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 $f(x)=a{x}^3+b{x}^2+cx+d$
Put $x=0$ and $x=1$
Then, we get $f(0)=-1$ and $f(1)=0$
$d=-1$ and $a+b+c+d=0$
$a+b+c=1$ ...(1)
It is given that $x=0$ is a stationary point of $f(x)$, but it is not a point of extremum.
Therefore, $f^{\prime }(0)=0=f^{\prime }(0)$ and $f^{\prime \prime }(0)=0$
Now $f(x)=a{x}^3+b{x}^2+cx+d$
$\Rightarrow f^{\prime }(x)=3a{x}^2+2bx+c,f^{\prime }(x)$
$=6ax+2b;f^{\prime }(x)=6a$'
$f^{\prime }(0),f^{\prime \prime }(0)=0$ amd $f^{\prime \prime \prime }(0)=0\equiv 0$
$c=0,b=0a\equiv 0$
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, $\int \frac{f(x)}{x^3-1}dx$
$=\int \frac{x^3-1}{x^3-1}dx=\int 1dx=x+C$