Integration of a function provided as an implicit function

Question

Let a differentiable function $f$ satisfies the functional rule $$f(xy) = f(x) + f(y) + xy - x -y $$ for all values of $x,y > 0$ and $f'(1)=4$.

Based on this statement there were three questions:

1. If $f(x_0)=0$, then $x_0$ lies in the interval $$ (a) (0,1) \\ (b) (1,e) \\ (c) (e, e^2) \\ (d) (e^2,e^3) $$

2. $\int\frac{f(x)}{x}dx$ is equal to: $$ (a) 3(\ln x)^2 + x + c \\ (b) 3(\ln x) + x + c \\ (c) 1.5(\ln x)^2 + x + c \\ (d) 1.5(\ln x) + x + c $$

3. If $\int{e^{f(x)}}dx = e^x(ax^3 + bx^2 + cx + d) + \lambda$, then the value of $(a+b+c+d)$ is equal to: $$ (a) -1 \\ (b) -2 \\ (c) 3 \\ (d) 6 $$

My attempt : I had no idea how to deal with implicit function. Any hints or suggestion to handle this kind of question would have been extremely helpful. Thanks for giving me the pointers. I will try and update the full solutions soon.

Answer 1

Differentiate $f(xy) = f(x) + f(y) + xy - x -y$ of the $y$ variable: $$ xf'(xy)=f'(y)+x-1. $$ Let $y=1$ then we get $$ xf'(x)=f'(1)+x-1 $$ Since $f'(1)=4$ we get $$ f'(x)=\frac{3}{x}+1 $$ Hence $$ f(x)=3\ln x +x+c $$ where $c$ is an arbitrary constant. Let $x=y=1$ in the original equation we get $f(1)=1$, hence we can get $c=0$, hence $$ f(x)=3\ln x+x. $$ Take this into the equation to examine it satisfies the condition. Now you can solve the question directly.

Answer 2

The equation itself may be solved without any regularity conditions by observing that $g$, $g(x):=f(x)-x$ satisfies the logarithmic equation $g(xy)=g(x)+g(y)$ for $g\colon (0,\infty)\to\mathbb{R}$. Thus there is some additive function $a\colon\mathbb{R}\to\mathbb{R}$, i.e., $a$ satisfies $a(x+y)=a(x)+a(y)$ for all real $x,y$, such that $g(x)=a(\ln x)$ for all $x>0$. Thus $f$ with $f(x):=a(\ln x)+x$ is the general solutions. If $f$ is assumed to be regular, for example continuous, the function $a$ has to be of the form $x\mapsto c x$ with some constant $c$. Thus in particular holde true If $f$ is differentiable. In that case the condition on $f'$ results in the form of $f$ as given in the other answer.