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  • Provide solution for RD Sharma math class 12 chapter Mean value theorem exercise 14.3 question 3 sub question (x)

Provide solution for RD Sharma math class 12 chapter Mean value theorem exercise 14.3 question 3 sub question (x)

Answers (1)

Answer:

               c=0 \in(-1,1)

Hint:

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

Given:

               f(x)=\log \left(x^{2}+2\right)-\log3on[-1,1]

Explanation:

We have

f(x)=\log \left(x^{2}+2\right)-\log3on[-1,1]

We know that logarithmic function is continuous and differentiable on its own domain.

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

\\ \Rightarrow \quad f(-1)=\log \left((-1)^{2}+2\right)-\log 3 \\\\ \Rightarrow \quad f(-1)=\log (1+2)-\log 3\\\\ \Rightarrow \quad f(-1)=\log 3-\log 3\\\\ \therefore \quad f(-1)=0\\\\ \Rightarrow \quad f(1)=\log \left(1^{2}+2\right)-\log 3\\\\ \Rightarrow \quad f(1)=\log (1+2)-\log 3\\\\ \Rightarrow \quad f(1)=\log 3-\log 3\\\\ \therefore \quad f(1)=0

We have  f(-1)=f(1) , so there exists at c \in(-1,1), such thatf'(c) = 0

Let’s find the derivative of f(x) 

f^{\prime}(x)=\frac{d\left(\log \left(x^{2}+2\right)-\log 3\right)}{d x}

\\ \Rightarrow \quad f^{\prime}(x)=\frac{1}{x^{2}+2} \frac{d\left(x^{2}+2\right)}{d x}=0 \ \ \ \ \ \ \left[\because \frac{d(\log x)}{d x}=\frac{1}{x}\right] \\\\ \Rightarrow \quad f^{\prime}(x)=\frac{2 x}{x^{2}+2}

We have f'(c) = 0

\\ \Rightarrow \quad \frac{2 c}{c^{2}+2}=0\\\\ \Rightarrow 2c = 0 \\\\ \Rightarrow \quad c=0 \in(-1,1)

Hence, Rolle’s Theorem is verified.

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