Question

For constant of integration $C,$ if $\int \frac{x(x+1)(2x^2-x+1)}{{(x^3+x^2+x-1)}^3}dx=\frac{1}{A\left(f\right(x){)}^2}+C,$ where $f\left(1\right)=2,$ then

• A. $A$ equals $-2$
• B. range of function $f$ is $(-\infty ,\infty )$
• C. $f$ has a point of inflection at $x=1$
• D. $f$ is many-one function

Answer 1

Answer: (A), (B), (C), (D)

$f$ is many-one function

Let $I=\int \frac{(2x^4+x^3+x)}{(x^3+x^2+x-1{)}^3}dx$
$=\int \frac{(2x^4+x^3+x)}{x^3{(x^2+x+1-\frac{1}{x})}^3}dx$
$=\int \frac{2x+1+\frac{1}{x^2}}{{(x^2+x+1-\frac{1}{x})}^3}dx$
Put $x^2+x+1-\frac{1}{x}=t\Rightarrow (2x+1+\frac{1}{x^2})dx=dt$
Then, $I=\int \frac{dt}{t^3}$
$=-\frac{1}{2{(x^2+x+1-\frac{1}{x})}^2}+C$
$\therefore A=-2$ and $f\left(x\right)=x^2+x+1-\frac{1}{x}$


$\underset{x\to 0^{+}}{lim}f\left(x\right)=-\infty$ and $\underset{x\to \infty }{lim}f\left(x\right)=\infty$
Also, $\underset{x\to -\infty }{lim}f\left(x\right)=\infty$
$\Rightarrow R_f=(-\infty ,\infty )$


We have $f\left(x\right)=x^2+x+1-\frac{1}{x}$
Differentiating w.r.t. $x,$ we get
$f^{\prime }\left(x\right)=2x+1+\frac{1}{x^2}$
Again differentiating w.r.t. $x,$ we get
$f^{\prime \prime }\left(x\right)=2-\frac{2}{x^3}=0$
$\Rightarrow x=1$
$f^{\prime \prime }\left(x\right)$ changes sign in neighbourhood of $x=1$
$\therefore f\left(x\right)$ has a point of inflection at $x=1$


Since $f\left(1\right)=2$ and $f(-1)=2,$
$\therefore f$ is many-one.