Solution: (1) f (x) + 2 f (1 / [1 − x]) = x Put x = 2, f (2) + 2 f (-1) = 2 -- (1) Put x = -1, f (-1) + 2 f (1 / 2) = -1 --- (2) Put x = 1... View Article
1) f (x + 2) - 2 f (x + 1) + f (x) = 0 2) f (x) + f (x + 1) = f{x(x + 1)} 3) f (x) + f (y) = f [(x + y) / (1 + xy)] 4) f (x + y) = f (x) f (y)... View Article
Solution: (2) Let f (x) = sx and g (x) = logex f {g(x)} = f (log ex) = eloge^x = ex g{ f(x)} = g (ex) = ex log e = ex f {g(x)} = g{ f(x)}... View Article
Solution: Let z = x+iy iz = -y+ix z+iz = (x+iy+i(x+iy) = (x-y)+i(x+y) So the vertices of the triangle are (x,y), (-y, x) and (x-y, x+y)... View Article
