Question

If $f\left(x\right)=\{\begin{array}{ccc}\frac{1-\cos 4x}{x^2},& if& x<0\\ a,& if& x=0\\ \frac{\sqrt{x}}{\sqrt{16+\sqrt{4}}-4},& if& x>0\end{array}$, then what value of $a,f$ is continuous at $x=0$?

• A. $2$
• B. $4$
• C. $6$
• D. $8$

Answer 1

The correct option is D $8$

Consider the following function

$f\left(x\right)=\frac{1-\cos 4x}{x^2}$ where $x<0$

L.H.L$={lim}_{x\to 0}f\left(x\right)={lim}_{x\to 0}\frac{1-\cos 4x}{x^2}$

Put $x=0-h$ as $x\to 0^{-}$ then $h\to 0$

L.H.L$={lim}_{h\to 0}\frac{1-\cos 4h}{h^2}$

$={lim}_{h\to 0}\frac{2{\sin }^{2}2h}{h^2}$

$=2{lim}_{h\to 0}\frac{{\sin }^{2}2h}{h^2}$

$=2{lim}_{h\to 0}\frac{\sin 2h}{2h}\times 2{lim}_{h\to 0}\frac{\sin 2h}{2h}\times 2$

$=2\times 2\times 2=8$

R.H.L$={lim}_{x\to 0}f\left(x\right)={lim}_{x\to 0}\frac{1-\cos 4x}{x^2}$

$f\left(x\right)=\frac{\sqrt{x}}{\sqrt{16+\sqrt{x}}-4}$

Put $x=0+h$ as $x\to 0^{+}$ then $h\to 0$

R.H.L$={lim}_{h\to 0}\frac{\sqrt{h}}{\sqrt{16+\sqrt{h}}-4}\times \frac{\sqrt{16+\sqrt{h}}+4}{\sqrt{16+\sqrt{h}}+4}$

$={lim}_{h\to 0}\frac{\sqrt{h}(\sqrt{16+\sqrt{h}}+4)}{16+\sqrt{h}-16}$

$=2{lim}_{h\to 0}\frac{\sqrt{h}(\sqrt{16}+\sqrt{h}+4)}{16+\sqrt{h}-16}$

$=\sqrt{16}+4=4+4=8$

Also $f\left(0\right)=a$

Since $f$ is continuous at $x=0$ then

L.H.L$=$R.H.L$=f\left(0\right)$

$\therefore a=8$