Question

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

• A. all real roots if a > 1
• B. two real roots if 1 < a < 2
• C. no real root if a < – 1
• D. all real roots if a < – 1

Answer 1

Answer: (B), (D)

The correct options are
B two real roots if 1 < a < 2
D all real roots if a < – 1

we have ${\left(\frac{x}{x+1}\right)}^2+{\left(\frac{x}{x-1}\right)}^2=a(a-1)$
$\Rightarrow {(\frac{x}{x+1}+\frac{x}{x-1})}^2-2\left(\frac{x}{x+1}\right)\left(\frac{x}{x-1}\right)=a(a-1)\Rightarrow {\left(\frac{2x^2}{x^2-1}\right)}^2-\frac{2x^2}{x^2-1}=a(a-1)$
$\Rightarrow z^2-z-a(a-1)=0,wherez=\frac{2x^2}{x^2-1}\Rightarrow z=aor1-a$
$When,z=a,\frac{2x^2}{x^2-1}=a\Rightarrow 2x^2=a{x}^2-a\Rightarrow x=\pm \sqrt{\frac{a}{a-2}}$
$When,z=1-a,\frac{2x^2}{x^2-1}=1-a\Rightarrow 2x^2=(1-a)x^2-1+a\Rightarrow x=\pm \sqrt{\frac{a-1}{a+1}}$
$\therefore x=\pm \sqrt{\frac{a}{a-2}},\pm \sqrt{\frac{a-1}{a+1}}$
If a < – 1 $\Rightarrow$ All roots are real.
$If1<a<2\Rightarrow x=\pm \sqrt{\frac{a}{2-a}}i,\pm \sqrt{\frac{a-1}{a+1}}\Rightarrow$ Only two roots are real.
If a > 2 $\Rightarrow$ All roots are real.