Răspuns:
Explicație pas cu pas:
[tex]\displaystyle\texttt{Consideram functia }g:{[1,x^2]}\rightarrow \mathbb{R},g(t)=\dfrac{e^t}{t}\texttt{ si fie G o primitiva}\\\texttt{a acesteia. Atunci: }\\f(x)=6\ln x-G(x^2)+G(1) \\\texttt{Prin derivare rezulta: }\\f'(x)=\dfrac{6}{x}-2xg(x^2)=\dfrac{6}{x}-2x\cdot\dfrac{e^{x^2}}{x^2}=\dfrac{6}{x}-\dfrac{2e^{x^2}}{x}=\dfrac{6-2e^{x^2}}{x}\\f'(x)=0\Leftrightarrow 6-2e^{x^2}=0\\~~~~~~~~~~~~~~~~~e^{x^2}=3\\~~~~~~~~~~~~~~~~~x=\sqrt{\ln 3}[/tex]
[tex]\displaystyle\texttt{Fara a face tabel sa observam ca functia este crescatoare}\\\texttt{pe }{[1,\sqrt{\ln 3}]}\texttt{ si descrescatoare pe }{[\sqrt{\ln 3},\infty)}. \texttt{Ne mai ramane sa}\\\texttt{calculam limita la infinit}\\\texttt{Mai intai sa observam ca }\lim_{x\to\infty}\int_1^{x^2}\dfrac{e^t}{t}dt=\infty.\texttt{ Motivul? Stim ca }\\\dfrac{e^t}{t}\geq1~\forall~t\geq 1,\texttt{ de unde rezulta din proprietatea de monotonie a }[/tex]
[tex]\displaystyle\texttt{integralei ca }\int_1^{x^2}\dfrac{e^t}{t}dt\geq\int_1^{x^2}dx=x^2-1. \texttt{Cum }\lim_{x\to\infty} x^2-1=\infty,\\\texttt{rezulta din criteriul majorarii ca }\lim_{x\to\infty}\int_1^{x^2}\dfrac{e^t}{t}dt=\infty.\\\texttt{Asta ne duce cu gandul la factor fortat( fiind in cazul }\infty-\infty)\\\lim_{x\to\infty}f(x)=\lim_{x\to\infty} \left(6\ln x-\int_1^{x^2}\dfrac{e^t}{t}dt\right)=\lim_{x\to\infty}\ln x\left(6-\dfrac{\int_1^{x^2}\frac{e^t}{t}dt}{\ln x}\right)[/tex]
[tex]\displaystyle\texttt{ Dar }\lim_{x\to\infty}\dfrac{\int_1^{x^2}\frac{e^t}{t}dt }{\ln x}\stackrel{\frac{\infty}{\infty}}{=}\lim_{x\to\infty} \dfrac{2xg(x^2)}{\frac{1}{x}}=\lim_{x\to\infty} 2x^2\cdot \dfrac{e^{x^2}}{x^2}=\lim_{x\to\infty}2e^{x^2}=\infty\\\texttt{Prin urmare }\lim_{x\to\infty}f(x)=\infty(6-\infty)=-\infty\\\texttt{Tinand cont si de faptul ca }f(1)=0,\texttt{ rezulta ca }\sqrt{\ln 3}\texttt{ este punct}\\\texttt{de maxim. In concluzie raspunsul corect este e).}[/tex]
