Question

Number of real solution $(x,y)$ of the equation $x^2+\frac{1}{x^2}=2^{1-y^2}$ is

• A. $0$
• B. $1$
• C. $2$
• D. $4$

Answer 1

The correct option is C $2$

$x^2+\frac{1}{x^2}\ge 2$ for all $x$ ($AM-GM$ inequality)
$2^{1-y^2}\ge 2$
$2y^2\le 1$
But $y^2\ge 0$ for real $y$
$\therefore 2y^2\ge 2^0=1$
$\therefore 2y^2=1$ i.e., $y=0$ is the only value;
$\text{if}y=0,x^2+\frac{1}{x^2}=2$
$\therefore x=\pm 1$
$(1,0),(-1,0)$ are the $2$ solutions.
Number of real solutions $=2$.
$x^5(x^3-1)+\frac{(1-x^3)}{x^4}=\frac{(x^3-1)}{x^4}(x^9-1)$
$=\frac{{(x^3-1)}^2(x^6+x^3+1)}{x^4}$