Continuity and Differentiability: CBSE Class 12 Previous Year Questions

33 different questions from Continuity and Differentiability (NCERT Chapter 5) asked in CBSE Class 12 Maths board exams 2026. Pick a mark group to practise, each with answers and step-by-step solutions.

1 mark questions12 questions 2 marks questions12 questions 3 marks questions3 questions 4 marks questions1 questions 5 marks questions5 questions

Most asked Continuity and Differentiability questions

Q71 markMCQContinuity and Differentiability CBSE 2026 · 65/3/1

The value of k for which the function $f (x) = {x^{2} sin \frac{1}{x}, k (x + 1), x \neq = 0 x = 0$ is a continuous function, is :

(A) $\frac{1}{4}$
(B) $2$
(C) $\frac{1}{2}$
(D) $0$

Show answer & solution

Answer: (D)

0

$∣ x^{2} sin \frac{1}{x} ∣ \leq x^{2} \to 0$ , so $lim_{x \to 0} f (x) = 0$ .
$f (0) = k (0 + 1) = k$ .
Continuity at 0 needs $k = 0$ .

Also asked in: 2026 65/3/2, 2026 65/3/3

Q51 markMCQContinuity and Differentiability CBSE 2026 · 65/4/1

Differential of $e^{e^{x}}$ with respect to x is :

(A) $lo g x$
(B) $e^{e^{x}}$
(C) $e^{x} e^{e^{x}}$
(D) $(e^{x})^{2}$

Show answer & solution

Answer: (C)

e^{x} e^{e^{x}}

By the chain rule, $\frac{d}{d x} e^{e^{x}} = e^{e^{x}} \cdot \frac{d}{d x} (e^{x})$ .
$= e^{e^{x}} \cdot e^{x}$ .

Also asked in: 2026 65/4/2, 2026 65/4/3

Q71 markMCQContinuity and Differentiability CBSE 2026 · 65/4/1

If $f (x) = {\frac{s i n x}{x} + cos x, k, x \neq = 0 x = 0$ is continuous at x = 0, then the value of k is :

(A)0
(B) $- 2$
(C) $- 1$
(D)2

Show answer & solution

Answer: (D) 2

$lim_{x \to 0} (\frac{s i n x}{x} + cos x) = 1 + 1 = 2$ .
For continuity at $x = 0$ , $k = f (0) = 2$ .

Also asked in: 2026 65/4/2, 2026 65/4/3

Q81 markMCQContinuity and Differentiability CBSE 2026 · 65/4/1

The greatest integer function, $f (x) = [x]$ , $0 < x < 3$ is not differentiable at how many points ?

(A)At only one point
(B)At only two points
(C)At no point
(D)At three points

Show answer & solution

Answer: (B) At only two points

$[x]$ jumps at every integer, so it is discontinuous (hence not differentiable) at integers.
In $0 < x < 3$ the integers are $x = 1$ and $x = 2$ .
Elsewhere $f$ is constant on each piece and differentiable. So it fails at exactly two points.

Also asked in: 2026 65/4/2, 2026 65/4/3

Q61 markMCQContinuity and Differentiability CBSE 2026 · 65/5/1

If $f (x) = {\frac{x ^{2} - 4 x - 5}{x + 1}, k, x \neq = - 1 x = - 1$
is continuous at $x = - 1$ , then the value of k is :

(A)Any real value
(B)6
(C) $- 1$
(D) $- 6$

Show answer & solution

Answer: (D)

- 6

For $x \neq = - 1$ , $\frac{x ^{2} - 4 x - 5}{x + 1} = \frac{( x - 5 ) ( x + 1 )}{x + 1} = x - 5$ .
$lim_{x \to - 1} f (x) = - 1 - 5 = - 6$ .
Continuity at $x = - 1$ needs $k = - 6$ .

Also asked in: 2026 65/5/2, 2026 65/5/3

Q81 markMCQContinuity and Differentiability CBSE 2026 · 65/5/1

Derivative of $cos^{- 1} (\frac{s i n x + c o s x}{2})$ , $- \frac{π}{4} < x < \frac{π}{4}$ with respect to x is :

(A) $- 1$
(B)1
(C) $\frac{π}{4}$
(D) $- \frac{π}{4}$

Show answer & solution

Answer: (A)

- 1

Let $u = \frac{s i n x + c o s x}{2}$ , so $u^{'} = \frac{c o s x - s i n x}{2}$ .
$1 - u^{2} = 1 - \frac{1 + s i n 2 x}{2} = \frac{( c o s x - s i n x ) ^{2}}{2}$ , and $cos x - sin x > 0$ for $- \frac{π}{4} < x < \frac{π}{4}$ .
$\frac{d}{d x} cos^{- 1} u = - \frac{u ^{'}}{1 - u ^{2}} = - \frac{( c o s x - s i n x ) / 2}{( c o s x - s i n x ) / 2} = - 1$ .

