Question

If $\left(f\right(x)-1){(x^2+x+1)}^2-\left(f\right(x)+1)(x^4+x^2+1)=0$ is true for all $x\in R-\left\{0\right\},$ then which of the following statements is (are) CORRECT ?

• A. $\left|f\right(x\left)\right|\ge 2,\forall x\in R-\left\{0\right\}$
• B. $f\left(x\right)$ has local maximum at $x=-1$
• C. $f\left(x\right)$ has local minimum at $x=1$
• D. $\underset{-\pi }{\overset{\pi }{\int }}\left(\cos x\right)f\left(x\right)dx=0$

Answer 1

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

$\underset{-\pi }{\overset{\pi }{\int }}\left(\cos x\right)f\left(x\right)dx=0$

$\left(f\right(x)-1){(x^2+x+1)}^2-\left(f\right(x)+1)(x^4+x^2+1)=0$
$\Rightarrow \left(f\right(x)-1){(x^2+x+1)}^2=\left(f\right(x)+1)(x^4+x^2+1)$
$\Rightarrow \frac{f\left(x\right)-1}{f\left(x\right)+1}=\frac{x^4+x^2+1}{{(x^2+x+1)}^2}$
$\Rightarrow \frac{f\left(x\right)-1}{f\left(x\right)+1}=\frac{x^2-x+1}{x^2+x+1}$

Using componendo and dividendo,
$f\left(x\right)=\frac{x^2+1}{x}=x+\frac{1}{x},x\ne 0$

$\Rightarrow |x+\frac{1}{x}|\ge 2$

Now, $f^{\prime }\left(x\right)=1-\frac{1}{x^2}=\frac{x^2-1}{x^2}=0$
Critical points are $x=\pm 1$
and $f^{\prime \prime }\left(x\right)=\frac{2}{x^3}$
$\Rightarrow f\left(x\right)$ has local maximum at $x=-1$
and local minimum at $x=1$

Since $f(-x)=-f\left(x\right),$ so $f\left(x\right)$ is odd.
$\Rightarrow \underset{-\pi }{\overset{\pi }{\int }}\left(\cos x\right)f\left(x\right)dx=0$