Matemática Ciência e aplicações, Vol 2, Cap 4

7212019 Matemaacutetica Ciecircncia e aplicaccedilotildees Vol 2 Cap 4

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Funccedilotildees trigonomeacutetricas 4

Exerciacutecios

1 1o quadrante 17π

4 4π 983083 π

4

2o quadrante 991251 5π

4 26π

3 8π 983083 2π

3 991251 19π

6 ndash2π ndash 7π

6

3o quadrante 991251 3π

4 22π

3 6π 983083 4π

3

4o quadrante 99125105 (observe que 99125105 lt 991251 π

2 cong 991251 157)

2

AC

B

D

A 40π (20 2π) 99125114π (7 983223 (9912512π))

B 17π2

8π 983083 π2

991251 11π2

9912514π 991251 3π2

C 13π 991251 21π

D 991251 5π2

9912512π 991251 π2

7π2

2π 983083 3π2

3 a)

PP

c)

P

36deg

b)

P

30deg

P

d)P

P

4

5π

3

4π

3

2π

3

π

3

0π

a) Hexaacutegono regular com lado de medida 1

b) periacutemetro = 6 aacuterea = 6 983223 12 middot 34

= 3

23

5 a) sen 991251 π2

983101 991251sen π

2 983101 9912511lt0

b) sen 991251 5π4

983101 sen 3π4

983101 22

gt0

5

4ndash

c) sen 10π3

983101 sen 2π 983083 4π3

983101 sen 4π3

983101 991251 sen π

3 983101

983101 991251 32

lt 0

d) sen 850983216 983101 sen (2 983223 360983216 983083 130983216) 983101 sen 130983216 gt 0

e) sen 3 816983216 983101 sen (10 983223 360983216 983083 216983216) 983101 sen 216983216 lt 0

6 a) sen 4π 983101 sen 0 983101 0

b) sen 17π2

983101 sen 8π 983083 π2

983101 sen π

2 983101 1

c) sen 19π3

983101 sen 6π 983083 π3

983101 sen π

3 983101 3

2

d) sen 1 290983216 983101 sen (3 983223 360983216 983083 210983216) 983101 sen 210983216 983101 983101 991251 sen 30983216 983101 991251 1

2

e) sen 991251 π3

983101 sen 5π3

983101 991251 sen π

3 983101 991251 3

2

7 a) sen 2π3

983083 κ 983223 2π 983101 sen 2π3

983101 sen π

3 983101 3

2 (V )

b) κ 983101 0rarr sen 0 983101 0

κ 983101 1rarr sen π 983101 0

κ 983101 2rarr sen 2π 983101 0

κ 983101 3rarr sen 3π 983101 0

(V )

c) sen 1

000983216 983101 sen(2 983223 360983216 983083280983216) 983101 sen 280983216 lt 0 (280983216

tem imagem no 4o Q) (F)

MATEMAacuteTICA CIEcircNC IA E APLICACcedilOtilde ES 2



d) Observe que o nuacutemero real 10 eacute maior que 3π (aproxi-

madamente 942) e menor que 7π2

(aproximadamente

1099)

seno

10

3

2

7

2

3

Desse modo o nuacutemero real 10 tem imagem no 3o

quadrante e assim sen 10 lt 0 (V)

8 periacuteodo 2π Im 983101 [ndash2 2]

3

2

2

0

x

y

2

ndash2

2

9 p 983101 2π Im 983101 [ndash1 1]

3

2

20

y

x

1

ndash1

2

10 p 983101 2π3

Im 983101 [ndash1 1]

2

3

2

3

6

0

y

x

1

ndash1

11 p = 2π Im 983101 [2 4]

3

2

2

0

y

x

2

3

4

2

12

0

y

x

1

2

3

2 3 4

p 983101 4π

Im 983101 [1 3]

13 Devemos ter 9912511 ⩽ sen α ⩽ 1 rArr 9912511 ⩽ t 983083 12

⩽ 1 multi-

plicando por 2

9912512 ⩽ t 983083 1 ⩽ 2

Somando (9912511) temos

9912513 ⩽ t ⩽ 1rArr R t isin ℝ | 991251 3 ⩽ t ⩽ 1

14

2

π

2 ⩽ α ⩽ π rArr 0 ⩽ sen α ⩽ 1 daiacute

0 ⩽ 2m 991251 3 ⩽ 1rArr 3 ⩽ 2m ⩽ 4rArr 32

⩽ m ⩽ 2

15 a) p 983101 2π∙4∙

983101 π

2

b) f maacutex

983101 3 983083 2 983223 1 983101 5

16 a) t 983101 0rarr h(0) 983101 6 983083 4 middot sen 0 983101 6 983083 0 983101 6 (m)

b) h (9) 983101 6 983083 4 983223 sen 9π12

983101 6 983083 4 983223 sen 3π4

983101

983101 6 983083 4 983223 22

983101 6 983083 2 983223 14 rArr h = 88 m

c) O maior valor possiacutevel de sen π

12 983223 t t isin 9831310 270] eacute

1

Assim hmaacutex

983101 6 983083 4 983223 1 983101 10 m

d) O periacuteodo de f eacute 2π

π

12

983101 2ππ

12

983101 24s

e) n 983101 27024

983101 1125

Assim satildeo 11 voltas completas

17 a) cos 11 π 983101 cos (10π 983083 π) 983101 cos π 983101 9912511

b) cos 10 π 983101 cos (5 983223 2π) 983101 cos 0 983101 1

c) cos 13π2

983101 cos 6π 983083 π2

983101 cos π2

983101 0

d) cos 27π2

983101 cos 12π 983083 3π2

983101 cos 3π2

983101 0

e) cos 991251 2π3 983101 cos

4π3 983101 991251 cos

π

3 983101 99125112

Ca piacutetulo 4 bull Funccedilotildees trigonomeacutetricas



18 a) cos 1 560983216 983101 cos (

1

440983216 983083 120deg)

360983216 983223 4

983101 cos 120983216 983101

983101 991251 cos 60983216 983101 99125112

b) cos 1

035983216 983101 cos (720983216 983083 315983216) 983101 cos 315983216 983101

983101 cos 45983216 983101 22

c) cos 19π2 983101 cos 2π 983083 7π6 983101 cos 7π6 983101 991251 cos π6 983101

983101 991251 32

d) cos 22π3

983101 cos 6π 983083 4π3

983101 cos 4π3

983101 991251 cos π3

983101

983101 99125112

e) cos(991251270983216) 983101 cos 90983216 983101 0

19 cos 9π2

983101 cos 4π 983083 π2

983101 cos π2

983101 0 sen 9π2

983101 sen π2

983101 1

cos 17π4 983101 cos 4π 983083 π4 983101 cos π4 983101 22

sen 17π4 983101

983101 sen π4

983101 22

y 983101 0 991251 1

22

983083 3 98322322

983101 9912511

2 2 983101 991251 2

4

20 x isin π 3π2

rArr 9912511 ⩽ cos x ⩽ 0rArr 9912511 ⩽ 2m5

⩽ 0rArr

rArr 9912515 ⩽ 2m ⩽ 0rArr 99125152

⩽ m ⩽ 0

21 ndash1⩽ cos x ⩽ 1rArrndash1⩽ 3m2 ndash 13 ⩽ 1 rArrndash 23 ⩽ 3m2 ⩽ 43 rArr

rArr ndash43

⩽ 3m ⩽ 83 rArr ndash

49

⩽ m ⩽ 89

O uacutenico nuacutemero inteiro pertencente ao intervalo

ndash

49

89

eacute o zero

22 p 983101 2π Im 983101 [ndash2 2]

3π

2

π

2

ndash2

2

y

x0 2ππ

23 p 983101 2π Im 983101 [1 3]

3

2

1

y

x0π 2π3π

2

π

2

24

1

1

y

x0

2π

3π 4ππ

p 983101 4π

Im 983101 [9912511 1]

25 p 983101 2π3

Im 983101 [ndash1 1]

1

1

y

x0 2π

3

π

2

π

3

π

6

26

0 x

3

y

6

3

2

2

3

2

1

p 983101 2π3

Im 983101 [1 3]

27 a) cos π

6 983083 κ 983223 2π 983101 cos π

6 983101 3

2 (F)

b) cos π

2 983083 κπ 983101 0 (V)

P

P

c) f miacuten

983101 2 983223 (9912511) 983101 9912512 (F)

d) f(x) 983101 x 983223 (9912511) 983101 991251x eacute uma funccedilatildeo linear (y 983101 ax) e

portanto natildeo eacute perioacutedica (F)

e) p 983101 2π

π

8 983101 16 (V)

28 a) 2

010 rArr f(0) 983101 400 983083 18 983223 cos 0 983101 418

2 015 rArr f(5) 983101 400 983083 18 983223 cos 5π3

983101

983101 400 983083 18 983223

1

2 983101 409




2

020 rArr f(10) 983101 400 983083 18 983223 cos 10π3

983101

983101 400 983083 18 983223 cos 4π3

983101 400 983083 18 983223 ndash

12

983101

983101 400 991251 9 983101 391

b) O menor valor possiacutevel de cos π

3 x eacute9912511 x isin 0 1hellip20

f miacuten

983101 400 983083 18 middot (9912511) 983101 400 991251 18 983101 382 (382 milhotildees

de doacutelares)Devemos ter

cos π

3 x 983101 cos (π 983083 κ 983223 2π)

ndash 1 κ isin ℤ

Isto eacute

π

3 x 983101 π 983083 κ middot 2π κ isin ℤ rArr x 983101 3 middot (1 983083 2κ) κ isin ℤ

κ 983101 0rArr x 983101 3 middot (1 983083 0) 983101 3 (2

013)

κ 983101 1rArr x 983101 3 middot (1 983083 2) 983101 9 (2

019)

κ 983101 2rArr x 983101 3 middot (1 983083 4) 983101 15 (2

025)

κ 983101 3rArr x 983101 3 middot (1 983083 6) 983101 21 (2

031) natildeo conveacutem

Assim em 3 vezes (2013 2009 2025) f atinge seu

menor valor

29 a) O conjunto imagem de f(x) 983101 tg x 991251 π6

eacute ℝ desse

modo eacute possiacutevel termos para x pertencente ao do-

miacutenio de f f(x) 983101 3 bem como f(x) 983101 99125110

b) tg x 991251 π6

983101 0rArr x 991251 π6

983101 κ π κ isin ℤ

κ 983101 0 x 983101 π

6

κ 983101 1 x 983101 π 983083 π

6 983101 7π

6

κ 983101 2 x 983101 π6 983083 2π 983101 13π6

e assim por diante

30 a) f(x) 983101 1 983083 tg x

x ne π2

983083 k π k isin 983130

D 983101 x isin ℝ | x 860697 π2

983083 k π k isin 983130

periacuteodo 0 ⩽ x ⩽ π rArr p = π

b) f(x) 983101 2 tg x

x 860697 π

2 983083 k π k isin 983130

D 983101 x isin ℝ | x 860697 π2

983083 k π k isin 983130

p 983101 π

c) f(x) 983101 tg 2x

2x 860697 π2

983083 k π rArr x 860697 π4

983083 k π2

k isin 983130

D 983101 x isin ℝ | x 860697 π4

983083 k π2

k isin 983130

periacuteodo 0 ⩽ 2x ⩽ π rArr 0 ⩽ x ⩽ π2

p 983101 π

2

d) f(x) 983101 tg x 983083 π

6

x 983083 π

6 1057305 π

2 983083 k π rArr x 1057305 π

2 ndash π

6 983083 k π rArr

rArr x 1057305 π

3 983083 k π k isin 983130

D 983101 x isin ℝ | x 1057305 π

3 983083 k π k isin 983130

periacuteodo 0 ⩽ x 983083 π

6 ⩽ π rArr ndash

π

6 ⩽ x ⩽ π ndash π

6 rArr

rArr ndash

π

6 ⩽ x ⩽ 5π

6 rArr p 983101 5π

6 ndash ndash π

6 rArr p 983101 π

31 a) tg x 991251 π

2 estaacute definida quando x 991251 π

2 ne π

2 983083 k π

k isin ℤ isto eacute x ne π 983083 k π k isin ℤ

Observe que os nuacutemeros reais descritos por

x isin 991769 x 983101 π 983083 k π k isin ℤ satildeo

k 983101 0 x 983101 π k 983101 1 x 983101 2π k 983101 2 x 983101 3π etc

Podemos entatildeo escrevecirc-los na forma

x isin ℝ 991769 x 983101 k π k isin ℤ

Daiacute o domiacutenio D eacute D 983101 x isin ℝ 991769 x ne k π k isin ℤ

b) p 983101

Faccedilamos t 983101 x991251 π2

para que tg t complete um periacuteodo

t deve variar entre 0 e π

0 ⩽ t⩽ π rArr 0 ⩽ x991251 π2

⩽ π rArrπ

2 ⩽ x⩽ 3π

2 e o periacuteodo

de f eacute 3π

2 991251 π

2 983101 π

c)

0

ndash1

1

y

4

2

ndash x

4ndash

2ndash

32 -

Periacuteodo

t 983101 x2

para que tg t complete um periacuteodo t deve

variar entre 0 e π

0 ⩽ t ⩽ π rArr 0 ⩽ x2

⩽ π rArr 0 ⩽ x ⩽ 2π o periacuteodo

eacute 2π

-Domiacutenio

tg x2

estaacute definida quando x2

ne π

2 983083 κπ rArr

rArr x ne π 983083 2 κπ rArr x ne (1 983083 2κ)π κ isin ℤ




-Graacutefico

1

0

y

x

ndash1

2

ndash

2

ndash

Exerciacutecios complementares

1 Lembre que sen(ndashx) = ndash sen x forallx isin

p = 2π Im = [ndash1 1]

