Matematic? M1, Varianta 1, BAC 2010

Scris de Andreea Neagu   
Vineri, 28 Ianuarie 2011 14:45
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Subiectul I
1. S? se determine num?rul natural x din egalitatea:
{tex}1+5+9+...+x=231{/tex}.
Rezolvare: Membrul stng al egalit??ii este o progresie aritmetic? cu primul termen 1 ?i ra?ia 4.
Suma primilor n termeni ai unei progresii aritmetice {tex}(a_{n})_{n\geq1}{/tex} este
{tex}S_{n}=\frac{n(a_{1}+a_{n})}{2}{/tex}.
A?adar
(1) {tex}1+5+9+...+x=231\Leftrightarrow{/tex}
{tex}\frac{n_{x}(1+x)}{2}=231{/tex},
Unde {tex inline}n_{x}{/tex} este num?rul termenilor sumei din membrul stng al egalit??ii date.
Dar al n-lea termen al unei progresii aritmetice {tex}(a_{n})_{n\geq1}{/tex} n func?ie de primul termen ?i ra?ia r, este dat de formula
{tex}a_{n}=a_{1}+(n-1)r, \foralln\geq1{/tex}.
Prin urmare
{tex}x=1+(n_{x}-1)\cdot4\Leftrightarrow{/tex}
(2) {tex}n_{x}=\frac{x+3}{4}{/tex}.
Din (1) ?i (2), rezult?
{tex}\frac{\frac{x+3}{4}(1+x)}{2}=231\Leftrightarrow{/tex}
{tex}\frac{x+3}{4}(1+x)=462\Leftrightarrow{/tex}
{tex}(x+3)(x+1)=1848\Leftrightarrow{/tex}
{tex}x^2+4x-1845=0{/tex}.
Avem
{tex}\Delta'=(\frac{b}{2})^2-ac=2^2-1\cdot(-1845)=4+1845=1849{/tex}
?i
{tex}x_{1}=\frac{\frac{-b}{2}+\sqrt{\Delta'}}{a}=\frac{-2+\sqrt{1849}}{1}=-2+43=41{/tex};
{tex}x_{2}=\frac{\frac{-b}{2}-\sqrt{\Delta'}}{a}=\frac{-2-\sqrt{1849}}{1}{/tex}, solu?ie care nu se accept?, nefiind num?r natural.
Deci, {tex}x=41{/tex}.

2. S? se rezolve n mul?imea numerelor reale inecua?ia:
{tex}2x^2-5x+3\leq0{/tex}.
Rezolvare: Rezolv?m ecua?ia
{tex}2^2-5x+3=0{/tex}.
Discriminantul ecua?iei este
{tex}\Delta=b^2-4ac=(-5)^2-4\cdot2\cdot3=25-24=1{/tex}.
Avem
{tex}x_{1}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{5+\sqrt{1}}{4}=\frac{6}{4}=\frac{3}{2}{/tex};
{tex}x_{2}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{5-\sqrt{1}}{4}=\frac{4}{4}=1{/tex}.
ntruct {tex}a=2>0{/tex}, func?ia de gradul al doilea
{tex}f:R\toR, f(x)=2x^2-5x+3{/tex}
iavalori de semn contrar lui a, adic? negative, ntre r?d?cini.
Prin urmare, solu?ia inecua?iei
{tex}2x^2-5x+3\leq0{/tex}
este
{tex}S=[1;\frac{3}{2}]{/tex}.

