If f(x) = Ax2 + Bx + C is such that f(a) = f(b), then write the value of c in Rolle’s theorem.
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
2Ac + B = 0
-------- (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,
{As a ≠ b}
From equation (i)
Hence the required value is .