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 answer
    B
    Both (i) and (ii) are true
  • C
    (i) is false, but (ii) is true
  • D
    Both (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'.

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