2

3π

2

ndash1

1

yy = sen x

x

0

π

2

2ππ

3π

2

ndash1

1

y

x0 π

22ππ

y = |sen x|

p = π Im = [0 1]

3(3o ) (1o ) (2o )

x 2x ndash π4

y = 1 + cos 2x ndash π4

π8 0 2

3π8

π2 1

5π8 π 0

7π8

3π2 1

9π8 2π 2

periacuteodo = 9π8

ndash π8

= π Im = [02]

y

2

1

0 xπ

8

3π

8

5π

8

7π

8

9π

8

4(3o ) (1o ) (2o )

x 3x ndash π y = 1 ndash 2 983223 sen(3x ndash π)

π3 0 1

π2

π2 ndash1

2π3 π 1

5π6

3π2 3

π 2π 1

periacuteodo = π ndash π3

= 2π3

Im = [ndash1 3]

y

3

1

ndash1

0 xπ

3

π

2

2π

3

5π

6

π

5

x

y = cos xy

1

ndash1

0 π

2

3π

2

π

2π

x

y = |cos x|y

1

ndash1

0 π

23π

2π 2π

x

y

1

ndash1

0

π

2

3π

2

π 2π

p = π Im = [ndash1 1]

6 f(x) = 0 rArr 2 middot ∙ cosx ∙ ndash 1 = 0 rArr | cos x ∙ = 1

2




Podemos tercosx = 1

2 rArr x = π

3 ou x = 5π

3

cosx = ndash 12

rArr x = 2π3

ou x = 4π3

7 a) Se x isin N (ndash1)x =1 se x eacute par

ndash1 se x eacute iacutempar

Desse modo o conjunto imagem de f eacute 1 3 poisbull x par rArr f(x) = 2 + 1 = 3

bull x iacutempar rArr f(x) = 2 + (ndash1) = 1

b) Note que f(0) = f(2) = f(4) = f(6) = hellip e

f(1) = f(3) = f(5) = f(7) = hellip

Sim p = 2

8 O periacuteodo de f eacute 2π

8π3

= 2π

8π3

= 075 s isto eacute o atleta

realiza uma oscilaccedilatildeo completa em 075 s Desse modo em

6 segundos ele realizaraacute

6

075 = 8 oscilaccedilotildees completas

9 a) tg x ndash 3π4

existe se e somente se x ndash 3π4

ne π2

+ k π

k isin ℤ isto eacute xne5π4

+ k π k isin ℤ

As imagens no ciclo correspondente aos nuacutemeros reais

x = 5π4

+ k π k isin ℤ satildeo A e B

A

B

π

4

Assim D = x isin ℝ ∙ x ne π4

+ k π k isin ℤ

b) Como o conjunto imagem de y = tg x ndash 3π4

eacute ℝ

basta garantir a existecircncia de m ndash 1m ndash 2

o que ocorre

para mne2

10 Como forallα isin ℝ ndash1 ⩽ cos α ⩽ 1 temos que

0 ⩽ cos2 α ⩽ 1 isto eacute

0 ⩽ 2m ndash 1 ⩽ 1 rArr 1 ⩽ 2m ⩽ 2 rArr 12

⩽ m ⩽ 1

11 f eacute a maacutexima quando cos x = ndash1 (observe que nesse caso

fmaacutex

= 3 ndash 2 middot (ndash1) = 5)

cos x = ndash1

Os nuacutemeros reais com extremidade em P satildeo da forma

π + k 983223 2π k isin ℤ

cos

P

2 a) p = 750 rArr 750 = 800 ndash 100 middot sen(t + 3)π

6 rArr

rArr 100 middot sen(t + 3)π

6 = 50 rArr sen

(t + 3)π6

= 12

As imagens dos nuacutemeros reais cujo seno vale 12

satildeo

os pontos P e Q da seguinte figura

P

30deg

Q

1

2

Daiacute

ou

(t + 3)π6

= π6

+ k 983223 2π 1

(t + 3)π6

= 5π6

+ k 983223 2π 2

De 1 t = ndash2 + 12k k isin ℤ

De 2 t = 2 + 12k k isin ℤ

Para k = 0

t = ndash 2 (natildeo serve)

out = 2 (mecircs de marccedilo)

Para k = 1t = 10 (novembro)

out = 14 (natildeo serve)

Observaccedilatildeo O aluno ainda natildeo formalizou a resolu-

ccedilatildeo de equaccedilotildees trigonomeacutetricas Uma outra possibi-

lidade de resoluccedilatildeo da equaccedilatildeo eacute ldquotestarrdquo valores

1

bull Para que sen(t + 3)π

6 = 1

2 podemos ter

(t + 3)π6 = π6 ou (t + 3)π6 = 5π6 t = ndash 2 (natildeo serve)

out = 2 (marccedilo)

bull Agora basta acrescentar um nuacutemero completo de

voltas a π6

e a 5π6

(t + 3)π6

= π6

+ 2π ou(t + 3)π

6 = 5π

6 + 2π

t = 10 (novembro)ou

t = 14 (natildeo serve pois 0 ⩽ t ⩽ 11)






Como os graacuteficos de f e g se interceptam em um uacutenico

ponto (P) a equaccedilatildeo f(x) = g(x) tem uma uacutenica soluccedilatildeo

real

21 bull sen (π x) cong 4x ndash 4x2

sen π4

cong 4 middot 14

ndash 4 middot 14

2

= 1 ndash 416

= 34

bull erro = 34

ndash 22

= 34

ndash 1412

= 3 ndash 2824

=

= |0045| = 0045

22 a) O maior valor de cos(2πt) eacute 1 (V)

Daiacute Pmaacutex

= 96 + 18 middot 1 = 114

b) O menor valor de cos(2πt) eacute ndash1

Daiacute Pmiacuten

= 96 + 18 middot (ndash1) = 78 (V)

c) O periacuteodo dessa funccedilatildeo eacute 2|2π|

= 1 s (V)π

d) P 13

= 96 + 18 cos 2π3

= 96 + 18 middot cos 120deg =

= 96 + 18 middot ndash 12

= 96 ndash 9 = 87 (F)

e) cos (2π t) = 1 rArr 2π t = k middot 2π k isin ℤ rArr t = k isin ℤ

isto eacute quando t = 0 t = 1 t = 2hellip P(t) eacute maacuteximo

cos (2π t) = ndash1 rArr 2π t = π + k middot 2π rArr

rArr 2t = 1 + 2k rArr t = 12

+ k

k isin ℤ isto eacute quando t = 12

t = 32

t = 52

hellip P(t) eacute

miacutenimo

O graacutefico natildeo pode ser o apresentado (F)

23 Como ndash1 ⩽ cosπ3 x ndash

4π3 ⩽ 1 o maior valor possiacutevel

para cosπ3 x ndash

4π3 = 1 hArr

π3 x ndash

4π3 = k middot 2π k isin ℤ

Daiacute x3

ndash 43

= 2k k isin ℤ

cos

x = 2k + 43

middot 3 k isin ℤ

x = 4 + 6k k isin ℤ

k = 0 rArr x = 4

k = 1 rArr x = 10

k = 2 rArr x = 16 (natildeo serve pois 1 ⩽ x ⩽ 12)

x = 4 (mecircs de abril) rArr f(4) = 200 + 54 middot 1 = 254 ton

x = 10 (mecircs de outubro) rArr f(10) = 200 + 60 middot 1 = 260 ton

Assim a produccedilatildeo maacutexima ocorreu em outubro (mecircs 10)

seu valor eacute 260 ton

24 a) ℝ

b) [ndash1 1]

c) 2π

d) bull Se x ⩾ 0 f(x) = cos |x| = cos x

y

1

2ππ0 xπ

23π

2ndash1

ndashπ

2

(1)

bull Se x ⩽ 0 f(x) = cos |x| = cos (ndashx) = cos x

y

1

ndash2π ndashπ 0 x

ndash1ndashndash

π

2

3π

2

(2)

bull Reunindo os dois casos concluiacutemos que o graacutefico de

f(x) = cos |x| coincide com o graacutefico de y = cos x

y

1

2π

π

0 xπ

2

π

2

3π

2

3π

2ndash2π

ndash ndash

ndash1

ndashπ

25 f(x) = 0 rArr sen 3π x2 = 0 ou ndash1 + x ndash 1 = 0

3 π x2

= k π ou x ndash 1 = 1 rArr

dArrx = 2π

3 k isin 2 x = 2 ()

k = 1 rArr x = 23

lt 1

k = 2 rArr x = 43

rArr P 43

0

k = 3 rArr x = 2 rArr()

Q (2 0)

k = 4 rArr x = 43

rArr R 83

0

k = 5 rArr x = 103

rArr S 103

0

26 a) p = 2π|2π|

= 1

Como ndash1 ⩽ sen 2πx ndash π2 ⩽ 1 temos ndash1 + 1 ⩽

⩽ 1 + sen 2πx ndash π2

⩽ 1 + 1 hArr 0 ⩽ f(x) ⩽ 2

Assim Im = [0 2]






k983101 6rArr cos 6π12

983101 0


983101 cos 105deg983101 minuscos 75deg


983101 cos 2π3

983101 minus 12


983101 cos 3π4

983101 minus 2

2

k983101 10rArr cos 10π12 983101 cos 5π6 983101 minus 32

k983101 11rArr cos 11π12


k983101 12rArr cos π 983101 minus1

cos

Para k⩾ 13 kisin ℤ comeccedila a haver repeticcedilatildeo de valores

o 3o quadrante repete os valores do 2o e o 4o repete os do

1o Assim temos 13 valores distintos

Resposta b

3 y 1 2 sen x= +

1 sen x 1 1 y 3minus le le rArr minus le le

Resposta d

4 Ao dar uma volta a partiacutecula faz 6 pausas (6 middot 60deg983101 360deg)

Como 512983101 85 middot 6983083 2 a partiacutecula realiza 85 voltas com-

pletas na 86a volta ela encerra o movimento na 2a parada

em um ponto do 4o quadrante

Resposta b

5 Observe que o periacuteodo de f eacute 5π4

minus π4

983101 π

Como Im983101 [0 4] f deve ser definida por y983101 2983083 2sen 2x

Resposta e

6 ( )x

f x 1 cos3

= +

( ) [ ]x

1 cos 1 0 f x 2 Im 0 23

minus le le rArr le le rArr =

2 p 6

1

3

π= = π

( )3

f 3 1 cos 1 1 03

ππ = + = minus =

3 3 f 1 cos 1 cos 1 0 1

2 6 2

π π π = + = + = + =

0

5 5 f 1 cos 1

2 12gt

π π = + gt

3 3 para x f 12 2

π π = =

( )2 1 1

para x 2 f 2 1 cos 13 2 2

π= π π = + = minus = entatildeo para

x ( )3

2 f x 02

π isin π gt

Resposta b

7 Do graacutefico ( ) ( )f 3 f 9 5 5π =

Resposta b

8 ( )y 2 sen 3x=

2p

3

π=

( ) ( )1 sen 3x 1 2 2 sen 3x 2minus le le rArr minus le le

[ ]Im 2 2= minus

Resposta b

9 Temos

f 12

983101 1rArr sen ω middot 12

983101 1rArr

rArr senω

2 983101 sen π

2 rArr ω 983101 π

f(x)983101 sen(ωx)983101 sen(πx)

f 16

983101 sen π6

983101 12

f

1

4 983101

sen

π

4 983101

2

2

1

2 983083

2

2 983083

1983101

3983083 2

2

f 12

983101 sen π2

983101 1

Resposta a

10 21 20

45 5 5 5

π π π π= + = π + rArr O menor valor natildeo

negativo cocircngruo ao arco de21

5

π eacute

5

π

Resposta a

| Ca piacutetulo 4 bull Funccedilotildees trigonomeacutetricas



11 sen x 2m 9= minus

1 2m 9 1minus le minus le rArr 8 2m 10le le rArr 4 m 5le le

Resposta e

12x983101

0rArr

f(0)983101

sen 0983101

0 e g(0)983101

cos 0983101

1

bull f (π)983101 1rArr sen π

k 983101 1rArr k 9831012

bull g π

4 983101 0rArr cos m π

4 983101 0rArr mπ

4 983101 π

2 983083 n middot 2π

nisin ℤ rArr

n983101 0 mπ

4 983101 π

2 rArr m983101 2

Resposta b

13 1 sen t 1

3

π minus le le rArr

1 3 3 sen t 1 3

3

π minus + le + le + rArr

2 3 sen t 43

π rArr le + le

Resposta e

14 f(x)= 3 sen983218 x ndash 5 (1 ndash sen983218 x)= 3 sen983218 x ndash 5+ 5 sen983218 x=

= 8 sen983218 x minus 5

f miacuten

= 8 middot 0 ndash 5= minus5

f maacutex

= 8 middot 1 ndash 5= 3

Resposta b

15 2 910ordm 8 360ordm 30ordm= sdot +

3195ordm 8 360ordm 315ordm 315ordm 45ordmminus = minus sdot minus rArr minus equiv

