SUBIECTUL I
1. Într-o progresie aritmetică, rația este
r = a 2 − a 1 = 15 − 4 = 11. r=a_2-a_1=15-4=11. r = a 2 − a 1 = 15 − 4 = 11. Atunci
a 3 = a 2 + r = 15 + 11 = 26. a_3=a_2+r=15+11=26. a 3 = a 2 + r = 15 + 11 = 26. Prin urmare, a 3 = 26 \boxed{a_3=26} a 3 = 26
2. Avem
f ( a ) = 3 a − 2 , f ( 2 ) = 3 ⋅ 2 − 2 = 4. f(a)=3a-2,\qquad f(2)=3\cdot2-2=4. f ( a ) = 3 a − 2 , f ( 2 ) = 3 ⋅ 2 − 2 = 4. Condiția din enunț devine
f ( a ) + f ( 2 ) = 2 a ⟺ 3 a − 2 + 4 = 2 a . f(a)+f(2)=2a
\Longleftrightarrow
3a-2+4=2a. f ( a ) + f ( 2 ) = 2 a ⟺ 3 a − 2 + 4 = 2 a . Deci
3 a + 2 = 2 a ⟺ a = − 2. 3a+2=2a \Longleftrightarrow a=-2. 3 a + 2 = 2 a ⟺ a = − 2. Prin urmare, a = − 2 \boxed{a=-2} a = − 2
3. Ecuația este
log 2 ( x 2 − 3 x + 2 ) = log 2 ( 2 + x ) . \log_2(x^2-3x+2)=\log_2(2+x). log 2 ( x 2 − 3 x + 2 ) = log 2 ( 2 + x ) . Condițiile de existență sunt
x 2 − 3 x + 2 > 0 , 2 + x > 0. x^2-3x+2>0,\qquad 2+x>0. x 2 − 3 x + 2 > 0 , 2 + x > 0. Prima condiție devine
( x − 1 ) ( x − 2 ) > 0 ⟺ x ∈ ( − ∞ , 1 ) ∪ ( 2 , + ∞ ) , (x-1)(x-2)>0 \Longleftrightarrow x\in(-\infty,1)\cup(2,+\infty), ( x − 1 ) ( x − 2 ) > 0 ⟺ x ∈ ( − ∞ , 1 ) ∪ ( 2 , + ∞ ) , iar a doua este x > − 2 x>-2 x > − 2 2 > 1 2>1 2 > 1
x 2 − 3 x + 2 = x + 2. x^2-3x+2=x+2. x 2 − 3 x + 2 = x + 2. Rezultă
x 2 − 4 x = 0 ⟺ x ( x − 4 ) = 0 , x^2-4x=0
\Longleftrightarrow
x(x-4)=0, x 2 − 4 x = 0 ⟺ x ( x − 4 ) = 0 , deci x = 0 x=0 x = 0 x = 4 x=4 x = 4
0 ∈ ( − 2 , 1 ) , 4 ∈ ( 2 , + ∞ ) . 0\in(-2,1),\qquad 4\in(2,+\infty). 0 ∈ ( − 2 , 1 ) , 4 ∈ ( 2 , + ∞ ) . Ambele valori sunt admise, așadar soluția este
x ∈ { 0 , 4 } . \boxed{x\in\{0,4\}}. x ∈ { 0 , 4 } . 4. Un număr par are cifra unităților pară. Din mulțimea
A = { 1 , 2 , 3 , 5 , 7 , 8 } , A=\{1,2,3,5,7,8\}, A = { 1 , 2 , 3 , 5 , 7 , 8 } , cifrele pare posibile pentru unități sunt 2 2 2 8 8 8
Pentru fiecare alegere a cifrei unităților, cifra zecilor se poate alege dintre celelalte 5 5 5
2 ⋅ 5 = 10. 2\cdot 5=10. 2 ⋅ 5 = 10. Se pot forma 10 \boxed{10} 10
