SUBIECTUL I
1. Folosim relația i 2 = − 1 i^2=-1 i 2 = − 1
2 i ( 6 − i ) = 12 i − 2 i 2 = 12 i + 2. 2i(6-i)=12i-2i^2=12i+2. 2 i ( 6 − i ) = 12 i − 2 i 2 = 12 i + 2. De asemenea,
3 ( 1 − 4 i ) = 3 − 12 i . 3(1-4i)=3-12i. 3 ( 1 − 4 i ) = 3 − 12 i . Prin urmare,
2 i ( 6 − i ) + 3 ( 1 − 4 i ) = 12 i + 2 + 3 − 12 i = 5. 2i(6-i)+3(1-4i)=12i+2+3-12i=5. 2 i ( 6 − i ) + 3 ( 1 − 4 i ) = 12 i + 2 + 3 − 12 i = 5. Am arătat că
2 i ( 6 − i ) + 3 ( 1 − 4 i ) = 5 . \boxed{2i(6-i)+3(1-4i)=5}. 2 i ( 6 − i ) + 3 ( 1 − 4 i ) = 5 . 2. Funcția este
f ( x ) = x + 5. f(x)=x+5. f ( x ) = x + 5. Compunerea funcției cu ea însăși este
( f ∘ f ) ( a ) = f ( f ( a ) ) = f ( a + 5 ) = a + 5 + 5 = a + 10. (f\circ f)(a)=f(f(a))=f(a+5)=a+5+5=a+10. ( f ∘ f ) ( a ) = f ( f ( a )) = f ( a + 5 ) = a + 5 + 5 = a + 10. Condiția din enunț devine
a + 10 = 2 a . a+10=2a. a + 10 = 2 a . Rezultă
a = 10. a=10. a = 10. Verificare:
( f ∘ f ) ( 10 ) = f ( 15 ) = 20 = 2 ⋅ 10. (f\circ f)(10)=f(15)=20=2\cdot 10. ( f ∘ f ) ( 10 ) = f ( 15 ) = 20 = 2 ⋅ 10. Deci
a = 10 . \boxed{a=10}. a = 10 . 3. Ecuația este
x 2 + 4 x − 4 = x 2 . \sqrt{x^2+4x-4}=x\sqrt2. x 2 + 4 x − 4 = x 2 . Deoarece membrul stâng este nenegativ, trebuie să avem
x 2 ≥ 0 ⟺ x ≥ 0. x\sqrt2\ge 0 \Longleftrightarrow x\ge 0. x 2 ≥ 0 ⟺ x ≥ 0. Pentru x ≥ 0 x\ge 0 x ≥ 0
x 2 + 4 x − 4 = 2 x 2 . x^2+4x-4=2x^2. x 2 + 4 x − 4 = 2 x 2 . Rezultă
x 2 − 4 x + 4 = 0 ⟺ ( x − 2 ) 2 = 0. x^2-4x+4=0
\Longleftrightarrow
(x-2)^2=0. x 2 − 4 x + 4 = 0 ⟺ ( x − 2 ) 2 = 0. Deci
x = 2. x=2. x = 2. Verificare:
2 2 + 4 ⋅ 2 − 4 = 8 = 2 2 . \sqrt{2^2+4\cdot2-4}=\sqrt8=2\sqrt2. 2 2 + 4 ⋅ 2 − 4 = 8 = 2 2 . Soluția este
x = 2 . \boxed{x=2}. x = 2 . 4. Mulțimea numerelor naturale de o cifră este
{ 0 , 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 } , \{0,1,2,3,4,5,6,7,8,9\}, { 0 , 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 } , deci sunt 10 10 10
Avem
16 = 2 4 . 16=2^4. 16 = 2 4 . Numărul 2 n 2^n 2 n 16 16 16
n ≥ 4. n\ge 4. n ≥ 4. Din mulțimea de mai sus, valorile favorabile sunt
n ∈ { 4 , 5 , 6 , 7 , 8 , 9 } . n\in\{4,5,6,7,8,9\}. n ∈ { 4 , 5 , 6 , 7 , 8 , 9 } . Sunt 6 6 6
P = 6 10 = 3 5 . P=\frac{6}{10}=\frac35. P = 10 6 = 5 3 . Răspuns:
3 5 . \boxed{\frac35}. 5 3 . 5. Avem
O ( 0 , 0 ) , A ( 3 , 1 ) , B ( 2 , 4 ) . O(0,0),\qquad A(3,1),\qquad B(2,4). O ( 0 , 0 ) , A ( 3 , 1 ) , B ( 2 , 4 ) . Pentru a arăta că triunghiul O A B OAB O A B A A A A O → \overrightarrow{AO} A O A B → \overrightarrow{AB} A B
