Question

The value of c form the Lagrange's mean value theorem for which $f\left(x\right)=\sqrt{25-x^2}$ in [1, 5], is

• A. 5
• B. 1
• C. $\sqrt{15}$
• D. None of these

Answer 1

The correct option is C $\sqrt{15}$

It is clear that f(x) has a definite and unique value of 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, $f^{\prime }\left(x\right)=\frac{-x}{\sqrt{25-x^2}}$, which clearly exists for all x in an open interval (1, 5).
So, $f^{\prime }\left(x\right)$ is differentiable in $(1,5)$.
So, there must be a value $c\in [1,5]$ such that
$f^{\prime }\left(c\right)=\frac{f\left(5\right)-f\left(1\right)}{5-1}=\frac{0-\sqrt{24}}{4}$
$=\frac{0-2\sqrt{6}}{4}=\frac{-\sqrt{6}}{2}$

But $f^{\prime }\left(c\right)=\frac{-c}{\sqrt{25-c^2}}$

$\Rightarrow \frac{-c}{\sqrt{25-c^2}}=-\frac{\sqrt{6}}{2}$

$\Rightarrow 4c^2=6(25-c^2)$
$\Rightarrow 4c^2=150-6c^2\Rightarrow 10c^2=150$
$\Rightarrow c^2=15\Rightarrow c=\pm \sqrt{15}$
$\therefore c=\sqrt{15}\in [1,5]$

Hence, option C is correct.