Also asked in: 2026 65/5/2, 2026 65/5/3

Q212 marksVery Short AnswerContinuity and Differentiability CBSE 2026 · 65/1/1

Check whether function f(x) defined as
$f (x) = {\frac{∣ x - 3∣}{2 ( x - 3 )}, \frac{x - 6}{6}, x < 3 x \geq 3$ is continuous at $x = 3$ or not ?

Show answer & solution

Answer:

f

is continuous at

x = 3

For $x < 3$ , $∣ x - 3∣ = - (x - 3)$ , so $f (x) = - \frac{1}{2}$ .
LHL $= lim_{x \to 3^{-}} f (x) = - \frac{1}{2}$ .
RHL $= lim_{x \to 3^{+}} \frac{x - 6}{6} = \frac{- 3}{6} = - \frac{1}{2}$ .
$f (3) = \frac{3 - 6}{6} = - \frac{1}{2}$ .
LHL = RHL = $f (3)$ , so $f$ is continuous at $x = 3$ .

Also asked in: 2026 65/1/2, 2026 65/1/3

Q21 (OR) (OR)2 marksVery Short AnswerContinuity and Differentiability CBSE 2026 · 65/1/1

If $3 (x^{2} + y^{2}) = 4 x y$ , then find $\frac{d y}{d x}$ at $(\frac{1}{2}, \frac{3}{2})$ .

Show answer & solution

Answer:

\frac{d y}{d x} = 3

Differentiate w.r.t. $x$ : $3 (2 x + 2 y \frac{d y}{d x}) = 4 y + 4 x \frac{d y}{d x}$ .
$\frac{d y}{d x} (23 y - 4 x) = 4 y - 23 x \Rightarrow \frac{d y}{d x} = \frac{2 y - 3 x}{3 y - 2 x}$ .
At $(\frac{1}{2}, \frac{3}{2})$ : numerator $= 3 - \frac{3}{2} = \frac{3}{2}$ , denominator $= \frac{3}{2} - 1 = \frac{1}{2}$ .
$\frac{d y}{d x} = 3$ .

Also asked in: 2026 65/1/2, 2026 65/1/3

Q222 marksVery Short AnswerContinuity and Differentiability CBSE 2026 · 65/4/1

Show that the function $f (x) = {\frac{c o s x}{- x + \frac{π}{2}}, 1, x \neq = \frac{π}{2} x = \frac{π}{2}$ is continuous at $x = \frac{π}{2}$ .

Show answer & solution

Answer: Proved.

Put $x = \frac{π}{2} - h$ ; as $x \to \frac{π}{2}$ , $h \to 0$ .
$lim_{x \to \frac{π}{2}} \frac{c o s x}{\frac{π}{2} - x} = lim_{h \to 0} \frac{c o s ( \frac{π}{2} - h )}{h} = lim_{h \to 0} \frac{s i n h}{h} = 1$ .
$f (\frac{π}{2}) = 1$ equals the limit, so f is continuous at $x = \frac{π}{2}$ .

Also asked in: 2026 65/4/2, 2026 65/4/3

Q22 (OR) (OR)2 marksVery Short AnswerContinuity and Differentiability CBSE 2026 · 65/4/1

Find whether the function $f (x) = {x - 1, 2 x - 3, x < 2 x \geq 2$ at x = 2 is differentiable or not.

Show answer & solution

Answer: Not differentiable at x = 2.

$f (2) = 1$ ; $lim_{x \to 2^{-}} (x - 1) = 1$ , so f is continuous at 2.
LHD $= lim_{h \to 0} \frac{f ( 2 - h ) - f ( 2 )}{- h} = lim_{h \to 0} \frac{( 1 - h ) - 1}{- h} = 1$ .
RHD $= lim_{h \to 0} \frac{f ( 2 + h ) - f ( 2 )}{h} = lim_{h \to 0} \frac{( 1 + 2 h ) - 1}{h} = 2$ .
LHD $\neq =$ RHD, so f is not differentiable at $x = 2$ .

Also asked in: 2026 65/4/2, 2026 65/4/3

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