If a,b,c are arbitrary constants such that the function defined by f(x)=casesax²+b,\ b≠0,\ x≤1\\ bx²+ax+c,\ x>1cases is differentiable at x=1, then which of the following could be a triplet for (a,b,c)? (A) (3,2,0) (B) (0,0,3) (C) (3/2,2,0 ) (D) (3,3/2,0 )

MathsCBSE Class 12 · Sample paper, session 2026-27 · 1 mark · MCQ

CBSE Class 12 Mathematics sample paper 2026-27: question 8

From the official CBSE sample question paper for 2026-27, section A, question 8, for 1 mark. Try it first; the hints and our step-by-step solution are below.

Q8 · 1 mark · MCQContinuity and Differentiability

If $a, b, c$ are arbitrary constants such that the function defined by $f (x) = {a x^{2} + b, b \neq = 0, x \leq 1 b x^{2} + a x + c, x > 1$ is differentiable at $x = 1$ , then which of the following could be a triplet for $(a, b, c)$ ?

(A) $(3, 2, 0)$
(B) $(0, 0, 3)$
(C) $(\frac{3}{2}, 2, 0)$
(D) $(3, \frac{3}{2}, 0)$

Source: official sample paper from cbseacademic.nic.in (PDF). See it in the full Sample paper 2026-27.

Hints

Open one hint at a time and try again before the next.

Hint 1

Differentiability at a point implies continuity there.

Hint 2

Equate the left-hand and right-hand derivatives at $x = 1$ .

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

Continuity at $x = 1$ : $a + b = b + a + c$ , so $c = 0$ .
Equal one-sided derivatives at $x = 1$ : $2 a = 2 b + a$ , so $a = 2 b$ . The triplet is $(2 b, b, 0)$ with $b \neq = 0$ , which matches $(3, \frac{3}{2}, 0)$ .
1 mark

Answer: Option (D)

Where the marks are

Gets $c = 0$ and $a = 2 b$ and picks the matching triplet (1 mark)

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Chapter: Continuity and Differentiability (notes, formulas, where marks are lost and every past question)

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CBSE Class 12 Mathematics sample paper 2026-27: question 8

Hints

Step-by-step solution

Revise the chapter

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