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Continuity and Differentiability — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsContinuity and Differentiability1 Mark
MCQ
The value of $k(k<0)$ for which the function $f$ defined as $f(x)=\left\{\begin{array}{cc}\frac{1-\cos k x}{x \sin x} & , x \neq 0 \\ \frac{1}{2} & , x=0\end{array}\right.$ is continuous at $x=0$ is
  • A
    $\pm 1$
  • ✓
    $-1$
  • C
    $\pm \frac{1}{2}$
  • D
    $\frac{1}{2}$
✓

Answer

Correct option: B.
$-1$
We have, $f(x)=\left\{\begin{array}{cc}\frac{1-\cos k x}{x \sin x}, & x \neq 0 \\ \frac{1}{2}, & x=0\end{array}\right.$
$\because f(x)$ is continuous at $x=0$.
$\therefore \lim _{x \rightarrow 0} \frac{1-\cos k x}{x \sin x}=\frac{1}{2} $
$\Rightarrow \lim _{x \rightarrow 0} \frac{2 \sin ^2 \frac{k x}{2}}{x^2 \frac{\sin x}{x}}=\frac{1}{2}$
$\Rightarrow \lim _{x \rightarrow 0} 2 \cdot \frac{k^2}{4}\left\{\frac{\sin \left(\frac{k x}{2}\right)}{\frac{k x}{2}}\right\}^2 \frac{1}{\frac{(\sin x)}{x}}=\frac{1}{2}$
$\Rightarrow \frac{k^2}{2}=\frac{1}{2} $
$\Rightarrow k= \pm 1$
$\text { But } k<0$
$ \therefore k=-1$

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