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 .


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