Download - 2. PROVA DE SMA301-CALCULO I - Professor Alexandre 29.04 ...conteudo.icmc.usp.br/pessoas/andcarva/sma301/Prova02_A_B_C_D.pdf · 2. PROVA DE SMA301-CALCULO I - Professor Alexandre

2. PROVA DE SMA301-CALCULO I - Professor Alexandre 29.04.2013 - A

NOME: No. USP: NOTA:

Questao Resposta Valor Questao Resposta Valor

01a. (a) (b) (c) (d) (e) 06a. (a) (b) (c) (d) (e)

02a. (a) (b) (c) (d) (e) 07a. (a) (b) (c) (d) (e)

03a. (a) (b) (c) (d) (e) 08a. (a) (b) (c) (d) (e)

04a. (a) (b) (c) (d) (e) 09a. (a) (b) (c) (d) (e)

05a. (a) (b) (c) (d) (e) 10a. (a) (b) (c) (d) (e)

Regras: 1 - UNICA SEM RASURA2 -3 -

01. Questao:x!0

x

4 + 5x- 1

2-px

2 + 4

x!1

x

3 - 1

(x- 1)2

+1-1

+1

02. Questao: f(x) = |x+ 1|-px

2 + 2

f [0, 1] y = 1 y = -1

f x = 1

2

y = 2

f [0, 1] x = 1 x = -1

f [0, 1]

f y = 1 y = -1

03. Questao:

f(x) =

�x+ 1, x ,

|x|+ 1, x .

g(x) =f(x)

|x|.

f Rf (-1, 0] g

f x = 0 g (0,1) g

f [0,1) g x = 0 g

f g R g

04. Questao: f : [a, b] ! R

f f

f M m (f) = [m,M]

c, d 2 [a, b] ⇠ 2 (min{f(c), f(d)},max{f(c), f(d)}) p 2 (c, d) f(p) = ⇠

f p 2 [a, b] f(p) = 0

(ii) (iii)

(i) (ii) (iii) (iv)

(i) (ii) (iv)

05. Questao: f : R ! R f(x) =

�a+ bx, x > 2

3, x = 2

b- ax

2

, x < 2

a, b 2 R

f a- b = 2

b- a = 2 f

a 2 (-1

2

, 0] b 2 [1, 2) f

f

a 2 (-1

2

, 0] b 2 [0, 1) f

06. Questao: f : R \ {0} ! R f(x) = x

1

|x|

x!-1f(x) =

u!0

-

u

u

f

f(0)

f

f

07. Questao:x!1

1

x

= 1

" > 0 � =

�1

2

,

"

2

�0 < |x- 1| < �

��1

x

- 1

�� < "

" > 0 � = {1, 2"} 0 < |x- 1| < �

��1

x

- 1

�� < "

" > 0 � =

�1

2

,

"

2

� ��1

x

- 1

�� < " 0 < |x- 1| < �

" > 0 � = {1, 2"}

��1

x

- 1

�� < " 0 < |x- 1| < �

x!1

1

x

= 1

x!1

+

1

x

= 1

x!1

-

1

x

= -1

08. Questao:x!0

⇣x

2 (3x)1- (x)

⌘

�x

2

�.

1

12

7

6

09. Questao: p(x) = 32x

5 - 80x

4 - 80x

3 + 200x

2 + 18x- 45

p(x)

p(x)

p(x)

p(x)

p(x)

10. Questao: f(x) =

✓2+ (

x

3 - 2

x

2 + 1

)

◆1+ (x2)

f

0(x) =�1+ (x2)

� ⇣2+

⇣x

3-2

x

2+1

⌘⌘ (x2)

f

0(x) =⇣2+

⇣x

3-2

x

2+1

⌘⌘1+ (x2)

1+ (x2)

2+⇣

x

3-2

x

2+1

⌘⇣

x

3-2

x

2+1

⌘x

4+3x

2+4x

(x2+1)2 - 2x (x2)⇣2+

⇣x

3-2

x

2+1

⌘⌘�

f

0(x) =h ⇣

x

2-2

x

2+1

⌘ �1+ (x2)

�+ [1+ (x2)]

