MCQ
If $\text{f(x)}=\begin{cases}\frac{1-\cos10\text{x}}{\text{x}^2},&\text{x}<0\\\text{a},&\text{x}=0\\\frac{\sqrt{\text{x}}}{\sqrt{625+\sqrt{\text{x}}}-25},&\text{x}>0\end{cases}$ then the value of so that $f(x)$ may be continuous at $x = 0$ is:
  • A
    $25$
  • $50$
  • C
    $-25$
  • D
    none of these

Answer

Correct option: B.
$50$
If $f(x)$ is continuous at $x = 0,$ then
$\lim\limits_{\text{x}\rightarrow0^-}\text{f(x)}=\text{f}(0)$
$\Rightarrow\lim\limits_{\text{h}\rightarrow0}\text{f}(-\text{h})=\text{f}(0)$
$\Rightarrow\lim\limits_{\text{h}\rightarrow0}\frac{(1-\cos(-10\text{h}))}{(-\text{h})^2}=\text{f}(0)$
$\Rightarrow\lim\limits_{\text{h}\rightarrow0}\frac{(1-\cos(10\text{h}))}{\text{h}^2}=\text{f}(0)$
$\Rightarrow\lim\limits_{\text{h}\rightarrow0}\frac{(2\sin^2(5\text{h}))}{\text{h}^2}=\text{a}$
$\Rightarrow\lim\limits_{\text{h}\rightarrow0}\frac{2\times25(\sin^2(5\text{h}))}{25\text{h}^2}=\text{a}$
$\Rightarrow50\lim\limits_{\text{h}\rightarrow0}\frac{(\sin^2(5\text{h}))}{(5\text{h})^2}=\text{a}$
$\Rightarrow50\lim\limits_{\text{h}\rightarrow0}\Big(\frac{\sin(5\text{h})}{5\text{h}}\Big)^2=\text{a}$
$\Rightarrow\text{a}=50$

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