[tex]f(x) = a(x+1)(x+2)(x+3)...(x+2014),\quad a\in \mathbb{R}^* \\ \\ \displaystyle \int_{x+1}^{x+2}\dfrac{f'(t)}{f(t)}\, dt =\ln\Big(f(t)\Big)\Bigg|_{x+1}^{x+2} = \ln\Big(f(x+2)\Big)-\ln\Big(f(x+1)\Big) = \\ \\ = \ln\Big(\dfrac{f(x+2)}{f(x+1)}\Big) = \ln\Big[\dfrac{a(x+3)(x+4)(x+5)...(x+2016)}{a(x+2)(x+3)(x+4)...(x+2015)}\Big] = \\ \\ = \ln\Big[\dfrac{x+2016}{x+2}\Big] = \ln(x+2016)-\ln(x+2)\\ \\ \Rightarrow \ln(x+2016)-\ln(x+2) = \ln(x+2016)-x^2 \\\\ \Rightarrow x^2 = \ln(x+2)[/tex]
Facem graficul lui x² si lui ln(x+2) in acelasi sistem de coordonate.
[tex]g(x) = \ln(x+2) \\ g'(x) = \dfrac{1}{x+2}[/tex]
=> Ecuatia are o solutie negativa si una pozitiva.
=> n = 1, m = 1
Am atașat imaginea cu graficul.
