BITSAT 2015 Paper

Let f(x) = ax3+bx2 + cx + d
Put x = 0 and x = 1
Then, we get f(0) = –1 and f(1) = 0
⇒ d = – 1 and a + b + c + d = 0
⇒ a + b + c = 1 ...(i)
It is given that x = 0 is a stationary point of
f(x), but it is not a point of extremum.
Therefore, f'(0) = 0 = f ''(0) and f '''(0) = 0
Now, f(x) = ax3+bx2 + cx + d
⇒ f'(x) = 3ax2 + 2bx + c,
f ''(x) = 6ax + 2b and f '''(x) = 6a
f' = 0, f''(0) = 0 and f '''(0) = 0 ≠ 0
⇒ c = 0, b = 0 and a ≠ 0
From Eqs. (i) and (ii), we get
a = 1, b = c = 0 and d = – 1
Put these values in f(x)
we get f(x) = x3 – 1
Hence, i

f(x)

x3−1

dx = ∫

x3−1

x3−1

dx = ∫ 1 dx = x + C