JEE Mathematics 2024 Question with Solution

Given:

$$f(x) = \begin{cases} \frac{1}{|x|}, & |x| \ge 2 \\ ax^2 + 2b, & |x| < 2 \end{cases}$$

The function is differentiable on $\mathbb{R}$.

Find: $48(a+b)$

The definition changes at $x=2$ and $x=-2$, so continuity and differentiability must hold at both points.

Write the function explicitly:

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

From continuity at $x=2$,

$$4a + 2b = \frac{1}{2}$$

From continuity at $x=-2$, the same equation is obtained.

Now differentiate each part:

$$f'(x) = \begin{cases} \frac{1}{x^2}, & x < -2 \\ 2ax, & -2 < x < 2 \\ -\frac{1}{x^2}, & x > 2 \end{cases}$$

From differentiability at $x=2$,

$$2a(2) = -\frac{1}{2^2}$$ $$4a = -\frac{1}{4}$$ $$a = -\frac{1}{16}$$

Using continuity,

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

Therefore,

$$a+b = -\frac{1}{16} + \frac{3}{8} = \frac{5}{16}$$

So,

$$48(a+b) = 48 \times \frac{5}{16} = 15$$

Therefore, the value of $48(a+b)$ is $15$.

Given: the piecewise function is differentiable on all real numbers.

Find: the numerical value of $48(a+b)$.

For differentiability of a piecewise function, first ensure continuity and then match derivatives at the joining points $x=\pm 2$.

At $x=2$, continuity gives

$$\lim_{x\to 2^-}(ax^2+2b)=\lim_{x\to 2^+}\frac{1}{x}$$ $$4a+2b=\frac{1}{2}$$

At $x=-2$,

$$\lim_{x\to -2^-}\left(-\frac{1}{x}\right)=\lim_{x\to -2^+}(ax^2+2b)$$ $$\frac{1}{2}=4a+2b$$

which is the same condition.

Now use derivatives. For $x<-2$,

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

For $-2<x<2$,

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

For $x>2$,

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

At $x=2$, differentiability gives

$$\lim_{x\to 2^-}2ax = \lim_{x\to 2^+}\left(-\frac{1}{x^2}\right)$$ $$4a=-\frac{1}{4}$$ $$a=-\frac{1}{16}$$

At $x=-2$,

$$\lim_{x\to -2^-}\frac{1}{x^2} = \lim_{x\to -2^+}2ax$$ $$\frac{1}{4}=-4a$$ $$a=-\frac{1}{16}$$

Again, the same value is obtained.

Substitute into the continuity equation:

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

Hence,

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

Therefore, the required value is $15$.