Calculăm:
A O → = O − A = ( 0 − 3 , 0 − 1 ) = ( − 3 , − 1 ) , \overrightarrow{AO}=O-A=(0-3,0-1)=(-3,-1), A O = O − A = ( 0 − 3 , 0 − 1 ) = ( − 3 , − 1 ) , A B → = B − A = ( 2 − 3 , 4 − 1 ) = ( − 1 , 3 ) . \overrightarrow{AB}=B-A=(2-3,4-1)=(-1,3). A B = B − A = ( 2 − 3 , 4 − 1 ) = ( − 1 , 3 ) . Produsul scalar este
A O → ⋅ A B → = ( − 3 ) ( − 1 ) + ( − 1 ) ⋅ 3 = 3 − 3 = 0. \overrightarrow{AO}\cdot\overrightarrow{AB}
=(-3)(-1)+(-1)\cdot3=3-3=0. A O ⋅ A B = ( − 3 ) ( − 1 ) + ( − 1 ) ⋅ 3 = 3 − 3 = 0. Deci
A O → ⊥ A B → . \overrightarrow{AO}\perp\overrightarrow{AB}. A O ⊥ A B . Prin urmare, triunghiul O A B OAB O A B A A A
6. Expresia este
E ( x ) = sin x + 2 cos 2 x + 2 sin 2 x 2 . E(x)=\sin x+2\cos 2x+2\sin^2\frac{x}{2}. E ( x ) = sin x + 2 cos 2 x + 2 sin 2 2 x . Pentru x = π 2 x=\dfrac{\pi}{2} x = 2 π
sin π 2 = 1 , cos π = − 1 , sin 2 π 4 = ( 2 2 ) 2 = 1 2 . \sin\frac{\pi}{2}=1,\qquad
\cos\pi=-1,\qquad
\sin^2\frac{\pi}{4}=\left(\frac{\sqrt2}{2}\right)^2=\frac12. sin 2 π = 1 , cos π = − 1 , sin 2 4 π = ( 2 2 ) 2 = 2 1 . Astfel,
E ( π 2 ) = 1 + 2 ( − 1 ) + 2 ⋅ 1 2 = 1 − 2 + 1 = 0. E\left(\frac{\pi}{2}\right)
=1+2(-1)+2\cdot\frac12
=1-2+1=0. E ( 2 π ) = 1 + 2 ( − 1 ) + 2 ⋅ 2 1 = 1 − 2 + 1 = 0. Am arătat că
E ( π 2 ) = 0 . \boxed{E\left(\frac{\pi}{2}\right)=0}. E ( 2 π ) = 0 . SUBIECTUL al II-lea
1. Se consideră
A ( x ) = ( x + 1 − x − 2 x 2 x + 1 ) , x ∈ R . A(x)=
\begin{pmatrix}
x+1&-x\\
-2x&2x+1
\end{pmatrix},
\qquad x\in\mathbb R. A ( x ) = ( x + 1 − 2 x − x 2 x + 1 ) , x ∈ R . 1.a) Pentru x = 1 x=1 x = 1
A ( 1 ) = ( 2 − 1 − 2 3 ) . A(1)=
\begin{pmatrix}
2&-1\\
-2&3
\end{pmatrix}. A ( 1 ) = ( 2 − 2 − 1 3 ) . Determinantul este
det ( A ( 1 ) ) = 2 ⋅ 3 − ( − 1 ) ( − 2 ) = 6 − 2 = 4. \det(A(1))=2\cdot3-(-1)(-2)=6-2=4. det ( A ( 1 )) = 2 ⋅ 3 − ( − 1 ) ( − 2 ) = 6 − 2 = 4. Deci
det ( A ( 1 ) ) = 4 . \boxed{\det(A(1))=4}. det ( A ( 1 )) = 4 . 1.b) Calculăm
A ( − 1 ) = ( 0 1 2 − 1 ) . A(-1)=
\begin{pmatrix}
0&1\\
2&-1
\end{pmatrix}. A ( − 1 ) = ( 0 2 1 − 1 ) . Atunci
A ( − 1 ) ⋅ A ( x ) = ( 0 1 2 − 1 ) ( x + 1 − x − 2 x 2 x + 1 ) . A(-1)\cdot A(x)
=
\begin{pmatrix}
0&1\\
2&-1
\end{pmatrix}
\begin{pmatrix}
x+1&-x\\
-2x&2x+1
\end{pmatrix}. A ( − 1 ) ⋅ A ( x ) = ( 0 2 1 − 1 ) ( x + 1 − 2 x − x 2 x + 1 ) . Rezultă
A ( − 1 ) ⋅ A ( x ) = ( − 2 x 2 x + 1 2 ( x + 1 ) + 2 x − 2 x − ( 2 x + 1 ) ) = ( − 2 x 2 x + 1 4 x + 2 − 4 x − 1 ) . A(-1)\cdot A(x)
=
\begin{pmatrix}
-2x&2x+1\\
2(x+1)+2x&-2x-(2x+1)
