Matematic? M1, Varianta 2, BAC 2010

Scris de Cristina Vu?can   
Sâmbătă, 19 Februarie 2011 11:56
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Subiectul I

1. S? se arate c? num?rul {tex}(1-i)^24{/tex} este real.
Rezolvare: Vom scrie num?rul sub form? trigonometric?. Fie
{tex}z=1-i{/tex}.
Atunci
{tex}r=|z|=\sqrt{a^2+b^2}=\sqrt{1^2+(-1)^2}=\sqrt{2}{/tex};
iar imaginea lui z este un punct din cadranul IV, prin urmare
{tex}\fi^*=2\pi-\arctan \left|\frac{b}{a}\right|=2\pi-\arctan 1=2\pi-\frac{\pi}{4}=\frac{7\pi}{4}{/tex}.
Forma trigonometric? a num?rului complex z este:
{tex}z=r(\cos \fi^*+i\sin \fi^*)=\sqrt{2}\left(\cos \frac{7\pi}{4}+i\sin \frac{7\pi}{4}\right){/tex}
iar
{tex}z^24=r^24(\cos 24\fi^*+i\sin 24\fi^*)={/tex}
{tex}=(\sqrt{2})^24\left[\cos \left(24\cdot \frac{7\pi}{4}\right)+i\sin \left(24\cdot \frac{7\pi}{4}\right)\right]=2^{12}(\cos 42\pi+i\sin 42\pi)=2^{12}[\cos (0+2\cdot21\pi)+i\sin (0+2\cdot21\pi)]=2^{12}\in R{/tex}.
Prin urmare, num?rul dat este egal cu {tex}2^{12}{/tex} ?i este un num?r real.

 

2. S? se rezolve n mul?imea numerelor reale ecua?ia {tex}\frac{3x-1}{x+1}+\frac{x+1}{2x-1}=3{/tex}.
Rezolvare: Domeniul de existen?? a solu?iilor ecua?iei este {tex}D=R\setminus \left\{-1;\frac{1}{2}\right\}{/tex}.
Atunci
{tex}\frac{3x-1}{x+1}+\frac{x+1}{2x-1}=3\Leftrightarrow{/tex}
{tex}(3x-1)(2x-1)+(x+1)^2=3(x+1)(2x-1)\Leftrightarrow{/tex}
{tex}6x^2-3x-2x+1+x^2+2x+1=3(2x^2+2x-x-1)\Leftrightarrow{/tex}
{tex}7x^2-3x+2=6x^2+3x-3\Leftrightarrow{/tex}
{tex}7x^2-3x+2-6x^2-3x+3=0\Leftrightarrow{/tex}
{tex}x^2-6x+5=0{/tex}.
Avem
{tex}\Delta=\left(\frac{b}{2}\right)^2-ac={/tex}
{tex}=(-3)^2-1\cdot5=9-5=4{/tex};
{tex}x_{1}=\frac{\frac{-b}{2}+\sqrt{\Delta}}{a}{/tex}; {tex}x_{2}=\frac{\frac{-b}{2}-\sqrt{\Delta}}{a}{/tex}.
Adic?
{tex}x_{1}=\frac{3+\sqrt{4}}{1}=3+2=5\in D{/tex}; {tex}x_{1}=\frac{3-\sqrt{4}}{1}=3-2=1\in D{/tex}.
Mul?imea solu?iilor ecua?iei este {tex}S=\{1;5\}{/tex}.

 

3. S? se determine inversa func?iei bijective {tex}f:R\to (1; +\infty){/tex}, {tex}f(x)=e^x+1{/tex}.
Rezolvare: Func?ia f este strict cresc?toare pe R, deci injectiv?, iar {tex}Im f=(1; +\infty){/tex}, adic? f este ?i surjectiv?. Prin urmare, func?ia f este bijectiv?.
Pentru a g?si legea func?iei inverse, rezolv?m n mul?imea numerelor reale ecua?ia
(*) {tex}f(x)=y{/tex}
n necunoscuta x ?i cu parametrul y, {tex}y\in (1; +\infty){/tex}.
Avem:
(*){tex}\Leftrightarrow e^x+1=y\Leftrightarrow{/tex}
{tex}e^x=y-1\Leftrightarrow{/tex}
{tex}x=\ln (y-1)\in R{/tex}.
A?adar, inversa func?iei f este func?ia
{tex}f^{-1}:(1; +\infty)\to R{/tex}
dat? de legea
{tex}f^{-1}(y)=\ln (y-1){/tex}, {tex}\forall y\in (1; +\infty){/tex}.

