GATE CS 2025 Set 1 Q21 — Calculus: Limits, Continuity, Differentiability

Consider the given function

f(x)

f(x)=\left\{\begin{array}{ll} a x+b & \text{ for } x<1 \\ x^{3}+x^{2}+1 & \text{ for } x \geq 1 \end{array}\right.

If the function is differentiable everywhere, the value of

b

must be ________. (rounded off to one decimal place)

View worked solution

Answer: -2.1:-1.9

For the function $f(x)$ to be differentiable everywhere, it must be both continuous and differentiable at the boundary point $x = 1$ .

For $f(x)$ to be continuous at $x = 1$ , the left-hand limit must equal the right-hand limit:

\lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x)

a(1) + b = 1^3 + 1^2 + 1

a + b = 3 \quad \text{--- (Equation 1)}

For $f(x)$ to be differentiable at $x = 1$ , the left-hand derivative (LHD) must equal the right-hand derivative (RHD):

\text{LHD} = \left. \frac{d}{dx}(ax + b) \right|_{x=1} = a

\text{RHD} = \left. \frac{d}{dx}(x^3 + x^2 + 1) \right|_{x=1} = \left. (3x^2 + 2x) \right|_{x=1} = 3(1)^2 + 2(1) = 5

Equating the LHD and RHD gives:

a = 5

Substitute $a = 5$ into Equation 1:

5 + b = 3 \implies b = -2

Thus, the value of $b$ must be -2.

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Published 10 Oct 2026

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