⇣2+

⇣x

3-2

x

2+1

⌘⌘i⇣2+

⇣x

3-2

x

2+1

⌘⌘ (x2)

f

0(x) = 6x

(x2+1)2

⇣x

2-2

x

2+1

⌘ �1+ (x2)

� ⇣2+

⇣x

3-2

x

2+1

⌘⌘ (x2)

f

0(x) = -2x (x2)�1+ (x2)

� ⇣2+

⇣x

3-2

x

2+1

⌘⌘ (x2)

2. PROVA DE SMA301-CALCULO I - Professor Alexandre 29.04.2013 - B



01a. (a) (b) (c) (d) (e) 06a. (a) (b) (c) (d) (e)

02a. (a) (b) (c) (d) (e) 07a. (a) (b) (c) (d) (e)

03a. (a) (b) (c) (d) (e) 08a. (a) (b) (c) (d) (e)

04a. (a) (b) (c) (d) (e) 09a. (a) (b) (c) (d) (e)

05a. (a) (b) (c) (d) (e) 10a. (a) (b) (c) (d) (e)


01. Questao:x!0

x

4 + 5x- 1

2-px

2 + 4

x!1

x

3 - 1

(x- 1)2

+1+1

-1


2 + 2

f x = 1

2

y = 2

f [0, 1] y = 1 y = -1

f [0, 1] x = 1 x = -1

f y = 1 y = -1

f [0, 1]

03. Questao:

f(x) =

�x+ 1, x ,

|x|+ 1, x .

g(x) =f(x)

|x|.

f g R g

f [0,1) g x = 0 g

f x = 0 g (0,1) g

f (-1, 0] g

f R


f f

f M m (f) = [m,M]


f p 2 [a, b] f(p) = 0

(i) (ii) (iii) (iv)

(ii) (iii)

(i) (ii) (iv)


�a+ bx, x > 2

3, x = 2

b- ax

2

, x < 2

a, b 2 R

f a- b = 2

b- a = 2 f

a 2 (-1

2

, 0] b 2 [0, 1) f

f

a 2 (-1

2

, 0] b 2 [1, 2) f

06. Questao: f : R \ {0} ! R f(x) = x

1

|x|f

f

f(0)

f

x!-1f(x) =

u!0

-

u

u

07. Questao:x!1

1

x

= 1

" > 0 � =

�1

2

,

"

2

� ��1

x

- 1

�� < " 0 < |x- 1| < �

" > 0 � = {1, 2"} 0 < |x- 1| < �

��1

x

- 1

�� < "

" > 0 � =

�1

2

,

"

2

�0 < |x- 1| < �

��1

x

- 1

�� < "

" > 0 � = {1, 2"}

��1

x

- 1

�� < " 0 < |x- 1| < �

x!1

1

x

= 1

x!1

+

1

x

= 1

x!1

-

1

x

= -1

08. Questao:x!0

⇣x

2 (3x)1- (x)

⌘

�x

2

�.

1

7

6

12


5 - 80x

4 - 80x

3 + 200x

2 + 18x- 45

p(x)

p(x)

p(x)

p(x)

p(x)

10. Questao: f(x) =

✓2+ (

x

3 - 2

x

2 + 1

)

◆1+ (x2)

f

0(x) =�1+ (x2)

� ⇣2+

⇣x

3-2

x

2+1

⌘⌘ (x2)

f

0(x) =h ⇣

x

2-2

x

2+1

⌘ �1+ (x2)

�+ [1+ (x2)]

⇣2+

⇣x

3-2

x

2+1

⌘⌘i⇣2+

⇣x

3-2

x

2+1

⌘⌘ (x2)

f

0(x) =⇣2+

⇣x

3-2

x

2+1

⌘⌘1+ (x2)

1+ (x2)

2+⇣

x

3-2

x

2+1

⌘⇣

x

3-2

x

2+1

⌘x

4+3x

2+4x

(x2+1)2 - 2x (x2)⇣2+

⇣x

3-2

x

2+1

⌘⌘�

f

0(x) = 6x

(x2+1)2

⇣x

2-2

x

2+1

⌘ �1+ (x2)