3. S? se determine inversa func?iei bijective {tex}f:(0;\infty)\to(1;\infty){/tex}, unde {tex}f(x)=x^2+1{/tex}.
Rezolvare: Func?ia {tex}f:(0;\infty)\to(1;\infty){/tex}, {tex}f(x)=x^2+1{/tex} este strict cresc?toare pe {tex}(0;\infty){/tex} ?i deci injectiv? iar {tex}Im f=(1;\infty){/tex}, adic? func?ia f este ?i surjectiv?.
Deci, f este bijectiv? ?i
{tex}f^{-1}:(1;\infty)\to(0;\infty){/tex}.
Pentru a determina legea func?iei inverse, vom rezolva ecua?ia
{tex}f(x)=y{/tex}
n mul?imea {tex}(0;\infty){/tex}, x fiind necunoscuta iar y fiind parametru.
{tex}f(x)=y, x\in(0;\infty)\Leftrightarrow{/tex}
{tex}x^2+1=y, x\in(0;\infty)\Leftrightarrow{/tex}
(1) {tex}x^2=y-1, x\in(0;\infty){/tex}.
Cum {tex}x\in(0;\infty){/tex}, rezult?
{tex}y-1>0\Leftrightarrow{/tex}
{tex}y>1\Leftrightarrow{/tex}
{tex}y\in(1;\infty){/tex}
Iar ecua?ia (1) este echivalent? cu
{tex}x=\sqrt{y-1}{/tex}.
A?adar,
{tex}f^{-1}:(1;\infty)\to(0;\infty){/tex}
iar
{tex}f^{-1}(y)=\sqrt{y-1}{/tex}.

4. Se consider? mul?imea A={1;2;3;...;10}. S? se determine num?rul submul?imilor cu trei elemente ale mul?imii A, care con?in elementul 1.
Rezolvare: Num?rul submul?imilor cu trei elemente ale mul?imii A, care con?in elementul 1 este acela?i cu num?rul submul?imilor cu dou? elemente ale mul?imii {2;3;4;...;10}, adic? este egal cu
{tex}C_{9}^{2}=\frac{9!}{2!(9-2)!}=\frac{9!}{2!7!}=\frac{7!\cdot8\cdot9}{2!7!}=\frac{8\cdot9}{2}=36{/tex}.
Deci, num?rul submul?imilor cu trei elemente ale mul?imii A, care con?in elementul 1 este egal cu 36.

5. S? se determine {tex}n\inR{/tex} astfel nct distan?a dintre punctele A(2;m) ?i B(m;-2) s? fie 4.
Rezolvare: Distan?a dintre dou? puncte {tex}A(x_{A};y_{A}){/tex} ?i {tex}B(x_{B};y_{B}){/tex} este dat? de formula
{tex}AB=\sqrt{(y_{b}-y_{A})^2+(x_{B}-x_{A})^2}{/tex}.
Pentru punctele A ?i B din enun?, distan?a AB este
{tex}AB=\sqrt{(-2-m)^2+((m-2)^2}=\sqrt{(m+2)^2+(m-2)^2}=\sqrt{2m^2+8}{/tex}.
A?adar
{tex}AB=4\Leftrightarrow{/tex}
{tex}\sqrt{2m^2+8}=4\Leftrightarrow{/tex}
{tex}2m^2+8=16\Leftrightarrow{/tex}
{tex}2m^2=8\Leftrightarrow{/tex}
{tex}m^2=4\Leftrightarrow{/tex}
{tex}m\in{-2;2}{/tex}.
Deci {tex}m\in{-2;2}{/tex}.

6. S? se calculeze {tex}\cos \frac{23\pi}{12}\sin \frac{\pi}{12}{/tex}.
Rezolvare:
{tex}\cos \frac{23\pi}{12}\sin \frac{\pi}{12}=\sin \frac{\pi}{12}\cos \frac{23\pi}{12}=\frac{\sin \left(\frac{\pi}{12}+\frac{23\pi}{12}\right)+\sin \left(\frac{\pi}{12}-\frac{23\pi}{12}\right)}{2}={/tex}
{tex}=\frac{\sin \frac{24\pi}{12}+\sin \frac{-22\pi}{12}}{2}=\frac{\sin 2\pi-\sin \frac{11\pi}{6}}{2}={/tex}
{tex}\frac{-1}{2}\sin \frac{11\pi}{6}=\frac{-1}{2}\sin \frac{12\pi-\pi}{6}=\frac{-1}{2}\sin (2\pi-\frac{\pi}{6})=\frac{1}{2}\sin \frac{\pi}{6}=\frac{1}{2}\cdot\frac{1}{2}=\frac{1}{4}{/tex}.
Prin urmare
{tex}\cos \frac{23\pi}{12}\sin \frac{\pi}{12}=\frac{1}{4}{/tex}.

