JEE Main 2024 — 30 January, Shift 1 — Question 29

Given

$$f(x)= \begin{cases} -\dfrac{1}{x}, & x \le -2,\\[6pt] ax^2+2b, & -2<\ x<2,\\[6pt] \dfrac{1}{x}, & x\ge 2 \end{cases}$$

Since the function is differentiable at $x=2$, it must first be continuous at $x=2$.

Step 1: Continuity at $x=2$

Right side value:

$$f(2)=\frac{1}{2}$$

Left limit:

$$\lim_{x\to2^-}f(x)=4a+2b$$

Equating,

$$4a+2b=\frac{1}{2}$$ $$8a+4b=1 \qquad (1)$$

Step 2: Differentiability at $x=2$

Derivative for $x>2$:

$$\frac{d}{dx}\left(\frac{1}{x}\right)=-\frac{1}{x^2}$$

So right derivative at $2$:

$$-\frac{1}{4}$$

Derivative for $-2<x<2$:

$$\frac{d}{dx}(ax^2+2b)=2ax$$

Left derivative at $2$:

$$4a$$

Equating derivatives,

$$4a=-\frac{1}{4}$$ $$a=-\frac{1}{16}$$

Step 3: Find $b$

Substitute in (1):

$$8\left(-\frac{1}{16}\right)+4b=1$$ $$-\frac{1}{2}+4b=1$$ $$4b=\frac{3}{2}$$ $$b=\frac{3}{8}$$

Step 4: Compute $48(a+b)$

$$a+b=-\frac{1}{16}+\frac{3}{8} =-\frac{1}{16}+\frac{6}{16} =\frac{5}{16}$$ $$48(a+b)=48\cdot\frac{5}{16} =3\times5 =15$$ $$\boxed{15}$$