The document contains 18 multi-part questions from an exam for the Brazilian Naval Academy in 2015. The questions cover topics such as calculus, geometry, trigonometry, complex numbers, and systems of equations.
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The US House of Representatives is deeply concerned by ongoing and pervasive acts of antisemitic
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Postsecondary education is a unique opportunity for students to learn and have their ideas and beliefs challenged. However, universities receiving hundreds of millions of federal funds annually have denied
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The Roman Empire, a vast and enduring power, stands as one of history's most remarkable civilizations, leaving an indelible imprint on the world. It emerged from the Roman Republic, transitioning into an imperial powerhouse under the leadership of Augustus Caesar in 27 BCE. This transformation marked the beginning of an era defined by unprecedented territorial expansion, architectural marvels, and profound cultural influence.
The empire's roots lie in the city of Rome, founded, according to legend, by Romulus in 753 BCE. Over centuries, Rome evolved from a small settlement to a formidable republic, characterized by a complex political system with elected officials and checks on power. However, internal strife, class conflicts, and military ambitions paved the way for the end of the Republic. Julius Caesar’s dictatorship and subsequent assassination in 44 BCE created a power vacuum, leading to a civil war. Octavian, later Augustus, emerged victorious, heralding the Roman Empire’s birth.
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The Roman Empire’s society was hierarchical, with a rigid class system. At the top were the patricians, wealthy elites who held significant political power. Below them were the plebeians, free citizens with limited political influence, and the vast numbers of slaves who formed the backbone of the economy. The family unit was central, governed by the paterfamilias, the male head who held absolute authority.
Culturally, the Romans were eclectic, absorbing and adapting elements from the civilizations they encountered, particularly the Greeks. Roman art, literature, and philosophy reflected this synthesis, creating a rich cultural tapestry. Latin, the Roman language, became the lingua franca of the Western world, influencing numerous modern languages.
Roman architecture and engineering achievements were monumental. They perfected the arch, vault, and dome, constructing enduring structures like the Colosseum, Pantheon, and aqueducts. These engineering marvels not only showcased Roman ingenuity but also served practical purposes, from public entertainment to water supply.
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Escola naval 2015
1. ESCOLA NAVAL 2015
1
01. (Esc. Naval 2015) Um gerador de corrente direta tem uma força eletromotriz de E volts e uma resistência interna de
r ohms. E e r são constantes. Se R ohms é a resistência externa, a resistência total é (r R)
+ ohms e, se P é a potência,
então
2
2
E R
P .
(r R)
=
+
Sendo assim, qual é a resistência externa que consumirá o máximo de potência?
a) 2r
b) r 1
+
c)
r
2
d) r
e) r(r 3)
+
02. (Esc. Naval 2015) O elemento químico Califórnio, 251
Cf , emite partículas alfa, transformando-se no elemento Cúrio,
247
Cm . Essa desintegração obedece à função exponencial t
0
N(t) N e ,
α
−
= onde N(t) é quantidade de partículas de 251
Cf
no instante t em determinada amostra; 0
N é a quantidade de partículas no instante inicial; e α é uma constante,
chamada constante de desintegração. Sabendo que em 898 anos a concentração de 251
Cf é reduzida à metade, pode-se
afirmar que o tempo necessário para que a quantidade de 251
Cf seja apenas 25% da quantidade inicial está entre
a) 500 e 1000 anos.
b) 1000 e 1500 anos.
c) 1500 e 2000 anos.
d) 2000 e 2500 anos.
e) 2500 e 3000 anos.
2. ESCOLA NAVAL 2015
2
03. (Esc. Naval 2015) Sejam f e g funções reais definidas por 2
4x 3,se x 0
f(x)
x 3x 2,se x 0
− ≥
=
− + <
e 2
x 1,se x 2
g(x)
1 x ,se x 2
+ >
=
− ≤
. Sendo
assim, pode-se dizer que (f g)(x)
é definida por
a) 2
4 2
4x 1, se x 2
(f g)(x) 1 4x , se 1 x 1
x x , se x 1ou 1 x 2
+ >
= − − ≤ ≤
+ < − < ≤
b) 2
4 2
4x 1, se x 2
(f g)(x) 1 4x , se 1 x 1
x x , se x 1ou 1 x 2
− >
= − − ≤ <
+ < − ≤ ≤
c) 2
4 2
4x 1, se x 2
(f g)(x) 1 4x , se 1 x 1
x x , se x 1ou 1 x 2
+ ≥
= − − < <
+ ≤ − ≤ <
d) 2
4 2
4x 1, se x 2
(f g)(x) 1 4x , se 1 x 1
x x , se x 1ou 1 x 2
+ ≥
= − − < ≤
+ < − < <
e) 2
4 2
4x 1, se x 2
(f g)(x) 1 4x , se 1 x 1
x x , se x 1ou 1 x 2
+ >
= − − − ≤ ≤
− < − ≤ ≤
04. (Esc. Naval 2015) A soma dos três primeiros termos de uma P.G. crescente vale 13 e a soma dos seus quadrados 91.