 

4. S? se determine probabilitatea ca alegnd un num?r {tex}\overline{ab}{/tex} din mul?imea numerelor naturale de dou? cifre s? avem {tex}a\neq b{/tex}.
Rezolvare: S? not?m cu A evenimentul
A: Alegnd un num?r {tex}\overline{ab}{/tex} din mul?imea numerelor naturale de dou? cifre, num?rul alesare cifrele diferite.
Atunci
{tex}P(A)=\frac{\mbox{nr. cazurilor favorabile}}{\mbox{nr. cazurilor posibile}}{/tex}.
Exist? 9 numere naturale de dou? cifre cu cifrele identice: 11;22;33;44;55;66;77;88 ?i 99. Avem {tex}(99-9=)90{/tex} de cazuri posibile (cte numere naturale de dou? cifre sunt) ?i (90-9=81) de cazuri favorabile (cte numere naturale de dou? cifre diferite sunt).
Prin urmare
{tex}P(A)=\frac{81}{90}=\frac{9}{10}{/tex}.

 

5. S? se calculeze lungimea medianei din A a triunghiului ABC, unde A(-2;-1), B(2;0) ?i C(0;6).
Rezolvare: Fie M mijlocul segmentului BC. Atunci
{tex}x_{M}=\frac{x_{B}+x_{C}}{2}=\frac{2+0}{2}=1{/tex};
{tex}y_{M}=\frac{y_{B}+y_{C}}{2}=\frac{0+6}{2}=3{/tex}.
Atunci
{tex}AM=\sqrt{(y_{M}-y_{A})^2+(x_{M}-x_{A})^2}=\sqrt{(3+1)^2+(1+2)^2}=\sqrt{16+9}=\sqrt{25}=5{/tex}.
Deci, {tex}AM=5{/tex}.

 

6. Fie vectorii {tex}\overrightarrow{u}=m\overrightarrow{i}+3\overrightarrow{j}{/tex} ?i {tex}\overrightarrow{v}=(m-2)\overrightarrow{i}-\overrightarrow{j}{/tex}. S? se determine m>0 astfel nct vectorii {tex}\overrightarrow{u}{/tex} ?i {tex}\overrightarrow{v}{/tex} s? fie perpendiculari.
Rezolvare: Vectorii {tex}\overrightarrow{u}{/tex} ?i {tex}\overrightarrow{v}{/tex} sunt perpendiculari dac? ?i numai dac? produsul lor scalar este nul, adic?
{tex}\overrightarrow{u}\cdot \overrightarrow{v}=0\Leftrightarrow{/tex}
{tex}u_{1}v_{1}+u_{2}v_{2}=0\Leftrightarrow{/tex}
{tex}m(m-2)+3\cdot (-1)=0\Leftrightarrow{/tex}
{tex}m(m-2)-3=0\Leftrightarrow{/tex}
{tex}m^2-2m-3=0{/tex}.
Avem
{tex}\Delta=\left(\frac{b}{2}\right)^2-ac={/tex}
{tex}=(-1)^2-1\cdot (-3)=1+3=4{/tex}
iar
{tex}m_{1}=\frac{\frac{-b}{2}+\sqrt{\Delta}}{a}=\frac{1+\sqrt{4}}{1}=1+2=3>0{/tex};
{tex}m_{2}=\frac{\frac{-b}{2}-\sqrt{\Delta}}{a}=\frac{1-\sqrt{4}}{1}=1-2=-1<0{/tex}.
Prin urmare, pentru {tex}m=3>0{/tex} vectorii {tex}\overrightarrow{u}{/tex} ?i {tex}\overrightarrow{v}{/tex} sunt perpendiculari.

