The number of roots of the equation `1/x+1/(sqrt((1-x^2)))=35/12` is

Options

A. 0
B. 1
C. 2
D. 3

Answer

D

Explanation

Let `(1)/(x)=u` and `(1)/(sqrt(1-x^(2)))=upsilon`. Then,
`u+upsilon=(35)/(12)` and `u^(2)+upsilon^(2)=u^(2)upsilon^(2)`
`rArr(u+upsilon)^(2)=((35)/(12))^(2)` and `u^(2)+upsilon^(2)=u^(2)upsilon^(2)`
`rArr u^(2)+upsilon^(2)+2"u" upsilon=((35)/(12))^(2)`
`rArr (u upsilon)^(2)+2(u upsilon)-((35)/(12))^(2)=0`
`rArr (u upsilon)^(2)+(49)/(12)u upsilon-(25)/(12)u upsilon-((35)/(12))^(2)=0`
`rArr (u upsilon+(49)/(12))(u upsilon-(25)/(12))^(2)=0 rArr u upsilon=-(49)/(12)" or ",u upsilon=(25)/(12)`
Now,`u upsilon=-(49)/(12)`
`rArr (1)/(xsqrt(1-x^(2)))=(-49)/(12)rArrx^(4)-x^(2)+(12^(2))/(49^(2))=0 rArr x=-(5+sqrt(73))/(14)`
and, `u upsilon=(25)/(12)`
`rArr (1)/(x sqrt(1-x^(2)))=(25)/(12)`
`rArr x^(4)-x^(2)+(12^(2))/(25^(2))=0`
`rArr x^(4)-((9)/(25)+(16)/(25))x^(2)+(12^(2))/(25^(2))=0`
`rArr (x^(2)-(9)/(25))(x^(2)-(16)/(25))=0 rArr x=(3)/(5),(4)/(5)`