VITEEE 2015 Paper

Consider the function f defined by
f (x) = a0

xn+1

n+1

+ an

xn

n

+ ... + an−1

x2

2

+ an x
Since, f(x) is a polynomial, so it is continuous and differentiable for all x. f(x) is continuous in the closed interval [0, 1] and differentiable in the open interval (0, 1).
Also, f(0) = 0
and
f (1) =

a0

n+1

+

a1

n

+ ... +

an−1

2

+ an = 0 [say]
i.e. f (0) = f (1)
Thus, all the three conditions of Rolle’s theorem are satisfied. Hence, there is atleast one value of x in the open interval
(0 , 1) where f ' (x) = 0
i.e. a0xn+a1xn−1 + ... + an = 0