5. Avem O ( 0 , 0 ) O(0,0) O ( 0 , 0 ) A ( 0 , 5 ) A(0,5) A ( 0 , 5 ) B ( 8 , 4 ) B(8,4) B ( 8 , 4 ) C C C O B OB O B
C ( 0 + 8 2 , 0 + 4 2 ) = C ( 4 , 2 ) . C\left(\frac{0+8}{2},\frac{0+4}{2}\right)=C(4,2). C ( 2 0 + 8 , 2 0 + 4 ) = C ( 4 , 2 ) . Calculăm lungimile:
A O = ( 0 − 0 ) 2 + ( 5 − 0 ) 2 = 5 , AO=\sqrt{(0-0)^2+(5-0)^2}=5, A O = ( 0 − 0 ) 2 + ( 5 − 0 ) 2 = 5 , A C = ( 4 − 0 ) 2 + ( 2 − 5 ) 2 = 16 + 9 = 5. AC=\sqrt{(4-0)^2+(2-5)^2}
=\sqrt{16+9}
=5. A C = ( 4 − 0 ) 2 + ( 2 − 5 ) 2 = 16 + 9 = 5. Prin urmare,
A O = A C . \boxed{AO=AC}. A O = A C . 6. Triunghiul A B C ABC A B C A A A B C BC B C B B B
cos B = cateta al a ˘ turat a ˘ ipotenuz a ˘ = A B B C . \cos B=\frac{\text{cateta alăturată}}{\text{ipotenuză}}=\frac{AB}{BC}. cos B = ipotenuz a ˘ cateta al a ˘ turat a ˘ = B C A B . Din enunț,
A B B C = 2 3 . \frac{AB}{BC}=\frac23. B C A B = 3 2 . Cum A B = 4 AB=4 A B = 4
4 B C = 2 3 ⟺ 2 B C = 12 ⟺ B C = 6. \frac{4}{BC}=\frac23
\Longleftrightarrow
2BC=12
\Longleftrightarrow
BC=6. B C 4 = 3 2 ⟺ 2 B C = 12 ⟺ B C = 6. Deci B C = 6 \boxed{BC=6} B C = 6
SUBIECTUL al II-lea
1. Se consideră
I 2 = ( 1 0 0 1 ) , A = ( − 2 8 − 1 3 ) . I_2=\begin{pmatrix}1&0\\0&1\end{pmatrix},
\qquad
A=\begin{pmatrix}-2&8\\-1&3\end{pmatrix}. I 2 = ( 1 0 0 1 ) , A = ( − 2 − 1 8 3 ) . 1.a) Determinantul matricei A A A
det A = ( − 2 ) ⋅ 3 − 8 ⋅ ( − 1 ) = − 6 + 8 = 2. \det A=(-2)\cdot3-8\cdot(-1)=-6+8=2. det A = ( − 2 ) ⋅ 3 − 8 ⋅ ( − 1 ) = − 6 + 8 = 2. Prin urmare, det A = 2 \boxed{\det A=2} det A = 2
1.b) Calculăm
I 2 − A = ( 1 0 0 1 ) − ( − 2 8 − 1 3 ) = ( 3 − 8 1 − 2 ) . I_2-A=
\begin{pmatrix}1&0\\0&1\end{pmatrix}
-
\begin{pmatrix}-2&8\\-1&3\end{pmatrix}
=
\begin{pmatrix}3&-8\\1&-2\end{pmatrix}. I 2 − A = ( 1 0 0 1 ) − ( − 2 − 1 8 3 ) = ( 3 1 − 8 − 2 ) . Deci
B = 1 2 ( I 2 − A ) = ( 3 2 − 4 1 2 − 1 ) . B=\frac12(I_2-A)
=
\begin{pmatrix}\frac32&-4\\[2pt]\frac12&-1\end{pmatrix}. B = 2 1 ( I 2 − A ) = ( 2 3 2 1 − 4 − 1 ) . Verificăm produsul:
A B = ( − 2 8 − 1 3 ) ( 3 2 − 4 1 2 − 1 ) = ( − 3 + 4 8 − 8 − 3 2 + 3 2 4 − 3 ) = ( 1 0 0 1 ) = I 2 . AB=
\begin{pmatrix}-2&8\\-1&3\end{pmatrix}
