Question

If $f\left(x\right)=\sqrt{x+2\sqrt{2x-4}}+\sqrt{x-2\sqrt{2x-4}}$, then the value of $10f^{\prime }\left({102}^{+}\right)$ is

• A. $-1$
• B. $0$
• C. $1$
• D. Does not exist

Answer 1

Answer: (C)

$1$

$f\left(x\right)=\sqrt{x+2\sqrt{2x-4}}+\sqrt{x-2\sqrt{2x-4}}$
$f\left(x\right)=\sqrt{{(\sqrt{x-2}+\sqrt{2})}^2}+\sqrt{{(\sqrt{x-2}-\sqrt{2})}^2}$
$f\left(x\right)=|\sqrt{x-2}+\sqrt{2}|+|\sqrt{x-2}-\sqrt{2}|$
For $\sqrt{x-2}$ to exist, $x\ge 2$
Also $\sqrt{x-1}+\sqrt{2}>0$.............(always true)
But $\sqrt{x-1}-\sqrt{2}\ge 0$ only if $x\ge 4$
$<0$ only if $x<4$
Now $f\left(x\right)$ becomes
$f\left(x\right)=\sqrt{x-2}+\sqrt{2}-\sqrt{x-2}+\sqrt{2}$ for $2\le x<4$
$f\left(x\right)=\sqrt{x-2}+\sqrt{2}+\sqrt{x-2}-\sqrt{2}$ for $x\ge 4$
$f\left(x\right)=2\sqrt{2,}$ for $2\le x<4$
$f\left(x\right)=2\sqrt{x-2}$ for $4\le x<\infty$

f is continous $[2,4]\cup [4,\infty ]$
$f^{\prime }\left(x\right)=0$ for $2\le x<4$

$f^{\prime }\left(x\right)=\frac{1}{\sqrt{x-2}}$ for $\le x<\infty$
$f^{\prime }\left({102}^{+}\right)=\frac{1}{\sqrt{102-2}}=\frac{1}{10}10f^{\prime }\left({102}^{+}\right)=1$