AIEEE 2004 Math Solved Paper

Let us define a function
f(x)=‌

ax3

3

+‌

bx2

2

+cx
Being polynomial, it is continuous and differentiable, also,
f(0)=0 and f(1)=‌

a

3

+‌

b

2

+c
⇒f(1)=‌

2a+3b+6c

6

=0 (given)
∴f(0)=f(1)
∴f(x) satisfies all conditions of Rolle's theorem
therefore f′(x)=0 has a root in (0,1)
i.e. ax2+bx+c=0 has at lease one root in (0,1)