A function f(x) is defined as f(x) = [ * 20 c x^m , ,1 x & x , , … - Vidyadip

MCQ

A function $f(x)$ is defined as $f(x) =\left[ {\begin{array}{*{20}{c}} {{x^m}\,\sin \,\tfrac{1}{x}}&{x\,\, \ne \,\,0\,,\,\,\,m\,\, \in \,\,N} \\ 0&{if\,\,\,\,\,x\,\, = \,\,0} \end{array}} \right. $ . The least value of $m$ for which $f ‘ (x)$ is continuous at $x = 0$ is

A
$1$
B
$2$
✓
$3$
D
none

✓

Answer

Correct option: C.

$3$

c
$f ‘ (0+) =\mathop {Lim}\limits_{h \to 0} \,\frac{{{h^m}\sin \frac{1}{x}}}{h}$

must exist $\Rightarrow m > 1$
for $m > 1 h ‘ (x) =\left[ \begin{gathered} \hfill \\ \hfill \\ \end{gathered} \right.$$\begin{gathered} m\,{x^{m - 1}}\sin \frac{1}{x} - {x^{m - 2}}\cos \frac{1}{x}\,\,x \ne 0 \hfill \\ \hfill \\ \,\,0\,\,\,\,if\,\,x = 0 \hfill \\ \end{gathered} $
now $\mathop {Lim}\limits_{h \to 0} \,\,h(x) = \mathop {Lim}\limits_{h \to 0} \,\,m\,{h^{m - 1}}\sin \frac{1}{h} - {h^{m - 2}}\cos \frac{1}{h}$
limit exist if $m > 2$
$\therefore$ $m \in N$ $\Rightarrow m = 3$

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