Subiectul II
1. Se consider? matricea

{tex}
A =
\left( {\begin{array}{cc}
a & b \\
b & a \\
\end{array} } \right)
{/tex},
cu {tex}a, b\inR{/tex} ?i {tex}b\neq0{/tex}.
a) S? se arate c? dac? matricea {tex}X\inM_{2}(R){/tex} verific? rela?ia AX=XA, atunci exist? {tex}u, v\inR{/tex} astfel nct
{tex}
X =
\left( {\begin{array}{cc}
u & v \\
v & u \\
\end{array} } \right)
{/tex}.
b) S? se arate c? {tex}\forall n \in N^*{/tex}, {tex}
A^n =
\left( {\begin{array}{cc}
x_{n} & y_{n} \\
y_{n} & x_{n} \\
\end{array} } \right)
{/tex},
Unde {tex}x_{n}=\frac{(a+b)^n+(a-b)^n}{2}{/tex} ?i {tex}y_{n}=\frac{(a+b)^n-(a-b)^n}{2}{/tex}.
c) S? se rezolve n mul?imea {tex}M_{2}(R){/tex} ecua?ia {tex}
X^3 =
\left( {\begin{array}{cc}
2 & 1 \\
1 & 2 \\
\end{array} } \right)
{/tex}.
Rezolvare: a) Fie o matrice {tex}X\inM_{2}(R){/tex} de forma
{tex}
X =
\left( {\begin{array}{cc}
x & y \\
z & t \\
\end{array} } \right){/tex},
Cu {tex}x, y, z, t\inR{/tex} care verific? rela?ia {tex}AX=XA{/tex}.
Atunci
{tex}AX=XA\Leftrightarrow{/tex}
{tex}
\left( {\begin{array}{cc}
a & b \\
b & a \\
\end{array} } \right)
\cdot
\left( {\begin{array}{cc}
x & y \\
z & t \\
\end{array} } \right)
=
\left( {\begin{array}{cc}
x & y \\
z & t \\
\end{array} } \right)
\cdot
\left( {\begin{array}{cc}
a & b \\
b & a \\
\end{array} } \right)
\Leftrightarrow{/tex}
{tex}
\left( {\begin{array}{cc}
ax+bz & ay+bt \\
bx+az & by+at \\
\end{array} } \right)
=
\left( {\begin{array}{cc}
xa+yb & xb+ya \\
za+tb & zb+ta \\
\end{array} } \right)
\Leftrightarrow{/tex}
{tex}\begin{cases}
ax+bz=ax+by \\
ay+bt=bx+ay \\
bx+az=az+bt \\
by+at=bz+at \\
\end{cases}\Leftrightarrow{/tex}
{tex}\begin{cases}
bz=by \\
bt=bx \\
\end{cases}\Leftrightarrow{/tex}
{tex}\begin{cases}
z=y \\
t=x\\
\end{cases}{/tex}, ntruct {tex}b\neq0{/tex}.
Notnd {tex}x=t=u{/tex} ?i {tex}y=z=v{/tex}, rezult? c? exist? {tex}u, v \in R{/tex} astfel nct
{tex}
X =
\left( {\begin{array}{cc}
u & v \\
v & u \\
\end{array} } \right)
{/tex}.

b) Fie P(n): {tex}
A^n =
\left( {\begin{array}{cc}
\frac{(a+b)^n+(a-b)^n}{2} & \frac{(a+b)^n-(a-b)^n}{2} \\
\frac{(a+b)^n-(a-b)^n}{2} & \frac{(a+b)^n+(a-b)^n}{2} \\
\end{array} } \right)
{/tex}, {tex}n \in N^*{/tex}.

