JEE Main 2026 Jan 24 Shift 1, Mathematics Q16: Differentiability Of A Function

JEE Main2026Jan 24, Shift 1Mathematics

Let $α, β \in R$ be such that the function

$f (x) = {2 α (x^{2} - 2) + 2 β x, x < 1 (α + 3) x + (α - β), x \geq 1$

be differentiable at all $x \in R$ . Then $34 (α + β)$ is equal to

Solution

Differentiability at $x = 1$ requires both continuity and matching derivatives.

Continuity at $x = 1$ :

$2 α (1 - 2) + 2 β = (α + 3) + (α - β)$

$- 2 α + 2 β = 2 α - β + 3 \Rightarrow 4 α - 3 β = - 3 \dots (1)$

Derivative match at $x = 1$ : $f_{-}^{'} (x) = 4 α x + 2 β$ , $f_{+}^{'} (x) = α + 3$ .

$4 α + 2 β = α + 3 \Rightarrow 3 α + 2 β = 3 \dots (2)$

Solving (1) and (2): multiply (1) by 2, (2) by 3, then add: $17 α = 3 \Rightarrow α = \frac{3}{17}$ .

From (2): $β = \frac{3 - 9/17}{2} = \frac{21}{17}$ .

$α + β = \frac{24}{17} \Rightarrow 34 (α + β) = 48$ .

Concept behind this question

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