 

Subiectul II

1. Se consider? matricea {tex}A\in M_{2}(R){/tex}, {tex}A =
\left( {\begin{array}{cc}
2 & 2 \\
1 & 1 \\
\end{array} } \right){/tex}.
a) S? se arate c? exist? {tex}a\in R{/tex} astfel nct {tex}A^2=a\cdot A{/tex}.
b) S? se calculeze {tex}(A-A^t)^{2009}{/tex}.
c) S? se rezolve ecua?ia {tex}X^5=A{/tex}, {tex}X\in M_{2}(R){/tex}.
Rezolvare:
a) Avem:
{tex}A^2=A\cdot A={/tex}
{tex}=\left( {\begin{array}{cc}
2 & 2 \\
1 & 1 \\
\end{array} } \right)\cdot \left( {\begin{array}{cc}
2 & 2 \\
1 & 1 \\
\end{array} } \right)={/tex}
{tex}=
\left( {\begin{array}{cc}
2\cdot2+2\cdot1 & 2\cdot2+2\cdot1 \\
1\cdot2+1\cdot1 & 1\cdot2+1\cdot1 \\
\end{array} } \right)=
\left( {\begin{array}{cc}
6 & 6 \\
3 & 3 \\
\end{array} } \right)={/tex}
{tex}=3\cdot \left( {\begin{array}{cc}
2 & 2 \\
1 & 1 \\
\end{array} } \right)=3\cdot A{/tex}.
A?adar, pentru {tex}a=3\in R{/tex}, avem {tex}A^2=a\cdot A{/tex}.
b) Avem:
{tex}A^t =
\left( {\begin{array}{cc}
2 & 1 \\
2 & 1 \\
\end{array} } \right){/tex}
?i
{tex}A-A^t =
\left( {\begin{array}{cc}
2 & 2 \\
1 & 1 \\
\end{array} } \right)- \left( {\begin{array}{cc}
2 & 1 \\
2 & 1 \\
\end{array} } \right)= \left( {\begin{array}{cc}
0 & 1 \\
-1 & 0 \\
\end{array} } \right){/tex}.
Apoi
{tex}(A-A^t)^2 =
\left( {\begin{array}{cc}
0 & 1 \\
-1 & 0 \\
\end{array} } \right)\cdot \left( {\begin{array}{cc}
0 & 1 \\
-1 & 0 \\
\end{array} } \right)={/tex}
{tex}\left( {\begin{array}{cc}
0\cdot0+1\cdot (-1) & 0\cdot1+1\cdot0 \\
(-1)\cdot0+0\cdot (-1) & (-1)\cdot1+0\cdot0 \\
\end{array} } \right)= \left( {\begin{array}{cc}
-1 & 0 \\
0 & -1 \\
\end{array} } \right)={/tex}
{tex}=(-1)\cdot I_{2}{/tex}.
Atunci
{tex}(A-A^t)^{2009}=(A-A^t)^{2008}\cdot (A-A^t)=[(A-A^t)^2]^{1004}\cdot (A-A^t)={/tex}
{tex}=[(-1)\cdot I_{2}]^{1004}\cdot (A-A^t)={/tex}
{tex}=I_{2}\cdot (A-A^t)={/tex}
{tex}= \left( {\begin{array}{cc}
0 & 1 \\
-1 & 0 \\
\end{array} } \right){/tex}.
c) Fie {tex}X\in M_{2}(R){/tex}, {tex}X =
\left( {\begin{array}{cc}
x & y \\
z & t \\
\end{array} } \right){/tex} o solu?ie a ecua?iei
{tex}X^5=A{/tex}.
Din egalitatea precedent? rezult? c?
{tex}det(X^5)=det(A)=0{/tex},
?i deci
{tex}det(X)=0\Leftrightarrow{/tex}
{tex}xt-yz=0\Leftrightarrowxt=yz=k{/tex}.
Dac? x ar fi egal cu 0, atunci y sau z ar fi 0, adic? X ar avea prima linie/coloan? nul?, ceea ce este imposibil, ntruct ar rezulta c? A are ?i ea prima linie/coloan? nul?. Prin urmare, x este diferit de 0. Analog se arat? c? t este diferit de 0 ?i atunci ?i y ?i z sunt diferite de 0.
Ob?inem atunci:
