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

f(x) = Ax² + Bx + C

Differentiating the above-mentioned function with respect to ‘x’,

f’(x) = 2Ax + B

f’(c) = 2Ac + B

f’(c) = 0

\(\therefore\) 2Ac + B = 0

c = \(-\frac{B}{2A}.....(i)\)

As f(a) = f(b),

f(a) = Aa² + Ba + C

f(b) = Ab² + Bb + C

Aa² + Ba + C = Ab² + Bb + C

Aa² + Ba = Ab² + Bb

A (a² – b²) + B (a – b) = 0

A (a + b) (a – b) + B (a – b) = 0

(a – b) {A (a + b) + B} = 0

a = b, A = \(-\frac{B}{a + b}\)

a + b = \(-\frac{B}{A}\) {As a ≠ b}

From equation (i)

c = \(\frac{a + b}{2}\)

Hence the required value is = \(\frac{a + b}{2}.\)