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  • Explain Solution RD Sharma Class 12 Chapter 14 Inverse Trigonometric Function Exercise Very Short Answer Question 5 Maths.

Explain Solution RD Sharma Class 12 Chapter 14 Inverse Trigonometric Function Exercise Very Short Answer Question 5 Maths.

Answers (1)

Answer:

                c=\sqrt{5}

Hint:

Find differentiability of f\left ( x \right ) and then apply the formula of {f}'\left ( c \right ).

Given:

                f\left ( x \right )=\sqrt{x^{2}-4},x\: \epsilon \left [ 2,3 \right ]

Solution:

                f\left ( x \right )=\sqrt{x^{2}-4}

            f\left ( x \right )=\left ( x^{2}-4 \right )^{\frac{1}{2}}

Differentiate,

\Rightarrow {f}'\left ( x \right )=\frac{1}{2}\left ( x^{2}-4 \right )^{\frac{1}{2}-1}\cdot 2x                                                                     \left [ \because \frac{d}{dx}\left ( x^{n} \right ) =n\left ( x \right )^{n-1}\right ]

  \begin{aligned} \Rightarrow &f^{\prime}(x)=\frac{1}{2}\left(x^{2}-4\right)^{\frac{-1}{2}} \cdot 2 x \\\\ \Rightarrow &f^{\prime}(x)=\frac{1}{2 \sqrt{x^{2}-4}} \cdot 2 x \\\\ \Rightarrow &f^{\prime}(x)=\frac{x}{\sqrt{x^{2}-4}} \\\\ \Rightarrow &f^{\prime}(c)=\frac{c}{\sqrt{c^{2}-4}} \end{aligned}    

 Use mean value theorem,

\begin{aligned} & f^{\prime}(c)=\frac{f(b)-f(a)}{b-a} \\\\ \Rightarrow & \frac{c}{\sqrt{c^{2}-4}}=\frac{\sqrt{5}-0}{3-2} \\\\ \Rightarrow & \frac{c}{\sqrt{c^{2}-4}}=\sqrt{5} \\\\ \Rightarrow & c=\sqrt{5}\left(\sqrt{c^{2}-4}\right) \\\\ \Rightarrow & c^{2}=5\left(c^{2}-4\right) \end{aligned}        

\Rightarrow c^{2}=5c^{2}-20        

\Rightarrow4 c^{2}=20        

\Rightarrow c^{2}=5        

\Rightarrow c=\pm \sqrt{5}                                                                                                                                    \left [ \because x\: \epsilon \left [ 2,3 \right ] \right ]

\Rightarrow c=\sqrt{5}        

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