Justapondo-se esses termos, obtém-se um número de três algarismos. Pode-se afirmar que o resto da divisão desse
número pelo inteiro 23 vale
a) 1
b) 4
c) 8
d) 9
e) 11
3. ESCOLA NAVAL 2015
3
05. (Esc. Naval 2015) Analise o sistema a seguir.
x y z 0
4x 2my 3z 0
2x 6y 4mz 0
+ + =
− + =
+ − =
Para o maior valor inteiro de m que torna o sistema acima possível e indeterminado, pode-se afirmar que a expressão
2
m 2 m
tg sec 1
4 3
π π
+ −
vale
a)
1
4
b)
9
4
c)
11
4
−
d)
7
4
e)
1
4
−
06. (Esc. Naval 2015) Três cones circulares 1 2
C , C e 3
C , possuem raios
R
R,
2
e
R
,
4
respectivamente. Sabe-se que
possuem a mesma altura e que 3 2 1.
C C C
⊂ ⊂ Escolhendo-se aleatoriamente um ponto de 1
C , a probabilidade de que
esse ponto esteja em 2
C e não esteja em 3
C é igual a
a)
1
4
b)
1
2
c)
3
4
d)
1
16
e)
3
16
4. ESCOLA NAVAL 2015
4
07. (Esc. Naval 2015) Em um polígono regular, cujos vértices A,B e C são consecutivos, a diagonal AC forma com o lado
BC um ângulo de 30 .
° Se o lado do polígono mede unidades de comprimento, o volume da pirâmide, cuja base é esse
polígono e cuja altura vale o triplo da medida do lado, é igual a
a)
3
3 3
2
b)
2
3 3
2
c)
3
3
2
d)
3 3
4
e)
3
3 3
3
08. (Esc. Naval 2015) Um prisma quadrangular regular tem área lateral 36 6 unidades de área. Sabendo que suas
diagonais formam um ângulo de 60° com suas bases, então a razão do volume de uma esfera de raio 1 6
24 unidades de
comprimento para o volume do prisma é
a)
8
81π
b)
81
8
π
c)
8
81
π
d)
8
27
π
e)
81
8π
5. ESCOLA NAVAL 2015
5
09. (Esc. Naval 2015) Uma reta r passa pelo ponto M(1,1,1) e é concorrente às seguintes retas 𝑟𝑟1: �
𝑥𝑥 = −1 + 3𝑡𝑡
𝑦𝑦 = −3 − 2𝑡𝑡
𝑧𝑧 = 2 − 𝑡𝑡
𝑡𝑡 ∈ ℝ
e
𝑟𝑟2: �
𝑥𝑥 = 4 − 𝑡𝑡
𝑦𝑦 = 2 − 5𝑡𝑡
𝑧𝑧 = −1 + 2𝑡𝑡
𝑡𝑡 ∈ ℝ
.
Pode-se dizer que as equações paramétricas dessa reta r são
a) �
𝑥𝑥 = 1 + 𝑡𝑡
𝑦𝑦 = 1 + 2𝑡𝑡
𝑧𝑧 = 1 − 𝑡𝑡
𝑡𝑡 ∈ ℝ
b) �
𝑥𝑥 = 1 + 𝑡𝑡
𝑦𝑦 = 1 + 2𝑡𝑡
𝑧𝑧 = 1
𝑡𝑡 ∈ ℝ
c) �
𝑥𝑥 = 1 − 𝑡𝑡
𝑦𝑦 = 1 + 2𝑡𝑡
𝑧𝑧 = 𝑡𝑡
𝑡𝑡 ∈ ℝ
d) �
𝑥𝑥 = 1 + 𝑡𝑡
𝑦𝑦 = 1 − 2𝑡𝑡
𝑧𝑧 = 1
𝑡𝑡 ∈ ℝ
e) �
𝑥𝑥 = 1 + 2𝑡𝑡
𝑦𝑦 = 1 − 𝑡𝑡
𝑧𝑧 = 1
𝑡𝑡 ∈ ℝ
6. ESCOLA NAVAL 2015
6
10. (Esc. Naval 2015) As retas 1
r : 2x y 1 0;
− + = 2
r : x y 3 0
+ + = e 3
r : x y 5 0
α + − = concorrem em um mesmo ponto P
para determinado valor de 𝛼𝛼 ∈ ℝ. Sendo assim, pode-se afirmar que o valor da expressão
3 ( 3 ) 5 3
cos 3sen tg
3 8 2 6
απ α π απ
− − −
− −
é
a)
2
3 1
4
+
b)
3 2
2
4
−
c)
2
2
8
+
d)
2
3
4
+
e)
2
3 1
4
−
11. (Esc. Naval 2015) Seja ABCD um quadrado de lado ,
em que AC e BD são suas diagonais. Seja O o ponto de
encontro dessas diagonais e sejam P e Q os pontos médios dos segmentos AO e BO, respectivamente. Pode-se dizer
que a área do quadrilátero que tem vértices nos pontos A, B, Q e P vale
a)
2
3
16
b)
2
16
c)
2
3
8
d)
2
8
e)
2
3
24
12. (Esc. Naval 2015) Considere os números complexos da forma n
z pcis (17 n), ,
50
π
= −
com 𝑛𝑛 ∈ ℕ ∗. O menor número
natural n, tal que o produto 1 2 n
z z z
⋅ ⋅ ⋅
