Continuity And Differentiability Question 4
Question 4 - 2024 (27 Jan Shift 2)
Consider the function $\mathrm{f}:(0,2) \rightarrow \mathrm{R}$ defined by $f(x)=\frac{x}{2}+\frac{2}{x}$ and the function $g(x)$ defined by $g(x)=\left{\begin{array}{cc}\min {f(t)}, & 0<\mathrm{t} \leq \mathrm{x} \text { and } 0<\mathrm{x} \leq 1 \ \frac{3}{2}+\mathrm{x}, & 1<\mathrm{x}<2\end{array}\right.$. Then
(1) $g$ is continuous but not differentiable at $x=1$
(2) $g$ is not continuous for all $x \in(0,2)$
(3) $g$ is neither continuous nor differentiable at $x=1$
(4) $g$ is continuous and differentiable for all $x \in(0,2)$
Show Answer
Answer (1)
Solution
$f:(0,2) \rightarrow R ; f(x)=\frac{x}{2}+\frac{2}{x}$
$f^{\prime}(x)=\frac{1}{2}-\frac{2}{x^{2}}$
$\therefore \mathrm{f}(\mathrm{x})$ is decreasing in domain.
$g(x)= \begin{cases}\frac{x}{2}+\frac{2}{x} & 0<x \leq 1 \ \frac{3}{2}+x & 1<x<2\end{cases}$
