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

all the above

We have,

$(\frac{x}{x+1})^2+(\frac{x}{x-1})^2=a(a-1)$

$⇒(\frac{x}{x+1}+\frac{x}{x-1})^2-\frac{2x^2}{x^2-1}=a(a-1)$

$⇒(\frac{2x^2}{x^2-1})^2-(\frac{2x^2}{x^2-1})=a(a-1)$

$⇒(\frac{2x^2}{x^2-1})(\frac{2x^2}{x^2-1}-1)=a(a-1)$

$⇒y (y-1)=a (a-1)$, where $y=\frac{2x^2}{x^2-1}$

$⇒y2-a2-y+a=0$

$⇒(y-a) (y+a-1)=0⇒y=a, y=1-a$

When $y = a$, we have

$\frac{2x^2}{x^2-1}=a⇒x=±\sqrt{\frac{a}{a-2}}$

Clearly, $x∈ R$ if $a∈ (-∞, 0) ∪ (2, ∞)$

When $y = 1-a$, we have

$\frac{2x^2}{x^2-1}=1-a⇒x=±\sqrt{\frac{a-1}{a+1}}$

Clearly, $x∈ R$ if $a∈ (-∞, -1) ∪ (1,∞)$

Thus, the given equation has four real roots if $a >2$ or $a <-1$ and exactly two real roots if $1 < a <2$.

Hence, all the options are true.