Let `f(x)+f(y)=f(xsqrt(1-y^2)+ysqrt(1-x^2))[f(x)` is not identically zero]. Then `a) f(4x^3-3x)+3f(x)=0` `b) f(4x^3-3x)=3f(x)` `c) f(2xsqrt(1-x^2)+2f(x)=0` `d) f(2xsqrt(1-x^2)=2f(x)`

Options

A. `f(4x^(3)-3x)+3f(x)=0`
B. `f(4x^(3)-3x)=3f(x)`
C. `f(2x sqrt(1-x^(2)))+2f(x)=0`
D. `f(2x sqrt(1-x^(2)))=2f(x)`

Answer

A, D

Explanation

Given `f(x)+f(y)=(xsqrt(1-y^(2))+ysqrt(1-x^(2))) " (1)" `
Replace y by x. Then `2f(x)=f(2xsqrt(1-x^(2)))`
`3f(x)=f(x)+2f(x)`
`=f(x)+f(2xsqrt(1-x^(2)))`
`=f(x sqrt(1-4x^(2)(1-x^(2)))+2xsqrt(1-x^(2))sqrt(1-x^(2)))`
`=f(xsqrt((2x^(2)-1)^(2))+2x(1-x^(2)))`
`=f(x|2x^(2)-1|+2x-2x^(3))`
`=f(2x^(3)-x+2x-2x^(3)) or f(x-2x^(3)+2x-2x^(3))`
`=f(x) or f(3-4x^(3))`
` :. f(x)=0 or 3f(x)=f(3x-4x^(3))`
Consider `3f(x)=f(3x-4x^(3))`
Replacing x by -x, we get
`3f(-x)=f(4x^(3)-3x) " (2)" `
Also, from (1),`f(x)+f(-x)=f(0)`.
Putting `x=y-0` in (1), we have `f(0)=0 " or " f(x)+f(-x)=0.`
Thus, f(x) is an odd function.
Now, from (2), `-3f(x) =f(4x^(3) -3x)`
or `f(4x^(3)-3x)=3f(x)=0`