Question

If $f\left(x\right)=\{\begin{array}{cc}\frac{1-\cos 4x}{x^2},& whenx<0\\ a,& whenx=0\\ \frac{\sqrt{x}}{\sqrt{(16+\sqrt{x})}-4},& whenx>0\end{array}$ is continuous at $x=0$, then the value of a will be.

• A. $8$
• B. $-8$
• C. $4$
• D. None of these

Answer 1

The correct option is A $8$

since the function is continues at x=0

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

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

$=\underset{x\to 0^{-}}{lim}\left(\frac{2{\sin }^{2}2x}{(2x{)}^2}\right)\cdot 4=8$

$\underset{x\to 0^{+}}{lim}f\left(x\right)=\underset{x\to 0^{+}}{lim}\frac{\sqrt{x}}{\sqrt{(16+\sqrt{x})}-4}$

$=\underset{x\to 0^{+}}{lim}\sqrt{(16+\sqrt{x})}+4=8$

also,$f\left(0\right)=8$

Hence, $a=8$.