2 580ordm 7 360ordm 60ordm= sdot +


A expressatildeo resulta em

sen 30ordm cos 45ordm sen 30ordm cos 45ordm

1cossec 60ordm tg30ordm tg30ordmsen60ordm

minus minus

= =+ +

1 2 1 2 1 2

2 2 2 26 3 32 3

3 3 333

minus minusminus

= = = =+

+

( )1 2 3 3 6

6 6

minus minus= =

Resposta e

sen16 x minus π

7 = minus1rArr sen x minus π

7 = sen 3π

2 + k middot 2π

kisin ℤ

Daiacute

x minus π

7 = 3π

2 + k middot 2π rArr x= π

7 + 3π

2 + k middot 2π rArr

rArr

x=

2π + 21π

14 +

kmiddot

2π

rArr

x=

23π

14 +

kmiddot

2π

kisin ℤ

Resposta c

17 bull O periacuteodo de f eacute 3π isso exclui a alternativa d

bull O conjunto imagem de f eacute [minus3 3] isso exclui a alter-

nativa b

bull f(0)= 0 isso exclui a alternativa c pois teriacuteamos

f(0)= 3 middot cos 0= 3 Exclui tambeacutem a alternativa e pois

teriacuteamos f(0)= 3 middot sen (0+ π2

)= 3 sen π

2 = 3

Resposta a

18 2

p3

π=

Resposta b

19 I Verdadeira pois se sen cos entatildeo eα = β α β

satildeo complementares

II Verdadeira pois2 2 2 2sen sen sen cos 1α + β = α + α =

III Falsa pois se por exemplo3

k 24π= α = β = e

sen cosα = minus β

IV Falsa pois por exemplo se 60ordmα = e

( ) ( )30ordm sen 60ordm cos 30ordmβ = minus = minus minus

Resposta a

20 2sec x cos x tg x sen x cos x cos xsdot minus sdot sdot minus =

1=cosx

cos xsdot senxcosxminus sen x cos xsdot sdot 2cos xminus =

( )2 2=1 sen x cos x 1 1 0minus + = minus =

a) 1 sen 1 0 1minus π = minus =

b) 1 cos 3 1 1 0+ π = minus =

c) ( )1 cos 1 1 2minus π = minus minus =

d) ( )1 2cos 1 2 1 1+ π = + minus = minus

e) ( )1 3 cos 1 3 1 4minus π = minus minus =

Resposta b

| MATEMAacuteTICA CIEcircNC IA E APLICACcedilOtilde ES 2



21 I f(x)= sen 2x periacuteodo de f = 2π2

= π

f(0)= sen 0= 0 f π

2 = 0 f(π)= 0 hellip

rarr B

II g(x)= sen |x|

x⩾ 0 g(x) = sen x g π

2 = 1

xlt 0 g(x) = sen(minusx)= minussen x g minus π

2 =

C

= minussenπ

2 = minus1III h(x)= sen(minusx)= minussen x

h π2

= minus1 h(π)= 0rarr A

Resposta d

22 bull yR = y

Q = 1

3 rArr 1

3 = cos x

1+ cos x rArr

rArr 3 cos x= 1+ cos xrArr cos x= 12

xisin [0π[

xQ= π

3

bull yP = 0rArr 0= cos x

1+ cos x rArr cos x= 0rArr x

P = π

2

bull Aacuterea do trapeacutezio= (OP+

RQ) middot OR2 =

=

π

2 + π

3 middot 1

3

2 = 5π

36

Resposta d

23 2655deg= 7 middot 360deg+ 135deg sen 2655deg= sen 135deg= 22

bull minus 41π4

= minus 40π4

minus π

4 = minus10π minus π

4

5 voltas

sec minus 40π4

= sec minus π

4 = 1

cos minus π

4

= 1

cos π

4

=

= 22

= 2

bull log2 256= log

2 2

8 = 8

6 middot 22

+ 5 middot 2 middot 8 = 8 2 middot 2 2 = 32

Resposta d

24 ( ) [ ]f x sen x h Im 2 0= + = minus

2 sen x h 0minus le + le

2 h sen x hminus minus le le minus

2 h 1 h 1minus minus = minus rArr = minus

Resposta c

25 2π|c|

= π2

rArr |c|= 4rArr c= plusmn 4

Como minus1⩽ sen cx⩽ 1 forallxisin ℝ f maacutex

= a+ b middot 1=

= a+ b e f miacuten

= a+ b middot (ndash1)= a ndash b

Devemos tera+ b= 9

andash

b=

ndash

7

rArr a= 1 e b = 8

bull c= 4 a= 1 b= 8rArr a+ b+ c= 13

bull c= ndash4 a= 1 e b = 8rArr a+ b+ c= 5

Resposta a

26 a) Falsa pois se ( )x 0 f 0 2 sen0 0= rArr = sdot = e pelo

graacutefico ( )f 0 2=

b) Falsa pois pelo graacutefico o periacuteodo de f eacute 4 π

c) Verdadeira pois no intervalo [ ]2 2 minus π π o graacutefico

cruza duas vezes o eixo nos pontos x = minusπ e

x = π

d) Falsa pois pelo graacutefico para ( )x 0 f x 0minusπ lt lt gt

e) Verdadeira pois pelo graacutefico ( )2 f x 2minus le le

I f(x)= g(x) apresenta duas soluccedilotildees no intervalo con-

siderado (F)

II f 9π10

gt g 9π10

pois f 9π10

gt 0 e g 9π10

lt 0 (V)

27

III exist x | g(x) lt 0 e f(x) lt 0 (F)

Resposta d

28 a) Se x cos 2x 1 cos 1 22

π= minus = π minus = minus

e pelo

graacutefico f 02

π

=

b) Se x 1 cos 2x 1 cos 2

2

π= minus = minus π = e pelo graacutefico

f 02

π =

c) Se

xx 2 cos 2 2 cos 2 2 2

2 2 4

π π= minus = minus = minus e pelo

graacutefico f 02

π =

3d) Se x 0 cos 2x 1 0

2 2

3 5 Se x cos 2x 1 1

4 4 4

π π = π rArr minus =

π π π = rArr minus = minus

de acordo com

o graacutefico

e) Se

xx cos 1 cos 1 1

2 2

π= π minus = minus = minus e pelo graacutefico

( )f 0π =

Resposta d




29 Observe que

bull f (π)= sen 2π + 5 cos 2π = 5

f (minusπ)= sen (minus2π)+ 5 cos (minus2π)= 5

bull f π

4 = sen 2π

4 + 5 cos 2π

4 = 1

rArr

rArr f ndash π4

= sen ndash2π4

+ 5 cos ndash 2π4

= minus1

rArr f natildeo eacute par nem iacutemparbull f (x+ π)= sen (2 middot (x+ π))+ 5 cos (2 middot (x+ π))

f (x+ π)= sen (2x+ 2π)+ 5 cos (2x+ 2π)=

= sen 2x+ 5 cos 2x= f (x)

O periacuteodo de f eacute π

Resposta c

30

x

y

2ππndashπ

ndash ndash

1 2

2

16

ndash2

ndash2 ndash1ndash3ndash4 3 4

0 π

2

π

2

3π

2

3π

2

Haacute dois pontos de interseccedilatildeo

Resposta c




d) Observe que o nuacutemero real 10 eacute maior que 3π (aproxi-

madamente 942) e menor que 7π2

(aproximadamente

1099)

seno

10

3

2

7

2

3

Desse modo o nuacutemero real 10 tem imagem no 3o

quadrante e assim sen 10 lt 0 (V)

8 periacuteodo 2π Im 983101 [ndash2 2]

3

2

2

0

x

y

2

ndash2

2

9 p 983101 2π Im 983101 [ndash1 1]

3

2

20

y

x

1

ndash1

2

10 p 983101 2π3

Im 983101 [ndash1 1]

2

3

2

3

6

0

y

x

1

ndash1

11 p = 2π Im 983101 [2 4]

3

2

2

0

y

x

2

3

4

2

12

0

y

x

1

2

3

2 3 4

p 983101 4π

Im 983101 [1 3]

13 Devemos ter 9912511 ⩽ sen α ⩽ 1 rArr 9912511 ⩽ t 983083 12

⩽ 1 multi-

plicando por 2

9912512 ⩽ t 983083 1 ⩽ 2

Somando (9912511) temos

9912513 ⩽ t ⩽ 1rArr R t isin ℝ | 991251 3 ⩽ t ⩽ 1

14

2

π

2 ⩽ α ⩽ π rArr 0 ⩽ sen α ⩽ 1 daiacute

0 ⩽ 2m 991251 3 ⩽ 1rArr 3 ⩽ 2m ⩽ 4rArr 32

⩽ m ⩽ 2

15 a) p 983101 2π∙4∙

983101 π

2

b) f maacutex

983101 3 983083 2 983223 1 983101 5

16 a) t 983101 0rarr h(0) 983101 6 983083 4 middot sen 0 983101 6 983083 0 983101 6 (m)

b) h (9) 983101 6 983083 4 983223 sen 9π12

983101 6 983083 4 983223 sen 3π4

983101

983101 6 983083 4 983223 22

983101 6 983083 2 983223 14 rArr h = 88 m

c) O maior valor possiacutevel de sen π

12 983223 t t isin 9831310 270] eacute

1

Assim hmaacutex

983101 6 983083 4 983223 1 983101 10 m

d) O periacuteodo de f eacute 2π

π

12

983101 2ππ

12

983101 24s

e) n 983101 27024

983101 1125

Assim satildeo 11 voltas completas

17 a) cos 11 π 983101 cos (10π 983083 π) 983101 cos π 983101 9912511

b) cos 10 π 983101 cos (5 983223 2π) 983101 cos 0 983101 1

c) cos 13π2

983101 cos 6π 983083 π2

983101 cos π2

983101 0

d) cos 27π2

983101 cos 12π 983083 3π2

983101 cos 3π2

983101 0

e) cos 991251 2π3 983101 cos

4π3 983101 991251 cos

π

3 983101 99125112




18 a) cos 1 560983216 983101 cos (

1

440983216 983083 120deg)

360983216 983223 4

983101 cos 120983216 983101

983101 991251 cos 60983216 983101 99125112

b) cos 1

035983216 983101 cos (720983216 983083 315983216) 983101 cos 315983216 983101

983101 cos 45983216 983101 22


983101 991251 32

d) cos 22π3

983101 cos 6π 983083 4π3

983101 cos 4π3

983101 991251 cos π3

983101

983101 99125112

e) cos(991251270983216) 983101 cos 90983216 983101 0

19 cos 9π2

983101 cos 4π 983083 π2

983101 cos π2

983101 0 sen 9π2

983101 sen π2

983101 1

cos 17π4 983101 cos 4π 983083 π4 983101 cos π4 983101 22

sen 17π4 983101

983101 sen π4

983101 22

y 983101 0 991251 1

22

983083 3 98322322

983101 9912511

2 2 983101 991251 2

4

20 x isin π 3π2

rArr 9912511 ⩽ cos x ⩽ 0rArr 9912511 ⩽ 2m5

⩽ 0rArr

rArr 9912515 ⩽ 2m ⩽ 0rArr 99125152

⩽ m ⩽ 0


rArr ndash43


49

⩽ m ⩽ 89


ndash

49

89

eacute o zero

22 p 983101 2π Im 983101 [ndash2 2]

3π

2

π

2

ndash2

2

y

x0 2ππ

23 p 983101 2π Im 983101 [1 3]

3

2

1

y

x0π 2π3π

2

π

2

24

1

1

y

x0

2π

3π 4ππ

p 983101 4π

Im 983101 [9912511 1]

25 p 983101 2π3

Im 983101 [ndash1 1]

1

1

y

x0 2π

3

π

2

π

3

π

6

26

0 x

3

y

6

3

2

2

3

2

1

p 983101 2π3

Im 983101 [1 3]

27 a) cos π

6 983083 κ 983223 2π 983101 cos π

6 983101 3

2 (F)

b) cos π

2 983083 κπ 983101 0 (V)

P

P

c) f miacuten

983101 2 983223 (9912511) 983101 9912512 (F)



e) p 983101 2π

π

8 983101 16 (V)

28 a) 2

010 rArr f(0) 983101 400 983083 18 983223 cos 0 983101 418

2 015 rArr f(5) 983101 400 983083 18 983223 cos 5π3

983101

983101 400 983083 18 983223

1

2 983101 409




2

020 rArr f(10) 983101 400 983083 18 983223 cos 10π3

983101

983101 400 983083 18 983223 cos 4π3

983101 400 983083 18 983223 ndash

12

983101

983101 400 991251 9 983101 391



f miacuten



cos π

3 x 983101 cos (π 983083 κ 983223 2π)

ndash 1 κ isin ℤ

Isto eacute

π


κ 983101 0rArr x 983101 3 middot (1 983083 0) 983101 3 (2

013)

κ 983101 1rArr x 983101 3 middot (1 983083 2) 983101 9 (2

019)

κ 983101 2rArr x 983101 3 middot (1 983083 4) 983101 15 (2

025)

κ 983101 3rArr x 983101 3 middot (1 983083 6) 983101 21 (2



menor valor


eacute ℝ desse



b) tg x 991251 π6

983101 0rArr x 991251 π6


κ 983101 0 x 983101 π

6

κ 983101 1 x 983101 π 983083 π

6 983101 7π

6

κ 983101 2 x 983101 π6 983083 2π 983101 13π6

e assim por diante

30 a) f(x) 983101 1 983083 tg x

x ne π2

983083 k π k isin 983130

D 983101 x isin ℝ | x 860697 π2

983083 k π k isin 983130


b) f(x) 983101 2 tg x

x 860697 π

2 983083 k π k isin 983130

D 983101 x isin ℝ | x 860697 π2

983083 k π k isin 983130

p 983101 π

c) f(x) 983101 tg 2x

2x 860697 π2

983083 k π rArr x 860697 π4

983083 k π2

k isin 983130

D 983101 x isin ℝ | x 860697 π4

983083 k π2

k isin 983130


p 983101 π

2

d) f(x) 983101 tg x 983083 π

6

x 983083 π

6 1057305 π

2 983083 k π rArr x 1057305 π

2 ndash π

6 983083 k π rArr

rArr x 1057305 π

3 983083 k π k isin 983130

D 983101 x isin ℝ | x 1057305 π

3 983083 k π k isin 983130


6 ⩽ π rArr ndash

π


6 rArr

rArr ndash

π

6 ⩽ x ⩽ 5π

6 rArr p 983101 5π

6 ndash ndash π

6 rArr p 983101 π

31 a) tg x 991251 π


2 ne π

2 983083 k π




k 983101 0 x 983101 π k 983101 1 x 983101 2π k 983101 2 x 983101 3π etc




b) p 983101




0 ⩽ t⩽ π rArr 0 ⩽ x991251 π2

⩽ π rArrπ

2 ⩽ x⩽ 3π

2 e o periacuteodo

de f eacute 3π

2 991251 π

2 983101 π

c)

