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Ministerul Educatiei al Republicii Moldova Directia Invatamant Preuniversitar Examenul de bacalaureat la matematica, iunie 2004 Profilul real

Timp alocat: 180 minute.

In itemii 1-3 incercuiti litera corespunzatoare variantei corecte de raspuns.

1. Functia f : [−4; 4] → ℝ este reprezentata grafic. Care propozitie este adevarata?

[figure: y; x; −6; −5; −4; −3; −2; −1; 0; 1; 2; 3; 4; 5; −1; −2; −3; −4; −5; −6; 1; 2; 3; .^{.................................}^{..............................}^{...........................}^{........................}^{.....................}^{..................}^{...............}^{..............}^{.............}^{............} . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .; .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .^{.............}^{..............}^{...............}^{................}^{...................}^{......................}^{.........................}^{............................}]
a) f′(0) = 0  b) f′(0) > 0  c) f′(0) < 0  d) f′(0) nu exista.

2. Dreapta definita de ecuatia ax + by = 1 este paralela cu axa absciselor daca

a) a = 0 si b = 0  b) a = 0 si b ≠ 0  c) a ≠ 0 si b = 0  d) a ≠ 0 si b ≠ 0.

3. Care ecuatie admite o singura solutie pe intervalul (0; 2π]?

a) tgx = 1  b) cos x = 0  c) ctgx = −3  d) sin x = 1.

4. Completati caseta libera cu unul dintre semnele <, =, >, astfel incat propozitia sa fie adevarata.

lg tg40^{o} + lg ctg40^{o} 0.

Argumentati raspunsul.

5. Calculati limita lim_{x→6} \frac{x^{2} − 36}{\sqrt{x + 3} − 3}.

6. Determinati termenul din mujloc al dezvoltarii binomului (\sqrt[3]{x} + \frac{1}{\sqrt[3]{x}})^{6}.

7. Rezolvati in ℝ inecuatia D(x) ≤ 0, unde D(x) = | \begin{matrix}1 & x & x^{2} & x^{3} \\ 1 & 1 & 1 & 1 \\ 1 & −1 & 1 & −1 \\ 1 & 0 & 0 & 0 \end{matrix} |.

Copyright\begin{matrix}c \\ ◯ \end{matrix}2004 ONG TCV Scoala Virtuala a Tanarului Matematician http://math.ournet.md 2

8. Baza unei piramide triunghiulare este triunghiul dreptunghic ABC, unde m(∠A) = 90^{o}, m(∠B) = 60^{o}, |AC| = 12\sqrt{3} cm. Determinati volumul piramidei, daca muchiile laterale sunt congruente si au lungimea egala cu 13 cm.

9. Determinati extremele locale ale functiei f : ℝ → ℝ, f(x) = x · e^{1−2x^{2}}.

10. Rezolvati in ℝ ecuatia log_{3} x^{2} − log_{3}^{2}(−x) + 3 = 0.

11. Unul dintre zerourile unei primitive a functiei f : ℝ → ℝ, f(x) = x^{2} − 4x + 1 este egal cu 2. Determinati celelalte zerouri ale acestei primitive.

12. Intr-un trapez isoscel bazele au lungimi egale cu 21 cm si 9 cm, iar lungimea inaltimii este egala cu 8 cm. Determinati raza cercului circumscris acestui trapez.

13. Fie polinomul P (X) = X^{4} − X^{3} − aX^{2} + (b − 2)X + a, unde a, b ∈ ℝ. Se stie ca α = 1 + i este radacina a polinomului P (X). Determinati valorile parametrilor reali a si b.

14. Rezolvati in ℝ inecuatia \frac{\sqrt{3^{2x+1} − 4 · 3^{x} + 1}}{x^{2} − x − 6} ≤ 0.

Solutii

1. Raspuns corect b) f′(0) > 0.

2. Raspuns corect b) a = 0 si b ≠ 0.

3. Raspuns corect d) sin x = 1.

4. lg tg40^{o} + lg ctg40^{o} = lg(tg40^{o} · ctg40^{o}) = lg 1 = 0.

5. lim_{x→6} \frac{x^{2} − 36}{\sqrt{x + 3} − 3} = lim_{x→6} \frac{(x − 6)(x + 6)(\sqrt{x + 3} + 3)}{(\sqrt{x + 3} − 3)(\sqrt{x + 3} + 3)} = lim_{x→6} \frac{(x − 6)(x + 6)(\sqrt{x + 3} + 3)}{(x + 3 − 9)} =

= lim_{x→6}(x + 6)(\sqrt{x + 3} + 3) = 12 · 6 = 72.

Raspuns: 72.