� ⇣2+

⇣x

3-2

x

2+1

⌘⌘ (x2)

f

0(x) = -2x (x2)�1+ (x2)

� ⇣2+

⇣x

3-2

x

2+1

⌘⌘ (x2)

2. PROVA DE SMA301-CALCULO I - Professor Alexandre 29.04.2013 - C



01a. (a) (b) (c) (d) (e) 06a. (a) (b) (c) (d) (e)

02a. (a) (b) (c) (d) (e) 07a. (a) (b) (c) (d) (e)

03a. (a) (b) (c) (d) (e) 08a. (a) (b) (c) (d) (e)

04a. (a) (b) (c) (d) (e) 09a. (a) (b) (c) (d) (e)

05a. (a) (b) (c) (d) (e) 10a. (a) (b) (c) (d) (e)


01. Questao:

f(x) =

�x+ 1, x ,

|x|+ 1, x .

g(x) =f(x)

|x|.

f g R g

f [0,1) g x = 0 g

f x = 0 g (0,1) g

f (-1, 0] g

f R


f f

f M m (f) = [m,M]


f p 2 [a, b] f(p) = 0

(i) (ii) (iii) (iv)

(ii) (iii)

(i) (ii) (iv)


�a+ bx, x > 2

3, x = 2

b- ax

2

, x < 2

a, b 2 R

f a- b = 2

b- a = 2 f

a 2 (-1

2

, 0] b 2 [0, 1) f

f

a 2 (-1

2

, 0] b 2 [1, 2) f

04. Questao:x!0

x

4 + 5x- 1

2-px

2 + 4

x!1

x

3 - 1

(x- 1)2

+1+1

-1


2 + 2

f x = 1

2

y = 2

f [0, 1] y = 1 y = -1

f [0, 1] x = 1 x = -1

f y = 1 y = -1

f [0, 1]

06. Questao:x!0

⇣x

2 (3x)1- (x)

⌘

�x

2

�.

1

7

6

12


5 - 80x

4 - 80x

3 + 200x

2 + 18x- 45

p(x)

p(x)

p(x)

p(x)

p(x)

08. Questao: f(x) =

✓2+ (

x

3 - 2

x

2 + 1

)

◆1+ (x2)

f

0(x) =�1+ (x2)

� ⇣2+

⇣x

3-2

x

2+1

⌘⌘ (x2)

f

0(x) =h ⇣

x

2-2

x

2+1

⌘ �1+ (x2)

�+ [1+ (x2)]

⇣2+

⇣x

3-2

x

2+1

⌘⌘i⇣2+

⇣x

3-2

x

2+1

⌘⌘ (x2)

f

0(x) =⇣2+

⇣x

3-2

x

2+1

⌘⌘1+ (x2)

1+ (x2)

2+⇣

x

3-2

x

2+1

⌘⇣

x

3-2

x

2+1

⌘x

4+3x

2+4x

(x2+1)2 - 2x (x2)⇣2+

⇣x

3-2

x

2+1

⌘⌘�

f

0(x) = 6x

(x2+1)2

⇣x

2-2

x

2+1

⌘ �1+ (x2)

� ⇣2+

⇣x

3-2

x

2+1

⌘⌘ (x2)

f

0(x) = -2x (x2)�1+ (x2)

� ⇣2+

⇣x

3-2

x

2+1

⌘⌘ (x2)

09. Questao: f : R \ {0} ! R f(x) = x

1

|x|f

f

f(0)

f

x!-1f(x) =

u!0

-

u

u

10. Questao:x!1

1

x

= 1

" > 0 � =

�1

2

,

"

2

� ��1

x

- 1

�� < " 0 < |x- 1| < �

" > 0 � = {1, 2"} 0 < |x- 1| < �

��1

x

- 1

�� < "

" > 0 � =

�1

2

,

"

2

�0 < |x- 1| < �

��1

x

- 1

�� < "

" > 0 � = {1, 2"}

��1

x

- 1

�� < " 0 < |x- 1| < �

x!1

1

x

= 1

x!1

+

1

x

= 1

x!1

-

1

x

= -1

2. PROVA DE SMA301-CALCULO I - Professor Alexandre 29.04.2013 - D



01a. (a) (b) (c) (d) (e) 06a. (a) (b) (c) (d) (e)

