Question

The equation ${\left(\frac{x}{x+1}\right)}^2+{\left(\frac{x}{x-1}\right)}^2=a(a-1)$ has

• A. four real roots if $a>2$
• B. four real roots if $a<-1$
• C. two real roots if $1<a<2$
• D. no real roots if $a<-1$

Answer 1

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

The correct options are
A four real roots if $a>2$
B four real roots if $a<-1$
C two real roots if $1<a<2$

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

Let $\frac{2x^2}{x^2-1}=t$
$\Rightarrow t^2-t-a(a-1)=0\Rightarrow t=a\text{or}t=1-a\Rightarrow \frac{2x^2}{x^2-1}=a\text{or}\frac{2x^2}{x^2-1}=1-a\Rightarrow x=\pm \sqrt{\frac{a}{a-2}}\text{or}x=\pm \sqrt{\frac{a-1}{a+1}}$

When $a<-1\Rightarrow$ all roots are real
$1<a<2\Rightarrow x=\pm \sqrt{\frac{a}{2-a}}i,\pm \sqrt{\frac{a-1}{a+1}}\Rightarrow$ two real roots
When $a>2\Rightarrow$ all roots are real