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  • need solution for rd sharma maths class 12 chapter 14 Mean Value Theoram exercise multiple choice question 3

need solution for rd sharma maths class 12 chapter 14 Mean Value Theoram exercise multiple choice question 3

Answers (1)

best_answer

Answer:

Option (b)

Hint:

You must know about Lagrange’s mean value theorem.

Given:

            f(x)=x+\frac{1}{x},\; x\in \left [ 1,3 \right ]

Solution:

            f(x)=x+\frac{1}{x},\; x\in \left [ 1,3 \right ]

            f(x)=\frac{x^{2}+1}{x}

Using mean value theorem,

            f^{'}(c)=\frac{f\left ( b \right )-f\left ( a \right )}{b-a}

            f^{'}(c)=\frac{f\left ( 3 \right )-f\left ( 1 \right )}{3-1}                    \left [ \because x=c,\; b=3,\; a=1 \right ]

            f^{'}(c)=\frac{\left ( 3+\frac{1}{3} \right )-\left ( 1+\frac{1}{1} \right )}{3-1}

\Rightarrow \; 1-\frac{1}{c^{2}}=\frac{\frac{9+1-6}{3}}{2}

\Rightarrow \; 1-\frac{1}{c^{2}}=\frac{4}{3}\times \frac{1}{2}

\Rightarrow \; 1-\frac{1}{c^{2}}=\frac{2}{3}

\Rightarrow \; 1-\frac{2}{3}=\frac{1}{c^{2}}

\Rightarrow \; 1-\frac{2}{3}=\frac{1}{c^{2}}

\Rightarrow \; \frac{1}{3}=\frac{1}{c^{2}}

\Rightarrow \; c^{2}=3

\Rightarrow \;c=\pm \sqrt{3}

but only c=\sqrt{3}\in \left [ 1,3 \right ]

Hence option (b) is correct.

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