{tex}t=\frac{k}{x}{/tex} ?i {tex}z=\frac{k}{y}{/tex}.
Prin urmare
{tex}X =
\left( {\begin{array}{cc}
x & y \\
\frac{k}{y} & \frac{k}{x} \\
\end{array} } \right){/tex}.
Efectund calculele ob?inem
{tex}X^2=\frac{x^2+k}{x}\cdot \left( {\begin{array}{cc}
x & y \\
\frac{k}{y} & \frac{k}{x} \\
\end{array} } \right){/tex},
adic?
{tex}X^2=\frac{x^2+k}{x}\cdot X{/tex}.
Analog
{tex}X^3=X^2\cdot X=\frac{x^2+k}{x}\cdot X^2=\left(\frac{x^2+k}{X}\right)^2\cdot X{/tex}
?i
{tex}X^5=X^3\cdot X^2={/tex}
{tex}=\left[\left(\frac{x^2+k}{x}\right)^2\cdot X\right]\cdot \left[\left(\frac{x^2+k}{x}\right)\cdot X\right]={/tex}
{tex}=\left(\frac{x^2+k}{x}\right)^3\cdot X^2={/tex}
{tex}=\left(\frac{x^2+k}{x}\right)^4\cdot X{/tex}.
Atunci
{tex}X^5=A\Leftrightarrow{/tex}
{tex}\left(\frac{x^2+k}{x}\right)^4\cdot \left( {\begin{array}{cc}
x & y \\
\frac{k}{y} & \frac{k}{x} \\
\end{array} } \right)= \left( {\begin{array}{cc}
2 & 2 \\
1 & 1 \\
\end{array} } \right)\Leftrightarrow{/tex}
{tex}\begin{cases}
\left(\frac{x^2+k}{x}\right)^4\cdot x=2 \\
\left(\frac{x^2+k}{x}\right)^4\cdot y=2 \\
\left(\frac{x^2+k}{x}\right)^4\cdot \frac{k}{y}=1 \\
\left(\frac{x^2+k}{x}\right)^4\cdot \frac{k}{x}=1 \\
\end{cases}{/tex}.
nlocuind {tex}\left(\frac{x^2+k}{x}\right)^4{/tex} din ultima ecua?ie n prima ecua?ie a sistemului, ob?inem:
{tex}\frac{x^2}{k}=2\Leftrightarrow{/tex}
{tex}x^2=2k\Leftrightarrow{/tex}
{tex}x=\sqrt{2k}{/tex}, ntruct x nu poate fi negativ, dup? cum rezult? din prima ecua?ie a sistemului.
nlocuind valoarea g?sit? pentru x n prima ecua?ie a sistemului, ob?inem:
{tex}\left(\frac{2k+k}{\sqrt{2k}}\right)^4\cdot \sqrt{2k}=2\Leftrightarrow{/tex}
{tex}\frac{81k^4}{4k^2}\cdot \sqrt{2k}=2\Leftrightarrow{/tex}
{tex}\sqrt{2k^5}=\frac{8}{81}|^2\Leftrightarrow{/tex}
{tex}2k^5=\frac{2^6}{3^8}\Leftrightarrow{/tex}
{tex}k=\frac{2}{3\sqrt[5]{27}}{/tex}.
Prin urmare
{tex}x=\sqrt{\frac{4}{\sqrt[5]{3^8}}}=\frac{2}{\sqrt[5]{\sqrt{3^8}}}=\frac{2}{\sqrt[5]{81}}{/tex}.
Din prima ?i a doua ecua?ie a sistemului rezult? c? {tex}y=x=\frac{2}{\sqrt[5]{81}}{/tex}.
A?adar, unica solu?ie a ecua?iei date este matricea
{tex}X =
\left( {\begin{array}{cc}
\frac{2}{\sqrt[5]{81}} & \frac{2}{\sqrt[5]{81}} \\
\frac{1}{\sqrt[5]{81}} & \frac{1}{\sqrt[5]{81}} \\
\end{array} } \right)=\frac{1}{\sqrt[5]{81}}\cdot A{/tex}.
Desigur, faptul c? matricea {tex}\frac{1}{\sqrt[5]{81}}\cdot A{/tex} este solu?ie a ecua?iei date rezult? imediat dac? utiliz?m pct. a), dar nu putem dovedi c? aceast? solu?ie este unic? dect urmnd calea de mai sus.

