Sample 2026-27 Class 12 Maths Q8 Solution: Triplet (a, b, c) that makes a piecewise function differentiable at x = 1

The question

If $a, b, c$ are arbitrary constants such that the function defined by

f(x) = \begin{cases} ax^2 + b,\ b \ne 0, & x \le 1 \\ bx^2 + ax + c, & x > 1 \end{cases}

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) $\left(\frac{3}{2}, 2, 0\right)$

(D) $\left(3, \frac{3}{2}, 0\right)$

Exam answer

$f$ is differentiable at $x = 1$ ⁠, so it is continuous there:

\begin{aligned} \text{LHL} = f(1) &= a(1)^2 + b = a + b \\ \text{RHL} &= b(1)^2 + a(1) + c \\ &= b + a + c \end{aligned}

a + b = b + a + c \Rightarrow c = 0

Why?

A function that is differentiable at a point is continuous there, so the left-hand limit, the right-hand limit and $f(1)$ must agree.

$f(1)$ and the left side use $ax^2 + b$ ⁠, because that piece has $x \le 1$ ⁠. The right side uses $bx^2 + ax + c$ ⁠. Cancelling $a + b$ from both sides leaves $c = 0$ ⁠.

For $x < 1$ ⁠, $f'(x) = 2ax$ ⁠, and for $x > 1$ ⁠, $f'(x) = 2bx + a$ ⁠. So at $x = 1$ ⁠,

LHD $= 2a(1) = 2a$ and RHD $= 2b(1) + a = 2b + a$ ⁠.

Why?

Each piece is a polynomial. Once the pieces join ( $c = 0$ ⁠), the one-sided derivatives at $x = 1$ are the derivatives of the pieces at $x = 1$ ⁠:

\begin{aligned} \tfrac{d}{dx}(ax^2 + b) &= 2ax \\ \tfrac{d}{dx}(bx^2 + ax + c) &= 2bx + a \end{aligned}

2a = 2b + a \Rightarrow a = 2b

Why?

Differentiable means LHD = RHD. Then take $a$ from both sides:

\begin{aligned} 2a &= 2b + a \\ 2a - a &= 2b \\ a &= 2b \end{aligned}

So $(a, b, c) = (2b, b, 0)$ ⁠, $b \ne 0$ ⁠, which is satisfied by $\left(3, \frac{3}{2}, 0\right)$ only.

\begin{aligned} \text{(A)}&: 3 \ne 2(2) \\ \text{(B)}&: b = 0,\ c = 3 \ne 0 \\ \text{(C)}&: \tfrac{3}{2} \ne 2(2) \\ \text{(D)}&: 3 = 2\left(\tfrac{3}{2}\right),\ c = 0 \end{aligned}

Why?

Test each triplet against all three conditions: $c = 0$ ⁠, $a = 2b$ and $b \ne 0$ ⁠.

\begin{aligned} \text{(A)}&: 2b = 2(2) = 4,\ a = 3 \ne 4 \\ \text{(C)}&: 2b = 2(2) = 4,\ a = \tfrac{3}{2} \ne 4 \\ \text{(D)}&: 2b = 2\left(\tfrac{3}{2}\right) = 3 = a \end{aligned}

Watch out: (B) fails twice. The question rules out $b = 0$ ⁠, and $c = 3$ would break continuity.

Answer: (D) $\left(3, \dfrac{3}{2}, 0\right)$

Where this comes from

Chapter 5, Continuity and Differentiability: Differentiability of a piecewise function.

Question type: Find $k$ so that $f$ is continuous + Is $f$ differentiable at a point? Compare LHD and RHD (combines 2 types)

How to spot it: click to see

Type 2. Find $k$ so that $f$ is continuous

Spot it: a function in pieces with an unknown $k$ (or $a$ and $b$ ⁠), and the words "find the value of $k$ ".

Here: First, differentiable means continuous: $ax^2 + b$ gives LHL $a + b$ ⁠, and $bx^2 + ax + c$ gives RHL $a + b + c$ ⁠. Equal only if $c = 0$ ⁠.

Type 3. Is $f$ differentiable at a point? Compare LHD and RHD

Spot it: a modulus such as $|x - 1|$ ⁠, or a function in pieces, and the word "differentiable".

Here: Then match the slopes at 1: LHD is $2a$ from $2ax$ ⁠, RHD is $2b + a$ from $2bx + a$ ⁠. So $a = 2b$ ⁠, which only $\left(3, \frac32, 0\right)$ fits.

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Q8: Triplet (a,b,c)(a, b, c)(a,b,c) that makes a piecewise function differentiable at x=1x = 1x=1

Q8: Triplet $(a, b, c)$ that makes a piecewise function differentiable at $x = 1$