2.f(x) is a differentiable funcn. such that f(x)f(y) +2 = f(x) + f(y) +f(xy);

f’(0)=0, f’(1)=2, then find f(x)

Given,
f(x) f(y) + 2 = f(x) + f(y) + f(xy) f’(0) = 0, f’(1)= 2

Differentiatiing above expression with respect to y,

f(x) f’(y) = 1 + f’(xy) x

For y=1, f(x) f’(1) = 1 + f’(x) x
2 f(x) = 1 + x f’(x) Put x=0, f(0) = 1/2

Differentiating this expression with respect to x,

2 f’(x) = f’(x) + x f’’(x)
f’(x) = x f’’(x) Put x=1, f’’(1) = 2

Again differentiating with respect to x,

f’’(x) = x f’’’(x) + f’’(x)
f’’’(x) = 0

Integrating this expession,

f’’(x) = x + c₁ (Since, f’’(1) = 2, c₁= 1)
f’’(x) = x + 1

Integrating again,

f’(x) = x² + x + c₂ (Since, f’(0) = 0, c₂= 0)
f’’(x) = (x²/2)+ x

Integrating again,

f(x) = (x³/6) + (x²/2) + c₃ (Since, f(0) = ½ , c₃= 1/2)

f(x) = (x³/6) + (x²/2) + (1/2)