Question

The value of $k$ for which the function $f\left(x\right)=\{\begin{array}{cc}\frac{1-\cos 4x}{8x^2}& ,x\ne 0\\ k& ,x=0\end{array}$ is continuous at $x=0$, is

• A. $k=0$
• B. $k=1$
• C. $k=-1$
• D. None of the above

Answer 1

The correct option is B $k=1$

Given, $f\left(x\right)=\{\begin{array}{cc}\frac{1-\cos 4x}{8x^2}& ,x\ne 0\\ k& ,x=0\end{array}$
LHL $=\underset{x\to 0^{-}}{lim}f\left(x\right)=\underset{h\to 0}{lim}f(0-h)=\underset{h\to 0}{lim}f\left(h\right)$
$=\underset{h\to 0}{lim}\frac{1-\cos 4h}{8h^2}$ ........ $\frac{0}{0}$ form
$=\underset{h\to 0}{lim}\frac{4\sin 4h}{16h}$ ........ (using L'Hospital's rule)

$=\underset{h\to 0}{lim}\frac{\sin 4h}{4h}$
$=1$ ....... $[\because \underset{x\to 0}{lim}\frac{\sin x}{x}=1]$
Since, $f\left(x\right)$ is continuous at $x=0$.
$\therefore f\left(0\right)=$ LHL $\Rightarrow k=1$

Hence, option B is correct.