\end{pmatrix}
=
\begin{pmatrix}
-2x&2x+1\\
4x+2&-4x-1
\end{pmatrix}. A ( − 1 ) ⋅ A ( x ) = ( − 2 x 2 ( x + 1 ) + 2 x 2 x + 1 − 2 x − ( 2 x + 1 ) ) = ( − 2 x 4 x + 2 2 x + 1 − 4 x − 1 ) . Pe de altă parte,
A ( − 2 x − 1 ) = ( − 2 x − ( − 2 x − 1 ) − 2 ( − 2 x − 1 ) 2 ( − 2 x − 1 ) + 1 ) = ( − 2 x 2 x + 1 4 x + 2 − 4 x − 1 ) . A(-2x-1)=
\begin{pmatrix}
-2x&-(-2x-1)\\
-2(-2x-1)&2(-2x-1)+1
\end{pmatrix}
=
\begin{pmatrix}
-2x&2x+1\\
4x+2&-4x-1
\end{pmatrix}. A ( − 2 x − 1 ) = ( − 2 x − 2 ( − 2 x − 1 ) − ( − 2 x − 1 ) 2 ( − 2 x − 1 ) + 1 ) = ( − 2 x 4 x + 2 2 x + 1 − 4 x − 1 ) . Prin urmare,
A ( − 1 ) ⋅ A ( x ) = A ( − 2 x − 1 ) , ∀ x ∈ R . \boxed{A(-1)\cdot A(x)=A(-2x-1)},\qquad \forall x\in\mathbb R. A ( − 1 ) ⋅ A ( x ) = A ( − 2 x − 1 ) , ∀ x ∈ R . 1.c) Din subpunctul anterior,
A ( − 1 ) ⋅ A ( m ) = A ( − 2 m − 1 ) , A(-1)\cdot A(m)=A(-2m-1), A ( − 1 ) ⋅ A ( m ) = A ( − 2 m − 1 ) , și
A ( − 1 ) ⋅ A ( n ) = A ( − 2 n − 1 ) . A(-1)\cdot A(n)=A(-2n-1). A ( − 1 ) ⋅ A ( n ) = A ( − 2 n − 1 ) . Prin distributivitate,
A ( − 1 ) ⋅ ( A ( m ) + A ( n ) ) = A ( − 2 m − 1 ) + A ( − 2 n − 1 ) . A(-1)\cdot(A(m)+A(n))=A(-2m-1)+A(-2n-1). A ( − 1 ) ⋅ ( A ( m ) + A ( n )) = A ( − 2 m − 1 ) + A ( − 2 n − 1 ) . Notăm
s = ( − 2 m − 1 ) + ( − 2 n − 1 ) = − 2 ( m + n ) − 2. s=(-2m-1)+(-2n-1)=-2(m+n)-2. s = ( − 2 m − 1 ) + ( − 2 n − 1 ) = − 2 ( m + n ) − 2. Pentru orice u , v ∈ R u,v\in\mathbb R u , v ∈ R
A ( u ) + A ( v ) = ( u + v + 2 − ( u + v ) − 2 ( u + v ) 2 ( u + v ) + 2 ) . A(u)+A(v)=
\begin{pmatrix}
u+v+2&-(u+v)\\
-2(u+v)&2(u+v)+2
\end{pmatrix}. A ( u ) + A ( v ) = ( u + v + 2 − 2 ( u + v ) − ( u + v ) 2 ( u + v ) + 2 ) . Astfel,
A ( − 2 m − 1 ) + A ( − 2 n − 1 ) = ( s + 2 − s − 2 s 2 s + 2 ) . A(-2m-1)+A(-2n-1)=
\begin{pmatrix}
s+2&-s\\
-2s&2s+2
\end{pmatrix}. A ( − 2 m − 1 ) + A ( − 2 n − 1 ) = ( s + 2 − 2 s − s 2 s + 2 ) . De asemenea,
2 A ( − 4 ) = 2 ( − 3 4 8 − 7 ) = ( − 6 8 16 − 14 ) . 2A(-4)=2
\begin{pmatrix}
-3&4\\
8&-7
\end{pmatrix}
=
\begin{pmatrix}
-6&8\\
16&-14
\end{pmatrix}. 2 A ( − 4 ) = 2 ( − 3 8 4 − 7 ) = ( − 6 16 8 − 14 ) . Egalitatea cerută este echivalentă cu
( s + 2 − s − 2 s 2 s + 2 ) = ( − 6 8 16 − 14 ) . \begin{pmatrix}
s+2&-s\\
-2s&2s+2
\end{pmatrix}
=
\begin{pmatrix}
-6&8\\
16&-14
\end{pmatrix}. ( s + 2 − 2 s − s 2 s + 2 ) = ( − 6 16 8 − 14 ) . Din prima poziție obținem
s + 2 = − 6 ⟺ s = − 8 , s+2=-6 \Longleftrightarrow s=-8, s + 2 = − 6 ⟺ s = − 8 , iar celelalte poziții dau aceeași condiție.
