CBSE Maths › Class 12 PYQs › Continuity and Differentiability

Continuity and Differentiability: 5 marks Questions (CBSE Class 12)

5 different 5 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)

Q345 marksLong AnswerContinuity and Differentiability CBSE 2026 · 65/1/1

If $x = cos t, y = cos m t$ , prove that $(1 - x^{2}) \frac{d ^{2} y}{d x ^{2}} - x \frac{d y}{d x} + m^{2} y = 0$ .

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Answer: Proved.

$\frac{d x}{d t} = - sin t$ , $\frac{d y}{d t} = - m sin m t$ , so $\frac{d y}{d x} = \frac{m s i n m t}{s i n t}$ .
Then $sin^{2} t (\frac{d y}{d x})^{2} = m^{2} sin^{2} m t$ , i.e. $(1 - x^{2}) (\frac{d y}{d x})^{2} = m^{2} (1 - y^{2})$ .
Differentiate w.r.t. $x$ : $- 2 x (\frac{d y}{d x})^{2} + 2 (1 - x^{2}) \frac{d y}{d x} \frac{d ^{2} y}{d x ^{2}} = - 2 m^{2} y \frac{d y}{d x}$ .
Divide by $2 \frac{d y}{d x}$ : $(1 - x^{2}) \frac{d ^{2} y}{d x ^{2}} - x \frac{d y}{d x} + m^{2} y = 0$ . Hence proved.

Q325 marksLong AnswerContinuity and Differentiability CBSE 2026 · 65/1/2

If $x = 3 sin t - sin 3 t, y = 3 cos t - cos 3 t$
find $\frac{d y}{d x}$ and prove that $\frac{d ^{2} y}{d x ^{2}} = \frac{- cosec ^{3} 2 t \cdot cosec t}{3}$

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Answer:

\frac{d y}{d x} = cot 2 t

; second part proved.

$\frac{d x}{d t} = 3 cos t - 3 cos 3 t$ , $\frac{d y}{d t} = - 3 sin t + 3 sin 3 t$ .
$\frac{d y}{d x} = \frac{s i n 3 t - s i n t}{c o s t - c o s 3 t} = \frac{2 c o s 2 t s i n t}{2 s i n 2 t s i n t} = cot 2 t$ .
$\frac{d ^{2} y}{d x ^{2}} = \frac{d}{d t} (cot 2 t) \cdot \frac{d t}{d x} = \frac{- 2 cosec ^{2} 2 t}{3 ( c o s t - c o s 3 t )}$ .
$cos t - cos 3 t = 2 sin 2 t sin t$ , so $\frac{d ^{2} y}{d x ^{2}} = \frac{- 2 cosec ^{2} 2 t}{6 s i n 2 t s i n t} = \frac{- cosec ^{3} 2 t \cdot cosec t}{3}$ . Hence proved.

Q355 marksLong AnswerContinuity and Differentiability CBSE 2026 · 65/1/3

If $x = a (sin t - t cos t)$ and $y = b (cos t + t sin t)$ , then find $\frac{d y}{d x}$ and $\frac{d ^{2} y}{d x ^{2}}$ .

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Answer:

\frac{d y}{d x} = \frac{b}{a} cot t

;

\frac{d ^{2} y}{d x ^{2}} = - \frac{b}{a ^{2} t} cosec^{3} t

$\frac{d x}{d t} = a (cos t - cos t + t sin t) = a t sin t$ .
$\frac{d y}{d t} = b (- sin t + sin t + t cos t) = b t cos t$ .
$\frac{d y}{d x} = \frac{b t c o s t}{a t s i n t} = \frac{b}{a} cot t$ .
$\frac{d ^{2} y}{d x ^{2}} = \frac{d}{d t} (\frac{b}{a} cot t) \cdot \frac{d t}{d x} = - \frac{b}{a} cosec^{2} t \cdot \frac{1}{a t s i n t} = - \frac{b}{a ^{2} t} cosec^{3} t$ .

Q335 marksLong AnswerContinuity and Differentiability CBSE 2026 · 65/5/1

If $y x^{2} + 1 = lo g x^{2} + 1 - x$ , show that
$(x^{2} + 1) \frac{d y}{d x} + x y + 1 = 0$ .

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Answer: Proved.

Read the right side as $lo g (x^{2} + 1 - x)$ .
Differentiate: $x^{2} + 1 \frac{d y}{d x} + \frac{x y}{x ^{2} + 1} = \frac{1}{x ^{2} + 1 - x} (\frac{x}{x ^{2} + 1} - 1)$ .
RHS $= \frac{1}{x ^{2} + 1 - x} \cdot \frac{- ( x ^{2} + 1 - x )}{x ^{2} + 1} = - \frac{1}{x ^{2} + 1}$ .
Multiply by $x^{2} + 1$ : $(x^{2} + 1) \frac{d y}{d x} + x y = - 1$ , i.e. $(x^{2} + 1) \frac{d y}{d x} + x y + 1 = 0$ .

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

Q33 (OR) (OR)5 marksLong AnswerContinuity and Differentiability CBSE 2026 · 65/5/1

Find the differential of $x^{c o t x} + \frac{2 x ^{2} - 3}{2 x ^{2} - x + 2}$ with respect to x.

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Answer:

x^{c o t x} (\frac{c o t x}{x} - cosec^{2} x lo g x) + \frac{- 2 x ^{2} + 20 x - 3}{( 2 x ^{2} - x + 2 ) ^{2}}

Let $u = x^{c o t x}$ : $lo g u = cot x lo g x$ , so $\frac{d u}{d x} = x^{c o t x} (\frac{c o t x}{x} - cosec^{2} x lo g x)$ .
Let $v = \frac{2 x ^{2} - 3}{2 x ^{2} - x + 2}$ : $\frac{d v}{d x} = \frac{4 x ( 2 x ^{2} - x + 2 ) - ( 2 x ^{2} - 3 ) ( 4 x - 1 )}{( 2 x ^{2} - x + 2 ) ^{2}}$ .
Numerator $= (8 x^{3} - 4 x^{2} + 8 x) - (8 x^{3} - 2 x^{2} - 12 x + 3) = - 2 x^{2} + 20 x - 3$ .
$\frac{d y}{d x} = x^{c o t x} (\frac{c o t x}{x} - cosec^{2} x lo g x) + \frac{- 2 x ^{2} + 20 x - 3}{( 2 x ^{2} - x + 2 ) ^{2}}$ .

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

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