Question

If $x$ is a non zero real number. Then the expression $(x^8-x^5-\frac{1}{x}+\frac{1}{x^4})$ is

• A. Always non negative
• B. Positive only for $7$ values of $x$
• C. Negative always
• D. Zero for exactly one value of $x>1.$

Answer 1

The correct option is A Always non negative

$=\frac{{(x^3-1)}^2}{x^4}\ge 0$ for all $x\ne 0\cdots \left(1\right)$
$x^6+x^3+1$ has only complex conjugate roots ( $3$ pairs ).
$\therefore x^6+x^3+1$ is always positive for all real $x\ne 0\cdots \left(2\right)$
Hence the given expression is non - negative from $\left(1\right)$ and $\left(2\right).$