\begin{pmatrix}\frac32&-4\\[2pt]\frac12&-1\end{pmatrix}
=
\begin{pmatrix}
-3+4&8-8\\[2pt]
-\frac32+\frac32&4-3
\end{pmatrix}
=
\begin{pmatrix}1&0\\0&1\end{pmatrix}
=I_2. A B = ( − 2 − 1 8 3 ) ( 2 3 2 1 − 4 − 1 ) = ( − 3 + 4 − 2 3 + 2 3 8 − 8 4 − 3 ) = ( 1 0 0 1 ) = I 2 . De asemenea,
B A = ( 3 2 − 4 1 2 − 1 ) ( − 2 8 − 1 3 ) = ( − 3 + 4 12 − 12 − 1 + 1 4 − 3 ) = I 2 . BA=
\begin{pmatrix}\frac32&-4\\[2pt]\frac12&-1\end{pmatrix}
\begin{pmatrix}-2&8\\-1&3\end{pmatrix}
=
\begin{pmatrix}
-3+4&12-12\\[2pt]
-1+1&4-3
\end{pmatrix}
=I_2. B A = ( 2 3 2 1 − 4 − 1 ) ( − 2 − 1 8 3 ) = ( − 3 + 4 − 1 + 1 12 − 12 4 − 3 ) = I 2 . Așadar, B B B A A A
B = A − 1 . \boxed{B=A^{-1}}. B = A − 1 . 1.c) Ecuația este
( A − I 2 ) X = 2 A . (A-I_2)X=2A. ( A − I 2 ) X = 2 A . Din punctul b), avem
A − 1 = 1 2 ( I 2 − A ) , A^{-1}=\frac12(I_2-A), A − 1 = 2 1 ( I 2 − A ) , de unde
I 2 − A = 2 A − 1 și A − I 2 = − 2 A − 1 . I_2-A=2A^{-1}
\quad\text{și}\quad
A-I_2=-2A^{-1}. I 2 − A = 2 A − 1 ș i A − I 2 = − 2 A − 1 . Astfel,
− 2 A − 1 X = 2 A . -2A^{-1}X=2A. − 2 A − 1 X = 2 A . Împărțind cu − 2 -2 − 2
A − 1 X = − A . A^{-1}X=-A. A − 1 X = − A . Înmulțim la stânga cu A A A
X = − A 2 . X=-A^2. X = − A 2 . Calculăm
A 2 = ( − 2 8 − 1 3 ) ( − 2 8 − 1 3 ) = ( 4 − 8 − 16 + 24 2 − 3 − 8 + 9 ) = ( − 4 8 − 1 1 ) . A^2=
\begin{pmatrix}-2&8\\-1&3\end{pmatrix}
\begin{pmatrix}-2&8\\-1&3\end{pmatrix}
=
\begin{pmatrix}
4-8&-16+24\\
2-3&-8+9
\end{pmatrix}
=
\begin{pmatrix}-4&8\\-1&1\end{pmatrix}. A 2 = ( − 2 − 1 8 3 ) ( − 2 − 1 8 3 ) = ( 4 − 8 2 − 3 − 16 + 24 − 8 + 9 ) = ( − 4 − 1 8 1 ) . Prin urmare,
X = − A 2 = ( 4 − 8 1 − 1 ) . X=-A^2
=
\boxed{\begin{pmatrix}4&-8\\1&-1\end{pmatrix}}. X = − A 2 = ( 4 1 − 8 − 1 ) . Verificare:
( A − I 2 ) X = ( − 3 8 − 1 2 ) ( 4 − 8 1 − 1 ) = ( − 4 16 − 2 6 ) = 2 A . (A-I_2)X
=
\begin{pmatrix}-3&8\\-1&2\end{pmatrix}
\begin{pmatrix}4&-8\\1&-1\end{pmatrix}
=
\begin{pmatrix}-4&16\\-2&6\end{pmatrix}
=2A. ( A − I 2 ) X = ( − 3 − 1 8 2 ) ( 4 1 − 8 − 1 ) = ( − 4 − 2 16 6 ) = 2 A . 2. Pe M = [ 0 , + ∞ ) M=[0,+\infty) M = [ 0 , + ∞ )
