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Mathematics · Limits, Continuity and Differentiability

JEE Main 2026 — 24 January, Morning Shift — Question 16

Let α,β∈R\alpha, \beta \in \mathbb{R} be such that the function f(x)={2α(x2−2)+2βx,x<1(α+3)x+(α−β),x≥1f(\mathrm{x})= \begin{cases}2 \alpha\left(\mathrm{x}^{2}-2\right)+2 \beta \mathrm{x} & , \mathrm{x}<1 \\(\alpha+3) \mathrm{x}+(\alpha-\beta) & , \mathrm{x} \geq 1\end{cases} be differentiable at all x∈R\mathrm{x} \in \mathbb{R}. Then 34(α+β)34(\alpha+\beta) is equal to :

  1. Option A:

    8484

  2. Option B:

    4848

    Correct
  3. Option C:

    3636

  4. Option D:

    2424

Answer: B

Step-by-step solution

f(x)={2αx2+2βx−4α;x<1(α+3)x+α−β;x≥1f(x)= \begin{cases}2 \alpha x^{2}+2 \beta x-4 \alpha ; & x<1 \\(\alpha+3) x+\alpha-\beta ; & x \geq 1\end{cases}

f(1+)=2α−β+3,f(1−)=−2α+2β\mathrm{f}\left(1^{+}\right)=2 \alpha-\beta+3, \mathrm{f}\left(1^{-}\right)=-2 \alpha+2 \beta

2α−β+3=2β−2α⇒4α−3β+3=0\begin{gathered} 2 \alpha-\beta+3=2 \beta-2 \alpha \Rightarrow 4 \alpha-3 \beta+3=0 \end{gathered}

f′(1+)=4α+2β,f′(1−)=α+3\mathrm{f}^{\prime}\left(1^{+}\right)=4 \alpha+2 \beta, \mathrm{f}^{\prime}\left(1^{-}\right)=\alpha+3

4α+2β=α+3⇒3α+2β−3=0\begin{gathered} 4 \alpha+2 \beta=\alpha+3 \Rightarrow 3 \alpha+2 \beta-3=0 \end{gathered}

Solving & (2) We get α=317,β=2117\alpha=\frac{3}{17}, \beta=\frac{21}{17}

⇒34(α+β)=34×2417=48\Rightarrow 34(\alpha+\beta)=34 \times \frac{24}{17}=48.

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
24 January, Morning Shift
Subject
Mathematics
Chapter
Limits, Continuity and Differentiability
Topic
Differentiability

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