Question

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

• A. $\frac{x^2}{2 } + C $
• B. $x + C $
• C. $\frac{x^3}{6} + C $
• D. None of these

Answer 1

Answer: (B)

Let $f(x)=a x^{3}+b x^{2}+c x+ d$
Put $x=0$ and $x=1$
Then, we get $f(0)=-1$ and $f(1)=0$
$d=-1$ and $a+b+c+d=0$
$a+b+c=1$ ...(1)
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 $f(x)=a x^{3}+b x^{2}+c x+ d$
$\Rightarrow f '( x )=3 ax ^{2}+2 bx + c , f'( x )$
$=6 ax +2 b ; f'( x )=6 a$'
$f'(0), f''(0)=0$ amd $f'''(0)=0 \equiv 0$
$c=0, b=0 a \equiv 0$
From Eqs (i) and (ii) we get
$a =1, b = c =0$ and $d =-1$
Put these values in $f(x)$
we get $f(x)=x^{3}-1$
Hence, $\int \frac{ f ( x )}{ x ^{3}-1} d x $
$=\int \frac{ x ^{3}-1}{ x ^{3}-1} dx =\int 1 dx = x + C$