Question

Let $f:R\to R$ be a function defined by
$f\left(x\right)=\{\begin{array}{cc}\frac{1-\cos ax}{x^2}& \text{if}x<0\\ b& \text{if}x=0\\ \frac{\sqrt{x}}{\sqrt{16+\sqrt{x}}-4}& \text{if}x>0\end{array}$

If $f$ is continuous in $R,$ then $b-a$ can be

• A. $8$
• B. $6$
• C. $2$
• D. $4$

Answer 1

The correct option is D $4$

$f$ is continuous in $R$
$\Rightarrow f$ is continuous at $x=0$ also.
$\underset{x\to 0^{-}}{lim}f\left(x\right)=\underset{x\to 0^{+}}{lim}f\left(x\right)=f\left(0\right)$

$\underset{x\to 0^{+}}{lim}f\left(x\right)=\underset{x\to 0^{+}}{lim}\frac{\sqrt{x}}{\sqrt{16+\sqrt{x}}-4}$
$=\underset{x\to 0^{+}}{lim}\frac{\sqrt{x}}{16+\sqrt{x}-16}\cdot (\sqrt{16+\sqrt{x}}+4)$
$=\underset{x\to 0^{+}}{lim}(\sqrt{16+\sqrt{x}}+4)=4+4=8$
$\therefore b=8$

$\underset{x\to 0^{-}}{lim}f\left(x\right)=\underset{x\to 0^{-}}{lim}\frac{1-\cos ax}{x^2}=8$
$\Rightarrow \underset{x\to 0^{-}}{lim}\frac{2{\sin }^2\frac{a}{2}x}{x^2}=8$
$\Rightarrow \underset{x\to 0^{-}}{lim}{\left(\frac{\sin \frac{a}{2}x}{x}\right)}^2=4$
$\Rightarrow \underset{x\to 0^{-}}{lim}{\left(\frac{\sin \frac{a}{2}x}{\frac{a}{2}x}\right)}^2\times {\left(\frac{a}{2}\right)}^2=4$
$\Rightarrow 1\times {\left(\frac{a}{2}\right)}^2=4$
$\Rightarrow a=\pm 4$

For $a=4,b=8,$ we have $b-a=4$