been verified to be continuous, there are no points of discontinuity.
🎯 Exam Significance
Exam Significance
›
  • For expressions containing absolute values, first identify where the arguments of the absolute values become zero.
  • These points are candidates for discontinuity, not necessarily actual discontinuities.
  • Always verify
    \[ \mathrm{LHL}=\mathrm{RHL}=f(a). \]
  • A piecewise constant or linear function is continuous within each open interval of its definition.
Significance for Competitive Entrance Examinations

For absolute-value functions, quickly identify the breakpoints by solving the equations inside the modulus:

\[ x=0,\qquad x+1=0. \]
Then construct the sign chart or piecewise form and check only those breakpoints. This technique is useful for more complicated functions involving multiple absolute values.

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  5 points

  1. The critical points are \(x=-1\) and \(x=0\).

  2. At \(x=-1\),

    \[ \mathrm{LHL}=\mathrm{RHL}=f(-1)=1. \]

  3. At \(x=0\),

    \[ \mathrm{LHL}=\mathrm{RHL}=f(0)=-1. \]

  4. The function is continuous at both critical points.

  5. Hence, the function is continuous on the entire real line.
← Q33
34 / 34  ·  100%
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ACADEMIA AETERNUM तमसो मा ज्योतिर्गमय · Est. 2025
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NCERT Class 12 Continuity: \(|x|-|x+1|\) Solution
NCERT Class 12 Continuity: \(|x|-|x+1|\) Solution — Complete Notes & Solutions · academia-aeternum.com
In NCERT Class 12 Mathematics Chapter 5: Continuity and Differentiability, understanding how absolute-value functions behave at their critical points is essential for mastering continuity problems. In this question, we examine the function \(f(x)=|x|-|x+1|\) and determine whether it has any points of discontinuity. Since absolute-value expressions can change their algebraic form when their arguments become zero, the points \(x=0\) and \(x=-1\) require special attention. The function is first…
🎓 Class 12 📐 Mathematics 📖 NCERT ✅ Free Access 🏆 CBSE · JEE
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    Frequently Asked Questions

    The function is \(f(x)=|x|-|x+1|\).

    The possible critical points are \(x=-1\) and \(x=0\), where the expressions inside the absolute values become zero.

    Yes. At \(x=-1\), the left-hand limit, right-hand limit, and function value are all equal to \(1\), so the function is continuous there.

    Yes. At \(x=0\), the left-hand limit, right-hand limit, and function value are all equal to \(-1\), so the function is continuous there.

    No. A change in algebraic form only identifies a possible point of discontinuity. The left-hand limit, right-hand limit, and function value must be compared.

    The piecewise form is \(f(x)=1\) for \(x<-1\), \(f(x)=-2x-1\) for \(-1\le x<0\), and \(f(x)=-1\) for \(x\ge0\).

    A function is continuous at \(x=a\) if \(\lim_{x\to a^-}f(x)=\lim_{x\to a^+}f(x)=f(a)\).

    No. The function is continuous at both critical points and hence has no points of discontinuity.

    Yes. The function is continuous for every \(x\in\mathbb R\).

    There are no points of discontinuity; \(f(x)=|x|-|x+1|\) is continuous on \(\mathbb R\).

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