Let $f, g: R \to R$ be two functions defined by $f(x)\, = \,\left\{ {\begin{array}{*{20}{c}}{x\,\sin \,\left( {\frac{1}{x}} \right),\,x\, \ne \,0\,\,\,\,\,\,\,\,\,\,}\\ {0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,,x\, = 0\,\,\,\,\,\,\,\,\,}\end{array}} \right.,$ and $g(x) =x\,f(x)$
Statement $I: f$ is a continuous function at $x = 0.$
Statement $II:$ $g$ is a differentiable function at $x = 0.$
- A
Both statement $I$ and $II$ are false.
- B
Both statement $I$ and $II$ are true.
- C
Statement $I$ is true, statement $II$ is false.
- D
Statement $I$ is false, statement $II$ is true.
✓
Answer
$f\left( x \right) = \left\{ \begin{array}{l} x\sin \left( {\frac{1}{x}} \right),\,\,\,x \ne 0\\ 0\,,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = 0 \end{array} \right.$ and $g(x)=xf(x)$
For $f(x)$
$\text{LHL} = \mathop {\lim }\limits_{h \to {0^ - }} \left\{ { - h\sin \left( { - \frac{1}{h}} \right)} \right\}$
$ = 0 \times a$ finite quatity between $-1$ and $1$
$=0$
$\text{RHL} = \mathop {\lim }\limits_{h \to {0^ + }} h\sin \frac{1}{h} = 0$
Also, $f(0)=0$
Thus $\text{LHL = RHL} = f\left( 0 \right)$
$\therefore $
$f(x)$ is continuous at $x=0$
$g\left( x \right) = \left\{{x^2}\sin \frac{1}{x},\,\,\,x \ne 0 \ 0 x = 0 \right\}.$
For $g(x)$
$\text{LHL} = \mathop {\lim }\limits_{h \to {0^ - }} \left\{ { - {h^2}\sin \left( {\frac{1}{h}} \right)} \right\}$
$ = {0^2} \times a$ a finite quantify between $-1$ and $1$
$=0$
$\text{RHL} = \mathop {\lim }\limits_{h \to {0^ + }} {h^2}\sin \left( {\frac{1}{h}} \right) = 0$
Also $g(0)=0$
$\therefore \ g(x)$ is continuous at $x=0$
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