x ∗ y = x + y − x y . x*y=x+y-\sqrt{xy}. x ∗ y = x + y − x y . Pentru x , y ∈ M x,y\in M x , y ∈ M x y ≥ 0 xy\ge 0 x y ≥ 0
x + y ≥ 2 x y ≥ x y , x+y\ge 2\sqrt{xy}\ge \sqrt{xy}, x + y ≥ 2 x y ≥ x y , pentru x , y ≥ 0 x,y\ge 0 x , y ≥ 0 x + y − x y ≥ 0 x+y-\sqrt{xy}\ge 0 x + y − x y ≥ 0 M M M
2.a) Calculăm direct:
1 ∗ 4 = 1 + 4 − 1 ⋅ 4 = 5 − 2 = 3. 1*4=1+4-\sqrt{1\cdot4}=5-2=3. 1 ∗ 4 = 1 + 4 − 1 ⋅ 4 = 5 − 2 = 3. Deci 1 ∗ 4 = 3 \boxed{1*4=3} 1 ∗ 4 = 3
2.b) Cerem x ∈ M x\in M x ∈ M x ≥ 0 x\ge 0 x ≥ 0
x ∗ ( 9 x ) = x + 9 x − x ⋅ 9 x . x*(9x)=x+9x-\sqrt{x\cdot 9x}. x ∗ ( 9 x ) = x + 9 x − x ⋅ 9 x . Cum x ≥ 0 x\ge 0 x ≥ 0
9 x 2 = 3 x . \sqrt{9x^2}=3x. 9 x 2 = 3 x . Prin urmare,
x ∗ ( 9 x ) = 10 x − 3 x = 7 x . x*(9x)=10x-3x=7x. x ∗ ( 9 x ) = 10 x − 3 x = 7 x . Condiția devine
7 x = x 2 ⟺ x 2 − 7 x = 0 ⟺ x ( x − 7 ) = 0. 7x=x^2
\Longleftrightarrow
x^2-7x=0
\Longleftrightarrow
x(x-7)=0. 7 x = x 2 ⟺ x 2 − 7 x = 0 ⟺ x ( x − 7 ) = 0. Rezultă
x = 0 sau x = 7. x=0 \quad \text{sau} \quad x=7. x = 0 sau x = 7. Ambele aparțin lui M M M
x ∈ { 0 , 7 } . \boxed{x\in\{0,7\}}. x ∈ { 0 , 7 } . 2.c) Deoarece 2 x > 0 2^x>0 2 x > 0 2 x + 2 > 0 2^{x+2}>0 2 x + 2 > 0 x ∈ R x\in\mathbb R x ∈ R
2 x ∗ 2 x + 2 = 2 x + 2 x + 2 − 2 x ⋅ 2 x + 2 . 2^x*2^{x+2}
=2^x+2^{x+2}-\sqrt{2^x\cdot 2^{x+2}}. 2 x ∗ 2 x + 2 = 2 x + 2 x + 2 − 2 x ⋅ 2 x + 2 . Avem
2 x + 2 = 4 ⋅ 2 x , 2^{x+2}=4\cdot 2^x, 2 x + 2 = 4 ⋅ 2 x , și
2 x ⋅ 2 x + 2 = 2 2 x + 2 = 2 x + 1 , \sqrt{2^x\cdot 2^{x+2}}
=\sqrt{2^{2x+2}}
=2^{x+1}, 2 x ⋅ 2 x + 2 = 2 2 x + 2 = 2 x + 1 , deoarece 2 x + 1 > 0 2^{x+1}>0 2 x + 1 > 0
2 x ∗ 2 x + 2 = 2 x + 4 ⋅ 2 x − 2 x + 1 = 5 ⋅ 2 x − 2 ⋅ 2 x = 3 ⋅ 2 x . 2^x*2^{x+2}
=2^x+4\cdot2^x-2^{x+1}
=5\cdot2^x-2\cdot2^x
=3\cdot2^x. 2 x ∗ 2 x + 2 = 2 x + 4 ⋅ 2 x − 2 x + 1 = 5 ⋅ 2 x − 2 ⋅ 2 x = 3 ⋅ 2 x . Ecuația devine
3 ⋅ 2 x = 6 x . 3\cdot2^x=6^x. 3 ⋅ 2 x = 6 x . Cum 6 x = 2 x ⋅ 3 x 6^x=2^x\cdot3^x 6 x = 2 x ⋅ 3 x 2 x > 0 2^x>0 2 x > 0 2 x 2^x 2 x
3 = 3 x . 3=3^x. 3 = 3 x . Rezultă
