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

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

Answers (1)

Answer:

               (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 onec \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)

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