Let f(x) = l x^p 1 x ,x 0 0 , , , , , , , , , , , , , ,,x = 0 . t… - Vidyadip

MCQ

Let $f(x) = \left\{ \begin{array}{l}{x^p}\sin \frac{1}{x},x \ne 0\\0\,\,\,\,\,\,\,\,\,\,\,\,\,\,,x = 0\end{array} \right.$ then $f(x)$ is continuous but not differential at $x = 0$ if

✓
$0 < p \le 1$
B
$1 \le p < \infty $
C
$ - \infty < p < 0$
D
$p = 0$

✓

Answer

Correct option: A.

$0 < p \le 1$

a
(a) $f(x) = {x^p}\sin \frac{1}{x},x \ne 0$ and $f(x) = 0,\;x = 0$

Since at $x = 0 , f(x)$ is a continuous function

$\therefore $ $\mathop {\lim }\limits_{x \to 0} f(x) = f(0) = 0$

==> $\mathop {\lim }\limits_{x \to 0} \,{x^p}\sin \frac{1}{x} = 0 \Rightarrow p > 0$.

$f(x)$ is differentiable at $x = 0$, if $\mathop {\lim }\limits_{x \to 0} \frac{{f(x) - f(0)}}{{x - 0}}$ exists

==> $\mathop {\lim }\limits_{x \to 0} \frac{{{x^p}\sin \frac{1}{x} - 0}}{{x - 0}}$ exists

==> $\mathop {\lim }\limits_{x \to 0} {x^{p - 1}}\sin \frac{1}{x}$ exists

==> $p - 1 > 0$ or $p > 1$

If $p \le 1$, then $\mathop {\lim }\limits_{x \to 0} {x^{p - 1}}\sin \left( {\frac{1}{x}} \right)$ does not exist and at $x = 0$ $f(x)$ is not differentiable.

$\therefore $ for $0 < p \le 1$ $f(x)$ is a continuous function at $x = 0$ but not differentiable.

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