JEE Main 2024 — 27 January, Shift 2 — Question 2

Given

$$f(x)=\frac{x}{2}+\frac{2}{x}, \qquad 0<\ x<2$$

First find behaviour of $f$.

$$f'(x)=\frac12-\frac{2}{x^2}$$

Set $f'(x)=0$:

$$\frac12=\frac{2}{x^2}$$ $$x^2=4 \Rightarrow x=2$$

Since domain is $(0,2)$, we check sign of derivative:

For $0\frac12 \Rightarrow f'(x)<0 ]

Hence $f$ is strictly decreasing on $(0,2)$.

Therefore on $(0,1]$,

$$g(x)=\min_{0<\ t\le x} f(t)=f(x)$$

because decreasing function attains minimum at right endpoint.

Thus

$$g(x)=f(x)=\frac{x}{2}+\frac{2}{x}, \quad 0<\ x\le1.$$

For $1<x<2$,

$$g(x)=\frac32+x.$$

Now evaluate behaviour at $x=1$.

$$g(1)=f(1)=\frac12+2=\frac52.$$

Right limit:

$$\lim_{x\to1^+}g(x)=\frac32+1=\frac52.$$

Hence continuous at $x=1$. Check differentiability at $x=1$.

Left derivative:

$$f'(1)=\frac12-2=-\frac32.$$

Right derivative:

$$\frac{d}{dx}\left(\frac32+x\right)=1.$$

Since

$$-\frac32 \ne 1,$$

$g$ is not differentiable at $x=1$.

$g$ is continuous at $x=1$ but not differentiable there.