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 the value of $\int \frac{f\left(x\right)}{x^3-1}dx$ is

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

Answer 1

Answer: (B)

$x+C$

Let $f\left(x\right)=a{x}^3+b{x}^2+cx+d$
Put $x=0$ and $x=1$
Then, we get $f\left(0\right)=-1$ and $f\left(1\right)=0$
$d=-1$ and $a+b+c+d=0$
$a+b+c=\mathrm{1....}\left(i\right)$
It is given that $x=0$ is a stationary point of $f\left(x\right)$, but it is not a point of extremum.
Therefore, $f^{\prime }\left(0\right)=0=f^{\prime }\left(0\right)$ and $f^{\prime \prime }\left(0\right)=0$
Now $f\left(x\right)=a{x}^3+b{x}^2+cx+d$
$\Rightarrow f^{\prime }\left(x\right)=3a{x}^2+2bx+c,f^{\prime }\left(x\right)=6ax+2b;f^{\prime \prime }\left(x\right)=6a$
$f^{\prime }\left(0\right),f^{\prime \prime }\left(0\right)=0$ amd $f^{\prime \prime \prime }\left(0\right)=0\ne 0$
$c=0,b=0$ $a\ne 0$
From Eqs (i) and (ii) we get
$a=1,b=c=0$ and $d=-1$
Put these values in $f\left(x\right)$
we get $f\left(x\right)=x^3-1$
Hence, $\int \frac{f\left(x\right)}{x^3-1}dx=\int \frac{x^3-1}{x^3-1}dx=\int 1dx=x+C$