Answer:
Option C
Explanation:
It is clear that f(x) has a definite and unique value for each x$\in [1,5]$
Thus, for every point in the interval [1,5] the value of f(x) exists
So, f(x) is continuous in the interval [1,5]
Also, $f'(x)=\frac{-x}{\sqrt{25-x^{2}}}$, which clearly exists for all x in an open interval (1,5)
So, f'(x) is differentiable in [1,5]
So, there must be a value $c \in [1,5] $ such that
$f'(c)= \frac{f(5)-f(1)}{5-1}=\frac{0-\sqrt{24}}{4}$
$=\frac{0-2\sqrt{6}}{4}=\frac{-\sqrt{6}}{2}$
but f'(c)$=\frac{-c}{\sqrt{25-c^{2}}}$
$\Rightarrow$ $\frac{-c}{\sqrt{25-c^{2}}}=-\frac{\sqrt{6}}{2}$
$\Rightarrow$ $4c^{2}=6(25-c^{2})$
$\Rightarrow$ $4c^{2}=150- 6c^{2} \Rightarrow 10c^{2}=150$
$\Rightarrow$ $c^{2}=15 \Rightarrow c= \pm \sqrt{15}$
$\therefore$ $c= \sqrt{15} \in [1,5]$
