If f (x) = Ax² + Bx + C is such that f (a) = f (b), then write the value of c in Rolle's theorem.

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Solution

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

Differentiating the given function with respect to x, we get
$f' (x) = 2 A x + B$
$\Rightarrow f' (c) = 2 A c + B$
$∴ f' (c) = 0 \Rightarrow 2 A c + B = 0 \Rightarrow c = \frac{- B}{2 A} . . . (1)$

$∵ f (a) = f (b) ∴ A a^{2} + B a + C = A b^{2} + b B + C \Rightarrow A a^{2} + B a = A b^{2} + b B \Rightarrow A (a^{2} - b^{2}) + B (a - b) = 0 \Rightarrow A (a - b) (a + b) + B (a - b) = 0 \Rightarrow (a - b) [A (a + b) + B] = 0 \Rightarrow a = b, A = \frac{- B}{(a + b)} \Rightarrow (a + b) = \frac{- B}{A} (∵ a \neq b)$

From (1), we have

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

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