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

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

Answers (1)

Answer:

                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}

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