Question

Let f be a differentiable function satisfying the functional rule

f(xy) = f(x) + f(y) + $\frac{x+y-1}{xy}$ $\forall$x, y>0

and f'(1) = 2. Find the value of [f(e$^{100}$)] .

Note : [k] denotes the greatest integer less than or equal to k.

Answer 1

99

Answer 2

1. Given:

   $f(xy) = f(x) + f(y) + \frac{x+y-1}{xy}$ for all $x,y>0$ and $f'(1)=2$.

2. Finding $f(1)$:

   Set $y=1$:

   $$f(x) = f(x) + f(1) + \frac{x+1-1}{x} \implies f(1) + \frac{x}{x} = 0 \implies f(1) + 1 = 0.$$

   Thus, $f(1)=-1$.

3. Guessing a form:

   Assume:

   $$f(x) = c\ln x + u(x).$$

   We desire $u(x)$ such that the extra term produces $\frac{x+y-1}{xy}$.

   Guess $u(x)= -\frac{1}{x}$. Then,

   $$f(x)= c\ln x -\frac{1}{x}.$$

4. Verification:

   Compute $f(xy)= c\ln(xy) -\frac{1}{xy} = c\ln x + c\ln y -\frac{1}{xy}$.

   Also,

   $$f(x) + f(y) = c\ln x - \frac{1}{x} + c\ln y - \frac{1}{y}.$$

   Then,

   $$f(x)+f(y) + \frac{x+y-1}{xy} = c\ln x + c\ln y - \left(\frac{1}{x}+\frac{1}{y}\right) + \frac{x+y-1}{xy}.$$

   Since

   $$\frac{x+y-1}{xy} - \left(\frac{1}{x}+\frac{1}{y}\right) = -\frac{1}{xy},$$

   we get

   $$f(x)+f(y)+\frac{x+y-1}{xy}= c\ln x + c\ln y -\frac{1}{xy},$$

   which matches $f(xy)$.

5. Determine $c$:

   Differentiate:

   $$f'(x)= \frac{c}{x} + \frac{1}{x^2}.$$

   At $x=1$,

   $$f'(1)= c + 1 = 2 \implies c=1.$$

6. Resulting Function:

   $$f(x)= \ln x - \frac{1}{x}.$$

7. Compute $f(e^{100})$:

   $$f(e^{100}) = \ln(e^{100}) - \frac{1}{e^{100}} = 100 - \frac{1}{e^{100}}.$$

   Since $\frac{1}{e^{100}}$ is positive yet extremely small,

   $$100 - \frac{1}{e^{100}} < 100,$$

   so the greatest integer less than or equal to $f(e^{100})$ is 99.