Question

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

• A. f is differentiable at all points of its domain except x=4
• B. f is differentiable on $(2,\infty )\sim \left\{4\right\}$
• C. f is differentiable on $(-\infty ,\infty )$
• D. $f^{\prime }\left(x\right)=0$ for all $xϵ[2,4)$

Answer 1

Answer: (A), (B), (D)

f is differentiable at all points of its domain except x=4
f is differentiable on $(2,\infty )\sim \left\{4\right\}$
$f^{\prime }\left(x\right)=0$ for all $xϵ[2,4)$

$f\left(x\right)=\sqrt{x+2\sqrt{2x-4}}+\sqrt{x-2\sqrt{2x-4}}$
Here, $2x-4\ge 0$
$\Rightarrow x\ge 2$
So, domain of f is $[2,\infty )$
Put $t=\sqrt{2x-4}$
$f\left(x\right)=\sqrt{\frac{t^2}{2}+2+2t}+\sqrt{\frac{t^2}{2}+2-2t}$
$=\frac{1}{\sqrt{2}}(t+2)+\frac{1}{\sqrt{2}}|t-2|$
$f\left(x\right)=\{\begin{array}{cc}\frac{1}{\sqrt{2}}\times 4& \text{if}t<2\\ \sqrt{2}t& \text{if}t\ge 2\end{array}$
$f\left(x\right)=\{\begin{array}{cc}2\sqrt{2}& \text{if}xϵ[2,4)\\ 2\sqrt{x-2}& \text{if}xϵ[4,\infty )\end{array}$
Hence $f^{\prime }\left(x\right)=\{\begin{array}{cc}0& \text{if}xϵ[2,4)\\ \frac{1}{\sqrt{x-2}}& \text{if}xϵ(4,\infty )\end{array}$
$f^{\prime }\left(4^{-}\right)=0$
$f^{\prime }\left(4^{+}\right)=\frac{1}{\sqrt{2}}$
Since, $LHD\ne RHD$ at $x=4$
So,$f\left(x\right)$ is not differentiable at $x=4$