Q105 / 160 · TS EAMCET 2020 (14 Sep Shift 1) · Mathematics
Let $f, g: \mathbf{R} \rightarrow \mathbf{R}$ be functions defined by
$f(x)=\left\{\begin{array}{cc}
x \sin \left(\frac{1}{x}\right), & \text { for } x \neq 0 \\
0, & \text { for } x=0
\end{array}\right.$
and $g(x)=x f(x)$
Consider the following statements
(i) $f(x)$ is continuous at $x=0$ but not differentiable at $x=0$
(ii) $g(x)$ is differentiable at $x=0$, but $g^1(x)$ is not continuous at $x=0$
Then, which one of the following is true?
- A(i) is true; but (ii) is false
- Correct answerBBoth (i) and (ii) are true
- C(i) is false, but (ii) is true
- DBoth (i) and (ii) are false
Solution
We have,
$\begin{aligned}
& f(x)=\left\{\begin{array}{cc}
x \sin \left(\frac{1}{x}\right), & \text { for } x \neq 0 \\
0, & \text { for } x=0
\end{array}\right. \\
& g(x)=x f(x)
\end{aligned}$
$\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} x \sin \left(\frac{1}{x}\right)=0 \sin \frac{1}{x} \in[-1,1]$
$\therefore f(x)$ is continuous at $x=0$
$f^{\prime}(x)=\left\{\begin{array}{cc}
\sin \left(\frac{1}{x}\right)-\frac{1}{x} \cos \left(\frac{1}{x}\right), & x \neq 0 \\
0, & x=0
\end{array}\right.$
$\therefore f(x)$ is not differentiable at $x=0$
Hence, the correct answer is option: 'B'.