Question

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

Answer 1

We have

\[f\left( x \right) = A x^2 + Bx + C\]

Differentiating the given function with respect to x, we get

\[f'\left( x \right) = 2Ax + B\] 

\[\Rightarrow f'\left( c \right) = 2Ac + B\]

\[\therefore f'\left( c \right) = 0 \Rightarrow 2Ac + B = 0 \Rightarrow c = \frac{- B}{2A} . . . \left( 1 \right)\]

\[\because f\left( a \right) = f\left( b \right)\]

\[ \therefore A a^2 + Ba + C = A b^2 + bB + C\]

\[ \Rightarrow A a^2 + Ba = A b^2 + bB\]

\[ \Rightarrow A\left( a^2 - b^2 \right) + B\left( a - b \right) = 0\]

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

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

\[ \Rightarrow a = b, A = \frac{- B}{\left( a + b \right)}\]

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

From (1), we have

\[c = \frac{a + b}{2}\]