x = 1. x=1. x = 1. Prin urmare, x = 1 \boxed{x=1} x = 1
SUBIECTUL al III-lea
1. Se consideră funcția
f : ( 0 , + ∞ ) → R , f ( x ) = 2 x x 2 + x + 2 . f:(0,+\infty)\to\mathbb R,\qquad
f(x)=\frac{2\sqrt{x}}{x^2+x+2}. f : ( 0 , + ∞ ) → R , f ( x ) = x 2 + x + 2 2 x . Pentru x > 0 x>0 x > 0 x \sqrt{x} x x 2 + x + 2 > 0 x^2+x+2>0 x 2 + x + 2 > 0
x 2 + x + 2 = ( x + 1 2 ) 2 + 7 4 > 0. x^2+x+2=\left(x+\frac12\right)^2+\frac74>0. x 2 + x + 2 = ( x + 2 1 ) 2 + 4 7 > 0. 1.a) Notăm
u ( x ) = 2 x , v ( x ) = x 2 + x + 2. u(x)=2\sqrt{x},\qquad v(x)=x^2+x+2. u ( x ) = 2 x , v ( x ) = x 2 + x + 2. Atunci
u ′ ( x ) = 1 x , v ′ ( x ) = 2 x + 1. u'(x)=\frac{1}{\sqrt{x}},\qquad v'(x)=2x+1. u ′ ( x ) = x 1 , v ′ ( x ) = 2 x + 1. Folosind regula derivării câtului,
f ′ ( x ) = u ′ ( x ) v ( x ) − u ( x ) v ′ ( x ) v 2 ( x ) . f'(x)=\frac{u'(x)v(x)-u(x)v'(x)}{v^2(x)}. f ′ ( x ) = v 2 ( x ) u ′ ( x ) v ( x ) − u ( x ) v ′ ( x ) . Deci
f ′ ( x ) = 1 x ( x 2 + x + 2 ) − 2 x ( 2 x + 1 ) ( x 2 + x + 2 ) 2 . f'(x)
=
\frac{\frac{1}{\sqrt{x}}(x^2+x+2)-2\sqrt{x}(2x+1)}
{(x^2+x+2)^2}. f ′ ( x ) = ( x 2 + x + 2 ) 2 x 1 ( x 2 + x + 2 ) − 2 x ( 2 x + 1 ) . Aducem numărătorul la factorul comun 1 x \frac{1}{\sqrt{x}} x 1
1 x ( x 2 + x + 2 ) − 2 x ( 2 x + 1 ) = x 2 + x + 2 − 2 x ( 2 x + 1 ) x . \frac{1}{\sqrt{x}}(x^2+x+2)-2\sqrt{x}(2x+1)
=
\frac{x^2+x+2-2x(2x+1)}{\sqrt{x}}. x 1 ( x 2 + x + 2 ) − 2 x ( 2 x + 1 ) = x x 2 + x + 2 − 2 x ( 2 x + 1 ) . Prin urmare,
f ′ ( x ) = x 2 + x + 2 − 4 x 2 − 2 x ( x 2 + x + 2 ) 2 x = − 3 x 2 − x + 2 ( x 2 + x + 2 ) 2 x . f'(x)
=
\frac{x^2+x+2-4x^2-2x}{(x^2+x+2)^2\sqrt{x}}
=
\frac{-3x^2-x+2}{(x^2+x+2)^2\sqrt{x}}. f ′ ( x ) = ( x 2 + x + 2 ) 2 x x 2 + x + 2 − 4 x 2 − 2 x = ( x 2 + x + 2 ) 2 x − 3 x 2 − x + 2 . Așadar,
f ′ ( x ) = − 3 x 2 − x + 2 ( x 2 + x + 2 ) 2 x , x ∈ ( 0 , + ∞ ) . \boxed{f'(x)=\frac{-3x^2-x+2}{(x^2+x+2)^2\sqrt{x}}},
\qquad x\in(0,+\infty). f ′ ( x ) = ( x 2 + x + 2 ) 2 x − 3 x 2 − x + 2 , x ∈ ( 0 , + ∞ ) . 1.b) Asimptota orizontală spre + ∞ +\infty + ∞ y = L y=L y = L
