Q103 / 180 · COMEDK 2020 · Mathematics
If the function $f(x)=\left\{\begin{array}{cl}\frac{1-\cos x}{x^{2}}, & \text { for } x \neq 0 \\ k, & \text { for } x=0\end{array}\right.$ is continuous at $x=0$, then the value of $k$ is
- A0
- B1
- C$-1$
- Correct answerD$1 / 2$
Solution
We have,
$$
f(x)=\left\{\begin{array}{cl}
\frac{1-\cos x}{x^{2}}, & x \neq 0 \\
k, & x=0
\end{array}\right.
$$
Since, $f(x)$ is continuous at $x=0$, then
$f(0)=\lim _{x \rightarrow 0} f(x) \Rightarrow k=\lim _{x \rightarrow 0} \frac{1-\cos x}{x^{2}}$
$\Rightarrow \quad k=\lim _{x \rightarrow 0} \frac{\sin x}{2 x} \quad$ [using L' Hospital Rule]
$\Rightarrow \quad k=\frac{1}{2}$
Hence, the correct answer is option: 'D'.