0

ndash1

1

y

4

2

ndash x

4ndash

2ndash

32 -

Periacuteodo

t 983101 x2


variar entre 0 e π

0 ⩽ t ⩽ π rArr 0 ⩽ x2


eacute 2π

-Domiacutenio

tg x2


ne π

2 983083 κπ rArr





-Graacutefico

1

0

y

x

ndash1

2

ndash

2

ndash




2

3π

2

ndash1

1

yy = sen x

x

0

π

2

2ππ

3π

2

ndash1

1

y

x0 π

22ππ

y = |sen x|

p = π Im = [0 1]

3(3o ) (1o ) (2o )

x 2x ndash π4


π8 0 2

3π8

π2 1

5π8 π 0

7π8

3π2 1

9π8 2π 2

periacuteodo = 9π8

ndash π8

= π Im = [02]

y

2

1

0 xπ

8

3π

8

5π

8

7π

8

9π

8

4(3o ) (1o ) (2o )


π3 0 1

π2

π2 ndash1

2π3 π 1

5π6

3π2 3

π 2π 1


= 2π3

Im = [ndash1 3]

y

3

1

ndash1

0 xπ

3

π

2

2π

3

5π

6

π

5

x

y = cos xy

1

ndash1

0 π

2

3π

2

π

2π

x

y = |cos x|y

1

ndash1

0 π

23π

2π 2π

x

y

1

ndash1

0

π

2

3π

2

π 2π



2




Podemos tercosx = 1

2 rArr x = π

3 ou x = 5π

3

cosx = ndash 12

rArr x = 2π3

ou x = 4π3






f(1) = f(3) = f(5) = f(7) = hellip

Sim p = 2


8π3

= 2π

8π3




6




ne π2

+ k π


+ k π k isin ℤ


x = 5π4


A

B

π

4


+ k π k isin ℤ


eacute ℝ


o que ocorre

para mne2




⩽ m ⩽ 1


fmaacutex


cos x = ndash1


π + k 983223 2π k isin ℤ

cos

P


6 rArr


6 = 50 rArr sen

(t + 3)π6

= 12


satildeo


P

30deg

Q

1

2

Daiacute

ou

(t + 3)π6

= π6

+ k 983223 2π 1

(t + 3)π6

= 5π6

+ k 983223 2π 2



Para k = 0








1


6 = 1

2 podemos ter




voltas a π6

e a 5π6

(t + 3)π6

= π6

+ 2π ou(t + 3)π

6 = 5π

6 + 2π

t = 10 (novembro)ou









real


sen π4

cong 4 middot 14

ndash 4 middot 14

2

= 1 ndash 416

= 34

bull erro = 34

ndash 22

= 34

ndash 1412

= 3 ndash 2824

=

= |0045| = 0045


Daiacute Pmaacutex

= 96 + 18 middot 1 = 114


Daiacute Pmiacuten

= 96 + 18 middot (ndash1) = 78 (V)


= 1 s (V)π

d) P 13

= 96 + 18 cos 2π3



= 96 ndash 9 = 87 (F)




rArr 2t = 1 + 2k rArr t = 12

+ k


t = 32

t = 52

hellip P(t) eacute

miacutenimo




para cosπ3 x ndash

4π3 = 1 hArr

π3 x ndash


Daiacute x3

ndash 43

= 2k k isin ℤ

cos

x = 2k + 43

middot 3 k isin ℤ


k = 0 rArr x = 4

k = 1 rArr x = 10






24 a) ℝ

b) [ndash1 1]

c) 2π


y

1

2ππ0 xπ

23π

2ndash1

ndashπ

2

(1)


y

1


ndash1ndashndash

π

2

3π

2

(2)



y

1

2π

π

0 xπ

2

π

2

3π

2

3π

2ndash2π

ndash ndash

ndash1

ndashπ


3 π x2


dArrx = 2π

3 k isin 2 x = 2 ()

k = 1 rArr x = 23

lt 1

k = 2 rArr x = 43

rArr P 43

0


Q (2 0)

k = 4 rArr x = 43

rArr R 83

0

k = 5 rArr x = 103

rArr S 103

0

26 a) p = 2π|2π|

= 1



⩽ 1 + 1 hArr 0 ⩽ f(x) ⩽ 2

Assim Im = [0 2]







983101 0




983101 cos 2π3

983101 minus 12


983101 cos 3π4

983101 minus 2

2


k983101 11rArr cos 11π12



cos




Resposta b

3 y 1 2 sen x= +


Resposta d





Resposta b


minus π4

983101 π


Resposta e

6 ( )x

f x 1 cos3

= +

( ) [ ]x

1 cos 1 0 f x 2 Im 0 23


2 p 6

1

3

π= = π

( )3

f 3 1 cos 1 1 03

ππ = + = minus =

3 3 f 1 cos 1 cos 1 0 1

2 6 2

π π π = + = + = + =

0

5 5 f 1 cos 1

2 12gt

π π = + gt

3 3 para x f 12 2

π π = =

( )2 1 1

para x 2 f 2 1 cos 13 2 2


x ( )3

2 f x 02

π isin π gt

Resposta b


Resposta b

8 ( )y 2 sen 3x=

2p

3

π=


[ ]Im 2 2= minus

Resposta b

9 Temos

f 12


983101 1rArr

rArr senω

2 983101 sen π

2 rArr ω 983101 π

f(x)983101 sen(ωx)983101 sen(πx)

f 16

983101 sen π6

983101 12

f

1

4 983101

sen

π

4 983101

2

2

1

2 983083

2

2 983083

1983101

3983083 2

2

f 12

983101 sen π2

983101 1

Resposta a

10 21 20

45 5 5 5



5

π eacute

5

π

Resposta a






Resposta e

12x983101

0rArr

f(0)983101

sen 0983101

0 e g(0)983101

cos 0983101

1


k 983101 1rArr k 9831012

bull g π


4 983101 0rArr mπ

4 983101 π


nisin ℤ rArr

n983101 0 mπ

4 983101 π

2 rArr m983101 2

Resposta b

13 1 sen t 1

3

π minus le le rArr

1 3 3 sen t 1 3

3


2 3 sen t 43

π rArr le + le

Resposta e


= 8 sen983218 x minus 5

f miacuten


f maacutex


Resposta b








minus minus

= =+ +

1 2 1 2 1 2

2 2 2 26 3 32 3

3 3 333

minus minusminus

= = = =+

+

( )1 2 3 3 6

6 6

minus minus= =

Resposta e

sen16 x minus π


7 = sen 3π

2 + k middot 2π

kisin ℤ

Daiacute

x minus π

7 = 3π


7 + 3π


rArr

x=

2π + 21π

14 +

kmiddot

2π

rArr

x=

23π

14 +

kmiddot

2π

kisin ℤ

Resposta c



nativa b




)= 3 sen π

2 = 3

Resposta a

18 2

p3

π=

Resposta b





k 24π= α = β = e




Resposta a


1=cosx




b) 1 cos 3 1 1 0+ π = minus =




Resposta b





= π

f(0)= sen 0= 0 f π


rarr B

II g(x)= sen |x|


2 = 1


2 =

C

= minussenπ


h π2


Resposta d

22 bull yR = y

Q = 1

3 rArr 1

3 = cos x

1+ cos x rArr


xisin [0π[

xQ= π

3



P = π

2


RQ) middot OR2 =

=

π

2 + π

3 middot 1

3

2 = 5π

36

Resposta d


bull minus 41π4

= minus 40π4

minus π


4

5 voltas

sec minus 40π4

= sec minus π

4 = 1

cos minus π

4

= 1

cos π

4

=

= 22

= 2

bull log2 256= log

2 2

8 = 8

6 middot 22


Resposta d

24 ( ) [ ]f x sen x h Im 2 0= + = minus




Resposta c

25 2π|c|

= π2



= a+ b middot 1=

= a+ b e f miacuten


Devemos tera+ b= 9

andash

b=

ndash

7

rArr a= 1 e b = 8



Resposta a






x = π




siderado (F)

II f 9π10

gt g 9π10

pois f 9π10

gt 0 e g 9π10

lt 0 (V)

27


Resposta d

28 a) Se x cos 2x 1 cos 1 22


e pelo

graacutefico f 02

π

=


2


f 02

π =

c) Se

xx 2 cos 2 2 cos 2 2 2

2 2 4


graacutefico f 02

π =

3d) Se x 0 cos 2x 1 0

2 2

3 5 Se x cos 2x 1 1

4 4 4



de acordo com

o graacutefico

e) Se

xx cos 1 cos 1 1

2 2


( )f 0π =

Resposta d




29 Observe que



bull f π

4 = sen 2π

4 + 5 cos 2π

4 = 1

rArr

rArr f ndash π4

= sen ndash2π4

+ 5 cos ndash 2π4

= minus1


f (x+ π)= sen (2x+ 2π)+ 5 cos (2x+ 2π)=



Resposta c

30

x

y

2ππndashπ

ndash ndash

1 2

2

16

ndash2


0 π

2

π

2

3π

2

3π

2


Resposta c




18 a) cos 1 560983216 983101 cos (

1

440983216 983083 120deg)

360983216 983223 4

983101 cos 120983216 983101

983101 991251 cos 60983216 983101 99125112

b) cos 1

035983216 983101 cos (720983216 983083 315983216) 983101 cos 315983216 983101

983101 cos 45983216 983101 22


983101 991251 32

d) cos 22π3

983101 cos 6π 983083 4π3

983101 cos 4π3

983101 991251 cos π3

983101

983101 99125112

e) cos(991251270983216) 983101 cos 90983216 983101 0

19 cos 9π2

983101 cos 4π 983083 π2

983101 cos π2

983101 0 sen 9π2

983101 sen π2

983101 1

cos 17π4 983101 cos 4π 983083 π4 983101 cos π4 983101 22

sen 17π4 983101

983101 sen π4

983101 22

y 983101 0 991251 1

22

983083 3 98322322

983101 9912511

2 2 983101 991251 2

4

20 x isin π 3π2

rArr 9912511 ⩽ cos x ⩽ 0rArr 9912511 ⩽ 2m5

⩽ 0rArr

rArr 9912515 ⩽ 2m ⩽ 0rArr 99125152

⩽ m ⩽ 0


rArr ndash43


49

⩽ m ⩽ 89


ndash

49

89

eacute o zero

22 p 983101 2π Im 983101 [ndash2 2]

3π

2

π

2

ndash2

2

y

x0 2ππ

23 p 983101 2π Im 983101 [1 3]

3

2

1

y

x0π 2π3π

2

π

2

24

1

1

y

x0

2π

3π 4ππ

p 983101 4π

Im 983101 [9912511 1]

25 p 983101 2π3

Im 983101 [ndash1 1]

1

1

y

x0 2π

3

π

2

π

3

π

6

26

0 x

3

y

6

3

2

2

3

2

1

p 983101 2π3

Im 983101 [1 3]

27 a) cos π

6 983083 κ 983223 2π 983101 cos π

6 983101 3

2 (F)

b) cos π

2 983083 κπ 983101 0 (V)

P

P

c) f miacuten

983101 2 983223 (9912511) 983101 9912512 (F)



e) p 983101 2π

π

8 983101 16 (V)

28 a) 2

010 rArr f(0) 983101 400 983083 18 983223 cos 0 983101 418

2 015 rArr f(5) 983101 400 983083 18 983223 cos 5π3

983101

983101 400 983083 18 983223

1

2 983101 409




2

020 rArr f(10) 983101 400 983083 18 983223 cos 10π3

983101

983101 400 983083 18 983223 cos 4π3

983101 400 983083 18 983223 ndash

12

983101

983101 400 991251 9 983101 391



f miacuten



cos π

3 x 983101 cos (π 983083 κ 983223 2π)

ndash 1 κ isin ℤ

Isto eacute

π


κ 983101 0rArr x 983101 3 middot (1 983083 0) 983101 3 (2

013)

κ 983101 1rArr x 983101 3 middot (1 983083 2) 983101 9 (2

019)

κ 983101 2rArr x 983101 3 middot (1 983083 4) 983101 15 (2

025)

κ 983101 3rArr x 983101 3 middot (1 983083 6) 983101 21 (2



menor valor


eacute ℝ desse



b) tg x 991251 π6

983101 0rArr x 991251 π6


κ 983101 0 x 983101 π

6

κ 983101 1 x 983101 π 983083 π

6 983101 7π

6

κ 983101 2 x 983101 π6 983083 2π 983101 13π6

e assim por diante

30 a) f(x) 983101 1 983083 tg x

x ne π2

983083 k π k isin 983130

D 983101 x isin ℝ | x 860697 π2

983083 k π k isin 983130


b) f(x) 983101 2 tg x

x 860697 π

2 983083 k π k isin 983130

D 983101 x isin ℝ | x 860697 π2

983083 k π k isin 983130

p 983101 π

c) f(x) 983101 tg 2x

2x 860697 π2

983083 k π rArr x 860697 π4

983083 k π2

k isin 983130

D 983101 x isin ℝ | x 860697 π4

983083 k π2

k isin 983130


p 983101 π

2

d) f(x) 983101 tg x 983083 π

6

x 983083 π

6 1057305 π

2 983083 k π rArr x 1057305 π

2 ndash π

6 983083 k π rArr

rArr x 1057305 π

3 983083 k π k isin 983130

D 983101 x isin ℝ | x 1057305 π

3 983083 k π k isin 983130


6 ⩽ π rArr ndash

π


6 rArr

rArr ndash

π

6 ⩽ x ⩽ 5π

6 rArr p 983101 5π

6 ndash ndash π

6 rArr p 983101 π

31 a) tg x 991251 π


2 ne π

2 983083 k π




k 983101 0 x 983101 π k 983101 1 x 983101 2π k 983101 2 x 983101 3π etc




b) p 983101




0 ⩽ t⩽ π rArr 0 ⩽ x991251 π2

⩽ π rArrπ

2 ⩽ x⩽ 3π

2 e o periacuteodo

de f eacute 3π

2 991251 π

2 983101 π

c)