L = lim x → + ∞ f ( x ) . L=\lim_{x\to+\infty}f(x). L = x → + ∞ lim f ( x ) . Calculăm
lim x → + ∞ 2 x x 2 + x + 2 = lim x → + ∞ 2 x 1 / 2 x 2 ( 1 + 1 x + 2 x 2 ) = lim x → + ∞ 2 x 3 / 2 ( 1 + 1 x + 2 x 2 ) = 0. \lim_{x\to+\infty}\frac{2\sqrt{x}}{x^2+x+2}
=
\lim_{x\to+\infty}
\frac{2x^{1/2}}{x^2\left(1+\frac1x+\frac2{x^2}\right)}
=
\lim_{x\to+\infty}
\frac{2}{x^{3/2}\left(1+\frac1x+\frac2{x^2}\right)}
=0. x → + ∞ lim x 2 + x + 2 2 x = x → + ∞ lim x 2 ( 1 + x 1 + x 2 2 ) 2 x 1/2 = x → + ∞ lim x 3/2 ( 1 + x 1 + x 2 2 ) 2 = 0. Prin urmare, asimptota orizontală spre + ∞ +\infty + ∞
y = 0 . \boxed{y=0}. y = 0 . 1.c) Tangenta la grafic este paralelă cu axa O x Ox O x 0 0 0
f ′ ( a ) = 0. f'(a)=0. f ′ ( a ) = 0. Pentru a > 0 a>0 a > 0
( a 2 + a + 2 ) 2 a (a^2+a+2)^2\sqrt{a} ( a 2 + a + 2 ) 2 a este strict pozitiv, deci trebuie ca numărătorul să fie 0 0 0
− 3 a 2 − a + 2 = 0. -3a^2-a+2=0. − 3 a 2 − a + 2 = 0. Echivalent,
3 a 2 + a − 2 = 0. 3a^2+a-2=0. 3 a 2 + a − 2 = 0. Factorizăm:
3 a 2 + a − 2 = ( 3 a − 2 ) ( a + 1 ) . 3a^2+a-2=(3a-2)(a+1). 3 a 2 + a − 2 = ( 3 a − 2 ) ( a + 1 ) . Astfel,
( 3 a − 2 ) ( a + 1 ) = 0 ⟺ a = 2 3 sau a = − 1. (3a-2)(a+1)=0
\Longleftrightarrow
a=\frac23 \quad \text{sau} \quad a=-1. ( 3 a − 2 ) ( a + 1 ) = 0 ⟺ a = 3 2 sau a = − 1. Cum a ∈ ( 0 , + ∞ ) a\in(0,+\infty) a ∈ ( 0 , + ∞ )
a = 2 3 . \boxed{a=\frac23}. a = 3 2 . 2. Se consideră funcția
f : R → R , f ( x ) = x 3 + 3 x + 3. f:\mathbb R\to\mathbb R,\qquad f(x)=x^3+3x+3. f : R → R , f ( x ) = x 3 + 3 x + 3. 2.a) Avem
f ( x ) − 3 x − 3 = x 3 + 3 x + 3 − 3 x − 3 = x 3 . f(x)-3x-3=x^3+3x+3-3x-3=x^3. f ( x ) − 3 x − 3 = x 3 + 3 x + 3 − 3 x − 3 = x 3 . Prin urmare,
∫ 0 2 ( f ( x ) − 3 x − 3 ) d x = ∫ 0 2 x 3 d x = [ x 4 4 ] 0 2 . \int_0^2 (f(x)-3x-3)\,dx
=
\int_0^2 x^3\,dx
=
\left[\frac{x^4}{4}\right]_0^2. ∫ 0 2 ( f ( x ) − 3 x − 3 ) d x = ∫ 0 2 x 3 d x = [ 4 x 4 ] 0 2 . Rezultă
[ x 4 4 ] 0 2 = 2 4 4 − 0 = 16 4 = 4. \left[\frac{x^4}{4}\right]_0^2
=\frac{2^4}{4}-0
=\frac{16}{4}
=4. [ 4 x 4 ] 0 2 = 4 2 4 − 0 = 4 16 = 4. Deci
