CBSE Class 12 · Chapter 5 · Calculus · 2026-27

Continuity and Differentiability Class 12: MCQ and case study questions

15 multiple-choice questions and 2 case-based questions on the current (rationalised) syllabus, in the style of the board paper's Sections A and E. Try each one first; the answer, a one-line reason and a worked solution open on a tap.

15 MCQs (1 mark each)
2 case studies (4 marks each)
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Multiple-choice questions

Tap an option to check it; the answer and a worked solution open. Section A of the board paper has 18 MCQs of 1 mark each, spread over all chapters.

Q1

·1 mark·Multiple choiceContinuity

The value of $k$ for which $f (x) = {\begin{cases} \frac{x^{2} - 9}{x - 3}, & x \neq 3 \\ k, & x = 3 \end{cases}$ is continuous at $x = 3$ is

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Answer: (a)

6

Why: For x ≠ 3, (x² − 9)/(x − 3) = x + 3, which tends to 6.

$lim_{x \to 3} \frac{(x - 3) (x + 3)}{x - 3} = lim_{x \to 3} (x + 3) = 6$ . Continuity needs $k = 6$ .

Q2

·1 mark·Multiple choiceDifferentiability

The function $f (x) = | x - 2 |$ at $x = 2$ is

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Answer: (b) continuous but not differentiable

Why: The graph has a corner at x = 2: left derivative −1, right derivative 1.

$lim_{x \to 2} | x - 2 | = 0 = f (2)$ , so $f$ is continuous. Left derivative $= - 1$ , right derivative $= 1$ ; they differ, so $f$ is not differentiable at $2$ .

Q3

·1 mark·Multiple choiceChain rule and standard derivatives

$\frac{d}{d x} [\cos (x^{2})]$ equals

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Answer: (c)

- 2 x \sin (x^{2})

Why: Chain rule: derivative of cos u is −sin u × u′, with u = x².

$\frac{d}{d x} \cos (x^{2}) = - \sin (x^{2}) \cdot 2 x = - 2 x \sin (x^{2})$ .

Q4

·1 mark·Multiple choiceChain rule and standard derivatives

$\frac{d}{d x} [e^{3 x} \sin x]$ equals

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Answer: (a)

e^{3 x} (3 \sin x + \cos x)

Why: Product rule: 3e^3x sin x + e^3x cos x.

$(e^{3 x})^{'} \sin x + e^{3 x} (\sin x)^{'} = 3 e^{3 x} \sin x + e^{3 x} \cos x = e^{3 x} (3 \sin x + \cos x)$ .

Q5

·1 mark·Multiple choiceChain rule and standard derivatives

$\frac{d}{d x} [\tan^{- 1} (2 x)]$ equals

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Answer: (d)

\frac{2}{1 + 4 x^{2}}

Why: d/dx tan⁻¹u = u′/(1 + u²) with u = 2x.

$\frac{1}{1 + (2 x)^{2}} \cdot 2 = \frac{2}{1 + 4 x^{2}}$ .

Q6

·1 mark·Multiple choiceChain rule and standard derivatives

For $0 < x < π$ , $\frac{d}{d x} [\log (\sin x)]$ equals (Here $\log$ means $\log_{e}$ .)

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Answer: (b)

\cot x

Why: Derivative of log u is u′/u = cos x/sin x.

$\frac{1}{\sin x} \cdot \cos x = \cot x$ .

Q7

·1 mark·Multiple choiceChain rule and standard derivatives

For $- 2 < x < 2$ , $\frac{d}{d x} [\sin^{- 1} \frac{x}{2}]$ equals

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Answer: (c)

\frac{1}{\sqrt{4 - x^{2}}}

Why: (1/√(1 − x²/4)) × 1/2 = 1/√(4 − x²).

$\frac{1}{\sqrt{1 - \frac{x^{2}}{4}}} \cdot \frac{1}{2} = \frac{1}{2} \cdot \frac{2}{\sqrt{4 - x^{2}}} = \frac{1}{\sqrt{4 - x^{2}}}$ .

