Question

If the function $ f(x) = \begin{cases} \frac {1-Cos x }{x^2} & \quad \, for \, x \neq 0\\ k & \quad \, for \, x=0\\ \end{cases} $ is continuous at $x = 0,$ then the value of $k$ is

• A. 0
• B. 1
• C. -1
• D. 1/2

Answer 1

Answer: (D)

Given, $f(x) =
\begin{cases}
\frac{1 - \cos \; x }{x^2} & , x \ne 0 \\
k & , x = 0
\end{cases}$
Since, f(x) is continuous
$\therefore \, \displaystyle\lim_{x \to 0} f(x) = f(0)$
$\Rightarrow \displaystyle\lim_{x \to0} \frac{1- \cos x}{x^{2}} =k$
[using L’ Hospital’s rule]
$ \Rightarrow \frac{1}{2} \displaystyle\lim_{x \to 0} \frac{-\left(-\sin x\right)}{2x} = k $
$\Rightarrow \frac{1}{2} .1=K$
$ \Rightarrow k = \frac{1}{2} $