Question

Let ƒ(x) be a polynomial of degree three satisfying ƒ(0) = –1 and ƒ(1) = 0. Also, 0 is a stationary point of ƒ(x) does not have an extremum at x = 0, then the value of the integral $\int_{}^{}\frac{ƒ(x)}{x^{3} - 1}$dx is –

• A. $\frac{x^{2}}{2} + c$
• B. x + c
• C. $\frac{x^{3}}{b} + c$
• D. None of these

Answer 1

Answer: (B)

x + c

Answer 2

Let ƒ(x) = ax³ + bx² + cx + d. Then,

ƒ(0) = –1 and ƒ(1) = 0

Ž d = –1 and a + b + c + d = 0

Ž d = –1 and a + b + c = 1… (1)

It is given that x = 0 is a stationary point of ƒ(x) but it is not a point of extremum. Therefore,

ƒ¢(0) = 0, ƒ¢¢(0) = 0 and ƒ¢¢¢(0) ¹ 0

Now ƒ(x) = ax³ + bx² + cx + d

Ž ƒ¢(x) = 3ax² + 2bx + c, ƒ¢¢(x) = 6ax + b and ƒ¢¢¢(x) = 6a

\ ƒ¢(0) = 0, ƒ¢¢(0) = 0 and ƒ¢¢¢(0) ¹ 0

Ž c = 0, b = 0 and a ¹ 0…(2)

From equations (1) and (2), we get a = 1, b = c = 0 and

d = –1.

\ ƒ(x) = x³ – 1

Hence, $\int_{}^{}\frac{ƒ(x)}{x^{3} - 1}$ dx = $\int_{}^{}1$dx = x + c.

Hence (2) is the correct answer.