Question

lf $f\left(x\right)=\sqrt{x+2\sqrt{2x-4}}+\sqrt{x-2\sqrt{2x-4}}$ then $f\left(x\right)$ is differentiable on

• A. $(-\infty ,\infty )$
• B. $[$2,$\infty$)-$\left\{4\right\}$
• C. $[2,\infty )$
• D. $(0,\infty )$

Answer 1

Answer: (B)

$[$2,$\infty$)-$\left\{4\right\}$

$f\left(x\right)=\sqrt{x+2\sqrt{2x-4}}+\sqrt{x-2\sqrt{2x-4}}$

On rearranging the equation we get,

$=\sqrt{{\left(\frac{\sqrt{2x-4}}{2}\right)}^2+2+\sqrt{2x-4}}+\sqrt{{\left(\frac{\sqrt{2x-4}}{2}\right)}^2+2+\sqrt{2x-4}}$

$=\frac{1}{\sqrt{2}}\sqrt{4+{\left(\sqrt{2x-4}\right)}^2+4\sqrt{2x-4}}+\frac{1}{\sqrt{2}}\sqrt{4+{\left(\sqrt{2x-4}\right)}^2-4\sqrt{2x-4}}$

$=\frac{1}{\sqrt{2}}[\left(\sqrt{2x-4}\right)+2]+(\sqrt{2x-4}-2)$

$\Rightarrow [1+\frac{dy}{dx}]=0\Rightarrow \frac{dy}{dx}=-1$

$\Rightarrow f\left(x\right)=\{\begin{array}{c}\begin{array}{cc}\frac{4}{\sqrt{2}},& \sqrt{2x-4}-2<0\Rightarrow \sqrt{2x-4}<0\Rightarrow x<4\end{array}\\ \begin{array}{cc}\sqrt{x-2},& \sqrt{2x-4}\ge 2\Rightarrow 2x\ge 8\Rightarrow x\ge 4\end{array}\\ \begin{array}{cc}2\sqrt{2},& 2\le x<4\end{array}\end{array}$

Now $f^{\prime }\left(x\right)=\{\begin{array}{c}\begin{array}{cc}\frac{1}{2\sqrt{x-2}},& 4\le x<\infty \end{array}\\ \begin{array}{cc}0,& 2\le x\le 4\end{array}\end{array}$

But when $x=4,f^{\prime }\left(4\right)=\frac{1}{0}$ not defined.

$\therefore f\left(x\right)$ is differentiable on $[2,4)\cup (4,\infty )\Rightarrow x\in [2,\infty )-\left\{4\right\}$