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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

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

Answers (1)

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}

 

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