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#### Please solve RD Sharma class 12 chapter Mean Value Theoram exercise 14.2 question 1 sub question (xi) maths textbook solution

Answer:  $\sqrt{3}$

Hint: You must know the formula of Lagrange’s Mean Value Theorem.

Given: $f(x)=x+\frac{1}{x} \text { on }[1,3]$

Solution:

$f(x)=x+\frac{1}{x}$

$f(x)$  is a polynomial function.

It is continuous in [1, 3]

$f^{\prime}(x)=1-\frac{1}{x^{2}}$

(Which is defined in [1, 3])

$\therefore f(x)$ is differentiable in [1, 3]. Hence it satisfies the condition of Lagrange’s Mean Value Theorem.

Now, there exists at least one value,

\begin{aligned} &c \in[1,3] \\ &f^{\prime}(c)=\frac{f(3)-f(1)}{3-1} \\ &1-\frac{1}{c^{2}}=\frac{3+\frac{1}{3}-1-\frac{1}{1}}{2} \end{aligned}

$1-\frac{1}{c^{2}}=\frac{3+\frac{1}{3}-1-1}{2}$

\begin{aligned} &1-\frac{1}{c^{2}}=\frac{9+1-3-3}{2 \times 3} \\ &1-\frac{1}{c^{2}}=\frac{10-6}{6} \end{aligned}

\begin{aligned} &1-\frac{1}{c^{2}}=\frac{4}{6} \\ &1-\frac{1}{c^{2}}=\frac{2}{3} \end{aligned}

\begin{aligned} &\frac{1}{c^{2}}=1-\frac{2}{3} \\ &\frac{1}{c^{2}}=\frac{3-2}{3} \end{aligned}

\begin{aligned} &\frac{1}{c^{2}}=\frac{1}{3} \\ &c^{2}=3 \\ &c=\sqrt{3} \end{aligned}