Cum
s = − 2 ( m + n ) − 2 , s=-2(m+n)-2, s = − 2 ( m + n ) − 2 , rezultă
− 2 ( m + n ) − 2 = − 8 ⟺ − 2 ( m + n ) = − 6 ⟺ m + n = 3. -2(m+n)-2=-8
\Longleftrightarrow
-2(m+n)=-6
\Longleftrightarrow
m+n=3. − 2 ( m + n ) − 2 = − 8 ⟺ − 2 ( m + n ) = − 6 ⟺ m + n = 3. Pentru m , n ∈ N m,n\in\mathbb N m , n ∈ N m < n m<n m < n
( m , n ) ∈ { ( 0 , 3 ) , ( 1 , 2 ) } . (m,n)\in\{(0,3),(1,2)\}. ( m , n ) ∈ {( 0 , 3 ) , ( 1 , 2 )} . Răspuns:
( m , n ) ∈ { ( 0 , 3 ) , ( 1 , 2 ) } . \boxed{(m,n)\in\{(0,3),(1,2)\}}. ( m , n ) ∈ {( 0 , 3 ) , ( 1 , 2 )} . 2. Pe R \mathbb R R
x ∘ y = x ⋅ 3 y + y ⋅ 3 x . x\circ y=x\cdot 3^y+y\cdot 3^x. x ∘ y = x ⋅ 3 y + y ⋅ 3 x . 2.a) Calculăm:
1 ∘ 2 = 1 ⋅ 3 2 + 2 ⋅ 3 1 = 9 + 6 = 15. 1\circ 2=1\cdot3^2+2\cdot3^1=9+6=15. 1 ∘ 2 = 1 ⋅ 3 2 + 2 ⋅ 3 1 = 9 + 6 = 15. Deci
1 ∘ 2 = 15 . \boxed{1\circ2=15}. 1 ∘ 2 = 15 . 2.b) Pentru orice x ∈ R x\in\mathbb R x ∈ R
x ∘ 0 = x ⋅ 3 0 + 0 ⋅ 3 x = x ⋅ 1 + 0 = x , x\circ0=x\cdot3^0+0\cdot3^x=x\cdot1+0=x, x ∘ 0 = x ⋅ 3 0 + 0 ⋅ 3 x = x ⋅ 1 + 0 = x , și
0 ∘ x = 0 ⋅ 3 x + x ⋅ 3 0 = 0 + x ⋅ 1 = x . 0\circ x=0\cdot3^x+x\cdot3^0=0+x\cdot1=x. 0 ∘ x = 0 ⋅ 3 x + x ⋅ 3 0 = 0 + x ⋅ 1 = x . Prin urmare, 0 0 0
e = 0 . \boxed{e=0}. e = 0 . 2.c) Condiția este
x ∘ ( 3 x ) = ( 2 x ) ∘ ( 2 x ) , x ≠ 0. x\circ(3x)=(2x)\circ(2x),\qquad x\ne0. x ∘ ( 3 x ) = ( 2 x ) ∘ ( 2 x ) , x = 0. Folosim definiția legii:
x ∘ ( 3 x ) = x ⋅ 3 3 x + 3 x ⋅ 3 x , x\circ(3x)=x\cdot3^{3x}+3x\cdot3^x, x ∘ ( 3 x ) = x ⋅ 3 3 x + 3 x ⋅ 3 x , și
( 2 x ) ∘ ( 2 x ) = 2 x ⋅ 3 2 x + 2 x ⋅ 3 2 x = 4 x ⋅ 3 2 x . (2x)\circ(2x)=2x\cdot3^{2x}+2x\cdot3^{2x}=4x\cdot3^{2x}. ( 2 x ) ∘ ( 2 x ) = 2 x ⋅ 3 2 x + 2 x ⋅ 3 2 x = 4 x ⋅ 3 2 x . Ecuația devine
x ⋅ 3 3 x + 3 x ⋅ 3 x = 4 x ⋅ 3 2 x . x\cdot3^{3x}+3x\cdot3^x=4x\cdot3^{2x}. x ⋅ 3 3 x + 3 x ⋅ 3 x = 4 x ⋅ 3 2 x . Deoarece x ≠ 0 x\ne0 x = 0 x x x
3 3 x + 3 ⋅ 3 x = 4 ⋅ 3 2 x . 3^{3x}+3\cdot3^x=4\cdot3^{2x}. 3 3 x + 3 ⋅ 3 x = 4 ⋅ 3 2 x . Notăm
t = 3 x . t=3^x. t = 3 x . Atunci t > 0 t>0 t > 0
t 3 + 3 t = 4 t 2 . t^3+3t=4t^2. t 3 + 3 t = 4 t 2 . Mutăm toți termenii în același membru:
t 3 − 4 t 2 + 3 t = 0 ⟺ t ( t 2 − 4 t + 3 ) = 0. t^3-4t^2+3t=0
\Longleftrightarrow
t(t^2-4t+3)=0. t 3 − 4 t 2 + 3 t = 0 ⟺ t ( t 2 − 4 t + 3 ) = 0. Cum t > 0 t>0 t > 0 t = 0 t=0 t = 0