Vom demonstra prin induc?ie dup? n c? P(n) este adev?rat? pentru orice {tex}n \in N^*{/tex}.

I. Pasul de verificare:
P(1): {tex}
A^1 =
\left( {\begin{array}{cc}
\frac{(a+b)^1+(a-b)^1}{2} & \frac{(a+b)^1-(a-b)^1}{2} \\
\frac{(a+b)^1-(a-b)^1}{2} & \frac{(a+b)^1+(a-b)^1}{2} \\
\end{array} } \right)
\Leftrightarrow{/tex}
{tex}
A =
\left( {\begin{array}{cc}
\frac{a+b+a-b}{2} & \frac{a+b-a+b}{2} \\
\frac{a+b-a+b}{2} & \frac{a+b+a-b}{2} \\
\end{array} } \right)
\Leftrightarrow{/tex}
{tex}
A =
\left( {\begin{array}{cc}
a & b \\
b & a \\
\end{array} } \right)
{/tex}, (A).

II. Presupunem c? P(n) este adev?rat? ?i vom demonstra atunci c? ?i P(n+1) este adev?rat?.
P(n+1): {tex}
A^{n+1} =
\left( {\begin{array}{cc}
\frac{(a+b)^{n+1}+(a-b)^{n+1}}{2} & \frac{(a+b)^n-(a-b)^n}{2} \\
\frac{(a+b)^{n+1}-(a-b)^{n+1}}{2} & \frac{(a+b)^{n+1}+(a-b)^{n+1}}{2} \\
\end{array} } \right)
{/tex}.
Avem:
{tex} A^{n+1}=A\cdot A^n=
\left( {\begin{array}{cc}
a & b \\
b & a \\
\end{array} } \right)
\cdot
\left( {\begin{array}{cc}
\frac{(a+b)^n+(a-b)^n}{2} & \frac{(a+b)^n-(a-b)^n}{2} \\
\frac{(a+b)^n-(a-b)^n}{2} & \frac{(a+b)^n+(a-b)^n}{2} \\
\end{array} } \right)
={/tex}

{tex}
=\left( {\begin{array}{cc}
a\cdot\frac{(a+b)^n+(a-b)^n}{2}+b\cdot\frac{(a+b)^n-(a-b)^n}{2} & a\cdot\frac{(a+b)^n-(a-b)^n}{2}+b\cdot\frac{(a+b)^n+(a-b)^n}{2} \\
b\cdot\frac{(a+b)^n+(a-b)^n}{2}+a\cdot\frac{(a+b)^n-(a-b)^n}{2} & b\cdot\frac{(a+b)^n-(a-b)^n}{2}+a\cdot\frac{(a+b)^n+(a-b)^n}{2} \\
\end{array} } \right) ={/tex}

{tex}=
\left( {\begin{array}{cc}
\frac{(a+b)^n}{2}\cdot(a+b)+\frac{(a-b)^n}{2}\cdot(a-b) & \frac{(a+b)^n}{2}\cdot(a+b)-\frac{(a-b)^n}{2}\cdot(a-b) \\
\frac{(a+b)^n}{2}\cdot(a+b)-\frac{(a-b)^n}{2}\cdot(a-b) & \frac{(a+b)^n}{2}\cdot(a+b)+\frac{(a-b)^n}{2}\cdot(a-b) \\
\end{array} } \right)={/tex}
{tex}=
\left( {\begin{array}{cc}
\frac{(a+b)^{n+1}+(a-b)^{n+1}}{2} & \frac{(a+b)^{n+1}-(a-b)^{n+1}}{2} \\
\frac{(a+b)^{n+1}-(a-b)^{n+1}}{2} & \frac{(a+b)^{n+1}+(a-b)^{n+1}}{2} \\
\end{array} } \right)
{/tex}.
Prin urmare din P(n) rezult? P(n+1) ?i deci P(n) este adev?rat? pentru orice {tex}n \in N^*{/tex}.

