28251: Difference between revisions
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'''Soluție.''' a) Funcția <math> f: [0,1] \longrightarrow \mathbb{R}, \quad f(x) = \ln\sqrt{1 + \frac{4x}{n^3}}</math> are toate proprietățile din enunț. | '''Soluție.''' a) Funcția <math display="block"> f: [0,1] \longrightarrow \mathbb{R}, \quad f(x) = \ln\sqrt{1 + \frac{4x}{n^3}}</math> are toate proprietățile din enunț. | ||
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b) Deoarece <math> e^t \geq t + 1</math> pentru orice <math> t \in \mathbb{R}</math>, avem | b) Deoarece <math> e^t \geq t + 1</math> pentru orice <math> t \in \mathbb{R}</math>, avem<math display="block"> 1 + \frac{2}{n^3} = \int_{0}^{1} e^{2f(x)} dx \ge \int_{0}^{1} \left(2f(x) + 1\right) dx = 2\int_{0}^{1} f(x)dx + 1,</math> | ||
de unde rezultă că <math display="block"> \int_{0}^{1} f(x)dx\leq \frac{1}{n^3}.</math>Cum <math> \int_{0}^{1} x^{n^3-1}dx = \dfrac{1}{n^3}</math>, deducem că <math> \int_{0}^{1} \left(f(x) - x^{n^3-1} \right) dx \leq 0 </math>, deci există <math> a \in [0,1] </math>, astfel încât <math> f(a) - a^{n^{3}-1} \leq 0 </math>. | |||
<math> 1 + \frac{2}{n^3} = \int_{0}^{1} e^{2f(x)} dx \ | |||
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Functia <math> g : [0,1] \longrightarrow \mathbb{R}, g(x) = f(x) - x^{n^{3}}</math> este continuă și <math> g(0) \cdot g(a) \leq 0</math>. | |||
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Rezultă că există <math> c \in [0,a] \subseteq [0,1]</math> astfel încât <math> g(c) = 0 </math>. | |||
Rezultă că | |||
Revision as of 18:58, 7 January 2024
28251 (Gheorghe Boroica)
Fie un număr natural și o funcție continuă astfel încât și Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int_{0}^{1} e^{2f(x)} dx = 1+\frac{2}{n^3}}
.
a) Dați un exemplu de o funcție Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f}
cu proprietățile din enunț.
b) Arătați că există Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c \in [0,1] }
astfel încât Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(c) = c^{n^{3}} - 1 }
.
![{\displaystyle f:[0,1]\longrightarrow \mathbb {R} }](https://wikimedia.org/api/rest_v1/media/math/render/svg/7e2685a8f42d36b37dbfb4ab625e6f2f9be12745)
Soluție. a) Funcția Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f: [0,1] \longrightarrow \mathbb{R}, \quad f(x) = \ln\sqrt{1 + \frac{4x}{n^3}}}
are toate proprietățile din enunț.
b) Deoarece Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle e^t \geq t + 1}
pentru orice Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle t \in \mathbb{R}}
, avemFailed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 1 + \frac{2}{n^3} = \int_{0}^{1} e^{2f(x)} dx \ge \int_{0}^{1} \left(2f(x) + 1\right) dx = 2\int_{0}^{1} f(x)dx + 1,}
de unde rezultă că Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int_{0}^{1} f(x)dx\leq \frac{1}{n^3}.}
Cum Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int_{0}^{1} x^{n^3-1}dx = \dfrac{1}{n^3}}
, deducem că Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int_{0}^{1} \left(f(x) - x^{n^3-1} \right) dx \leq 0 }
, deci există Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a \in [0,1] }
, astfel încât Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(a) - a^{n^{3}-1} \leq 0 }
.
Functia Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g : [0,1] \longrightarrow \mathbb{R}, g(x) = f(x) - x^{n^{3}}}
este continuă și Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g(0) \cdot g(a) \leq 0}
.
Rezultă că există Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c \in [0,a] \subseteq [0,1]}
astfel încât Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g(c) = 0 }
.