Q8

·1 mark·Multiple choiceImplicit, parametric and logarithmic differentiation

If $x^{2} + y^{2} = 25$ , then $\frac{d y}{d x}$ at the point $(3, 4)$ is

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Answer: (a)

- \frac{3}{4}

Why: 2x + 2y y′ = 0 gives y′ = −x/y.

Differentiating: $2 x + 2 y \frac{d y}{d x} = 0 \Rightarrow \frac{d y}{d x} = - \frac{x}{y} = - \frac{3}{4}$ at $(3, 4)$ .

Q9

·1 mark·Multiple choiceImplicit, parametric and logarithmic differentiation

If $x = 2 t^{2}$ and $y = 4 t$ , then $\frac{d y}{d x}$ at $t = 2$ is

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Answer: (b)

\frac{1}{2}

Why: dy/dx = (dy/dt)/(dx/dt) = 4/(4t) = 1/t.

$\frac{d x}{d t} = 4 t$ , $\frac{d y}{d t} = 4$ , so $\frac{d y}{d x} = \frac{1}{t} = \frac{1}{2}$ at $t = 2$ .

Q10

·1 mark·Multiple choiceImplicit, parametric and logarithmic differentiation

For $x > 0$ , if $y = x^{x}$ , then $\frac{d y}{d x}$ equals (Here $\log$ means $\log_{e}$ .)

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Answer: (d)

x^{x} (1 + \log x)

Why: Take logs: log y = x log x, so y′/y = log x + 1.

$\log y = x \log x \Rightarrow \frac{1}{y} \frac{d y}{d x} = \log x + 1 \Rightarrow \frac{d y}{d x} = x^{x} (1 + \log x)$ .

Q11

·1 mark·Multiple choiceSecond-order derivatives

If $y = e^{2 x} + x^{3}$ , then $\frac{d^{2} y}{d x^{2}}$ equals

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Answer: (c)

4 e^{2 x} + 6 x

Why: Differentiate twice: 2e^2x + 3x², then 4e^2x + 6x.

$y^{'} = 2 e^{2 x} + 3 x^{2}$ , $y^{″} = 4 e^{2 x} + 6 x$ .

Q12

·1 mark·Multiple choiceSecond-order derivatives

If $y = 3 \cos 2 x$ , then $\frac{d^{2} y}{d x^{2}} + 4 y$ equals

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Answer: (a)

0

Why: y″ = −12 cos 2x = −4y.

$y^{'} = - 6 \sin 2 x$ , $y^{″} = - 12 \cos 2 x = - 4 y$ . So $y^{″} + 4 y = 0$ .

Q13

·1 mark·Multiple choiceContinuity

The function $f (x) = {\begin{cases} k x^{2}, & x \leq 2 \\ 3, & x > 2 \end{cases}$ is continuous at $x = 2$ when $k$ equals

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Answer: (d)

\frac{3}{4}

Why: Match the two pieces at x = 2: 4k = 3.

$f (2) = 4 k$ and $lim_{x \to 2^{+}} 3 = 3$ . So $4 k = 3 \Rightarrow k = \frac{3}{4}$ .

Q14

·1 mark·Multiple choiceChain rule and standard derivatives

For a constant $a > 0$ , $\frac{d}{d x} (a^{x})$ equals (Here $\log$ means $\log_{e}$ .)

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Answer: (b)

a^{x} \log a

Why: a^x = e^{x log a}; differentiate by the chain rule.

$a^{x} = e^{x \log a}$ , so $\frac{d}{d x} a^{x} = e^{x \log a} \cdot \log a = a^{x} \log a$ . ( $x a^{x - 1}$ is the power rule, which does not apply: the variable is in the exponent.)

Q15

·1 mark·Multiple choiceDifferentiability

The function $f (x) = | x - 1 | + | x + 2 |$ is not differentiable at

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Answer: (c)

x = 1

and

x = - 2

Why: Each modulus term has a corner where its inside is zero.

$| x - 1 |$ has a corner at $x = 1$ and $| x + 2 |$ at $x = - 2$ ; the other term is differentiable there, so the sum has corners at both points. (E.g. at $x = 1$ : left derivative $0$ , right derivative $2$ .)