c) ntruct
{tex}X^4=X^3 \cdot X=X \cdot X^3{/tex},
unde {tex}
X^3 =
\left( {\begin{array}{cc}
2 & 1 \\
1 & 2 \\
\end{array} } \right)
{/tex}, rezult? c? matricea X comut? cu o matrice de tipul A, n care {tex}a=2{/tex} ?i {tex}b=1{/tex}. Atunci, conform pct. a) rezult? c? X este o matrice de forma {tex}
X =
\left( {\begin{array}{cc}
x & y \\
y & x \\
\end{array} } \right)
{/tex}, unde {tex}x,y \in R{/tex}, iar {tex}y\neq0{/tex}.
Atunci, din b) rezult? c? {tex}
X^3 =
\left( {\begin{array}{cc}
\frac{(x+y)^3+(x-y)^3}{2} & \frac{(x+y)^3-(x-y)^3}{2} \\
\frac{(x+y)^3-(x-y)^3}{2} & \frac{(x+y)^3+(x-y)^3}{2} \\
\end{array} } \right)
{/tex}.
A?adar
{tex}
X^3 =
\left( {\begin{array}{cc}
2 & 1 \\
1 & 2 \\
\end{array} } \right)
\Leftrightarrow{/tex}
{tex}
\left( {\begin{array}{cc}
\frac{(x+y)^3+(x-y)^3}{2} & \frac{(x+y)^3-(x-y)^3}{2} \\
\frac{(x+y)^3-(x-y)^3}{2} & \frac{(x+y)^3+(x-y)^3}{2} \\
\end{array} } \right) ={/tex}
{tex}
=\left( {\begin{array}{cc}
2 & 1 \\
1 & 2 \\
\end{array} } \right)
\Leftrightarrow{/tex}
{tex}
\begin{cases}
\frac{(x+y)^3+(x-y)^3}{2}=2 \\
\frac{(x+y)^3-(x-y)^3}{2}=1 \\
\end{cases}\Leftrightarrow{/tex}
{tex}\begin{cases}
(x+y)^3+(x-y)^3=4 \\
(x+y)^3-(x-y)^3=2 \\
\end{cases}{/tex}.
Adunnd, respectiv sc?znd cele dou? rela?ii ob?inem:
{tex}\begin{cases}
(x+y)^3=3 \\
(x-y)^3=1 \\
\end{cases}\Leftrightarrow{/tex}
{tex}\begin{cases}
x+y=\sqrt[3]{3} \\
x-y=1 \\
\end{cases}{/tex}.
Adunnd, respectiv sc?znd ultimele dou? rela?ii ob?inem:
{tex}x=\frac{\sqrt[3]{3}+1}{2}{/tex}, {tex}y=\frac{\sqrt[3]{3}-1}{2}{/tex}.
Prin urmare, solu?ia ecua?iei date este matricea
{tex}
X =
\left( {\begin{array}{cc}
\frac{\sqrt[3]{3}+1}{2} & \frac{\sqrt[3]{3}-1}{2} \\
\frac{\sqrt[3]{3}-1}{2} & \frac{\sqrt[3]{3}+1}{2} \\
\end{array} } \right)
{/tex}.

