Question

If Rolle's theorem holds for the function $f\left(x\right)=\ln (2-x^2),x\in [-1,1]$ at the point $x=c$. Then the value of $c$ is

• A. $0$
• B. $2$
• C. $3$
• D. $6$

Answer 1

The correct option is A $0$

First, we determine the domain of the function:
$2-x^2>0\Rightarrow D=(-\sqrt{2},\sqrt{2}).$
Since the interval $(-1,1)$ belongs the domain of the function, the function is continuous and differentiable on $(-1,1).$
The function is even, so $f(-1)=f\left(1\right).$
Henec, Rolle's theorem is applicable to this function.
we find the derivative by the chain rule.
$f^{\prime }\left(x\right)=\left(\ln \right(2-x^2){)}^{\prime }=\frac{1}{2-x^2}\cdot (2-x^2{)}^{\prime }=-\frac{2x}{2-x^2}.$
Equating the derivative to zero, we get the value of $c:$
$f^{\prime }\left(c\right)=0\Rightarrow -\frac{2c}{2-c^2}=0,$
$\Rightarrow$$\{\begin{array}{c}-2c=0\\ 2-c^2\ne 0\end{array}$, $\Rightarrow c=0.$