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

Answer:  $\pm \frac{1}{\sqrt{2}}$

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

Given:$f(x)=\sqrt{25-x^{2}} \text { on }[-3,4]$

Solution:

$f(x)=\sqrt{25-x^{2}}$

$f(x)$  is a polynomial function.

It is continuous in [-3, 4]

\begin{aligned} &f^{\prime}(x)=\frac{1}{2}\left(25-x^{2}\right)^{\frac{-1}{2}}(-2 x) \\ &f^{\prime}(x)=\frac{-x}{\sqrt{25-x^{2}}} \end{aligned}

(Which is defined in [-3, 4])

$\therefore f(x)$ is differentiable in [-3, 4]. Hence it satisfies the condition of Lagrange’s Mean Value Theorem.

Now, there exists at least one value,

\begin{aligned} &c \in[-3,4] \\ &f^{\prime}(c)=\frac{f(4)-f(-3)}{4-(-3)} \end{aligned}

\begin{aligned} &\frac{-c}{\sqrt{25-c^{2}}}=\frac{\sqrt{25-(4)^{2}}-\sqrt{25-(-3)^{2}}}{4+3} \\ &\frac{-c}{\sqrt{25-c^{2}}}=\frac{\sqrt{25-16}-\sqrt{25-9}}{7} \end{aligned}

\begin{aligned} &\frac{-c}{\sqrt{25-c^{2}}}=\frac{\sqrt{9}-\sqrt{16}}{7} \\ &\frac{-c}{\sqrt{25-c^{2}}}=\frac{3-4}{7} \\ &\frac{-c}{\sqrt{25-c^{2}}}=\frac{-1}{7} \end{aligned}

\begin{aligned} &-7 c=-\sqrt{25-c^{2}} \\ &7 c=\sqrt{25-c^{2}} \end{aligned}

Squaring on both sides,

\begin{aligned} &49 c^{2}=25-c^{2} \\ &49 c^{2}+c^{2}-25=0 \\ &50 c^{2}-25=0 \end{aligned}

\begin{aligned} &2 c^{2}-1=0 \\ &2 c^{2}=1 \\ &c^{2}=\frac{1}{2} \\ &c^{2}=\pm \frac{1}{\sqrt{2}} \end{aligned}

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