CBSE Maths › Class 12 PYQs › Continuity and Differentiability

Continuity and Differentiability: 2 marks Questions (CBSE Class 12)

12 different 2 marks questions on Continuity and Differentiability from CBSE Class 12 Maths board exams 2026, newest first.

1 mark (12)2 marks (12)3 marks (3)4 marks (1)5 marks (5)

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 · 65/2/1

If $x = a sin^{3} t$ , $y = b cos^{3} t$ , then find $\frac{d y}{d x}$ at $t = \frac{π}{4}$ .

Show answer & solution

Answer:

- \frac{b}{a}

$\frac{d x}{d t} = 3 a sin^{2} t cos t$ , $\frac{d y}{d t} = - 3 b cos^{2} t sin t$ .
$\frac{d y}{d x} = \frac{- 3 b c o s ^{2} t s i n t}{3 a s i n ^{2} t c o s t} = - \frac{b}{a} cot t$ .
At $t = \frac{π}{4}$ : $\frac{d y}{d x} = - \frac{b}{a}$ .

Q252 marksVery Short AnswerContinuity and Differentiability CBSE 2026 · 65/2/2

If $x = sin t - cos t$ , $y = sin t cos t$ , find $\frac{d y}{d x}$ at $t = \frac{π}{4}$ .

Show answer & solution

Answer: 0

$\frac{d x}{d t} = cos t + sin t$ ; $y = \frac{1}{2} sin 2 t \Rightarrow \frac{d y}{d t} = cos 2 t$ .
$\frac{d y}{d x} = \frac{c o s 2 t}{c o s t + s i n t}$ .
At $t = \frac{π}{4}$ : $\frac{d y}{d x} = \frac{c o s \frac{π}{2}}{2} = 0$ .

Q232 marksVery Short AnswerContinuity and Differentiability CBSE 2026 · 65/2/3

If $x = e^{s i n^{- 1} t}$ , $y = e^{c o s^{- 1} t}$ ,
find $\frac{d y}{d x}$ at $t = \frac{1}{2}$

Show answer & solution

Answer:

- 1

$\frac{d x}{d t} = \frac{e ^{s i n^{- 1} t}}{1 - t ^{2}}$ , $\frac{d y}{d t} = - \frac{e ^{c o s^{- 1} t}}{1 - t ^{2}}$ .
$\frac{d y}{d x} = - \frac{e ^{c o s^{- 1} t}}{e ^{s i n^{- 1} t}}$ .
At $t = \frac{1}{2}$ : $sin^{- 1} t = cos^{- 1} t = \frac{π}{4}$ , so $\frac{d y}{d x} = - 1$ .

Q212 marksVery Short AnswerContinuity and Differentiability CBSE 2026 · 65/3/1

If $x = t + \frac{1}{t}$ and $y = t - \frac{1}{t}$ , find $\frac{d y}{d x}$ at $t = 2$ .

Show answer & solution

Answer:

\frac{5}{3}

$\frac{d x}{d t} = 1 - \frac{1}{t ^{2}}$ , $\frac{d y}{d t} = 1 + \frac{1}{t ^{2}}$ .
$\frac{d y}{d x} = \frac{1 + \frac{1}{t ^{2}}}{1 - \frac{1}{t ^{2}}} = \frac{t ^{2} + 1}{t ^{2} - 1}$ .
At $t = 2$ : $\frac{d y}{d x} = \frac{5}{3}$ .

Q242 marksVery Short AnswerContinuity and Differentiability CBSE 2026 · 65/3/2

If $x = e^{t + \frac{1}{t}}$ and $y = e^{t - \frac{1}{t}}$ , then find $\frac{d y}{d x}$ at $t = - 2$ .

Show answer & solution

Answer:

\frac{5 e}{3}

$\frac{d x}{d t} = e^{t + \frac{1}{t}} (1 - \frac{1}{t ^{2}})$ , $\frac{d y}{d t} = e^{t - \frac{1}{t}} (1 + \frac{1}{t ^{2}})$ .
$\frac{d y}{d x} = e^{- \frac{2}{t}} \cdot \frac{t ^{2} + 1}{t ^{2} - 1}$ .
At $t = - 2$ : $\frac{d y}{d x} = e^{1} \cdot \frac{5}{3} = \frac{5 e}{3}$ .

Q222 marksVery Short AnswerContinuity and Differentiability CBSE 2026 · 65/3/3

If $x = e^{s i n^{- 1} t}$ and $y = e^{c o s^{- 1} t}$ , then find $\frac{d y}{d x}$ at $t = \frac{1}{2}$ .

Show answer & solution

Answer:

- 1

$\frac{d x}{d t} = \frac{e ^{s i n^{- 1} t}}{1 - t ^{2}}$ , $\frac{d y}{d t} = \frac{- e ^{c o s^{- 1} t}}{1 - t ^{2}}$ .
$\frac{d y}{d x} = - \frac{e ^{c o s^{- 1} t}}{e ^{s i n^{- 1} t}} = - e^{c o s^{- 1} t - s i n^{- 1} t}$ .
At $t = \frac{1}{2}$ : $sin^{- 1} t = cos^{- 1} t = \frac{π}{4}$ , so $\frac{d y}{d x} = - e^{0} = - 1$ .

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

Q222 marksVery Short AnswerContinuity and Differentiability CBSE 2026 · 65/5/1

Differentiate $x^{x}$ with respect to $x lo g x$ .

Show answer & solution

Answer:

x^{x}

Let $u = x^{x}$ , $v = x lo g x$ .
$lo g u = x lo g x \Rightarrow \frac{d u}{d x} = x^{x} (1 + lo g x)$ .
$\frac{d v}{d x} = lo g x + 1$ .
$\frac{d u}{d v} = \frac{x ^{x} ( 1 + l o g x )}{1 + l o g x} = x^{x}$ .

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

Q22 (OR) (OR)2 marksVery Short AnswerContinuity and Differentiability CBSE 2026 · 65/5/1

If $y = P cos ux + Q sin ux$ , show that $\frac{d ^{2} y}{d x ^{2}} + u^{2} y = 0$ .

Show answer & solution

Answer: Proved.

$\frac{d y}{d x} = - P u sin ux + Q u cos ux$ .
$\frac{d ^{2} y}{d x ^{2}} = - P u^{2} cos ux - Q u^{2} sin ux = - u^{2} (P cos ux + Q sin ux) = - u^{2} y$ .
Hence $\frac{d ^{2} y}{d x ^{2}} + u^{2} y = 0$ .

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

← Determinants Application of Derivatives →

Practise smarter: chapter-wise revision, formula sheets and step-by-step NCERT solutions on MonoMath CBSE →