0

ndash1

1

y

4

2

ndash x

4ndash

2ndash

32 -

Periacuteodo

t 983101 x2


variar entre 0 e π

0 ⩽ t ⩽ π rArr 0 ⩽ x2


eacute 2π

-Domiacutenio

tg x2


ne π

2 983083 κπ rArr





-Graacutefico

1

0

y

x

ndash1

2

ndash

2

ndash




2

3π

2

ndash1

1

yy = sen x

x

0

π

2

2ππ

3π

2

ndash1

1

y

x0 π

22ππ

y = |sen x|

p = π Im = [0 1]

3(3o ) (1o ) (2o )

x 2x ndash π4


π8 0 2

3π8

π2 1

5π8 π 0

7π8

3π2 1

9π8 2π 2

periacuteodo = 9π8

ndash π8

= π Im = [02]

y

2

1

0 xπ

8

3π

8

5π

8

7π

8

9π

8

4(3o ) (1o ) (2o )


π3 0 1

π2

π2 ndash1

2π3 π 1

5π6

3π2 3

π 2π 1


= 2π3

Im = [ndash1 3]

y

3

1

ndash1

0 xπ

3

π

2

2π

3

5π

6

π

5

x

y = cos xy

1

ndash1

0 π

2

3π

2

π

2π

x

y = |cos x|y

1

ndash1

0 π

23π

2π 2π

x

y

1

ndash1

0

π

2

3π

2

π 2π



2




Podemos tercosx = 1

2 rArr x = π

3 ou x = 5π

3

cosx = ndash 12

rArr x = 2π3

ou x = 4π3






f(1) = f(3) = f(5) = f(7) = hellip

Sim p = 2


8π3

= 2π

8π3




6




ne π2

+ k π


+ k π k isin ℤ


x = 5π4


A

B

π

4


+ k π k isin ℤ


eacute ℝ


o que ocorre

para mne2




⩽ m ⩽ 1


fmaacutex


cos x = ndash1


π + k 983223 2π k isin ℤ

cos

P


6 rArr


6 = 50 rArr sen

(t + 3)π6

= 12


satildeo


P

30deg

Q

1

2

Daiacute

ou

(t + 3)π6

= π6

+ k 983223 2π 1

(t + 3)π6

= 5π6

+ k 983223 2π 2



Para k = 0








1


6 = 1

2 podemos ter




voltas a π6

e a 5π6

(t + 3)π6

= π6

+ 2π ou(t + 3)π

6 = 5π

6 + 2π

t = 10 (novembro)ou









real


sen π4

cong 4 middot 14

ndash 4 middot 14

2

= 1 ndash 416

= 34

bull erro = 34

ndash 22

= 34

ndash 1412

= 3 ndash 2824

=

= |0045| = 0045


Daiacute Pmaacutex

= 96 + 18 middot 1 = 114


Daiacute Pmiacuten

= 96 + 18 middot (ndash1) = 78 (V)


= 1 s (V)π

d) P 13

= 96 + 18 cos 2π3



= 96 ndash 9 = 87 (F)




rArr 2t = 1 + 2k rArr t = 12

+ k


t = 32

t = 52

hellip P(t) eacute

miacutenimo




para cosπ3 x ndash

4π3 = 1 hArr

π3 x ndash


Daiacute x3

ndash 43

= 2k k isin ℤ

cos

x = 2k + 43

middot 3 k isin ℤ


k = 0 rArr x = 4

k = 1 rArr x = 10






24 a) ℝ

b) [ndash1 1]

c) 2π


y

1

2ππ0 xπ

23π

2ndash1

ndashπ

2

(1)


y

1


ndash1ndashndash

π

2

3π

2

(2)



y

1

2π

π

0 xπ

2

π

2

3π

2

3π

2ndash2π

ndash ndash

ndash1

ndashπ


3 π x2


dArrx = 2π

3 k isin 2 x = 2 ()

k = 1 rArr x = 23

lt 1

k = 2 rArr x = 43

rArr P 43

0


Q (2 0)

k = 4 rArr x = 43

rArr R 83

0

k = 5 rArr x = 103

rArr S 103

0

26 a) p = 2π|2π|

= 1



⩽ 1 + 1 hArr 0 ⩽ f(x) ⩽ 2

Assim Im = [0 2]







983101 0




983101 cos 2π3

983101 minus 12


983101 cos 3π4

983101 minus 2

2


k983101 11rArr cos 11π12



cos




Resposta b

3 y 1 2 sen x= +


Resposta d





Resposta b


minus π4

983101 π


Resposta e

6 ( )x

f x 1 cos3

= +

( ) [ ]x

1 cos 1 0 f x 2 Im 0 23


2 p 6

1

3

π= = π

( )3

f 3 1 cos 1 1 03

ππ = + = minus =

3 3 f 1 cos 1 cos 1 0 1

2 6 2

π π π = + = + = + =

0

5 5 f 1 cos 1

2 12gt

π π = + gt

3 3 para x f 12 2

π π = =

( )2 1 1

para x 2 f 2 1 cos 13 2 2


x ( )3

2 f x 02

π isin π gt

Resposta b


Resposta b

8 ( )y 2 sen 3x=

2p

3

π=


[ ]Im 2 2= minus

Resposta b

9 Temos

f 12


983101 1rArr

rArr senω

2 983101 sen π

2 rArr ω 983101 π

f(x)983101 sen(ωx)983101 sen(πx)

f 16

983101 sen π6

983101 12

f

1

4 983101

sen

π

4 983101

2

2

1

2 983083

2

2 983083

1983101

3983083 2

2

f 12

983101 sen π2

983101 1

Resposta a

10 21 20

45 5 5 5



5

π eacute

5

π

Resposta a






Resposta e

12x983101

0rArr

f(0)983101

sen 0983101

0 e g(0)983101

cos 0983101

1


k 983101 1rArr k 9831012

bull g π


4 983101 0rArr mπ

4 983101 π


nisin ℤ rArr

n983101 0 mπ

4 983101 π

2 rArr m983101 2

Resposta b

13 1 sen t 1

3

π minus le le rArr

1 3 3 sen t 1 3

3


2 3 sen t 43

π rArr le + le

Resposta e


= 8 sen983218 x minus 5

f miacuten


f maacutex


Resposta b








minus minus

= =+ +

1 2 1 2 1 2

2 2 2 26 3 32 3

3 3 333

minus minusminus

= = = =+

+

( )1 2 3 3 6

6 6

minus minus= =

Resposta e

sen16 x minus π


7 = sen 3π

2 + k middot 2π

kisin ℤ

Daiacute

x minus π

7 = 3π


7 + 3π


rArr

x=

2π + 21π

14 +

kmiddot

2π

rArr

x=

23π

14 +

kmiddot

2π

kisin ℤ

Resposta c



nativa b




)= 3 sen π

2 = 3

Resposta a

18 2

p3

π=

Resposta b





k 24π= α = β = e




Resposta a


1=cosx




b) 1 cos 3 1 1 0+ π = minus =




Resposta b





= π

f(0)= sen 0= 0 f π


rarr B

II g(x)= sen |x|


2 = 1


2 =

C

= minussenπ


h π2


Resposta d

22 bull yR = y

Q = 1

3 rArr 1

3 = cos x

1+ cos x rArr


xisin [0π[

xQ= π

3



P = π

2


RQ) middot OR2 =

=

π

2 + π

3 middot 1

3

2 = 5π

36

Resposta d


bull minus 41π4

= minus 40π4

minus π


4

5 voltas

sec minus 40π4

= sec minus π

4 = 1

cos minus π

4

= 1

cos π

4

=

= 22

= 2

bull log2 256= log

2 2

8 = 8

6 middot 22


Resposta d

24 ( ) [ ]f x sen x h Im 2 0= + = minus




Resposta c

25 2π|c|

= π2



= a+ b middot 1=

= a+ b e f miacuten


Devemos tera+ b= 9

andash

b=

ndash

7

rArr a= 1 e b = 8



Resposta a






x = π




siderado (F)

II f 9π10

gt g 9π10

pois f 9π10

gt 0 e g 9π10

lt 0 (V)

27


Resposta d

28 a) Se x cos 2x 1 cos 1 22


e pelo

graacutefico f 02

π

=


2


f 02

π =

c) Se

xx 2 cos 2 2 cos 2 2 2

2 2 4


graacutefico f 02

π =

3d) Se x 0 cos 2x 1 0

2 2

3 5 Se x cos 2x 1 1

4 4 4



de acordo com

o graacutefico

e) Se

xx cos 1 cos 1 1

2 2


( )f 0π =

Resposta d




29 Observe que



bull f π

4 = sen 2π

4 + 5 cos 2π

4 = 1

rArr

rArr f ndash π4

= sen ndash2π4

+ 5 cos ndash 2π4

= minus1


f (x+ π)= sen (2x+ 2π)+ 5 cos (2x+ 2π)=



Resposta c

30

x

y

2ππndashπ

ndash ndash

1 2

2

16

ndash2


0 π

2

π

2

3π

2

3π

2


Resposta c




2

020 rArr f(10) 983101 400 983083 18 983223 cos 10π3

983101

983101 400 983083 18 983223 cos 4π3

983101 400 983083 18 983223 ndash

12

983101

983101 400 991251 9 983101 391



f miacuten



cos π

3 x 983101 cos (π 983083 κ 983223 2π)

ndash 1 κ isin ℤ

Isto eacute

π


κ 983101 0rArr x 983101 3 middot (1 983083 0) 983101 3 (2

013)

κ 983101 1rArr x 983101 3 middot (1 983083 2) 983101 9 (2

019)

κ 983101 2rArr x 983101 3 middot (1 983083 4) 983101 15 (2

025)

κ 983101 3rArr x 983101 3 middot (1 983083 6) 983101 21 (2



menor valor


eacute ℝ desse



b) tg x 991251 π6

983101 0rArr x 991251 π6


κ 983101 0 x 983101 π

6

κ 983101 1 x 983101 π 983083 π

6 983101 7π

6

κ 983101 2 x 983101 π6 983083 2π 983101 13π6

e assim por diante

30 a) f(x) 983101 1 983083 tg x

x ne π2

983083 k π k isin 983130

D 983101 x isin ℝ | x 860697 π2

983083 k π k isin 983130


b) f(x) 983101 2 tg x

x 860697 π

2 983083 k π k isin 983130

D 983101 x isin ℝ | x 860697 π2

983083 k π k isin 983130

p 983101 π

c) f(x) 983101 tg 2x

2x 860697 π2

983083 k π rArr x 860697 π4

983083 k π2

k isin 983130

D 983101 x isin ℝ | x 860697 π4

983083 k π2

k isin 983130


p 983101 π

2

d) f(x) 983101 tg x 983083 π

6

x 983083 π

6 1057305 π

2 983083 k π rArr x 1057305 π

2 ndash π

6 983083 k π rArr

rArr x 1057305 π

3 983083 k π k isin 983130

D 983101 x isin ℝ | x 1057305 π

3 983083 k π k isin 983130


6 ⩽ π rArr ndash

π


6 rArr

rArr ndash

π

6 ⩽ x ⩽ 5π

6 rArr p 983101 5π

6 ndash ndash π

6 rArr p 983101 π

31 a) tg x 991251 π


2 ne π

2 983083 k π




k 983101 0 x 983101 π k 983101 1 x 983101 2π k 983101 2 x 983101 3π etc




b) p 983101




0 ⩽ t⩽ π rArr 0 ⩽ x991251 π2

⩽ π rArrπ

2 ⩽ x⩽ 3π

2 e o periacuteodo

de f eacute 3π

2 991251 π

2 983101 π

c)

0

ndash1

1

y

4

2

ndash x

4ndash

2ndash

32 -

Periacuteodo

t 983101 x2


variar entre 0 e π

0 ⩽ t ⩽ π rArr 0 ⩽ x2


eacute 2π

-Domiacutenio

tg x2


ne π

2 983083 κπ rArr





-Graacutefico

1

0

y

x

ndash1

2

ndash

2

ndash




2

3π

2

ndash1

1

yy = sen x

x

0

π

2

2ππ

3π

2

ndash1

1

y

x0 π

22ππ

y = |sen x|

p = π Im = [0 1]

3(3o ) (1o ) (2o )

x 2x ndash π4


π8 0 2

3π8

π2 1

5π8 π 0

7π8

3π2 1

9π8 2π 2

periacuteodo = 9π8

ndash π8

= π Im = [02]

y

2

1

0 xπ

8

3π

8

5π

8

7π

8

9π

8

4(3o ) (1o ) (2o )


