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Question:

If $f(x)=A x^{2}+B x+C$ is such that $f(a)=f(b)$, then write the value of $c$ in Rolle’s theorem.

Solution:

We have

$f(x)=A x^{2}+B x+C$

Differentiating the given function with respect to x, we get

$f^{\prime}(x)=2 A x+B$

$\Rightarrow f^{\prime}(c)=2 A c+B$

$\therefore f^{\prime}(c)=0 \Rightarrow 2 A c+B=0 \Rightarrow c=\frac{-B}{2 A}$         …..(1)

$\because f(a)=f(b)$

$\therefore A a^{2}+B a+C=A b^{2}+b B+C$

$\Rightarrow A a^{2}+B a=A b^{2}+b B$

$\Rightarrow A\left(a^{2}-b^{2}\right)+B(a-b)=0$

$\Rightarrow A(a-b)(a+b)+B(a-b)=0$

$\Rightarrow(a-b)[A(a+b)+B]=0$

$\Rightarrow a=b, A=\frac{-B}{(a+b)}$

$\Rightarrow(a+b)=\frac{-B}{A} \quad(\because a \neq b)$

From (1), we have

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