Question

$f\left(x\right)=\{\begin{array}{c}\begin{array}{cc}\frac{{\cos }^{2}x-{\sin }^{2}x-1}{\sqrt{x^2+4}-2},& x\ne 0\end{array}\\ \begin{array}{cc}a,& x=0\end{array}\end{array}$ then the value of $a$ in order that $f\left(x\right)$ may be continuous at $x=0$ is

• A. $-8$
• B. $8$
• C. $-4$
• D. $4$

Answer 1

The correct option is A $-8$

Given: $f\left(x\right)=\{\begin{array}{c}\begin{array}{cc}\frac{{\cos }^{2}x-{\sin }^{2}x-1}{\sqrt{x^2+4}-2},& (x\ne 0)\end{array}\\ \begin{array}{cc}a,& (x\ne 0)\end{array}\end{array}$ is continuous at $x=0$, then

$\underset{x\to 0}{lim}f\left(x\right)=\underset{x\to 0}{lim}\frac{{\cos }^{2}x-{\sin }^{2}x-1}{\sqrt{x^2+4}-2}$

$=\underset{x\to 0}{lim}\frac{\cos 2x-1}{\sqrt{x^2+4}-2}$

As th function is of the $\frac{0}{0}$ form,applying L-Hospital's rule,

$=\underset{x\to 0}{lim}\left[\frac{\frac{-2\sin 2x}{x}}{\sqrt{x^2+4}}\right]$

$=\underset{x\to 0}{lim}2\frac{-\sin 2x\sqrt{x^2+4}}{x}$

$=-\underset{x\to 0}{lim}2[2\cos 2x\sqrt{x^2+4}+\sin 2x\frac{x}{\sqrt{x^2+4}}]$

$=-\left[4\sqrt{4}\right]=-8$

Thus, $\underset{x\to 0}{lim}f\left(x\right)=-8$

For function to be continuous at $x=0$.

$f\left(0\right)=\underset{x\to 0}{lim}f\left(x\right)$ $\Rightarrow a=-8$