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STD 12 - 5. continuity and differentiation — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 5. continuity and differentiation1 Mark
MCQ
Let $f(x) = \left\{ {\begin{array}{*{20}{c}} {{x^p}\,\sin \left( {\frac{1}{x}} \right) + x|{x^3}|,\,\,x\, \ne 0}\\{0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = 0} \end{array}} \right.$  then complete set of values of $p$ for which $f"(x)$ is continuous at $x = 0$ is
  • A
    $[2,\infty)$
  • B
    $[3, \infty )$
  • ✓
    $(4,\infty)$
  • D
    $[-2,\infty)$
✓

Answer

Correct option: C.
$(4,\infty)$
c
$g(x)=x|x|^{3}$ has $4$ repeated roots

$\therefore \mathrm{g}^{\prime \prime}(\mathrm{x})$ is cont. and diff. at $\mathrm{x}=0$

$\therefore$ consider $f(x) = {x^p}\sin \frac{1}{x}$         $x \ne 0$

                                         $0$                $x=0$

${f^\prime }\left( {{0^ + }} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{{h^p}\sin \frac{1}{h}}}{h} = 0\quad $

$\therefore f^{\prime}(0)=0$ for $p>1$

$\therefore {f^\prime }(x) = p{x^{p - 1}}\sin \frac{1}{x} - {x^{p - 2}}\cos \frac{1}{x}$         $x \ne 0$

                                                   $0$                           ${\rm{x}} = 0$

${f^{\prime \prime }}\left( {{0^ + }} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{p{h^{p - 1}}\sin \frac{1}{h} - {h^{p - 2}}\cos \frac{1}{h}}}{h} = 0$

if $p>3$

$\therefore f^{\prime \prime}(x)=\left(p(p-1) x^{p-2}-x^{p-4}\right) \sin \frac{1}{x}+\left(2 x^{p-3}\right) \cos \frac{1}{x}$

for $x \neq 0$

$=0 \quad \text { for } x=0$

$\therefore \mathrm{f}^{\prime \prime}(\mathrm{x})$ to be continuous $p \in(4, \infty)$

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