Question

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

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

Answer 1

The correct option is B $x+c$

Let $f\left(x\right)=a{x}^3+b{x}^2+cx+d$
Since $f\left(0\right)=-1$ and $f\left(1\right)=0$
$\Rightarrow d=-1$ and $a+b+c+d=0$
$\Rightarrow d=-1$ and $a+b+c=1$ ...(1)
Since $0$ is a stationary point of $f\left(x\right)$, $\therefore f^{\prime }\left(0\right)=0$
$\Rightarrow 3a{\left(0\right)}^2+2b\left(0\right)+c=0\Rightarrow c=0$
Since $f\left(x\right)$ does not have an extremum at $x=0$,
$\therefore f^{\prime \prime }\left(0\right)=0\Rightarrow b=0$ and therefore from (1) $a=1$
So, $f\left(x\right)=x^3-1$
$\therefore \int \frac{f\left(x\right)}{x^3-1}dx=\int dx=x+c$