t 2 − 4 t + 3 = 0 ⟺ ( t − 1 ) ( t − 3 ) = 0. t^2-4t+3=0
\Longleftrightarrow
(t-1)(t-3)=0. t 2 − 4 t + 3 = 0 ⟺ ( t − 1 ) ( t − 3 ) = 0. Deci
t = 1 sau t = 3. t=1 \quad \text{sau}\quad t=3. t = 1 sau t = 3. Din 3 x = 1 3^x=1 3 x = 1 x = 0 x=0 x = 0 3 x = 3 3^x=3 3 x = 3
x = 1. x=1. x = 1. Verificare:
1 ∘ 3 = 1 ⋅ 27 + 3 ⋅ 3 = 36 , 1\circ3=1\cdot27+3\cdot3=36, 1 ∘ 3 = 1 ⋅ 27 + 3 ⋅ 3 = 36 , 2 ∘ 2 = 2 ⋅ 9 + 2 ⋅ 9 = 36. 2\circ2=2\cdot9+2\cdot9=36. 2 ∘ 2 = 2 ⋅ 9 + 2 ⋅ 9 = 36. Soluția este
x = 1 . \boxed{x=1}. x = 1 . SUBIECTUL al III-lea
1. Se consideră
f : R → R , f ( x ) = e x x 2 + 3 x + 3 . f:\mathbb R\to\mathbb R,\qquad
f(x)=\frac{e^x}{x^2+3x+3}. f : R → R , f ( x ) = x 2 + 3 x + 3 e x . Numitorul este
x 2 + 3 x + 3 = ( x + 3 2 ) 2 + 3 4 > 0 , x^2+3x+3=\left(x+\frac32\right)^2+\frac34>0, x 2 + 3 x + 3 = ( x + 2 3 ) 2 + 4 3 > 0 , pentru orice x ∈ R x\in\mathbb R x ∈ R R \mathbb R R
1.a) Notăm
q ( x ) = x 2 + 3 x + 3. q(x)=x^2+3x+3. q ( x ) = x 2 + 3 x + 3. Atunci
q ′ ( x ) = 2 x + 3. q'(x)=2x+3. q ′ ( x ) = 2 x + 3. Aplicând regula derivării unui raport, obținem
f ′ ( x ) = e x q ( x ) − e x q ′ ( x ) q ( x ) 2 = e x ( q ( x ) − q ′ ( x ) ) q ( x ) 2 . f'(x)=\frac{e^x q(x)-e^x q'(x)}{q(x)^2}
=\frac{e^x(q(x)-q'(x))}{q(x)^2}. f ′ ( x ) = q ( x ) 2 e x q ( x ) − e x q ′ ( x ) = q ( x ) 2 e x ( q ( x ) − q ′ ( x )) . Calculăm:
q ( x ) − q ′ ( x ) = x 2 + 3 x + 3 − ( 2 x + 3 ) = x 2 + x . q(x)-q'(x)=x^2+3x+3-(2x+3)=x^2+x. q ( x ) − q ′ ( x ) = x 2 + 3 x + 3 − ( 2 x + 3 ) = x 2 + x . Prin urmare,
f ′ ( x ) = e x ( x 2 + x ) ( x 2 + 3 x + 3 ) 2 , x ∈ R . \boxed{
f'(x)=\frac{e^x(x^2+x)}{(x^2+3x+3)^2}
},\qquad x\in\mathbb R. f ′ ( x ) = ( x 2 + 3 x + 3 ) 2 e x ( x 2 + x ) , x ∈ R . 1.b) Avem o formă de tip + ∞ + ∞ \dfrac{+\infty}{+\infty} + ∞ + ∞
lim x → + ∞ e x x 2 + 3 x + 3 = lim x → + ∞ e x 2 x + 3 . \lim_{x\to+\infty}\frac{e^x}{x^2+3x+3}
=
\lim_{x\to+\infty}\frac{e^x}{2x+3}. x → + ∞ lim x 2 + 3 x + 3 e x = x → + ∞ lim 2 x + 3 e x . Din nou,
lim x → + ∞ e x 2 x + 3 = lim x → + ∞ e x 2 = + ∞ . \lim_{x\to+\infty}\frac{e^x}{2x+3}
=
\lim_{x\to+\infty}\frac{e^x}{2}
=+\infty. x → + ∞ lim 2 x + 3 e x = x → + ∞ lim 2 e x = + ∞. Așadar,
