The equation $(\frac{x}{x+1})^2+(\frac{x}{x-1})^2=a(a-1)$ has |
four real roots, if $a > 2$ two real roots, if $1< a< 2$ four real roots for all $a < -1$ all the above |
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. |