Please solve RD Sharma class 12 chapter Mean value theorem exercise 14.1 question 3 sub question (ix) maths textbook solution

$c=0 \in[-1,1]$

Hint:

$f(-1)=f(1)$ , so there exists at$c \in(-1,1)$

Given:

$f(x)=e^{1-x^{2}}{ }_{\text {on }}[-1,1]$

Explanation:

We have

$f(x)=e^{1-x^{2}}{ }_{\text {on }}[-1,1]$

We know that exponential are continuous and differentiable on R.

Let’s find the values of the function at an extreme

$\\ \Rightarrow \quad f(-1)=e^{1-(-1)^{2}} \\\\ \Rightarrow \quad f(-1)=e^{1-1}$

$\\ \Rightarrow f(-1)=e^{0} \\\\ \therefore \quad f(-1)=1\\\\ \Rightarrow f(1)=e^{1-(1)^{2}}\\\\ \Rightarrow f(1)=e^{1-1}\\\\ \Rightarrow \quad f(1)=e^{0}\\\\ \therefore \quad f(1)=1$

We have$f(-1)=f(1)$ , so there exists at $c \in(-1,1)$ , such that$f^{\prime}(c)=0$

Let’s find the derivative of f(x)

$\\ \Rightarrow \quad f^{\prime}(x)=\frac{d\left(e^{1-x^{2}}\right)}{d x}\\\\ \Rightarrow \quad f^{\prime}(x)=\frac{e^{1-x^{2}} d\left(1-x^{2}\right)}{d x}\\\\ \Rightarrow \quad f^{\prime}(x)=e^{1-x^{2}}(-2 x)$

We have $f^{\prime}(c)=0$

$\\ \Rightarrow \quad e^{1-c^{2}}(-2 c)=0\\\\ \Rightarrow \quad 2 c=0\\\\ \Rightarrow \quad c=0 \in[-1,1]$

Hence, Rolle’s Theorem is verified.