∫ 0 2 ( f ( x ) − 3 x − 3 ) d x = 4 . \boxed{\int_0^2 (f(x)-3x-3)\,dx=4}. ∫ 0 2 ( f ( x ) − 3 x − 3 ) d x = 4 . 2.b) Observăm că
f ( x ) − x 3 = x 3 + 3 x + 3 − x 3 = 3 x + 3 = 3 ( x + 1 ) . f(x)-x^3=x^3+3x+3-x^3=3x+3=3(x+1). f ( x ) − x 3 = x 3 + 3 x + 3 − x 3 = 3 x + 3 = 3 ( x + 1 ) . Pe intervalul [ 0 , 1 ] [0,1] [ 0 , 1 ] 3 ( x + 1 ) > 0 3(x+1)>0 3 ( x + 1 ) > 0
∫ 0 1 1 ( f ( x ) − x 3 ) 2 d x = ∫ 0 1 1 9 ( x + 1 ) 2 d x = 1 9 ∫ 0 1 ( x + 1 ) − 2 d x . \int_0^1 \frac{1}{(f(x)-x^3)^2}\,dx
=
\int_0^1 \frac{1}{9(x+1)^2}\,dx
=
\frac19\int_0^1 (x+1)^{-2}\,dx. ∫ 0 1 ( f ( x ) − x 3 ) 2 1 d x = ∫ 0 1 9 ( x + 1 ) 2 1 d x = 9 1 ∫ 0 1 ( x + 1 ) − 2 d x . O primitivă este
∫ ( x + 1 ) − 2 d x = − ( x + 1 ) − 1 + C . \int (x+1)^{-2}\,dx=-(x+1)^{-1}+C. ∫ ( x + 1 ) − 2 d x = − ( x + 1 ) − 1 + C . Prin urmare,
1 9 ∫ 0 1 ( x + 1 ) − 2 d x = 1 9 [ − 1 x + 1 ] 0 1 = 1 9 ( − 1 2 + 1 ) = 1 18 . \frac19\int_0^1 (x+1)^{-2}\,dx
=
\frac19\left[-\frac{1}{x+1}\right]_0^1
=
\frac19\left(-\frac12+1\right)
=
\frac{1}{18}. 9 1 ∫ 0 1 ( x + 1 ) − 2 d x = 9 1 [ − x + 1 1 ] 0 1 = 9 1 ( − 2 1 + 1 ) = 18 1 . Așadar,
∫ 0 1 1 ( f ( x ) − x 3 ) 2 d x = 1 18 . \boxed{\int_0^1 \frac{1}{(f(x)-x^3)^2}\,dx=\frac{1}{18}}. ∫ 0 1 ( f ( x ) − x 3 ) 2 1 d x = 18 1 . 2.c) Avem
f ( x ) − 3 = x 3 + 3 x , f(x)-3=x^3+3x, f ( x ) − 3 = x 3 + 3 x , deci, pentru x ∈ [ 1 , e ] x\in[1,e] x ∈ [ 1 , e ]
f ( x ) − 3 x 2 = x 3 + 3 x x 2 = x + 3 x . \frac{f(x)-3}{x^2}
=
\frac{x^3+3x}{x^2}
=
x+\frac3x. x 2 f ( x ) − 3 = x 2 x 3 + 3 x = x + x 3 . Integrala devine
∫ 1 e f ( x ) − 3 x 2 ln x d x = ∫ 1 e ( x + 3 x ) ln x d x . \int_1^e \frac{f(x)-3}{x^2}\ln x\,dx
=
\int_1^e \left(x+\frac3x\right)\ln x\,dx. ∫ 1 e x 2 f ( x ) − 3 ln x d x = ∫ 1 e ( x + x 3 ) ln x d x . Calculăm separat:
∫ x ln x d x = x 2 2 ln x − ∫ x 2 2 ⋅ 1 x d x = x 2 2 ln x − x 2 4 + C . \int x\ln x\,dx
=\frac{x^2}{2}\ln x-\int \frac{x^2}{2}\cdot\frac1x\,dx
=\frac{x^2}{2}\ln x-\frac{x^2}{4}+C. ∫ x ln x d x = 2 x 2 ln x − ∫ 2 x 2 ⋅ x 1 d x = 2 x 2 ln x − 4 x 2 + C . Astfel,
∫ 1 e x ln x d x = [ x 2 2 ln x − x 2 4 ] 1 e . \int_1^e x\ln x\,dx
