Manipal Entrance Test 2014 Math Paper

Here, we observe that
(a) f(x) is a polynomial, so it is continuous in the interval [0, 2].
(b) f′(x)=3x2−6x+2 exists for all x∈(0,2)
So, f(x) is differentiable for all x∈(0,2) and
(c) f(0)=0,f(2)=23−3(2)2+2(2)=0
⇒f(0)=f(2)
Thus, all the three conditions of Rolle's theorem are satisfied.
So, there must exist c∈[0,2] such that f′(c)=0
⇒ f′(c)=3c2−6c+2=0
⇒ c=1±

1

√3

∈[0,2]