lim x → + ∞ f ( x ) = + ∞ . \boxed{\displaystyle\lim_{x\to+\infty} f(x)=+\infty}. x → + ∞ lim f ( x ) = + ∞ . 1.c) Trebuie să demonstrăm că
f ( x ) − f ( y ) ≤ 3 − e 3 e , x ≤ 0 ≤ y . f(x)-f(y)\le \frac{3-e}{3e},
\qquad x\le0\le y. f ( x ) − f ( y ) ≤ 3 e 3 − e , x ≤ 0 ≤ y . Din formula derivatei,
f ′ ( x ) = e x x ( x + 1 ) ( x 2 + 3 x + 3 ) 2 . f'(x)=\frac{e^x x(x+1)}{(x^2+3x+3)^2}. f ′ ( x ) = ( x 2 + 3 x + 3 ) 2 e x x ( x + 1 ) . Deoarece
e x > 0 și ( x 2 + 3 x + 3 ) 2 > 0 , e^x>0
\quad \text{și}\quad
(x^2+3x+3)^2>0, e x > 0 ș i ( x 2 + 3 x + 3 ) 2 > 0 , semnul lui f ′ ( x ) f'(x) f ′ ( x ) x ( x + 1 ) x(x+1) x ( x + 1 )
Astfel,
f ′ ( x ) > 0 pe ( − ∞ , − 1 ) , f'(x)>0 \text{ pe } (-\infty,-1), f ′ ( x ) > 0 pe ( − ∞ , − 1 ) , f ′ ( x ) < 0 pe ( − 1 , 0 ) , f'(x)<0 \text{ pe } (-1,0), f ′ ( x ) < 0 pe ( − 1 , 0 ) , și
f ′ ( x ) > 0 pe ( 0 , + ∞ ) . f'(x)>0 \text{ pe } (0,+\infty). f ′ ( x ) > 0 pe ( 0 , + ∞ ) . Rezultă că, pe intervalul ( − ∞ , 0 ] (-\infty,0] ( − ∞ , 0 ] x = − 1 x=-1 x = − 1 x ≤ 0 x\le0 x ≤ 0
f ( x ) ≤ f ( − 1 ) . f(x)\le f(-1). f ( x ) ≤ f ( − 1 ) . Pe intervalul [ 0 , + ∞ ) [0,+\infty) [ 0 , + ∞ ) y ≥ 0 y\ge0 y ≥ 0
f ( y ) ≥ f ( 0 ) . f(y)\ge f(0). f ( y ) ≥ f ( 0 ) . Prin urmare,
f ( x ) − f ( y ) ≤ f ( − 1 ) − f ( 0 ) . f(x)-f(y)\le f(-1)-f(0). f ( x ) − f ( y ) ≤ f ( − 1 ) − f ( 0 ) . Calculăm:
f ( − 1 ) = e − 1 ( − 1 ) 2 + 3 ( − 1 ) + 3 = e − 1 1 = 1 e , f(-1)=\frac{e^{-1}}{(-1)^2+3(-1)+3}
=\frac{e^{-1}}{1}=\frac1e, f ( − 1 ) = ( − 1 ) 2 + 3 ( − 1 ) + 3 e − 1 = 1 e − 1 = e 1 , și
f ( 0 ) = 1 3 . f(0)=\frac{1}{3}. f ( 0 ) = 3 1 . Rezultă
f ( − 1 ) − f ( 0 ) = 1 e − 1 3 = 3 − e 3 e . f(-1)-f(0)=\frac1e-\frac13
=\frac{3-e}{3e}. f ( − 1 ) − f ( 0 ) = e 1 − 3 1 = 3 e 3 − e . Deci
f ( x ) − f ( y ) ≤ 3 − e 3 e , x ≤ 0 ≤ y . \boxed{
f(x)-f(y)\le \frac{3-e}{3e}
},
\qquad x\le0\le y. f ( x ) − f ( y ) ≤ 3 e 3 − e , x ≤ 0 ≤ y . 2. Se consideră
f : ( 0 , + ∞ ) → R , f ( x ) = 4 x + 1 + 3 x ln x . f:(0,+\infty)\to\mathbb R,\qquad
f(x)=4x+1+3x\ln x. f : ( 0 , + ∞ ) → R , f ( x ) = 4 x + 1 + 3 x ln x . 2.a) Pe intervalul [ 1 , 2 ] [1,2] [ 1 , 2 ] x > 0 x>0 x > 0
f ( x ) − 3 x ln x = 4 x + 1. f(x)-3x\ln x=4x+1. f ( x ) − 3 x ln x = 4 x + 1. Prin urmare,