6. Cum n = 6, dezvoltarea binomului contine 7 termeni si termenul din mijloc este T_{4}.

T_{4} = T_{3+1} = C_{6}^{3}(x^{\frac{1}{3}} )^{6−3}(x^{− \frac{1}{3}} )^{3} = \frac{6!}{3!3!}x^{\frac{1}{3} ·3}x^{− \frac{1}{3} ·3} = \frac{3! · 4 · 5 · 6}{3! · 1 · 2 · 3}x^{0} = 20.

Raspuns: T_{4} = 20.

7. Descompunem determinantul dupa linia a 4 si obtinem:

D(x) = 1·(−1)^{4+1}·| \begin{matrix}x & x^{2} & x^{3} \\ 1 & 1 & 1 \\ −1 & 1 & −1 \end{matrix} | = −x | \begin{matrix}1 & x & x^{2} \\ 1 & 1 & 1 \\ 0 & 1 & 0 \end{matrix} | = −2x·(−1)^{2+3}·| \begin{matrix}1 & x^{2} \\ 1 & 1 \end{matrix} | = 2x(1−x^{2}) =

= 2x(1 − x)(1 + x).

Utilizand metoda intervalelor, pentru inecuatia D(x) ≤ 0 ⇔ 2x(1 − x)(1 + x) ≤ 0, se obtine x ∈ [−1; 0] ∪ [1; +∞).

[figure: −1; 0; 1; +; −; +; −]

Raspuns: x ∈ [−1; 0] ∪ [1; +∞).

8.

[figure: S; A; B; C; D]

Cum muchiile laterale sunt congruente, piciorul D al inaltimii SD = h se afla in mijlocul ipotenuzei BC. Determinam ipotenuza BC:

BC = \frac{AC}{sin ∠B} = \frac{12\sqrt{3}}{\frac{\sqrt{3}}{2}} = 24(cm).

Rezulta BD = \frac{24}{2} = 12. Din △SDB (dreptunghic in D) aflam inaltimea piramidei h:

h = SD = \sqrt{SB^{2} − BD^{2}} = \sqrt{13^{2} − 12^{2}} = \sqrt{1 · 25} = 5(cm).

Aflam aria bazei:

S_{△ABC} = \frac{1}{2}AC · BC · sin C = \frac{1}{2}12\sqrt{3} · 24 · sin(90^{o} − 60^{o}) = \frac{1}{2}12\sqrt{3} · 24 · \frac{1}{2} = 72\sqrt{3} (cm^{2}).

Aflam volumul piramidei:

V = \frac{1}{3}S_{△ABC} · SD = \frac{1}{3}72\sqrt{3} · 5 = 120\sqrt{3} (cm^{3}).

Raspuns: V = 120\sqrt{3} (cm^{3}).

9. Aflam derivata functiei f:

f′(x) = x′e^{1−2x^{2}} + xe^{1−2x^{2}}(1 − 2x^{2})′ = e^{1−2x^{2}} + xe^{1−2x^{2}}(−4x) = = e^{1−2x^{2}}(1 − 4x^{2}).

Aflam punctele critice, rezolvand ecuatia f′(x) = 0:

f′(x) = 0 ⇔ e^{1−2x^{2}}(1 − 4x^{2}) = 0 ⇔ x_{1} = −\frac{1}{2} si x_{2} = \frac{1}{2}.

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Determinam semnul functiei f′ pe intervale (−∞; −\frac{1}{2}), (−\frac{1}{2}, \frac{1}{2}) si (\frac{1}{2}; +∞)

[figure: − \frac{1}{2}; \frac{1}{2}; −; +; −; ↘; ↗; ↘]

si obtinem extremele locale: x = −\frac{1}{2} punct de minim, f_{min} = f(−\frac{1}{2}) = −\frac{\sqrt{e}}{2} si x = \frac{1}{2} punct de maxim, f_{max} = f(\frac{1}{2}) = \frac{\sqrt{e}}{2}.

Raspuns: f_{min} = −\frac{\sqrt{e}}{2}, f_{max} = \frac{\sqrt{e}}{2}.

10. DVA: x < 0. log_{3} x^{2} − log_{3}^{2}(−x) + 3 = 0 ⇔ 2 log_{3} |x| − log_{3}^{2}(−x) + 3 = 0 ⇔

⇔ log_{3}^{2}(−x)−2 log_{3} |x|−3 = 0 ⇔ [ \begin{matrix}log_{3}(−x) = −1, \\ log_{3}(−x) = 3, \end{matrix} ⇔ [ \begin{matrix}−x = 3^{−1}, \\ −x = 3^{3}, \end{matrix} ⇔ [ \begin{matrix}x = −\frac{1}{3} ∈ DVA, \\ x = −27 ∈ DVA. \end{matrix}

Raspuns: x ∈ \{−\frac{1}{3}; −27\}.

