JEE Main · 2024 · 27 Jan 2024 · Shift 2 · Continuity and Differentiability

1. (A)g is continuous but not differentiable at $x=1$✓ Correct
2. (B)g is not continuous for all $x\in (0,2)$
3. (C)g is neither continuous nor differentiable at $x=1$
4. (D)$g$ is continuous and differentiable for all $x\in (0,2)$

Answer: (A)

Solution

$f(x)=\frac{x}{2}+\frac{2}{x}$ on $(0,2)$; $f^{\prime }(x)=\frac{1}{2}-\frac{2}{x^2}$. For $0<x<2$, $x^2<4$ so $\frac{2}{x^2}>\frac{1}{2}$, giving $f^{\prime }(x)<0$: $f$ is strictly decreasing on $(0,2)$.

Since $f$ is decreasing, $\min \{f(t):0<t\le x\}=f(x)$ for $0<x\le 1$. So

$$g(x)=\{\begin{array}{ll}\frac{x}{2}+\frac{2}{x},& 0<x\le 1\\ \frac{3}{2}+x,& 1<x<2.\end{array}$$

At $x=1$: $f(1)=\frac{1}{2}+2=\frac{5}{2}$, and $\underset{x\to 1^{+}}{\lim }\left(\frac{3}{2}+x\right)=\frac{5}{2}$. The two pieces agree, so $g$ is continuous at $x=1$.

Left derivative at $1$: $f^{\prime }(1)=\frac{1}{2}-2=-\frac{3}{2}$. Right derivative: $\frac{d}{dx}\left(\frac{3}{2}+x\right)=1$. These differ, so $g$ is not differentiable at $x=1$.

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

JEE Main · 2024 · 27 Jan 2024 · Shift 2 · Single correct option