π3 0 1

π2

π2 ndash1

2π3 π 1

5π6

3π2 3

π 2π 1


= 2π3

Im = [ndash1 3]

y

3

1

ndash1

0 xπ

3

π

2

2π

3

5π

6

π

5

x

y = cos xy

1

ndash1

0 π

2

3π

2

π

2π

x

y = |cos x|y

1

ndash1

0 π

23π

2π 2π

x

y

1

ndash1

0

π

2

3π

2

π 2π



2




Podemos tercosx = 1

2 rArr x = π

3 ou x = 5π

3

cosx = ndash 12

rArr x = 2π3

ou x = 4π3






f(1) = f(3) = f(5) = f(7) = hellip

Sim p = 2


8π3

= 2π

8π3




6




ne π2

+ k π


+ k π k isin ℤ


x = 5π4


A

B

π

4


+ k π k isin ℤ


eacute ℝ


o que ocorre

para mne2




⩽ m ⩽ 1


fmaacutex


cos x = ndash1


π + k 983223 2π k isin ℤ

cos

P


6 rArr


6 = 50 rArr sen

(t + 3)π6

= 12


satildeo


P

30deg

Q

1

2

Daiacute

ou

(t + 3)π6

= π6

+ k 983223 2π 1

(t + 3)π6

= 5π6

+ k 983223 2π 2



Para k = 0








1


6 = 1

2 podemos ter




voltas a π6

e a 5π6

(t + 3)π6

= π6

+ 2π ou(t + 3)π

6 = 5π

6 + 2π

t = 10 (novembro)ou









real


sen π4

cong 4 middot 14

ndash 4 middot 14

2

= 1 ndash 416

= 34

bull erro = 34

ndash 22

= 34

ndash 1412

= 3 ndash 2824

=

= |0045| = 0045


Daiacute Pmaacutex

= 96 + 18 middot 1 = 114


Daiacute Pmiacuten

= 96 + 18 middot (ndash1) = 78 (V)


= 1 s (V)π

d) P 13

= 96 + 18 cos 2π3



= 96 ndash 9 = 87 (F)




rArr 2t = 1 + 2k rArr t = 12

+ k


t = 32

t = 52

hellip P(t) eacute

miacutenimo




para cosπ3 x ndash

4π3 = 1 hArr

π3 x ndash


Daiacute x3

ndash 43

= 2k k isin ℤ

cos

x = 2k + 43

middot 3 k isin ℤ


k = 0 rArr x = 4

k = 1 rArr x = 10






24 a) ℝ

b) [ndash1 1]

c) 2π


y

1

2ππ0 xπ

23π

2ndash1

ndashπ

2

(1)


y

1


ndash1ndashndash

π

2

3π

2

(2)



y

1

2π

π

0 xπ

2

π

2

3π

2

3π

2ndash2π

ndash ndash

ndash1

ndashπ


3 π x2


dArrx = 2π

3 k isin 2 x = 2 ()

k = 1 rArr x = 23

lt 1

k = 2 rArr x = 43

rArr P 43

0


Q (2 0)

k = 4 rArr x = 43

rArr R 83

0

k = 5 rArr x = 103

rArr S 103

0

26 a) p = 2π|2π|

= 1



⩽ 1 + 1 hArr 0 ⩽ f(x) ⩽ 2

Assim Im = [0 2]







983101 0




983101 cos 2π3

983101 minus 12


983101 cos 3π4

983101 minus 2

2


k983101 11rArr cos 11π12



cos




Resposta b

3 y 1 2 sen x= +


Resposta d





Resposta b


minus π4

983101 π


Resposta e

6 ( )x

f x 1 cos3

= +

( ) [ ]x

1 cos 1 0 f x 2 Im 0 23


2 p 6

1

3

π= = π

( )3

f 3 1 cos 1 1 03

ππ = + = minus =

3 3 f 1 cos 1 cos 1 0 1

2 6 2

π π π = + = + = + =

0

5 5 f 1 cos 1

2 12gt

π π = + gt

3 3 para x f 12 2

π π = =

( )2 1 1

para x 2 f 2 1 cos 13 2 2


x ( )3

2 f x 02

π isin π gt

Resposta b


Resposta b

8 ( )y 2 sen 3x=

2p

3

π=


[ ]Im 2 2= minus

Resposta b

9 Temos

f 12


983101 1rArr

rArr senω

2 983101 sen π

2 rArr ω 983101 π

f(x)983101 sen(ωx)983101 sen(πx)

f 16

983101 sen π6

983101 12

f

1

4 983101

sen

π

4 983101

2

2

1

2 983083

2

2 983083

1983101

3983083 2

2

f 12

983101 sen π2

983101 1

Resposta a

10 21 20

45 5 5 5



5

π eacute

5

π

Resposta a






Resposta e

12x983101

0rArr

f(0)983101

sen 0983101

0 e g(0)983101

cos 0983101

1


k 983101 1rArr k 9831012

bull g π


4 983101 0rArr mπ

4 983101 π


nisin ℤ rArr

n983101 0 mπ

4 983101 π

2 rArr m983101 2

Resposta b

13 1 sen t 1

3

π minus le le rArr

1 3 3 sen t 1 3

3


2 3 sen t 43

π rArr le + le

Resposta e


= 8 sen983218 x minus 5

f miacuten


f maacutex


Resposta b








minus minus

= =+ +

1 2 1 2 1 2

2 2 2 26 3 32 3

3 3 333

minus minusminus

= = = =+

+

( )1 2 3 3 6

6 6

minus minus= =

Resposta e

sen16 x minus π


7 = sen 3π

2 + k middot 2π

kisin ℤ

Daiacute

x minus π

7 = 3π


7 + 3π


rArr

x=

2π + 21π

14 +

kmiddot

2π

rArr

x=

23π

14 +

kmiddot

2π

kisin ℤ

Resposta c



nativa b




)= 3 sen π

2 = 3

Resposta a

18 2

p3

π=

Resposta b





k 24π= α = β = e




Resposta a


1=cosx




b) 1 cos 3 1 1 0+ π = minus =




Resposta b





= π

f(0)= sen 0= 0 f π


rarr B

II g(x)= sen |x|


2 = 1


2 =

C

= minussenπ


h π2


Resposta d

22 bull yR = y

Q = 1

3 rArr 1

3 = cos x

1+ cos x rArr


xisin [0π[

xQ= π

3



P = π

2


RQ) middot OR2 =

=

π

2 + π

3 middot 1

3

2 = 5π

36

Resposta d


bull minus 41π4

= minus 40π4

minus π


4

5 voltas

sec minus 40π4

= sec minus π

4 = 1

cos minus π

4

= 1

cos π

4

=

= 22

= 2

bull log2 256= log

2 2

8 = 8

6 middot 22


Resposta d

24 ( ) [ ]f x sen x h Im 2 0= + = minus




Resposta c

25 2π|c|

= π2



= a+ b middot 1=

= a+ b e f miacuten


Devemos tera+ b= 9

andash

b=

ndash

7

rArr a= 1 e b = 8



Resposta a






x = π




siderado (F)

II f 9π10

gt g 9π10

pois f 9π10

gt 0 e g 9π10

lt 0 (V)

27


Resposta d

28 a) Se x cos 2x 1 cos 1 22


e pelo

graacutefico f 02

π

=


2


f 02

π =

c) Se

xx 2 cos 2 2 cos 2 2 2

2 2 4


graacutefico f 02

π =

3d) Se x 0 cos 2x 1 0

2 2

3 5 Se x cos 2x 1 1

4 4 4



de acordo com

o graacutefico

e) Se

xx cos 1 cos 1 1

2 2


( )f 0π =

Resposta d




29 Observe que



bull f π

4 = sen 2π

4 + 5 cos 2π

4 = 1

rArr

rArr f ndash π4

= sen ndash2π4

+ 5 cos ndash 2π4

= minus1


f (x+ π)= sen (2x+ 2π)+ 5 cos (2x+ 2π)=



Resposta c

30

x

y

2ππndashπ

ndash ndash

1 2

2

16

ndash2


0 π

2

π

2

3π

2

3π

2


Resposta c




-Graacutefico

1

0

y

x

ndash1

2

ndash

2

ndash




2

3π

2

ndash1

1

yy = sen x

x

0

π

2

2ππ

3π

2

ndash1

1

y

x0 π

22ππ

y = |sen x|

p = π Im = [0 1]

3(3o ) (1o ) (2o )

x 2x ndash π4


π8 0 2

3π8

π2 1

5π8 π 0

7π8

3π2 1

9π8 2π 2

periacuteodo = 9π8

ndash π8

= π Im = [02]

y

2

1

0 xπ

8

3π

8

5π

8

7π

8

9π

8

4(3o ) (1o ) (2o )


π3 0 1

π2

π2 ndash1

2π3 π 1

5π6

3π2 3

π 2π 1


= 2π3

Im = [ndash1 3]

y

3

1

ndash1

0 xπ

3

π

2

2π

3

5π

6

π

5

x

y = cos xy

1

ndash1

0 π

2

3π

2

π

2π

x

y = |cos x|y

1

ndash1

0 π

23π

2π 2π

x

y

1

ndash1

0

π

2

3π

2

π 2π



2




Podemos tercosx = 1

2 rArr x = π

3 ou x = 5π

3

cosx = ndash 12

rArr x = 2π3

ou x = 4π3






f(1) = f(3) = f(5) = f(7) = hellip

Sim p = 2


8π3

= 2π

8π3




6




ne π2

+ k π


+ k π k isin ℤ


x = 5π4


A

B

π

4


+ k π k isin ℤ


eacute ℝ


o que ocorre

para mne2




⩽ m ⩽ 1


fmaacutex


cos x = ndash1


π + k 983223 2π k isin ℤ

cos

P


6 rArr


6 = 50 rArr sen

(t + 3)π6

= 12


satildeo


P

30deg

Q

1

2

Daiacute

ou

(t + 3)π6

= π6

+ k 983223 2π 1

(t + 3)π6

= 5π6

+ k 983223 2π 2



Para k = 0








1


6 = 1

2 podemos ter




voltas a π6

e a 5π6

(t + 3)π6

= π6

+ 2π ou(t + 3)π

6 = 5π

6 + 2π

t = 10 (novembro)ou









real


sen π4

cong 4 middot 14

ndash 4 middot 14

2

= 1 ndash 416

= 34

bull erro = 34

ndash 22

= 34

ndash 1412

= 3 ndash 2824

=

= |0045| = 0045


Daiacute Pmaacutex

= 96 + 18 middot 1 = 114


Daiacute Pmiacuten

= 96 + 18 middot (ndash1) = 78 (V)


= 1 s (V)π

d) P 13

= 96 + 18 cos 2π3



= 96 ndash 9 = 87 (F)




rArr 2t = 1 + 2k rArr t = 12

+ k


t = 32

t = 52

hellip P(t) eacute

miacutenimo




para cosπ3 x ndash

4π3 = 1 hArr

π3 x ndash


Daiacute x3

ndash 43

= 2k k isin ℤ

cos

x = 2k + 43

middot 3 k isin ℤ


k = 0 rArr x = 4

k = 1 rArr x = 10






24 a) ℝ

b) [ndash1 1]

c) 2π


y

1

2ππ0 xπ

23π

2ndash1

ndashπ

2

(1)


y

1


ndash1ndashndash

π

2

3π

2

(2)



y

1

2π

π

0 xπ

2

π

2

3π

2

3π

2ndash2π

ndash ndash

ndash1

ndashπ


3 π x2


dArrx = 2π

3 k isin 2 x = 2 ()

k = 1 rArr x = 23

lt 1

k = 2 rArr x = 43

rArr P 43

0


Q (2 0)

k = 4 rArr x = 43

rArr R 83

0

k = 5 rArr x = 103

rArr S 103

0

26 a) p = 2π|2π|

= 1



⩽ 1 + 1 hArr 0 ⩽ f(x) ⩽ 2

Assim Im = [0 2]







983101 0




983101 cos 2π3

983101 minus 12


983101 cos 3π4

983101 minus 2

2


k983101 11rArr cos 11π12



cos




Resposta b

3 y 1 2 sen x= +


Resposta d





Resposta b


minus π4

983101 π


Resposta e

6 ( )x

f x 1 cos3

= +

( ) [ ]x

1 cos 1 0 f x 2 Im 0 23


2 p 6

1

3

π= = π

( )3

f 3 1 cos 1 1 03

ππ = + = minus =

3 3 f 1 cos 1 cos 1 0 1

2 6 2

π π π = + = + = + =

0

5 5 f 1 cos 1

2 12gt

π π = + gt

3 3 para x f 12 2

π π = =

( )2 1 1

para x 2 f 2 1 cos 13 2 2


x ( )3

2 f x 02

π isin π gt

Resposta b


Resposta b

8 ( )y 2 sen 3x=

2p

3

π=


[ ]Im 2 2= minus

Resposta b

9 Temos

f 12


983101 1rArr

rArr senω

2 983101 sen π

2 rArr ω 983101 π

f(x)983101 sen(ωx)983101 sen(πx)

f 16

983101 sen π6

983101 12

f

1

4 983101

sen

π

4 983101

2

2

1

2 983083

2

2 983083

1983101

3983083 2

2

f 12

983101 sen π2

983101 1

Resposta a

10 21 20

45 5 5 5



5

π eacute

5

π

Resposta a






Resposta e

12x983101

0rArr

f(0)983101

sen 0983101

0 e g(0)983101

cos 0983101

1


k 983101 1rArr k 9831012

bull g π


4 983101 0rArr mπ

4 983101 π


nisin ℤ rArr

n983101 0 mπ

4 983101 π

2 rArr m983101 2

Resposta b

13 1 sen t 1

3

π minus le le rArr

1 3 3 sen t 1 3

3


2 3 sen t 43

π rArr le + le

Resposta e


= 8 sen983218 x minus 5

f miacuten


f maacutex


Resposta b








minus minus

= =+ +

1 2 1 2 1 2

2 2 2 26 3 32 3

3 3 333

minus minusminus

= = = =+

+

( )1 2 3 3 6

6 6

minus minus= =

Resposta e

sen16 x minus π


7 = sen 3π

2 + k middot 2π

kisin ℤ

Daiacute

x minus π

7 = 3π


7 + 3π


rArr

x=

2π + 21π

14 +

kmiddot

2π

rArr

x=

23π

14 +

kmiddot

2π

kisin ℤ

Resposta c



nativa b




)= 3 sen π

2 = 3

Resposta a

18 2

p3

π=

Resposta b





k 24π= α = β = e




Resposta a


1=cosx




b) 1 cos 3 1 1 0+ π = minus =




Resposta b





= π

f(0)= sen 0= 0 f π


rarr B

II g(x)= sen |x|


2 = 1


2 =

C

= minussenπ


h π2


Resposta d

22 bull yR = y

Q = 1

3 rArr 1

3 = cos x

1+ cos x rArr


xisin [0π[

xQ= π

3



P = π

2


RQ) middot OR2 =

=

π

2 + π

3 middot 1

3

2 = 5π

36

Resposta d


bull minus 41π4

= minus 40π4

minus π


4

5 voltas

sec minus 40π4

= sec minus π

4 = 1

cos minus π

4

= 1

cos π

4

=

= 22

= 2

bull log2 256= log

2 2

8 = 8

6 middot 22


Resposta d

24 ( ) [ ]f x sen x h Im 2 0= + = minus




Resposta c

25 2π|c|

= π2



= a+ b middot 1=

= a+ b e f miacuten


Devemos tera+ b= 9

andash

b=

ndash

7

rArr a= 1 e b = 8



Resposta a






x = π




siderado (F)

