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

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               c=0 \in[-1,1]


               f(-1)=f(1) , so there exists atc \in(-1,1) 


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


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 havef(-1)=f(1) , so there exists at c \in(-1,1) , such thatf^{\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.

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