 

2. Pentru {tex}a, b\in M=[0; \infty){/tex}, se define?te opera?ia {tex}a*b=\ln (e^a+e^b-1){/tex}.
a) S? se arate c? dac? {tex}a, b\in M{/tex}, atunci {tex}a*b\in M{/tex}.
b) S? se arate c? legea de compozi?ie * este asociativ?.
c) Pentru {tex}m\in N{/tex}, {tex}m\geq2{/tex}, s? se determine {tex}a\in M{/tex} astfel nct {tex}a^m=a*...*a=2a{/tex}.
Rezolvare:
Fie {tex}a, b\in M{/tex}. Atunci:
{tex}a\geq0, b\geq0{/tex}
?i
{tex}e^a\geq1, e^b\geq1{/tex}.
Rezult? atunci c?
{tex}e^a+e^b\geq2\Leftrightarrow{/tex}
{tex}e^a+e^b-1\geq1\Leftrightarrow{/tex}
{tex}\ln (e^a+e^b-1)\geq0\Leftrightarrow{/tex}
{tex}a*b\in M{/tex}.
b) * este asociativ? dac? ?i numai dac? {tex}\forall a, b, c \in M, (a*b)*c=a*(b*c)\Leftrightarrow{/tex}
{tex}\forall a, b, c \in M, [\ln(e^a+e^b-1)]*c=a*[\ln (e^b+e^c-1)]\Leftrightarrow{/tex}
{tex}\forall a, b, c \in M, \ln (e^{\ln (e^a+e^b-1)}+e^c-1)=\ln (e^a+e^{\ln (e^b+e^c-1)}-1)\Leftrightarrow{/tex}
{tex}\forall a, b, c \in M, \ln (e^a+e^b-1+e^c-1)=\ln (e^a+e^b+e^c-1-1)\Leftrightarrow{/tex}
{tex}\forall a, b, c \in M, \ln (e^a+e^b+e^c-2)=\ln (e^a+e^b+e^c-2){/tex} (A).
c) Fie {tex}m\in N, m\geq2{/tex} ?i {tex}a\in M{/tex}. Atunci
{tex}a^2=a*a=\ln (e^a+e^a-1)=\ln (2e^a-1){/tex};
{tex}a^3=a^2*a={/tex}
{tex}=[\ln (2e^a-1)]*a=\ln (e^{\ln (2e^a-1)}+e^a-1)=\ln (2e^a-1+e^a-1)=\ln (3e^a-2){/tex};
{tex}a^4=a^3*a={/tex}
{tex}=[\ln (3e^a-2)]*a=\ln (e^{\ln (3e^a-2)}+e^a-1)=\ln (4e^a-3){/tex};
...
{tex}a^m=\ln (me^a-(m-1)){/tex}.
A?adar, avem
{tex}a^m=2a\Leftrightarrow{/tex}
{tex}\ln (me^a-(m-1))=2a\Leftrightarrow{/tex}
{tex}\ln (me^a-m+1)=\ln e^{2a}\Leftrightarrow{/tex}
{tex}me^a-m+1=e^{2a}\Leftrightarrow{/tex}
{tex}(e^{a})^2-me^a+m-1=0{/tex}.
Fie {tex}e^{a}=t, t\geq1{/tex}.
Ultima ecua?ie devine atunci
{tex}t^2-mt+m-1=0{/tex}
{tex}\Delta=m^2-4(m-1)=m^2-4m+4=(m-2)^2{/tex}.
Atunci
{tex}t_{1}=\frac{m+(m-2)}{2}=m-1\geq1{/tex};
{tex}t_{2}=\frac{m-(m-2)}{2}=1\geq1{/tex}.
Vom avea de rezolvat dou? ecua?ii n a:
{tex}e^{a}=m-1{/tex} ?i {tex}e^{a}=1{/tex}.
Ob?inem {tex}a=\ln (m-1)\in M{/tex} ?i {tex}a=0\in M{/tex}.
Valorile lui a pentru care {tex}a^m=2a{/tex} sunt {tex}a=\ln (m-1){/tex} ?i {tex}a=0{/tex}.

 

