Question

If LMVT holds for the function $f\left(x\right)=\sqrt{25-x^2}$ on the interval $[0,4]$ at $x=c,$ then which of the following is/are true

• A. $c$ have two values
• B. $c$ has only one value
• C. $c^4-5c^2=5$
• D. $c^4-5c^2=0$

Answer 1

Answer: (B), (D)

$c^4-5c^2=0$

Given, $f\left(x\right)=\sqrt{25-x^2}$
$\Rightarrow f^{\prime }\left(x\right)=\frac{-x}{\sqrt{25-x^2}}$
$\Rightarrow f\left(x\right)$ is continuous in $[0,4]$ and differentiable in $(0,4)$
So, $LMVT$ is applicable.
$\therefore f^{\prime }\left(c\right)=\frac{f\left(4\right)-f\left(0\right)}{4-0}$
$\Rightarrow \frac{-c}{\sqrt{25-c^2}}=-\frac{2}{4}=-\frac{1}{2}$
$\therefore c^2=5\Rightarrow c=\pm \sqrt{5}$
But $c\in (0,4),$ then $c=\sqrt{5}$
$\Rightarrow c^4-5c^2=0$