=
\left[\frac{x^2}{2}\ln x-\frac{x^2}{4}\right]_1^e. ∫ 1 e x ln x d x = [ 2 x 2 ln x − 4 x 2 ] 1 e . Deoarece ln e = 1 \ln e=1 ln e = 1 ln 1 = 0 \ln1=0 ln 1 = 0
( e 2 2 − e 2 4 ) − ( 0 − 1 4 ) = e 2 4 + 1 4 = e 2 + 1 4 . \left(\frac{e^2}{2}-\frac{e^2}{4}\right)
-
\left(0-\frac14\right)
=
\frac{e^2}{4}+\frac14
=
\frac{e^2+1}{4}. ( 2 e 2 − 4 e 2 ) − ( 0 − 4 1 ) = 4 e 2 + 4 1 = 4 e 2 + 1 . De asemenea,
∫ 1 e 3 x ln x d x = 3 [ ( ln x ) 2 2 ] 1 e = 3 2 ( 1 2 − 0 2 ) = 3 2 . \int_1^e \frac3x\ln x\,dx
=
3\left[\frac{(\ln x)^2}{2}\right]_1^e
=
\frac32(1^2-0^2)
=
\frac32. ∫ 1 e x 3 ln x d x = 3 [ 2 ( ln x ) 2 ] 1 e = 2 3 ( 1 2 − 0 2 ) = 2 3 . Prin urmare,
∫ 1 e ( x + 3 x ) ln x d x = e 2 + 1 4 + 3 2 = e 2 + 1 + 6 4 = e 2 + 7 4 . \int_1^e \left(x+\frac3x\right)\ln x\,dx
=
\frac{e^2+1}{4}+\frac32
=
\frac{e^2+1+6}{4}
=
\frac{e^2+7}{4}. ∫ 1 e ( x + x 3 ) ln x d x = 4 e 2 + 1 + 2 3 = 4 e 2 + 1 + 6 = 4 e 2 + 7 . Conform enunțului,
e 2 + 7 4 = e 2 + m 4 . \frac{e^2+7}{4}=\frac{e^2+m}{4}. 4 e 2 + 7 = 4 e 2 + m . Rezultă
m = 7. m=7. m = 7. Deci m = 7 \boxed{m=7} m = 7
Autoevaluare pentru punctaj maxim
Toate cele 18 subpuncte au fost rezolvate în ordinea enunțului.
Am verificat condițiile de existență pentru logaritm, radicalul din legea de compoziție, radicalul funcției f f f
Calculele de progresie aritmetică, ecuație liniară, ecuație logaritmică, numărare, geometrie analitică, trigonometrie, matrice, lege de compoziție, derivare, asimptotă și integrare sunt dezvoltate explicit.
Rezultatele de tip „Arătați” sunt demonstrate prin calcule complete, nu doar prin enunțarea rezultatului.
Rezultatele finale au fost reverificate: a 3 = 26 a_3=26 a 3 = 26 a = − 2 a=-2 a = − 2 x ∈ { 0 , 4 } x\in\{0,4\} x ∈ { 0 , 4 } 10 10 10 B C = 6 BC=6 B C = 6 X = ( 4 − 8 1 − 1 ) X=\begin{pmatrix}4&-8\\1&-1\end{pmatrix} X = ( 4 1 − 8 − 1 ) x ∈ { 0 , 7 } x\in\{0,7\} x ∈ { 0 , 7 } x = 1 x=1 x = 1 y = 0 y=0 y = 0 a = 2 3 a=\frac23 a = 3 2 m = 7 m=7 m = 7
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