Subiectul III
1. Se consider? ?irul {tex}(a_{n})_{n\in N^*}{/tex} dat de {tex}a_{1}\in (0;1){/tex} ?i {tex}a_{n+1}=a_{n}(1-\sqrt{a_{n}}), \forall n\in N^*{/tex}.
a) S? se arate c? {tex}a_{n}\in (0;1), \forall n\in N^*{/tex}.
b) S? se demonstreze c? ?irul {tex}(a_{n})_{n \in N^*}{/tex} este strict descresc?tor.
c) S? se arate c? ?irul {tex}(b_{n})_{n\in N^*}{/tex} dat de {tex}b_{n}=a_{1}^2+a_{2}^2+...+a_{n}^2, \forall n\in N^*{/tex} este m?rginit superior de {tex}a_{1}{/tex}.
Rezolvare:
a) Fie predicatul unar P, dat de {tex}P(n): a_{n}\in (0;1), n\in N^*{/tex}.
Vom demonstra prin induc?ie matematic? dup? n c? P(n) este o propozi?ie adev?rat? pentru orice n, num?r natural nenul.
I. {tex}P(1): a_{1}\in (0;1){/tex}, (A).
II. {tex}P(n)\Rightarrow P(n+1), n\in N^* \Leftrightarrow{/tex}
{tex}a_{n}\in (0;1)\Rightarrow a_{n+1} \in (0;1), n \in N^* \Leftrightarrow {/tex}
{tex} a_{n} \in (0;1) \Rightarrow a_{n}(1-\sqrt{a_{n}}) \in (0;1), n \in N^* {/tex}.
ntr-adev?r, dac? {tex}a_{n} \in (0;1), n \in N^* {/tex}, atunci
{tex}0<a_{n}<1, n \in N^* {/tex}
?i
{tex} 0<1-\sqrt{a_{n}}<1, n \in N^* {/tex}
?i prin urmare
{tex}0<a_{n}(1-\sqrt{a_{n}})<1, n \in N^* {/tex}.
Deci,
{tex}P(n) \Rightarrow P(n+1), n \in N^* {/tex}.
Din I ?i II, rezult? conform principiului induc?iei matematice c? P(n) este adev?rat? pentru orice n, num?r natural nenul, adic? {tex} a_{n} \in (0;1), \forall n \in N^*{/tex}.
b) ?irul {tex} (a_{n})_{n \in N^*}{/tex} este strict descresc?tor dac? ?i numai dac?
{tex} a_{n+1} < a_{n}, \forall n \in N^* \Leftrightarrow {/tex}
{tex} a_{n}(1-\sqrt{a_{n}}) < a_{n}, \forall n \in N^* \Leftrightarrow {/tex}
{tex} a_{n}(1-\sqrt{a_{n}}) - a_{n} < 0, \forall n \in N^* {/tex}
{tex} -a_{n}\sqrt{a_{n}} < 0, \forall n \in N^* {/tex}, (A dac? ?inem cont de pct. a)).
c) Vom demonstra c?:
{tex} a_{n}^2 < a_{n} - a_{n+1}, \forall n \in N^*{/tex}.
ntr-adev?r,
{tex} a_{n}^2 < a_{n} - a_{n+1}, \forall n \in N^* \Leftrightarrow{/tex}
{tex} a_{n}^2 < a_{n} - a_{n}(1-\sqrt{a_{n}}), \forall n \in N^* \Leftrightarrow{/tex}
{tex} a_{n}^2 < a_{n}\sqrt{a_{n}}, \forall n \in N^* |:a_{n}>0 \Leftrightarrow{/tex}
{tex} a_{n} < \sqrt{a_{n}}, \forall n \in N^*{/tex}
?i ntruct ambii membri ai inegalit??ii sunt pozitivi, avem
{tex} a_{n} < \sqrt{a_{n}}, \forall n \in N^* |^2 \Leftrightarrow{/tex}
{tex} a_{n}^2 < a_{n}, \forall n \in N^* |:a_{n}>0 \Leftrightarrow{/tex}
{tex} a_{n} < 1, \forall n \in N^*{/tex} (A, cf. pct. a)).
Prin urmare
{tex} a_{n}^2 < a_{n} - a_{n+1}, \forall n \in N^*{/tex}.
Din aceast? inegalitate, rezult? urm?torul ?ir de inegalit??i:
{tex} a_{1}^2 < a_{1} - a_{2}{/tex};
{tex} a_{2}^2 < a_{2} - a_{3}{/tex};
{tex} a_{3}^2 < a_{3} - a_{4}{/tex};
...
{tex} a_{n}^2 < a_{n} - a_{n+1}{/tex}.
Adunnd membru cu membru aceste inegalit??i, ob?inem:
{tex} b_{n}=a_{1}^2+a_{2}^2+...+a_{n}^2 < a_{1} - a_{n+1} < a_{1}, \forall n \in N^*{/tex}.
A?adar, ?irul {tex}(b_{n})_{n \in N^*}{/tex} este m?rginit superior de {tex}a_{1}{/tex}.