2. Se consider? {tex}a \in Z_{7}{/tex} ?i polinomul {tex}f=X^6+aX+\widehat{5}\in Z_{7}[X]{/tex}.
a) S? se verifice c? pentru orice {tex}b\in Z_{7}{/tex}, {tex}b\neq\widehat{0}{/tex}, are loc rela?ia {tex}b^6=\widehat{1}{/tex}.
b) S? se arate c? {tex}x^6+\widehat{5}=(x^3-\widehat{4})(x^3+\widehat{4}){/tex}, {tex}\forall x \in Z_{7}{/tex}.
c) S? se demonstreze c? pentru orice {tex}a \in Z_{7}{/tex}, polinomul f este reductibil n {tex}Z_{7}[X]{/tex}.
Rezolvare:
a) Fie {tex}b \in Z_{7}{/tex}, {tex}b\neq\widehat{0}{/tex}.
Atunci
{tex}(\widehat{1})^6=\widehat{1^6}=\widehat{1}{/tex};
{tex}(\widehat{2})^6=((\widehat{2})^2)^3=(\widehat{2^2})^3=\widehat{4^3}=\widehat{64}=\widehat{1}{/tex};
{tex}(\widehat{3})^6=((\widehat{3})^2)^3=(\widehat{3^2})^3=(\widehat{2})^3=\widehat{2^3}=\widehat{8}=\widehat{1}{/tex};
{tex}(\widehat{4})^6=((\widehat{4})^2)^3=(\widehat{4^2})^3=\widehat{2^3}=\widehat{8}=\widehat{1}{/tex};
{tex}(\widehat{5})^6=((\widehat{5})^2)^3=(\widehat{5^2})^3=\widehat{4^3}=\widehat{64}=\widehat{1}{/tex};
{tex}(\widehat{6})^6=((\widehat{6})^2)^3=(\widehat{6^2})^3=\widehat{1^3}=\widehat{1}{/tex}.
A?adar, pentru orice {tex}b \in Z_{7}{/tex}, {tex}b\neq\widehat{0}{/tex}, are loc rela?ia {tex}b^6=\widehat{1}{/tex}.
b) Fie {tex}x \in Z_{7}{/tex}.
Atunci
{tex}(x^3-\widehat{4})(x^3+\widehat{4})=(x^3)^2-(\widehat{4})^2=x^6-\widehat{16}=x^6-\widehat{2}=x^6+\widehat{5}{/tex}.
c) Pentru {tex}a=\widehat{0}{/tex}, polinomul f este
{tex}f=X^6+\widehat{5}=(X^3-\widehat{4})(X^3+\widehat{4}){/tex}, conform pct. b)
Prin urmare, pentru {tex}a=\widehat{0}{/tex}, polinomul f este reductibil n {tex}Z_{7}{/tex}, descompunndu-se ca produs de dou? polinoame de gradul 3, cu coeficien?i n {tex}Z_{7}{/tex}.
ntruct {tex}Z_{7}{/tex} este corp, orice element {tex}a \in Z_{7}{/tex}, {tex}a\neq\widehat{0}{/tex}, este inversabil ?i pentru orice {tex}a \in Z_{7}{/tex}, {tex}a\neq\widehat{0}{/tex}, elementul {tex}a^{-1} \in Z_{7}{/tex}este o r?d?cin? a lui f. ntr-adev?r,
{tex}f(a^{-1})=(a^{-1})^6+a \cdot a^{-1}+\widehat{5}=\widehat{1}+\widehat{1}+\widehat{5}=\widehat{7}=\widehat{0}{/tex}.
(Am folosit c? {tex}(a^{-1})^6=\widehat{1}{/tex}, conform pct. a)
Atunci, conform teoremei lui Bezout, polinomul {tex}X-a^{-1}{/tex} divide pe f, ?i prin urmare, f este reductibil n {tex}Z_{7}[X]{/tex}, el putndu-se descompune ca produs dintre un polinom de gradul 1 ?i un polinom de gradul 5.
A?adar, pentru orice {tex}a \in Z_{7}{/tex}, polinomul f este reductibil n {tex}Z_{7}[X]{/tex}.