II f 9π10

gt g 9π10

pois f 9π10

gt 0 e g 9π10

lt 0 (V)

27


Resposta d

28 a) Se x cos 2x 1 cos 1 22


e pelo

graacutefico f 02

π

=


2


f 02

π =

c) Se

xx 2 cos 2 2 cos 2 2 2

2 2 4


graacutefico f 02

π =

3d) Se x 0 cos 2x 1 0

2 2

3 5 Se x cos 2x 1 1

4 4 4



de acordo com

o graacutefico

e) Se

xx cos 1 cos 1 1

2 2


( )f 0π =

Resposta d




29 Observe que



bull f π

4 = sen 2π

4 + 5 cos 2π

4 = 1

rArr

rArr f ndash π4

= sen ndash2π4

+ 5 cos ndash 2π4

= minus1


f (x+ π)= sen (2x+ 2π)+ 5 cos (2x+ 2π)=



Resposta c

30

x

y

2ππndashπ

ndash ndash

1 2

2

16

ndash2


0 π

2

π

2

3π

2

3π

2


Resposta c




Podemos tercosx = 1

2 rArr x = π

3 ou x = 5π

3

cosx = ndash 12

rArr x = 2π3

ou x = 4π3






f(1) = f(3) = f(5) = f(7) = hellip

Sim p = 2


8π3

= 2π

8π3




6




ne π2

+ k π


+ k π k isin ℤ


x = 5π4


A

B

π

4


+ k π k isin ℤ


eacute ℝ


o que ocorre

para mne2




⩽ m ⩽ 1


fmaacutex


cos x = ndash1


π + k 983223 2π k isin ℤ

cos

P


6 rArr


6 = 50 rArr sen

(t + 3)π6

= 12


satildeo


P

30deg

Q

1

2

Daiacute

ou

(t + 3)π6

= π6

+ k 983223 2π 1

(t + 3)π6

= 5π6

+ k 983223 2π 2



Para k = 0








1


6 = 1

2 podemos ter




voltas a π6

e a 5π6

(t + 3)π6

= π6

+ 2π ou(t + 3)π

6 = 5π

6 + 2π

t = 10 (novembro)ou









real


sen π4

cong 4 middot 14

ndash 4 middot 14

2

= 1 ndash 416

= 34

bull erro = 34

ndash 22

= 34

ndash 1412

= 3 ndash 2824

=

= |0045| = 0045


Daiacute Pmaacutex

= 96 + 18 middot 1 = 114


Daiacute Pmiacuten

= 96 + 18 middot (ndash1) = 78 (V)


= 1 s (V)π

d) P 13

= 96 + 18 cos 2π3



= 96 ndash 9 = 87 (F)




rArr 2t = 1 + 2k rArr t = 12

+ k


t = 32

t = 52

hellip P(t) eacute

miacutenimo




para cosπ3 x ndash

4π3 = 1 hArr

π3 x ndash


Daiacute x3

ndash 43

= 2k k isin ℤ

cos

x = 2k + 43

middot 3 k isin ℤ


k = 0 rArr x = 4

k = 1 rArr x = 10






24 a) ℝ

b) [ndash1 1]

c) 2π


y

1

2ππ0 xπ

23π

2ndash1

ndashπ

2

(1)


y

1


ndash1ndashndash

π

2

3π

2

(2)



y

1

2π

π

0 xπ

2

π

2

3π

2

3π

2ndash2π

ndash ndash

ndash1

ndashπ


3 π x2


dArrx = 2π

3 k isin 2 x = 2 ()

k = 1 rArr x = 23

lt 1

k = 2 rArr x = 43

rArr P 43

0


Q (2 0)

k = 4 rArr x = 43

rArr R 83

0

k = 5 rArr x = 103

rArr S 103

0

26 a) p = 2π|2π|

= 1



⩽ 1 + 1 hArr 0 ⩽ f(x) ⩽ 2

Assim Im = [0 2]







983101 0




983101 cos 2π3

983101 minus 12


983101 cos 3π4

983101 minus 2

2


k983101 11rArr cos 11π12



cos




Resposta b

3 y 1 2 sen x= +


Resposta d





Resposta b


minus π4

983101 π


Resposta e

6 ( )x

f x 1 cos3

= +

( ) [ ]x

1 cos 1 0 f x 2 Im 0 23


2 p 6

1

3

π= = π

( )3

f 3 1 cos 1 1 03

ππ = + = minus =

3 3 f 1 cos 1 cos 1 0 1

2 6 2

π π π = + = + = + =

0

5 5 f 1 cos 1

2 12gt

π π = + gt

3 3 para x f 12 2

π π = =

( )2 1 1

para x 2 f 2 1 cos 13 2 2


x ( )3

2 f x 02

π isin π gt

Resposta b


Resposta b

8 ( )y 2 sen 3x=

2p

3

π=


[ ]Im 2 2= minus

Resposta b

9 Temos

f 12


983101 1rArr

rArr senω

2 983101 sen π

2 rArr ω 983101 π

f(x)983101 sen(ωx)983101 sen(πx)

f 16

983101 sen π6

983101 12

f

1

4 983101

sen

π

4 983101

2

2

1

2 983083

2

2 983083

1983101

3983083 2

2

f 12

983101 sen π2

983101 1

Resposta a

10 21 20

45 5 5 5



5

π eacute

5

π

Resposta a






Resposta e

12x983101

0rArr

f(0)983101

sen 0983101

0 e g(0)983101

cos 0983101

1


k 983101 1rArr k 9831012

bull g π


4 983101 0rArr mπ

4 983101 π


nisin ℤ rArr

n983101 0 mπ

4 983101 π

2 rArr m983101 2

Resposta b

13 1 sen t 1

3

π minus le le rArr

1 3 3 sen t 1 3

3


2 3 sen t 43

π rArr le + le

Resposta e


= 8 sen983218 x minus 5

f miacuten


f maacutex


Resposta b








minus minus

= =+ +

1 2 1 2 1 2

2 2 2 26 3 32 3

3 3 333

minus minusminus

= = = =+

+

( )1 2 3 3 6

6 6

minus minus= =

Resposta e

sen16 x minus π


7 = sen 3π

2 + k middot 2π

kisin ℤ

Daiacute

x minus π

7 = 3π


7 + 3π


rArr

x=

2π + 21π

14 +

kmiddot

2π

rArr

x=

23π

14 +

kmiddot

2π

kisin ℤ

Resposta c



nativa b




)= 3 sen π

2 = 3

Resposta a

18 2

p3

π=

Resposta b





k 24π= α = β = e




Resposta a


1=cosx




b) 1 cos 3 1 1 0+ π = minus =




Resposta b





= π

f(0)= sen 0= 0 f π


rarr B

II g(x)= sen |x|


2 = 1


2 =

C

= minussenπ


h π2


Resposta d

22 bull yR = y

Q = 1

3 rArr 1

3 = cos x

1+ cos x rArr


xisin [0π[

xQ= π

3



P = π

2


RQ) middot OR2 =

=

π

2 + π

3 middot 1

3

2 = 5π

36

Resposta d


bull minus 41π4

= minus 40π4

minus π


4

5 voltas

sec minus 40π4

= sec minus π

4 = 1

cos minus π

4

= 1

cos π

4

=

= 22

= 2

bull log2 256= log

2 2

8 = 8

6 middot 22


Resposta d

24 ( ) [ ]f x sen x h Im 2 0= + = minus




Resposta c

25 2π|c|

= π2



= a+ b middot 1=

= a+ b e f miacuten


Devemos tera+ b= 9

andash

b=

ndash

7

rArr a= 1 e b = 8



Resposta a






x = π




siderado (F)

II f 9π10

gt g 9π10

pois f 9π10

gt 0 e g 9π10

lt 0 (V)

27


Resposta d

28 a) Se x cos 2x 1 cos 1 22


e pelo

graacutefico f 02

π

=


2


f 02

π =

c) Se

xx 2 cos 2 2 cos 2 2 2

2 2 4


graacutefico f 02

π =

3d) Se x 0 cos 2x 1 0

2 2

3 5 Se x cos 2x 1 1

4 4 4



de acordo com

o graacutefico

e) Se

xx cos 1 cos 1 1

2 2


( )f 0π =

Resposta d




29 Observe que



bull f π

4 = sen 2π

4 + 5 cos 2π

4 = 1

rArr

rArr f ndash π4

= sen ndash2π4

+ 5 cos ndash 2π4

= minus1


f (x+ π)= sen (2x+ 2π)+ 5 cos (2x+ 2π)=



Resposta c

30

x

y

2ππndashπ

ndash ndash

1 2

2

16

ndash2


0 π

2

π

2

3π

2

3π

2


Resposta c








real


sen π4

cong 4 middot 14

ndash 4 middot 14

2

= 1 ndash 416

= 34

bull erro = 34

ndash 22

= 34

ndash 1412

= 3 ndash 2824

=

= |0045| = 0045


Daiacute Pmaacutex

= 96 + 18 middot 1 = 114


Daiacute Pmiacuten

= 96 + 18 middot (ndash1) = 78 (V)


= 1 s (V)π

d) P 13

= 96 + 18 cos 2π3



= 96 ndash 9 = 87 (F)




rArr 2t = 1 + 2k rArr t = 12

+ k


t = 32

t = 52

hellip P(t) eacute

miacutenimo




para cosπ3 x ndash

4π3 = 1 hArr

π3 x ndash


Daiacute x3

ndash 43

= 2k k isin ℤ

cos

x = 2k + 43

middot 3 k isin ℤ


k = 0 rArr x = 4

k = 1 rArr x = 10






24 a) ℝ

b) [ndash1 1]

c) 2π


y

1

2ππ0 xπ

23π

2ndash1

ndashπ

2

(1)


y

1


ndash1ndashndash

π

2

3π

2

(2)



y

1

2π

π

0 xπ

2

π

2

3π

2

3π

2ndash2π

ndash ndash

ndash1

ndashπ


3 π x2


dArrx = 2π

3 k isin 2 x = 2 ()

k = 1 rArr x = 23

lt 1

k = 2 rArr x = 43

rArr P 43

0


Q (2 0)

k = 4 rArr x = 43

rArr R 83

0

k = 5 rArr x = 103

rArr S 103

0

26 a) p = 2π|2π|

= 1



⩽ 1 + 1 hArr 0 ⩽ f(x) ⩽ 2

Assim Im = [0 2]