∫ 1 2 ( f ( x ) − 3 x ln x ) d x = ∫ 1 2 ( 4 x + 1 ) d x . \int_1^2 (f(x)-3x\ln x)\,dx
=
\int_1^2 (4x+1)\,dx. ∫ 1 2 ( f ( x ) − 3 x ln x ) d x = ∫ 1 2 ( 4 x + 1 ) d x . Calculăm:
∫ 1 2 ( 4 x + 1 ) d x = [ 2 x 2 + x ] 1 2 = ( 8 + 2 ) − ( 2 + 1 ) = 10 − 3 = 7. \int_1^2 (4x+1)\,dx
=
\left[2x^2+x\right]_1^2
=(8+2)-(2+1)=10-3=7. ∫ 1 2 ( 4 x + 1 ) d x = [ 2 x 2 + x ] 1 2 = ( 8 + 2 ) − ( 2 + 1 ) = 10 − 3 = 7. Așadar,
∫ 1 2 ( f ( x ) − 3 x ln x ) d x = 7 . \boxed{\displaystyle\int_1^2 (f(x)-3x\ln x)\,dx=7}. ∫ 1 2 ( f ( x ) − 3 x ln x ) d x = 7 . 2.b) Pe intervalul [ 1 , e ] [1,e] [ 1 , e ] x > 0 x>0 x > 0
f ( x ) − 4 x − 1 = 3 x ln x . f(x)-4x-1=3x\ln x. f ( x ) − 4 x − 1 = 3 x ln x . Atunci
∫ 1 e f ( x ) − 4 x − 1 x d x = ∫ 1 e 3 x ln x x d x = 3 ∫ 1 e ln x d x . \int_1^e \frac{f(x)-4x-1}{x}\,dx
=
\int_1^e \frac{3x\ln x}{x}\,dx
=
3\int_1^e \ln x\,dx. ∫ 1 e x f ( x ) − 4 x − 1 d x = ∫ 1 e x 3 x ln x d x = 3 ∫ 1 e ln x d x . Folosim primitiva
∫ ln x d x = x ln x − x + C . \int \ln x\,dx=x\ln x-x+C. ∫ ln x d x = x ln x − x + C . Deci
3 ∫ 1 e ln x d x = 3 [ x ln x − x ] 1 e . 3\int_1^e \ln x\,dx
=3\left[x\ln x-x\right]_1^e. 3 ∫ 1 e ln x d x = 3 [ x ln x − x ] 1 e . Calculăm:
[ x ln x − x ] 1 e = ( e ln e − e ) − ( 1 ln 1 − 1 ) = ( e − e ) − ( 0 − 1 ) = 1. \left[x\ln x-x\right]_1^e
=(e\ln e-e)-(1\ln1-1)
=(e-e)-(0-1)=1. [ x ln x − x ] 1 e = ( e ln e − e ) − ( 1 ln 1 − 1 ) = ( e − e ) − ( 0 − 1 ) = 1. Prin urmare,
∫ 1 e f ( x ) − 4 x − 1 x d x = 3 . \boxed{\displaystyle\int_1^e \frac{f(x)-4x-1}{x}\,dx=3}. ∫ 1 e x f ( x ) − 4 x − 1 d x = 3 . 2.c) Pe intervalul [ 2 , 4 ] [2,4] [ 2 , 4 ] x > 0 x>0 x > 0 ln x > 0 \ln x>0 ln x > 0
f ( x ) − 1 = 4 x + 3 x ln x = x ( 4 + 3 ln x ) . f(x)-1=4x+3x\ln x=x(4+3\ln x). f ( x ) − 1 = 4 x + 3 x ln x = x ( 4 + 3 ln x ) . Astfel,
f ( x ) − 1 x 2 ln x = x ( 4 + 3 ln x ) x 2 ln x = 4 + 3 ln x x ln x = 4 x ln x + 3 x . \frac{f(x)-1}{x^2\ln x}
=
\frac{x(4+3\ln x)}{x^2\ln x}
=
\frac{4+3\ln x}{x\ln x}
=
\frac{4}{x\ln x}+\frac3x. x 2 ln x f ( x ) − 1 = x 2 ln x x ( 4 + 3 ln x ) = x ln x 4 + 3 ln x = x ln x 4 + x 3 . Integrala devine
∫ 2 4 f ( x ) − 1 x 2 ln x d x = 4 ∫ 2 4 1 x ln x d x + 3 ∫ 2 4 1 x d x . \int_2^4 \frac{f(x)-1}{x^2\ln x}\,dx
=
