Let α, β ∈ ℝ be such that the function f(x)= cases 2 α(x^ 2 -2)+2 β x &, x be differentiable at all x ∈ ℝ . Then 34(α+β) is equal to

Correct answer: D. 48 f(x) = cases 2α x^2 + 2β x - 4α & ; x f(1^-) = 2α - β + 3 , f(1) = -2α + 2β 2α - β + 3 = 2β - 2α 4α - 3β + 3 = 0 ...(1) f'(1^+) = 4α + 2β , f'(1^-) = α + 3 4α + 2β = α + 3 3α + 2β - 3 = 0 ...(2) Solving (1) & (2) We get α = (3)/(17) , β = (21)/(17) 34(α + β) = 34 × (27)/(17) = 48

JEE Main 202624 JanuaryMorning Shift

Exam: JEE Main
Year: 2026
Date: 24 January
Shift: Morning Shift
Subject: Mathematics
Chapter: Continuity and Differentiability

JEE Main 2026 · 24 January · Morning Shift · Mathematics · Continuity and Differentiability

Let α, β ∈ ℝ be such that the function f(x)= cases 2 α(x^ 2 -2)+2 β...

Let

\alpha, \beta \in \mathbb{R}

be such that the function

f(x)=

\begin{cases}2 \alpha\left(x^{2}-2\right)+2 \beta x &, x<1 \\ (\alpha+3) x+(\alpha-\beta) &, x \geq 1\end{cases}$

<br> be differentiable at all

x \in \mathbb{R}

. Then

34(\alpha+\beta)$ is equal to

Options

Correct answer

D. 48

Step-by-step solution

f(x) =

\begin{cases} 2\alpha x^2 + 2\beta x - 4\alpha & ; \quad x < 1 \\ (\alpha + 3)x + \alpha - \beta & ; \quad x \geq 1 \end{cases}$

<br>

f(1^-) = 2\alpha - \beta + 3

,

f(1) = -2\alpha + 2\beta

<br>

2\alpha - \beta + 3 = 2\beta - 2\alpha \Rightarrow 4\alpha - 3\beta + 3 = 0

...(1)<br>

f'(1^+) = 4\alpha + 2\beta

,

f'(1^-) = \alpha + 3

<br>

4\alpha + 2\beta = \alpha + 3 \Rightarrow 3\alpha + 2\beta - 3 = 0

...(2)<br>Solving (1) & (2)<br>We get

\alpha = \dfrac{3}{17}

,

\beta = \dfrac{21}{17}

<br>

\Rightarrow 34(\alpha + \beta) = 34 \times \dfrac{27}{17} = 48$

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