#### Provide solution for RD Sharma math class 12 chapter Mean value theorem exercise 14.1 question 8 sub question (ii)

(0,e)

Hint:

All condition of Rolle’s Theorem satisfied.

Given:

$y=e^{1-x^{2}}, x \in[-1,1]$

Explanation:

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

1. Being polynomial f(x) is continuous for all x and hence continuous in [-1,1].
2. $f^{\prime}(x)=-2 x e^{1-x^{2}}$ , which exist (-1,1)

$\therefore f(x )$ is derivable in (-1,1)

3.

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

Thus all conditions of Rolle’s Theorem are satisfied.

Therefore, there exists at least one$c \in(-1,1)$ such that f'(c)=0

Nowf'(c)=0

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

Hence by geometrical transformation of Rolle’s Theorem (0,e)