 

2. Se consider? func?ia {tex}f: R \to R, f(x)=\frac{1}{x^2+x+1}{/tex}.
a) S? se arate c? func?ia {tex}F: R \to R, F(x)=\frac{2\sqrt{3}}{3} \cdot \arctan \left(\frac{2x+1}{\sqrt{3}}\right), x \in R{/tex} este o primitiv? a lui f.
b) S? se calculeze aria suprafe?ei delimitate de dreptele {tex}x=0, x=1{/tex}, Ox ?i graficul func?iei {tex}g: R \to R, g(x)=(2x+1)f(x){/tex}.
c) S? se calculeze:
{tex}\lim_{n \to \infty} \int_{-n}^{n} f(x)dx{/tex}, unde {tex}n \in N^*{/tex}.
Rezolvare:
a) Func?ia F este o primitiv? a lui f dac? ?i numai dac?
{tex}F(x)=f(x), \forall x \in R \Leftrightarrow{/tex}
{tex} \left[\left(\frac{2\sqrt{3}}{3}\right) \cdot \arctan \left(\frac{2x+1}{\sqrt{3}}\right)\right]=\frac{1}{x^2+x+1}, \forall x \in R \Leftrightarrow{/tex}
{tex} \left(\frac{2\sqrt{3}}{3}\right) \cdot \frac{1}{\left(\frac{2x+1}{\sqrt{3}}\right)^2+1} \cdot \left(\frac{2x+1}{\sqrt{3}}\right)=\frac{1}{x^2+x+1}, \forall x \in R \Leftrightarrow{/tex}
{tex} \left(\frac{2\sqrt{3}}{3}\right) \cdot \frac{3}{4x^2+4x+4} \cdot \frac{2}{\sqrt{3}}=\frac{1}{x^2+x+1}, \forall x \in R \Leftrightarrow{/tex}
{tex} \frac{4}{3} \cdot \frac{3}{4(x^2+x+1)}=\frac{1}{x^2+x+1}, \forall x \in R{/tex}, (A).
b) Aria cerut? este:
{tex} A=\int _{0}^{1} g(x)dx={/tex}
{tex}=\int_{0}^{1} (2x+1)f(x)dx={/tex}
{tex}\int_{0}^{1} \frac{2x+1}{x^2+x+1}dx={/tex}
{tex}=[\ln (x^2+x+1)] |_{0}^{1}={/tex}
{tex} =\ln (1^2+1+1)-\ln (0^2+0+1)={/tex}
{tex}=\ln 3-0=\ln 3{/tex}.
c) Calcul?m mai nti
{tex}I_{n}=\int_{-n}^{n} f(x)dx, n \in N^*{/tex}.
Avem:
{tex}I_{n}=\int_{-n}^{n} \frac{1}{x^2+x+1}dx={/tex}
{tex}=\int_{-n}^{n} \frac{1}{(x+\frac{1}{2})^2+1-\frac{1}{4}}dx={/tex}
{tex}=\int_{-n}^{n} \frac{1}{(x+\frac{1}{2})^2+(\frac{\sqrt{3}}{2})^2}dx={/tex}
{tex}=\left[\frac{2}{\sqrt{3}} \arctan \frac{2}{\sqrt{3}}\left(x+\frac{1}{2}\right)\right] \right|_{-n}^{n}={/tex}
{tex}=\left[\frac{2\sqrt{3}}{3} \arctan \frac{2\sqrt{3}(2x+1)}{3}\right|_{-n}^{n}={/tex}
{tex}=\frac{2\sqrt{3}}{3}\left(\arctan \frac{\sqrt{3}(2n+1)}{3}-\arctan \frac{\sqrt{3}(-2n+1)}{3}\right)={/tex}
{tex}\frac{2\sqrt{3}}{3}\left(\arctan \frac{\sqrt{3}(2n+1)}{3}+\arctan \frac{\sqrt{3}(2n-1)}{3}\right){/tex}.
Atunci
{tex}\lim_{n \to \infty} I_{n}={/tex}
{tex}=\lin_{n \to \infty} \frac{2\sqrt{3}}{3}\left(\arctan \frac{\sqrt{3}(2n+1)}{3}+\arctan \frac{\sqrt{3}(2n-1)}{3}\right)={/tex}
{tex}=\frac{2\sqrt{3}}{3}\left(\frac{\pi}{2}+\frac{\pi}{2}\right)={/tex}
{tex}=\frac{2\pi \sqrt{3}}{3}{/tex}.

Ultima actualizare ( Luni, 28 Februarie 2011 10:43 )