Q119 / 180 · KCET 2020 · Mathematics
If $f(x)=\left\{\begin{array}{cl}\frac{1-\cos K x}{x \sin x}, & \text { if } x \neq 0 \\ \frac{1}{2}, & \text { if } x=0\end{array}\right.$ is continuous at $x=0$, then the value of $K$ is
- A$\pm \frac{1}{2}$
- B0
- C$\pm 2$
- Correct answerD$\pm 1$
Solution
We have,
$f(x)=\left\{\begin{array}{cc}
\frac{1-\cos k x}{x \sin x} & , x \neq 0 \\
\frac{1}{2} & , x=0
\end{array}\right.$
$f(x)$ is continuous at $x=0$
$\begin{aligned}
&\therefore \quad \lim _{x \rightarrow 0} \frac{1-\cos k x}{x \sin x}=\frac{1}{2} \\
&\Rightarrow \quad \lim _{x \rightarrow 0} \frac{2 \sin ^{2} \frac{k}{2} x}{x \sin x}=\frac{1}{2} \\
&\Rightarrow \lim _{x \rightarrow 0} 2\left(\frac{\sin \frac{k x}{2}}{\frac{k x}{2}}\right)^{2} \times \lim _{x \rightarrow 0} \frac{1}{\sin x} \times \frac{k^{2}}{4}=\frac{1}{2} \\
&\Rightarrow \quad 2 \times \frac{k^{2}}{4}=\frac{1}{2}
\end{aligned}$
$\Rightarrow \quad \begin{aligned}
k^{2} &=1 \\
k &=\pm 1
\end{aligned}$
Hence, the correct answer is option: 'D'.