Subiectul III
1. Se consider? num?rul real a>0 ?i func?ia {tex}f:R \to R{/tex}, {tex}f(x)=e^x-ax{/tex}.
a) S? se determine asimptota oblic? la graficul func?iei f c?tre {tex}\infty{/tex}.
b) S? se determine punctele de extrem local ale func?iei f.
c) S? se determine {tex}a \in (0;\infty){/tex} ?tiind c? {tex}f(x) \geq 1{/tex}, {tex}\forall x \in R{/tex}.
Rezolvare: Ecua?ia asimptotei oblice la graficul func?iei f c?tre {tex}-\infty{/tex} este
{tex}y=mx+n{/tex},
Unde {tex}m=\lim_{x \to -\infty}\frac{f(x)}{x}{/tex}, iar {tex}n=\lim_{x \to -\infty}[f(x)-mx]{/tex}.
A?adar
{tex}m=\lim_{x \to -\infty}\frac{f(x)}{x}=\lim_{x \to -\infty}\frac{e^x-ax}{x}=\lim_{x \to -\infty}(\frac{e^x}{x}-a)=-a{/tex};
{tex}n=\lim_{x \to -\infty}[f(x)-mx]=\lim_{x \to -\infty}(e^x-ax+ax)=\lim_{x \to -\infty}e^x=0{/tex}.
Deci, ecua?ia asimptotei oblice la graficul func?iei f c?tre {tex}-\infty{/tex} este
{tex}y=-ax{/tex}.
b) Func?ia {tex}f:R \to R{/tex}, {tex}f(x)=e^x-ax{/tex}, este derivabil? pe R ?i
{tex}f':R \to R{/tex}, {tex}f'(x)=e^x-a{/tex}, {tex}\forall x\in R{/tex}.
n plus,
{tex}f'(x)=0\Leftrightarrow{/tex}
{tex}e^x-a=0\Leftrightarrow{/tex}
{tex}x=\ln a{/tex}.
Prin urmare, {tex}\ln a{/tex} este un punct critic al func?iei f ?i n plus
{tex}f'(x)>0{/tex}, {tex}\forall x \in (\ln a;+\infty){/tex}
?i
{tex}f'(x)<0{/tex}, {tex}\forall x \in (-\infty;\ln a){/tex}.
Deci, {tex}\ln a{/tex} este un punct de minim global al func?iei f iar func?ia f nu mai are alte puncte de extrem. Minimul func?iei f este
{tex}m=f(\ln a)=e^{\ln a}-a\ln a=a-a\ln a{/tex}.
c) Am ar?tat c?
{tex}f(x) \geq a-a\ln a{/tex}, {tex}\forall x \in R{/tex}.
Atunci
{tex}f(x) \geq 1{/tex}, {tex}\forall x \in R\Leftrightarrow{/tex}
{tex}a-a\ln a\geq 1\Leftrightarrow{/tex}
{tex}a-a\ln a-1\geq 0{/tex}.
Vom rezolva inecua?ia
{tex}x-x\ln x-1{/tex} n mul?imea {tex}(0;\infty){/tex}.
Consider?m func?ia {tex}g:(0;\infty)\to R{/tex}, {tex}g(x)=x-x\ln x-1{/tex}.
Func?ia g este derivabil? pe {tex}(0;\infty){/tex} ?i
{tex}g':(0;\infty) \to R{/tex}, {tex}g'(x)=1-\ln x-x \cdot \frac{1}{x}=1-\ln x-1=-\ln x{/tex}, {tex}\forall x \in(0;\infty){/tex}.
Avem
{tex}g'(1)=0{/tex}
?i
{tex}g'(x)>0{/tex}, {tex}\forall x\in(0;1){/tex}
?i
{tex}g'(x)<0{/tex}, {tex}\forall x\in(1;\infty){/tex}.
Cu alte cuvinte, 1 este punct de maxim global al func?iei g pe {tex}(0;\infty){/tex}, adic?
{tex}g(x)\leq g(1)=0{/tex}, {tex}\forall x\in(0;\infty){/tex}, valoarea 0 fiind luat? n 1.
Prin urmare, inecua?ia
{tex}g(x)\geq 0{/tex}
Are o singur? solu?ie n intervalul {tex}(0;\infty){/tex}, anume pe 1.
Deci
{tex}a-a\ln a-1\geq 0\Leftrightarrow{/tex}
{tex}a=1{/tex}
sau
{tex}f(x)\geq 1{/tex}, {tex}\forall x \in R{/tex} pentru {tex}a=1{/tex}.

