The equation `(x/(x+1))^2+(x/(x-1))^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`

Options

A. four real roots, if `a gt 2`
B. two real roots, if `1 lt a lt 2`
C. four real roots for all `a lt-1`
D. all the above

Answer

D

Explanation

We have,
`((x)/(x+1))^(2)+((x)/(x-1))^(2)=a(a-1)`
`implies((x)/(x+1)+(x)/(x-1))^(2)-(2x^(2))/(x^(2)-1)=a(a-1)`
`implies ((2x^(2))/(x^(2)-1))^(2)-((2x^(2))/(x^(2)-1))=a(a-1)`
`implies ((2x^(2))/(x^(2)-1))((2x^(2))/(x^(2)-1)-1)=a(a-1)`
` implies y(y-1)=a(a-1)," where "y=(2x^(2))/(x^(2)-1)`
`implies y^(2)-a^(2)-y+a=0`
`implies (y-a)(y+a-1)=0 impliesy=a,y=1-a`
When y=a, we have
`(2x^(2))/(x^(2)-1)=a impliesx=+-sqrt((a)/(a-2))`
Clearly, `x in R`if `a in (-oo,0) cup(2,oo)`
When y=1-a, we have
`(2x^(2))/(x^(2)-1)=a impliesx=+-sqrt((a-1)/(a-2))`
Clearly, `x in R` if `a in (-oo,-1)cup (1,oo)`.
Thus,the given equation has four real rootsif `a gt2` or a lt-1`
and exactly two real roots if `1 lt a lt 2`.
Hence,all the options are true .