From the first relation,
dy/dx = f(x)
Equation of tangent at (0,1/2):
(y-0.5)/(x-0) = f(x)
or, x.f(x) = y-1/2
From the second relation,
dy/dx = d(f(x))
Equation of tangent at (0,1):
(y-1)/(x) = d(f(x))
x.d(f(x)) = y-1
Note (0,1/2) and (0,1) has same abscissae, so,
On the x-axis, let the common point be (h,0)
Both the equations should satisfy this point.
h.f(h) = -0.5 ............(i)
h.d(f(h)) = -1 ...........(ii)
dividing (i) and (ii),
d(f(h))/f(h) = 2
Integrating both sides,
ln (f(h)) = 2h + c
f(h) = e^(2h+c)
the function is f(x)=e^(2x+c)
Given the y=f(x) passes through (0,1), putting the values, in the above relation,
1 = e^(c)
or, c = 0
therefore the funtion is, f(x) = e^(2x)
Now,
f(x) = 2e^(x)
or, e^(2x) = 2.e^(x)
or, e^(x) = 2
or, x = ln2
so, i get just one solution. And i m really curious to know the answer. :) ..