2. Se consider? func?ia {tex}f:(0;\infty) \to R{/tex}, {tex}f(x)=\frac{\ln x}{\sqrt{x}}{/tex}.
a) S? se arate c? func?ia {tex}F:(0;\infty) \to R{/tex}, {tex}F(x)=2\sqrt{x}(\ln x-2){/tex} este o primitiv? a func?iei f.
b) S? se arate c? orice primitiv? G a func?iei f este cresc?toare pe {tex}[1;\infty){/tex}.
c) S? se calculeze aria suprafe?ei plane cuprinse ntre graficul func?iei f, axa Ox ?i dreptele de ecua?ii {tex}x=\frac{1}{e}{/tex} ?i {tex}x=e{/tex}.
Rezolvare:
a) Vom ar?ta c?
{tex}F'(x)=f(x){/tex}, {tex}\forall x\in (0;\infty){/tex}.
ntr-adev?r, pentru orice {tex}x \in (0;\infty){/tex}
{tex}F'(x)=2 \cdot \frac{1}{2\sqrt{x}} \cdot (\ln x - 2)+2\sqrt{x} \cdot \frac{1}{x}=\frac{1}{\sqrt{x}}+\frac{2}{\sqrt{x}}=\frac{1}{\sqrt{x}}(\ln x-2+2)=\frac{\ln x}{\sqrt{x}}}{b}.{/tex}
b) Conform pct. a, o primitiv? oarecare G a func?iei f este de forma
{tex}G:(0;\infty) \to R{/tex}, {tex}G(x)=2\sqrt{x}(\ln x-2)+C{/tex}, {tex}\forall x\in (0;\infty){/tex}, unde C este o constant? real? oarecare.
G este cresc?toare pe {tex}[1;\infty){/tex} dac? derivata ei, adic? func?ia f, este pozitiv? pe {tex}[1;\infty){/tex}, ceea ce este evident.
c) Aria cerut? este
{tex}A=\int_{\frac{1}{e}}^{e}f(x)dx=\int_{\frac{1}{e}}^{e}\frac{\ln x}{\sqrt{x}}dx=\int_{\frac{1}{e}}^{e}(2\sqrt{x})' \cdot \ln x dx{/tex}.
Aplic?m metoda de integrare prin p?r?i pentru func?iile f ?i g, unde {tex}f(x)=2\sqrt{x}{/tex} ?i {tex}g(x)=\ln x{/tex}:
{tex}A=[2\sqrt{x}\ln x]|_{\frac{1}{e}}^{e} - \int_{\frac{1}{e}}^{e} 2\sqrt{x} \cdot \frac{1}{x} dx=2\sqrt{e}\ln e - 2\sqrt{\frac{1}{e}} \ln (\frac{1}{e}) - \int_{\frac{1}{e}}^{e}\frac{2}{\sqrt{x}}dx=2\sqrt{e}+\frac{2}{\sqrt{e}}-4\sqrt{x}|_{\frac{1}{e}}^{e}=2\sqrt{e}+\frac{2}{\sqrt{e}}-(4\sqrt{e}-\frac{4}{\sqrt{e}})=2\sqrt{e}+\frac{2}{\sqrt{e}}-4\sqrt{e}+\frac{4}{\sqrt{e}}=-2\sqrt{e}+\frac{6}{\sqrt{e}}{/tex}.
A?adar
{tex}A=-2\sqrt{e}+\frac{6}{\sqrt{e}}=\frac{6-2e}{\sqrt{e}}{/tex}.

Ultima actualizare ( Duminică, 27 Martie 2011 17:28 )