Limits, Continuity and Differentiability – Differentiability – JEE Main 24 Jan 2026 Shift 1

Solution:

For $f(x)$ to be differentiable at $x=1$, it must be continuous at $x=1$.

**Continuity at $x=1$:**
$$\lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x)$$
$$2\alpha(1^2 – 2) + 2\beta(1) = (\alpha + 3)(1) + (\alpha – \beta)$$
$$2\alpha(-1) + 2\beta = \alpha + 3 + \alpha – \beta$$
$$-2\alpha + 2\beta = 2\alpha – \beta + 3$$
$$4\alpha – 3\beta = -3 \quad \dots(1)$$

**Differentiability at $x=1$:**
$$LHD = RHD$$
$$f'(x) = \begin{cases} 4\alpha x + 2\beta &, x < 1 \\ \alpha + 3 &, x > 1 \end{cases}$$
$$4\alpha(1) + 2\beta = \alpha + 3$$
$$3\alpha + 2\beta = 3 \quad \dots(2)$$

Solving equations (1) and (2):
From (2), $2\beta = 3 – 3\alpha \Rightarrow \beta = \frac{3-3\alpha}{2}$.
Substitute into (1):
$$4\alpha – 3(\frac{3-3\alpha}{2}) = -3$$
$$8\alpha – 9 + 9\alpha = -6$$
$$17\alpha = 3 \Rightarrow \alpha = \frac{3}{17}$$

Substitute $\alpha$ back to find $\beta$:
$$3(\frac{3}{17}) + 2\beta = 3$$
$$2\beta = 3 – \frac{9}{17} = \frac{51 – 9}{17} = \frac{42}{17}$$
$$\beta = \frac{21}{17}$$

We need to find $34(\alpha + \beta)$:
$$34(\frac{3}{17} + \frac{21}{17}) = 34(\frac{24}{17})$$
$$= 2 \times 24 = 48$$

Ans. (2)