Q45 / 75 · JEE Main 2026 (24 January Shift 1) · Mathematics
Let $\alpha, \beta \in \mathbb{R}$ be such that the function $f(x)= \begin{cases}2 \alpha\left(x^{2}-2\right)+2 \beta x &, x<1 \\ (\alpha+3) x+(\alpha-\beta) &, x \geq 1\end{cases}$
be differentiable at all $x \in \mathbb{R}$. Then $34(\alpha+\beta)$ is equal to
be differentiable at all $x \in \mathbb{R}$. Then $34(\alpha+\beta)$ is equal to
- A36
- B24
- C84
- Correct answerD48
Solution
$f(x) = \begin{cases} 2\alpha x^2 + 2\beta x - 4\alpha & ; \quad x < 1 \\ (\alpha + 3)x + \alpha - \beta & ; \quad x \geq 1 \end{cases}$
$f(1^-) = 2\alpha - \beta + 3$, $f(1) = -2\alpha + 2\beta$
$2\alpha - \beta + 3 = 2\beta - 2\alpha \Rightarrow 4\alpha - 3\beta + 3 = 0$ ...(1)
$f'(1^+) = 4\alpha + 2\beta$, $f'(1^-) = \alpha + 3$
$4\alpha + 2\beta = \alpha + 3 \Rightarrow 3\alpha + 2\beta - 3 = 0$ ...(2)
Solving (1) & (2)
We get $\alpha = \dfrac{3}{17}$, $\beta = \dfrac{21}{17}$
$\Rightarrow 34(\alpha + \beta) = 34 \times \dfrac{27}{17} = 48$
$f(1^-) = 2\alpha - \beta + 3$, $f(1) = -2\alpha + 2\beta$
$2\alpha - \beta + 3 = 2\beta - 2\alpha \Rightarrow 4\alpha - 3\beta + 3 = 0$ ...(1)
$f'(1^+) = 4\alpha + 2\beta$, $f'(1^-) = \alpha + 3$
$4\alpha + 2\beta = \alpha + 3 \Rightarrow 3\alpha + 2\beta - 3 = 0$ ...(2)
Solving (1) & (2)
We get $\alpha = \dfrac{3}{17}$, $\beta = \dfrac{21}{17}$
$\Rightarrow 34(\alpha + \beta) = 34 \times \dfrac{27}{17} = 48$
Hence, the correct answer is option: 'D'.