Question

If the function $f(x)=\{\begin{array}{ll}\frac{1-Cosx}{x^2}& forx\ne 0\\ k& forx=0\end{array}$ 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, $f(x)=<br/>\{\begin{array}{ll}<br/>\frac{1-\cos x}{x^2}& ,x\ne 0\\ <br/>k& ,x=0<br/>\end{array}$
Since, f(x) is continuous
$\therefore \underset{x\to 0}{\lim }f(x)=f(0)$
$\Rightarrow \underset{x\to 0}{\lim }\frac{1-\cos x}{x^2}=k$
[using L’ Hospital’s rule]
$\Rightarrow \frac{1}{2}\underset{x\to 0}{\lim }\frac{-\left(-\sin x\right)}{2x}=k$
$\Rightarrow \frac{1}{2}.1=K$
$\Rightarrow k=\frac{1}{2}$