02a. (a) (b) (c) (d) (e) 07a. (a) (b) (c) (d) (e)

03a. (a) (b) (c) (d) (e) 08a. (a) (b) (c) (d) (e)

04a. (a) (b) (c) (d) (e) 09a. (a) (b) (c) (d) (e)

05a. (a) (b) (c) (d) (e) 10a. (a) (b) (c) (d) (e)


01. Questao:

f(x) =

�x+ 1, x ,

|x|+ 1, x .

g(x) =f(x)

|x|.

f Rf (-1, 0] g

f x = 0 g (0,1) g

f [0,1) g x = 0 g

f g R g


f f

f M m (f) = [m,M]


f p 2 [a, b] f(p) = 0

(ii) (iii)

(i) (ii) (iii) (iv)

(i) (ii) (iv)

03. Questao:x!0

x

4 + 5x- 1

2-px

2 + 4

x!1

x

3 - 1

(x- 1)2

+1-1

+1


2 + 2

f [0, 1] y = 1 y = -1

f x = 1

2

y = 2

f [0, 1] x = 1 x = -1

f [0, 1]

f y = 1 y = -1

05. Questao:x!0

⇣x

2 (3x)1- (x)

⌘

�x

2

�.

1

12

7

6


5 - 80x

4 - 80x

3 + 200x

2 + 18x- 45

p(x)

p(x)

p(x)

p(x)

p(x)

07. Questao: f(x) =

✓2+ (

x

3 - 2

x

2 + 1

)

◆1+ (x2)

f

0(x) =�1+ (x2)

� ⇣2+

⇣x

3-2

x

2+1

⌘⌘ (x2)

f

0(x) =⇣2+

⇣x

3-2

x

2+1

⌘⌘1+ (x2)

1+ (x2)

2+⇣

x

3-2

x

2+1

⌘⇣

x

3-2

x

2+1

⌘x

4+3x

2+4x

(x2+1)2 - 2x (x2)⇣2+

⇣x

3-2

x

2+1

⌘⌘�

f

0(x) =h ⇣

x

2-2

x

2+1

⌘ �1+ (x2)

�+ [1+ (x2)]

⇣2+

⇣x

3-2

x

2+1

⌘⌘i⇣2+

⇣x

3-2

x

2+1

⌘⌘ (x2)

f

0(x) = 6x

(x2+1)2

⇣x

2-2

x

2+1

⌘ �1+ (x2)

� ⇣2+

⇣x

3-2

x

2+1

⌘⌘ (x2)

f

0(x) = -2x (x2)�1+ (x2)

� ⇣2+

⇣x

3-2

x

2+1

⌘⌘ (x2)


�a+ bx, x > 2

3, x = 2

b- ax

2

, x < 2

a, b 2 R

f a- b = 2

b- a = 2 f

a 2 (-1

2

, 0] b 2 [1, 2) f

f

a 2 (-1

2

, 0] b 2 [0, 1) f

09. Questao: f : R \ {0} ! R f(x) = x

1

|x|

x!-1f(x) =

u!0

-

u

u

f

f(0)

f

f

10. Questao:x!1

1

x

= 1

" > 0 � =

�1

2

,

"

2

�0 < |x- 1| < �

��1

x

- 1

�� < "

" > 0 � = {1, 2"} 0 < |x- 1| < �

��1

x

- 1

�� < "

" > 0 � =

�1

2

,

"

2

� ��1

x

- 1

�� < " 0 < |x- 1| < �

" > 0 � = {1, 2"}

��1

x

- 1

�� < " 0 < |x- 1| < �

x!1

1

x

= 1

x!1

+

1

x

= 1

x!1

-

1

x

= -1

Download - 2. PROVA DE SMA301-CALCULO I - Professor Alexandre 29.04 ...conteudo.icmc.usp.br/pessoas/andcarva/sma301/Prova02_A_B_C_D.pdf · 2. PROVA DE SMA301-CALCULO I - Professor Alexandre

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