é um número real positivo, é igual a
a) 8
b) 16
c) 25
d) 33
e) 50
7. ESCOLA NAVAL 2015
7
13. (Esc. Naval 2015) Em uma P.G.,
2 2
4
2(k 1)
a
5k
+
= e
2
1 2
25k
a ,
4(k 1)
=
+
onde 𝑘𝑘 ∈ ℝ+
∗
. Para o valor médio M de k, no intervalo
onde a P.G. é decrescente, o resto da divisão do polinômio 5 4 2
5 5
P(x) x x 25x 10
4 2
= − + − pelo binômio
15
Mx
8
−
é
a)
1039
32
b)
1231
16
c)
1103
32
d)
1487
32
e)
1103
16
14. (Esc. Naval 2015) Uma função y f(x)
= é definida pelo determinante da matriz
2
3
x x 1 x 2
x x x 1 x
A
1 0 0 0
x 1 0 1
− −
−
=
−
em cada 𝑥𝑥 ∈
ℝ tal que A é invertível. É correto afirmar que o conjunto imagem de f é igual a
a) ( , 4]
−∞
b) ℝ − {0, 4}
c) ( , 4] {0}
−∞ −
d) ( , 4)
−∞
e) [4, )
+∞
15. (Esc. Naval 2015) No limite 2
x 0
1 x (1 2ax)
lim ,
x
→
+ − −
o valor de a pode ser determinado para que tal limite exista. Nesse
caso, o valor do limite é
a)
1
4
−
b)
1
4
c)
1
8
d)
1
8
−
e) 0
8. ESCOLA NAVAL 2015
8
16. (Esc. Naval 2015) Calculando 3
x 0
tgx x x senx
lim
x senx tg x
→
− −
+
−
a)
7
3
b)
13
6
c)
5
2
d)
13
3
e)
7
6
17. (Esc. Naval 2015) Resolvendo
4
4
2
2tgx 2
sen (2x)
tg(2x)cos (2x)
cotg(2x)
sec (x)dx
e cos(4x) 1 sec (2x)
−
−
∫ encontra-se
a) 2x
1
e sen(2x) c
2
− +
b) 2tgx
1
e c
2
−
− +
c) 2x
1
e sen(2x) c
2
−
+
d) 2x
1
e cosx c
2
− +
e) 2x
1
e sec(4x) c
2
−
− +
18. (Esc. Naval 2015) O ângulo que a reta normal à curva C, definida por x 1
f(x) x ,
−
= no ponto P(2, 2), faz com a reta
r : 3x 2y 5 0
+ − = é
a) 2 1 2
arc cos((5 4 In2)(13(2 4 In2 4 In 2) ))
θ −
= + + +
b) 2 1 2
arc cos((5 4 In2)(13(2 4 In2 4 In 2) ))
θ −
= + − +
c) 2 1 2
arc cos((5 4 In2)(13(2 4 In2 4 In 2) ))
θ −
= + + −
d) 2 1 2
arc cos((5 4 In2)(13(2 4 In2 4 In 2)) )
θ −
= + + +
e) 2 1 2
arc cos((5 4 In2)(13(2 4 In2 4 In 2)))
θ −
= + + +
9. ESCOLA NAVAL 2015
9
19. (Esc. Naval 2015) As curvas representantes dos gráficos de duas funções de variável real y f(x)
= e y g(x)
=
interceptam-se em um ponto 0 0 0
P (x ,y ), sendo 0
x D(f) D(g).
∈ ∩ É possível definir o ângulo formado por essas duas curvas
no ponto 0
P como sendo o menor ângulo formado pelas retas tangentes àquelas curvas no ponto 0
P . Se 2
f(x) x 1,
= −
2
g(x) 1 x
= − e θ é o ângulo entre as curvas na interseção de abscissa positiva, então, pode-se dizer que o valor da
expressão
1 2
5 7
( 6 2) sen cos2 cossec
12 6
π π
θ
− + −
é
a)
82
5
b)
2
3
5
c)
68
25
d)
7
25
e)
17
2
5
20. (Esc. Naval 2015) Um plano 1
π contém os pontos M ( 1, 3, 2)
− e N ( 2, 0,1).
− Se 1
π é perpendicular ao plano
2 : 3x 2y z 15 0,
π − + − = é possível dizer que o ângulo entre 1
π e o plano 3 : x y 2z 7 0
π + + − = vale
a)
8 2
arccos
15
b)
4 2
arccot
15
c)
8 2
arccos
15
−
d)
61
arccos
45 2
e)
194
arc tg
16
−
10. ESCOLA NAVAL 2015
10
GABARITO
1 - D 2 - C 3 - A 4 - A 5 - ANULADA
6 - ANULADA 7 - A 8 - C 9 - ANULADA 10 - E
11 - A 12 - A 13 - ANULADA 14 - C 15 - D
16 - B 17 - ANULADA 18 - D 19 - E 20 - ANULADA