11. F (x) = ∫ (x^{2} − 4x + 1) dx = \frac{x^{3}}{3} − 2x^{2} + x + C.

Cum F (2) = 0, rezulta: \frac{8}{3} − 8 + 2 + C = 0, de unde C = \frac{10}{3}. Prin urmare, F (x) = \frac{x^{3}}{3} − 2x^{2} + x + \frac{10}{3}. Determinam, celelalte radacini: F (x) = 0 ⇔ x^{3} − 6x^{2} + 3x + 10 = 0 ⇔ x^{3} − 2x^{2} − 4x^{2} + 8x − 5x + 10 = 0 ⇔ ⇔ x^{2}(x−2)−4x(x−2)−5(x−2) = 0 ⇔ (x−2)(x^{2}−4x−5) = 0 ⇔ (x−2)(x+1)(x−5) = 0, de unde x_{2} = −1, x_{3} = 5.

Raspuns: x_{2} = −1, x_{3} = 5.

12.

[figure: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .^{.......................}^{........................}^{........................}^{........................}^{.......................}^{......................}^{......................}^{.......................}^{........................}^{........................}^{........................}^{.......................}^{.......................}^{........................}^{........................}^{........................}^{.......................}^{......................}^{......................}^{.......................}^{........................}^{........................}^{........................}^{.......................} . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . .; C; D; B; A; E; ^{.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .}]

Fie AB = 9 cm, CD = 21 cm, AE⊥DC, AE = 8 cm.

Cum trapezul ABCD este isoscel CE = \frac{CD − AB}{2} = \frac{21 − 9}{2} = 6 (cm), atunci DE = 21−6−15 (cm), AD = \sqrt{AE^{2} + DE^{2}} = \sqrt{64 + 225} = 17 (cm) (din △AED dreptuinghic in E). Din △AEC (dreptunghic in E) aflam AC: AC = \sqrt{CE^{2} + AE^{2}} = \sqrt{64 + 36} = 10 (cm).

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Atunci

sin ∠ACE = \frac{AE}{AC} = \frac{8}{10} = \frac{4}{5}.

Conform teoremei sinusurilor \frac{AD}{sin ∠ACE} = 2R, de unde

R = \frac{17 · 5}{2 · 4} = \frac{85}{8} = 10\frac{5}{8}.

Ramane de observat ca raza cercului circumscris △ACD coincide cu raza cercului circumscris trapezului.

Raspuns: R = 10\frac{5}{8} cm.

13. Cum α = 1 + i radacina a polinomului P (X), rezulta P (α) = 0 si (1 + i)^{4} − (1 + i)^{3} − a(1 + i)^{2} + (b − 2)(1 + i) + a = 0 ⇔ ⇔ 1 + 4i^{3} + 6i^{2} + 4i^{3} + i^{4} − 1 − 3i − 3i^{2} − i^{3} − a(1 + 2i + i^{2}) + (b − 2) + i(b − 2) + a = 0 ⇔ ⇔ 1 + 4i − 6 − 4i + 1 − 1 − 3i + 3 + i − 2ai + (b − 2) + i(b − 2) + a = 0 ⇔ ⇔ a − 2 + b − 2 + i(b − 2 − 2 − 2a) = 0 ⇔ a + b − 4 + i(b − 2a − 4) = 0.

Utilizand definitia egalitatii a doua numere complexe, obtinem \{ \begin{matrix}a + b − 4 = 0, \\ b − 2a − 4 = 0, \end{matrix} de unde

\{ \begin{matrix}a = 0, \\ b = 4. \end{matrix}

Raspuns: a = 0, b = 4.

14.

\frac{\sqrt{3^{2x+1} − 4 · 3^{x} + 1}}{x^{2} − x − 6} ≤ 0 ⇔ [ \begin{matrix}\{ & \begin{matrix}3^{2x+1} − 4 · 3^{x} + 1 = 0, \\ x ∈ ℝ\\{−2; 3\}, \end{matrix} \\ \{ & \begin{matrix}3^{2x+1} − 4 · 3^{x} + 1 > 0, \\ x^{2} − x − 6 < 0, \end{matrix} \end{matrix} ⇔ [ \begin{matrix}\{ & \begin{matrix}[ & \begin{matrix}3^{x} = 1, \\ 3^{x} = \frac{1}{3}, \end{matrix} \\ x ∈ ℝ\\{−2; 3\}, \end{matrix} \\ \{ & \begin{matrix}(3^{x} − 1) (3^{x} − \frac{1}{3}) > 0, \\ −2 < x < 3, \end{matrix} \end{matrix} ⇔

⇔ [ \begin{matrix}\{ & x = 0, x = −1, \\ \{ & \begin{matrix}[ & \begin{matrix}x > 0, \\ x < −1, \end{matrix} \\ −2 < x < 3, \end{matrix} \end{matrix} ⇔ x ∈ (−2; −1] ∪ [0; 3).

Raspuns: x ∈ (−2; −1] ∪ [0; 3).