#### Explain Solution RD Sharma Class 12 Chapter 14 Inverse Trigonometric Function Exercise Very Short Answer Question 1 Maths.

$C=\frac{a+b}{2}$

Hint:

Find differentiation of $f\left ( x \right )$ and apply Rolle’s Theorem.

Given:

$f\left ( x \right )=Ax^{2}+Bx+C$ and $f\left ( a \right )=f\left ( b \right )$

Solution:

$f\left ( a \right )=f\left ( b \right )$

$\begin{array}{ll} \Rightarrow & A a^{2}+B a+C=A b^{2}+B b+C \\\\ \Rightarrow & A a^{2}+B a=A b^{2}+B b \\\\ \Rightarrow & A\left(a^{2}-b^{2}\right)+B(a-b)=0 \\\\ \Rightarrow & A(a-b)(a+b)+B(a-b)=0 \\\\ \Rightarrow & (a-b)[A(a+b)+B]=0 \\\\ \Rightarrow & A(a+b)+B=0 \text { and } a-b=0 \end{array}$

$\Rightarrow A=\frac{-B}{a+b}$   and $a-b$

$\Rightarrow A=\frac{-B}{a+b}$                                                                                                                                      $\left [ \because a\neq b \right ]$

Now,     $f\left ( x \right )=Ax^{2}+Bx+C$

${f}'\left ( x \right )=2xA+B$

${f}'\left ( C \right )=2AC+B$

By Rolle’s Theorem,

$\Rightarrow$         ${f}'\left ( C \right )=0$

$\Rightarrow$         $2AC+B=0$

$\Rightarrow$         $C=\frac{-B}{2A}$

$\Rightarrow$         $C=\frac{-B}{2\left ( \frac{-B}{a+b}\right )}$                                                                                                        $\left [ \because A=\frac{-B}{a+b} \right ]$

$\Rightarrow$         $C=\frac{-B}{2}\times\frac{a+b}{\left ( -B \right )}$

$\Rightarrow$         $C=\frac{a+b}{2}$