Manipal Entrance Test 2016 Math Paper

Consider the function f(x)=

ax3

3

+

bx2

2

+cx+d
We have, f(0)=d
and f(1)=

a

3

+

b

2

+c+d =

2a+3b+6c

6

+d
=

0

6

+d [∵2a+3b+6c=0]
Therefore, 0 and 1 are roots of the polynomial f(x).
Hence, according to Rolle's theorem, there exists atleast one root of the polynomial f′(x)=ax2+bx+c lying between 0 and 1.