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Please Solve RD Sharma Class 12 Chapter 14 Inverse Trigonometric Functions Exercise Fill in the blanks Question 1 Maths Textbook Solution.

Answers (1)

Answer:

                \sqrt{3}

Hint:

Use mean value theorem’s formula.

Given:

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

Solution:

                f\left ( x \right )=x+\frac{1}{x}

On differentiating we get,

\Rightarrow         {f}'\left ( x \right )=1-\frac{1}{x^{2}}

\Rightarrow         {f}'\left ( c\right )=1-\frac{1}{c^{2}}

Apply mean value theorem,

    \begin{aligned} & f^{\prime}(c)=\frac{f(b)-f(a)}{b-a} \\ \Rightarrow & 1-\frac{1}{c^{2}}=\frac{3+\frac{1}{3}-(1+1)}{3-1} \\ \Rightarrow & 1-\frac{1}{c^{2}}=\frac{\frac{10}{3}-2}{2} \\ \Rightarrow & 1-\frac{1}{c^{2}}=\frac{4}{6} \end{aligned}

   \Rightarrow \frac{1}{c^{2}}=1-\frac{4}{6}

   \Rightarrow \frac{1}{c^{2}}=\frac{2}{6}=\frac{1}{3}

   \Rightarrow c^{2}=3

        c=\pm \sqrt{3}

   \Rightarrow c=\sqrt{3}                                                                                                                             \left [ \because x\: \epsilon \left [ 1,3 \right ] \right ]

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