Limits, Continuity & Differentiability
Continuity and Differentiability of Min-defined Function
nta_pyq_2024_jan
Grade 12
Question:
Consider the function $f:(0,2)\to\mathbb{R}$ defined by $f(x)=\frac{x}{2}+\frac{2}{x}$ and the function $g(x)$ defined by
$$g(x)=\begin{cases}\min\{f(t)\},\; 0<t\le x & ,\; 0<x\le 1\\\frac{3}{2}+x & ,\; 1<x<2\end{cases}$$
Then
$g$ is continuous but not differentiable at $x=1$
$g$ is not continuous for all $x\in(0,2)$
$g$ is neither continuous nor differentiable at $x=1$
$g$ is continuous and differentiable for all $x\in(0,2)$
Step-by-Step Solution
Key Concept: $f(x)=\frac{x}{2}+\frac{2}{x}$ is strictly decreasing on $(0,2)$ (since $f'(x)=\frac{1}{2}-\frac{2}{x^2}<0$ for $x<2$). So $\min_{0<t\le x}f(t)=f(x)=\frac{x}{2}+\frac{2}{x}$ for $0<x\le1$. Check continuity and differentiability at $x=1$: $g(1^-)=f(1)=\frac{5}{2}$, $g(1^+)=\frac{3}{2}+1=\frac{5}{2}$. ✓ continuous. LHD at $x=1$: $f'(1)=\frac{1}{2}-2=-\frac{3}{2}$; RHD: $1$. Not differentiable.
$f'(x)=\frac{1}{2}-\frac{2}{x^2}<0$ for $x\in(0,2)$, so $f$ is decreasing. For $0<x\le1$: $g(x)=f(x)=\frac{x}{2}+\frac{2}{x}$.
At $x=1$: $g(1^-)=\frac{5}{2}$, $g(1^+)=\frac{3}{2}+1=\frac{5}{2}$. Continuous.
LHD $=f'(1)=-\frac{3}{2}$, RHD $=1$. Not differentiable at $x=1$.
Correct Answer: 1