Case study questions

Section E of the board paper has three case-based questions of 4 marks: a real-life passage, then parts of 1, 1 and 2 marks, with a choice (OR) on the 2-mark part. All Class 12 case study questions by chapter.

Case study 1: Taxi fare (4 marks)

A city taxi charges ₹ $F (d)$ for a trip of $d$ km, where $F (d) = 50 + 12 d$ for $0 \leq d \leq 5$ and $F (d) = k + 10 d$ for $d > 5$ . The transport authority insists that the fare must not jump at $5$ km.

(i) Find

k

so that

F

is continuous at

d = 5

. [1 mark]

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Answer:

k = 60

F (5) = 110

and

lim_{d \to 5^{+}} (k + 10 d) = k + 50

; so

k + 50 = 110 \Rightarrow k = 60

. A1

(ii) With this

k

, find the fare for an

8

km trip. [1 mark]

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Answer: ₹

140

F (8) = 60 + 80 = ₹ 140

. A1

(iii) Is

F

differentiable at

d = 5

? Justify. [2 marks]

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Answer: No

Left derivative at

5

= 12

, right derivative

= 10

M1; they differ, so

F

is not differentiable at

d = 5

(though continuous). A1

OR Find the rate at which the fare increases per km for a

3

km trip and for a

7

km trip. [2 marks]

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Answer: ₹

12

per km; ₹

10

per km

F^{'} (d) = 12

for

0 < d < 5

M1 and

F^{'} (d) = 10

for

d > 5

: ₹

12

/km at

3

km, ₹

10

/km at

7

km. A1

Case study 2: Cooling cup of tea (4 marks)

Priya pours a cup of tea and leaves it on the kitchen table while she answers the door. The room is at $25^{\circ}$ C, and the tea cools towards room temperature. Its temperature (in °C) $t$ minutes after pouring is $T (t) = 25 + 60 e^{- t / 10}$ .

(i) What is the temperature of the tea at

t = 0

? [1 mark]

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Answer:

85^{\circ}

T (0) = 25 + 60 = 85^{\circ}

C. A1

(ii) Find

\frac{d T}{d t}

. [1 mark]

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Answer:

- 6 e^{- t / 10}

\frac{d T}{d t} = 60 \cdot (- \frac{1}{10}) e^{- t / 10} = - 6 e^{- t / 10}

. A1

(iii) Find

\frac{d T}{d t}

t = 10

and hence the rate at which the tea is cooling then (exact form). State the sign of

\frac{d^{2} T}{d t^{2}}

. [2 marks]

Show answer

Answer: cooling at

\frac{6}{e}^{\circ}

C/min; positive

T^{'} (10) = - 6 e^{- 1} = - \frac{6}{e}

, so the tea is cooling at

\frac{6}{e}^{\circ}

C per minute A1.

T^{″} (t) = \frac{3}{5} e^{- t / 10} > 0

: the cooling slows down. A1

OR Show that

\frac{d T}{d t} = - \frac{1}{10} (T - 25)

. [2 marks]

Show answer

Answer: Shown

T - 25 = 60 e^{- t / 10}

M1, so

- \frac{1}{10} (T - 25) = - 6 e^{- t / 10} = \frac{d T}{d t}

. A1

Next steps for Continuity and Differentiability

This free set is separate from the chapter's question bank. On the Continuity and Differentiability chapter page the revision notes and the first step of the Route to 95 are free for everyone. CBSE Pro adds all 73 questions in the chapter bank (short and long answers, assertion–reason and more case studies) with their full step-marking schemes, the Route to 95 checkpoints with your progress saved, Skill Builders, the full common-mistakes library and AI marking of your written working.

Continuity and Differentiability: important questions What Pro includes

Original questions written by CBSE Math Revision for the CBSE 2026-27 syllabus; not taken from NCERT, NCERT Exemplar or CBSE papers. Every answer was checked twice, the second time by an independent reviewer. CBSE Math Revision is independent and not affiliated with CBSE or NCERT. Spotted a slip? Tell us and it goes in the corrections log.