Question

The number of real solution of $x-\frac{1}{x^2-4}=2-\frac{1}{x^2-4}$, is:

• A. $0$
• B. $1$
• C. $2$
• D. infinite

Answer 1

The correct option is A $0$

$x-\frac{1}{x^2-4}=2-\frac{1}{x^2-4}$
Is defined when $x^2-4\ne 0\Rightarrow (x-2)(x+2)\ne 0\Rightarrow x\ne \pm 2$
Now
$x-\frac{1}{x^2-4}=2-\frac{1}{x^2-4}\Rightarrow x^3-4x-1=2x^2-8-1\Rightarrow x^3-2x^2-4x+8=0\Rightarrow (x-2)(x-2)(x+2)=0$
Hence number of real solution is $0$