983101 0




983101 cos 2π3

983101 minus 12


983101 cos 3π4

983101 minus 2

2


k983101 11rArr cos 11π12



cos




Resposta b

3 y 1 2 sen x= +


Resposta d





Resposta b


minus π4

983101 π


Resposta e

6 ( )x

f x 1 cos3

= +

( ) [ ]x

1 cos 1 0 f x 2 Im 0 23


2 p 6

1

3

π= = π

( )3

f 3 1 cos 1 1 03

ππ = + = minus =

3 3 f 1 cos 1 cos 1 0 1

2 6 2

π π π = + = + = + =

0

5 5 f 1 cos 1

2 12gt

π π = + gt

3 3 para x f 12 2

π π = =

( )2 1 1

para x 2 f 2 1 cos 13 2 2


x ( )3

2 f x 02

π isin π gt

Resposta b


Resposta b

8 ( )y 2 sen 3x=

2p

3

π=


[ ]Im 2 2= minus

Resposta b

9 Temos

f 12


983101 1rArr

rArr senω

2 983101 sen π

2 rArr ω 983101 π

f(x)983101 sen(ωx)983101 sen(πx)

f 16

983101 sen π6

983101 12

f

1

4 983101

sen

π

4 983101

2

2

1

2 983083

2

2 983083

1983101

3983083 2

2

f 12

983101 sen π2

983101 1

Resposta a

10 21 20

45 5 5 5



5

π eacute

5

π

Resposta a






Resposta e

12x983101

0rArr

f(0)983101

sen 0983101

0 e g(0)983101

cos 0983101

1


k 983101 1rArr k 9831012

bull g π


4 983101 0rArr mπ

4 983101 π


nisin ℤ rArr

n983101 0 mπ

4 983101 π

2 rArr m983101 2

Resposta b

13 1 sen t 1

3

π minus le le rArr

1 3 3 sen t 1 3

3


2 3 sen t 43

π rArr le + le

Resposta e


= 8 sen983218 x minus 5

f miacuten


f maacutex


Resposta b








minus minus

= =+ +

1 2 1 2 1 2

2 2 2 26 3 32 3

3 3 333

minus minusminus

= = = =+

+

( )1 2 3 3 6

6 6

minus minus= =

Resposta e

sen16 x minus π


7 = sen 3π

2 + k middot 2π

kisin ℤ

Daiacute

x minus π

7 = 3π


7 + 3π


rArr

x=

2π + 21π

14 +

kmiddot

2π

rArr

x=

23π

14 +

kmiddot

2π

kisin ℤ

Resposta c



nativa b




)= 3 sen π

2 = 3

Resposta a

18 2

p3

π=

Resposta b





k 24π= α = β = e




Resposta a


1=cosx




b) 1 cos 3 1 1 0+ π = minus =




Resposta b





= π

f(0)= sen 0= 0 f π


rarr B

II g(x)= sen |x|


2 = 1


2 =

C

= minussenπ


h π2


Resposta d

22 bull yR = y

Q = 1

3 rArr 1

3 = cos x

1+ cos x rArr


xisin [0π[

xQ= π

3



P = π

2


RQ) middot OR2 =

=

π

2 + π

3 middot 1

3

2 = 5π

36

Resposta d


bull minus 41π4

= minus 40π4

minus π


4

5 voltas

sec minus 40π4

= sec minus π

4 = 1

cos minus π

4

= 1

cos π

4

=

= 22

= 2

bull log2 256= log

2 2

8 = 8

6 middot 22


Resposta d

24 ( ) [ ]f x sen x h Im 2 0= + = minus




Resposta c

25 2π|c|

= π2



= a+ b middot 1=

= a+ b e f miacuten


Devemos tera+ b= 9

andash

b=

ndash

7

rArr a= 1 e b = 8



Resposta a






x = π




siderado (F)

II f 9π10

gt g 9π10

pois f 9π10

gt 0 e g 9π10

lt 0 (V)

27


Resposta d

28 a) Se x cos 2x 1 cos 1 22


e pelo

graacutefico f 02

π

=


2


f 02

π =

c) Se

xx 2 cos 2 2 cos 2 2 2

2 2 4


graacutefico f 02

π =

3d) Se x 0 cos 2x 1 0

2 2

3 5 Se x cos 2x 1 1

4 4 4



de acordo com

o graacutefico

e) Se

xx cos 1 cos 1 1

2 2


( )f 0π =

Resposta d




29 Observe que



bull f π

4 = sen 2π

4 + 5 cos 2π

4 = 1

rArr

rArr f ndash π4

= sen ndash2π4

+ 5 cos ndash 2π4

= minus1


f (x+ π)= sen (2x+ 2π)+ 5 cos (2x+ 2π)=



Resposta c

30

x

y

2ππndashπ

ndash ndash

1 2

2

16

ndash2


0 π

2

π

2

3π

2

3π

2


Resposta c







983101 0




983101 cos 2π3

983101 minus 12


983101 cos 3π4

983101 minus 2

2


k983101 11rArr cos 11π12



cos




Resposta b

3 y 1 2 sen x= +


Resposta d





Resposta b


minus π4

983101 π


Resposta e

6 ( )x

f x 1 cos3

= +

( ) [ ]x

1 cos 1 0 f x 2 Im 0 23


2 p 6

1

3

π= = π

( )3

f 3 1 cos 1 1 03

ππ = + = minus =

3 3 f 1 cos 1 cos 1 0 1

2 6 2

π π π = + = + = + =

0

5 5 f 1 cos 1

2 12gt

π π = + gt

3 3 para x f 12 2

π π = =

( )2 1 1

para x 2 f 2 1 cos 13 2 2


x ( )3

2 f x 02

π isin π gt

Resposta b


Resposta b

8 ( )y 2 sen 3x=

2p

3

π=


[ ]Im 2 2= minus

Resposta b

9 Temos

f 12


983101 1rArr

rArr senω

2 983101 sen π

2 rArr ω 983101 π

f(x)983101 sen(ωx)983101 sen(πx)

f 16

983101 sen π6

983101 12

f

1

4 983101

sen

π

4 983101

2

2

1

2 983083

2

2 983083

1983101

3983083 2

2

f 12

983101 sen π2

983101 1

Resposta a

10 21 20

45 5 5 5



5

π eacute

5

π

Resposta a






Resposta e

12x983101

0rArr

f(0)983101

sen 0983101

0 e g(0)983101

cos 0983101

1


k 983101 1rArr k 9831012

bull g π


4 983101 0rArr mπ

4 983101 π


nisin ℤ rArr

n983101 0 mπ

4 983101 π

2 rArr m983101 2

Resposta b

13 1 sen t 1

3

π minus le le rArr

1 3 3 sen t 1 3

3


2 3 sen t 43

π rArr le + le

Resposta e


= 8 sen983218 x minus 5

f miacuten


f maacutex


Resposta b








minus minus

= =+ +

1 2 1 2 1 2

2 2 2 26 3 32 3

3 3 333

minus minusminus

= = = =+

+

( )1 2 3 3 6

6 6

minus minus= =

Resposta e

sen16 x minus π


7 = sen 3π

2 + k middot 2π

kisin ℤ

Daiacute

x minus π

7 = 3π


7 + 3π


rArr

x=

2π + 21π

14 +

kmiddot

2π

rArr

x=

23π

14 +

kmiddot

2π

kisin ℤ

Resposta c



nativa b




)= 3 sen π

2 = 3

Resposta a

18 2

p3

π=

Resposta b





k 24π= α = β = e




Resposta a


1=cosx




b) 1 cos 3 1 1 0+ π = minus =




Resposta b





= π

f(0)= sen 0= 0 f π


rarr B

II g(x)= sen |x|


2 = 1


2 =

C

= minussenπ


h π2


Resposta d

22 bull yR = y

Q = 1

3 rArr 1

3 = cos x

1+ cos x rArr


xisin [0π[

xQ= π

3



P = π

2


RQ) middot OR2 =

=

π

2 + π

3 middot 1

3

2 = 5π

36

Resposta d


bull minus 41π4

= minus 40π4

minus π


4

5 voltas

sec minus 40π4

= sec minus π

4 = 1

cos minus π

4

= 1

cos π

4

=

= 22

= 2

bull log2 256= log

2 2

8 = 8

6 middot 22


Resposta d

24 ( ) [ ]f x sen x h Im 2 0= + = minus




Resposta c

25 2π|c|

= π2



= a+ b middot 1=

= a+ b e f miacuten


Devemos tera+ b= 9

andash

b=

ndash

7

rArr a= 1 e b = 8



Resposta a






x = π




siderado (F)

II f 9π10

gt g 9π10

pois f 9π10

gt 0 e g 9π10

lt 0 (V)

27


Resposta d

28 a) Se x cos 2x 1 cos 1 22


e pelo

graacutefico f 02

π

=


2


f 02

π =

c) Se

xx 2 cos 2 2 cos 2 2 2

2 2 4


graacutefico f 02

π =

3d) Se x 0 cos 2x 1 0

2 2

3 5 Se x cos 2x 1 1

4 4 4



de acordo com

o graacutefico

e) Se

xx cos 1 cos 1 1

2 2


( )f 0π =

Resposta d




29 Observe que



bull f π

4 = sen 2π

4 + 5 cos 2π

4 = 1

rArr

rArr f ndash π4

= sen ndash2π4

+ 5 cos ndash 2π4

= minus1


f (x+ π)= sen (2x+ 2π)+ 5 cos (2x+ 2π)=



Resposta c

30

x

y

2ππndashπ

ndash ndash

1 2

2

16

ndash2


0 π

2

π

2

3π

2

3π

2


Resposta c






Resposta e

12x983101

0rArr

f(0)983101

sen 0983101

0 e g(0)983101

cos 0983101

1


k 983101 1rArr k 9831012

bull g π


4 983101 0rArr mπ

4 983101 π


nisin ℤ rArr

n983101 0 mπ

4 983101 π

2 rArr m983101 2

Resposta b

13 1 sen t 1

3

π minus le le rArr

1 3 3 sen t 1 3

3


2 3 sen t 43

π rArr le + le

Resposta e


= 8 sen983218 x minus 5

f miacuten


f maacutex


Resposta b








minus minus

= =+ +

1 2 1 2 1 2

2 2 2 26 3 32 3

3 3 333

minus minusminus

= = = =+

+

( )1 2 3 3 6

6 6

minus minus= =

Resposta e

sen16 x minus π


7 = sen 3π

2 + k middot 2π

kisin ℤ

Daiacute

x minus π

7 = 3π


7 + 3π


rArr

x=

2π + 21π

14 +

kmiddot

2π

rArr

x=

23π

14 +

kmiddot

2π

kisin ℤ

Resposta c



nativa b




)= 3 sen π

2 = 3

Resposta a

18 2

p3

π=

Resposta b





k 24π= α = β = e




Resposta a


1=cosx




b) 1 cos 3 1 1 0+ π = minus =




Resposta b





= π

f(0)= sen 0= 0 f π


rarr B

II g(x)= sen |x|


2 = 1


2 =

C

= minussenπ


h π2


Resposta d

22 bull yR = y

Q = 1

3 rArr 1

3 = cos x

1+ cos x rArr


xisin [0π[

xQ= π

3



P = π

2


RQ) middot OR2 =

=

π

2 + π

3 middot 1

3

2 = 5π

36

Resposta d


bull minus 41π4

= minus 40π4

minus π


4

5 voltas

sec minus 40π4

= sec minus π

4 = 1

cos minus π

4

= 1

cos π

4

=

= 22

= 2

bull log2 256= log

2 2

8 = 8

6 middot 22


Resposta d

24 ( ) [ ]f x sen x h Im 2 0= + = minus




Resposta c

25 2π|c|

= π2



= a+ b middot 1=

= a+ b e f miacuten


Devemos tera+ b= 9

andash

b=

ndash

7

rArr a= 1 e b = 8



Resposta a






x = π




siderado (F)

II f 9π10

gt g 9π10

pois f 9π10

gt 0 e g 9π10

lt 0 (V)

27


Resposta d

28 a) Se x cos 2x 1 cos 1 22


e pelo

graacutefico f 02

π

=


2


f 02

π =

c) Se

xx 2 cos 2 2 cos 2 2 2

2 2 4


graacutefico f 02

π =

3d) Se x 0 cos 2x 1 0

2 2

3 5 Se x cos 2x 1 1

4 4 4



de acordo com

o graacutefico

e) Se

xx cos 1 cos 1 1

2 2


( )f 0π =

Resposta d




29 Observe que



bull f π

4 = sen 2π

4 + 5 cos 2π

4 = 1

rArr

rArr f ndash π4

= sen ndash2π4

+ 5 cos ndash 2π4

= minus1


f (x+ π)= sen (2x+ 2π)+ 5 cos (2x+ 2π)=



Resposta c

30

x

y

2ππndashπ

ndash ndash

1 2

2

16

ndash2


0 π

2

π

2

3π

2

3π

2


Resposta c





= π

f(0)= sen 0= 0 f π


rarr B

II g(x)= sen |x|


2 = 1


2 =

C

= minussenπ


h π2


Resposta d

22 bull yR = y

Q = 1

3 rArr 1

3 = cos x

1+ cos x rArr


xisin [0π[

xQ= π

3



P = π

2


RQ) middot OR2 =

=

π

2 + π

3 middot 1

3

2 = 5π

36

Resposta d


bull minus 41π4

= minus 40π4

minus π


4

5 voltas

sec minus 40π4

= sec minus π

4 = 1

cos minus π

4

= 1

cos π

4

=

= 22

= 2

bull log2 256= log

2 2

8 = 8

6 middot 22


Resposta d

24 ( ) [ ]f x sen x h Im 2 0= + = minus




Resposta c

25 2π|c|

= π2



= a+ b middot 1=

= a+ b e f miacuten


Devemos tera+ b= 9

andash

b=

ndash

7

rArr a= 1 e b = 8



Resposta a






x = π




siderado (F)

II f 9π10

gt g 9π10

pois f 9π10

gt 0 e g 9π10

lt 0 (V)

27


Resposta d

28 a) Se x cos 2x 1 cos 1 22


e pelo

graacutefico f 02

π

=


2


f 02

π =

c) Se

xx 2 cos 2 2 cos 2 2 2

2 2 4


graacutefico f 02

π =

3d) Se x 0 cos 2x 1 0

2 2

3 5 Se x cos 2x 1 1

4 4 4



de acordo com

o graacutefico

e) Se

xx cos 1 cos 1 1

2 2


( )f 0π =

Resposta d




29 Observe que



bull f π

4 = sen 2π

4 + 5 cos 2π

4 = 1

rArr

rArr f ndash π4

= sen ndash2π4

+ 5 cos ndash 2π4

= minus1


f (x+ π)= sen (2x+ 2π)+ 5 cos (2x+ 2π)=



Resposta c

30

x

y

2ππndashπ

ndash ndash

1 2

2

16

ndash2


0 π

2

π

2

3π

2

3π

2


Resposta c




29 Observe que



bull f π

4 = sen 2π

4 + 5 cos 2π

4 = 1

rArr

rArr f ndash π4

= sen ndash2π4

+ 5 cos ndash 2π4

= minus1


f (x+ π)= sen (2x+ 2π)+ 5 cos (2x+ 2π)=



Resposta c

30

x

y

2ππndashπ

ndash ndash

1 2

2

16

ndash2


0 π

2

π

2

3π

2

3π

2


Resposta c


Matemática Ciência e aplicações, Vol 2, Cap 4

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Transcript of Matemática Ciência e aplicações, Vol 2, Cap 4