Răspuns:
Explicație pas cu pas:
[tex]\displaystyle 14.\texttt{Fie functia }g:[x,x+1]\rightarrow\mathbb{R},g(x)=\dfrac{x^2}{\sqrt{x^4+x^2+1}},\texttt{ continua si }\\\texttt{strict crescatoare.Conforma teoremei de medie integral edition}\\\texttt{rezulta ca }\exists~c_x\in[x,x+1]\texttt{ astfel incat }\\g(c_x)=\dfrac{1}{x+1-x}\int_x^{x+1}g(t)dt=\int_x^{x+1}g(t)dt.\\x\leq c_x\leq x+1\\g(x)\leq g(c_x)\leq g(x+1)\\\dfrac{x^2}{\sqrt{x^4+x^2+1}}\leq\int_x^{x+1}\dfrac{t^2}{\sqrt{t^4+t^2+1}}dt\leq\dfrac{(x+1)^2}{\sqrt{(x+1)^4+(x+1)^2+1}}[/tex]
[tex]\displaystyle\texttt{Cum }\lim_{x\to\infty}\dfrac{x^2}{\sqrt{x^4+x^2+1}}=\lim_{x\to\infty}\dfrac{(x+1)^2}{\sqrt{(x+1)^4+(x+1)^2+1}}=1,\texttt{rezulta }\\\texttt{din criteriul clestelui ca }\lim_{x\to\infty}\int_x^{x+1}\dfrac{x^2}{\srt{x^4+x^2+1}}dx=1.\\\texttt{Raspunsul crect este b).}[/tex]
[tex]\displaystyle15. \texttt{Evident } 2\leq 3+\cos t\leq 4\texttt{ de unde }\dfrac{1}{2}\geq\dfrac{1}{3+\cos t}\geq\dfrac{1}{4}~.\texttt{Din }\\\texttt{proprietatea de monotonie a integralei rezulta }\\\int_0^x\dfrac{1}{2}dt\geq\int_0^x\dfrac{dt}{3+\cos t}\geq\int_0^x\dfrac{1}{4}dt\\\dfrac{x}{2}\geq\int_0^x\dfrac{dt}{3+\cos t}\geq\dfrac{x}{4}|:x^2\\\dfrac{1}{2x}\geq\dfrac{1}{x^2}\int_0^x\dfrac{dt}{3+\cos t}\geq\dfrac{1}{4x}[/tex]
[tex]\displaystyle\texttt{Cum }\lim_{x\to\infty}\dfrac{1}{2x}=\lim_{x\to\infty}\dfrac{1}{4x}=0~\texttt{rezulta din criteriul clestelui}\\\texttt{ca }\lim_{x\to\infty}\int_0^x\dfrac{dt}{3+\cos t}=0.\texttt{ Raspunsul corect este e) }[/tex]