4\int_2^4 \frac{1}{x\ln x}\,dx
+3\int_2^4 \frac1x\,dx. ∫ 2 4 x 2 ln x f ( x ) − 1 d x = 4 ∫ 2 4 x ln x 1 d x + 3 ∫ 2 4 x 1 d x . Pentru prima integrală folosim substituția u = ln x u=\ln x u = ln x d u = 1 x d x du=\dfrac1x\,dx d u = x 1 d x
∫ 2 4 1 x ln x d x = [ ln ( ln x ) ] 2 4 = ln ( ln 4 ) − ln ( ln 2 ) . \int_2^4 \frac{1}{x\ln x}\,dx
=
\left[\ln(\ln x)\right]_2^4
=
\ln(\ln4)-\ln(\ln2). ∫ 2 4 x ln x 1 d x = [ ln ( ln x ) ] 2 4 = ln ( ln 4 ) − ln ( ln 2 ) . Cum
ln 4 = ln ( 2 2 ) = 2 ln 2 , \ln4=\ln(2^2)=2\ln2, ln 4 = ln ( 2 2 ) = 2 ln 2 , rezultă
ln ( ln 4 ) − ln ( ln 2 ) = ln ( 2 ln 2 ) − ln ( ln 2 ) = ln 2. \ln(\ln4)-\ln(\ln2)
=
\ln(2\ln2)-\ln(\ln2)
=
\ln2. ln ( ln 4 ) − ln ( ln 2 ) = ln ( 2 ln 2 ) − ln ( ln 2 ) = ln 2. De asemenea,
∫ 2 4 1 x d x = [ ln x ] 2 4 = ln 4 − ln 2 = ln 2. \int_2^4 \frac1x\,dx
=
\left[\ln x\right]_2^4
=
\ln4-\ln2
=
\ln2. ∫ 2 4 x 1 d x = [ ln x ] 2 4 = ln 4 − ln 2 = ln 2. Prin urmare,
∫ 2 4 f ( x ) − 1 x 2 ln x d x = 4 ln 2 + 3 ln 2 = 7 ln 2. \int_2^4 \frac{f(x)-1}{x^2\ln x}\,dx
=4\ln2+3\ln2=7\ln2. ∫ 2 4 x 2 ln x f ( x ) − 1 d x = 4 ln 2 + 3 ln 2 = 7 ln 2. Din condiția
∫ 2 4 f ( x ) − 1 x 2 ln x d x = a ln 2 \int_2^4 \frac{f(x)-1}{x^2\ln x}\,dx=a\ln2 ∫ 2 4 x 2 ln x f ( x ) − 1 d x = a ln 2 obținem
a ln 2 = 7 ln 2. a\ln2=7\ln2. a ln 2 = 7 ln 2. Deoarece ln 2 ≠ 0 \ln2\ne0 ln 2 = 0
a = 7 . \boxed{a=7}. a = 7 . Autoevaluare pentru punctaj maxim
Toate cele 18 subpuncte au fost rezolvate în ordinea enunțului oficial.
Am verificat notațiile pierdute în textul extras din PDF: la I.3 ecuația este cu radical, la I.4 numărul este 2 n 2^n 2 n x ∘ y = x ⋅ 3 y + y ⋅ 3 x x\circ y=x\cdot3^y+y\cdot3^x x ∘ y = x ⋅ 3 y + y ⋅ 3 x
Au fost precizate condițiile de domeniu pentru radical, puteri, logaritmi, numitori și integrale.
Calculele de algebră, probabilitate, geometrie analitică, trigonometrie, matrice, lege de compoziție, derivare, monotonie, limită și integrare sunt dezvoltate explicit.
Rezultatele finale au fost reverificate: a = 10 a=10 a = 10 x = 2 x=2 x = 2 3 5 \frac35 5 3 ( 0 , 3 ) (0,3) ( 0 , 3 ) ( 1 , 2 ) (1,2) ( 1 , 2 ) x = 1 x=1 x = 1 3 − e 3 e \frac{3-e}{3e} 3 e 3 − e a = 7 a=7 a = 7
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