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Let f(x)= r x \ 2r2 [(f(r))2-f(x)f(r) ]r2-x2-r3e f(r)r \ be differentiable in (- , 0) (0, ) and f(1)=1. Then the value of ea, such that f(a)=0, is equal to .
2
f2(x)= r x \ 2r2f(r)r+x f(r)-f(x)r-x-r3ef(r)/r \
y2=xyy'-x3ey/x (with y=f(x))
yx= dydx- x2yey/x — homogeneous; put y=vx
x dvdx= xvev ve-vdv=dx
-(1+v)e-v=x+c; x=1, v=1: c=-1- 2e
f(a)=0 v=0: -1=a-1- 2e a= 2e
ea=2
Answer: 2
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