bx2+cx Then, f(x) is a polynomial. So, it is continuous in R. Now, ‌‌f(0)=0 and
‌‌f(1)=‌
a
3
+‌
b
2
+c=‌
2a+3b+6c
6
⇒‌‌f(1)=0‌‌[∵2a+3b+6c=0, given ] ∵f(x) is a polynomial, so it is differentiable in R, so in (0,1). Hence, by Rolle's theorem, there exists atleast one point x∈(0,1), there exists such that f′(x)=0 ⇒‌‌ax2+bx+c=0 Hence, required interval is (0,1).