The number of integers a in the interval [1, 2014] for which the system of equations
`x+y=a,(x^(2))/(x-1)+(y^(2))/(y-1)=4`
has finitely many solutions is-

Options

A. 0
B. 1007
C. 2013
D. 2014

Answer

C

Explanation

`x+y=a,(x^(2))/(x-1)+(y^(2))/(y-1)=4`
`ain[1, 2014]`
`x+1+1/(x-1)+y+1+1/(y-1)=4`
`(x-1)+1/(x-1)+(y-1)+1/(y-1)=0`
`(a-2)+1/(x-1)+1/(y-1)=0`
`(a-2)+((a-2))/((x-1)(y-1))=0`
`(a-2)[1+1/(xy+1-a)]=`
`ane2` [for a = 2 infinitely many solutions]
`xy+1-a+1=0`
`x(a-x)-a+2=0`
`rArrx^(2)+ax-(2-a)=0`
`D=a^(2)+4(2+a)=a^(2)-4a+8`
alway +ve
there two real solutions.
`ane2" and " ain [1,2014]`
Total 2014 - 1 = 2013 values