AP® Exam Practice Questions for Chapter 1 · PDF fileAP® Exam Practice Questions for...

AP® Exam Practice Questions for Chapter 1 1

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AP® Exam Practice Questions for Chapter 1

1. sin sin 0lim 0x

x

xπ

ππ π→

= = =

So, the answer is A.

2.

( ) ( )( )

2

2

2

3 5 7lim

4

3 2 5 2 7

2 4

9

6

3

2

x

x x

x→ −

+ +−

− + − +=

− −

=−

= −

So, the answer is B.

3. Evaluate each statement.

I: As x approaches 3 from the left and right, the function approaches 1. So,

3lim 2 1x

x→

− = is

a true statement.

II: As x approaches 3 from the left and right, the function approaches 0. So, ( )

3lim 6 2 0x

x→

− = is

a true statement.

III: As x approaches 3 from the left, the function approaches 0 and as x approaches 3 from the right, the function approaches 1. So, the limit ( )

3limx

f x→

does not exist is a true statement.

Because I, II, and III are true statements, the answer is D.

4. Evaluate each limit.

I: Using a graphing utility, 3

1

1lim does not exist.

1x

x

x→

+−

II: ( )0 0

lim lim ,x x

xf x

x→ →= where ( ) 1, 0

1, 0

xf x

x

− <= ≥

does not exist because the limits on each side of 0x = do not agree.

III: ( ) ( )2

3, 2lim , where

0, 2x

xf x f x

x→

≤= >

does not exist

because the limits on each side of 2x = do not

agree.

Because the limits of I, II, and III do not exist, the answer is D.

5. The domain of

( ) 2is 1 0 1.

1f x x x

x= − > >

−

Because f is not continuous at 1,x = the answer is C.

6. ( ) ( ) ( ) ( )( ) ( )

( ) ( )

5 5 5

5 5

lim 5 lim 5 lim

5lim lim

5 10 1 49

x x x

x x

f x g x f x g x

f x g x

→ → →

→ →

− = −

= −

= − =

So, the answer is D.


I. Because ( )2

lim 1 andx

g x−→

= ( )2

lim 1,x

g x+→

=

( )2

lim 1.x

g x→

=

The statement is true.

II. ( ) ( )2

lim 1 2 3x

g x g→

= ≠ =

The statement is false.

III. g is continuous at 3.x =

The statement is true.

Because I and III are true, the answer is B.


A: Because ( ) ( )11 1

lim and lim , lim does not exist.xx x

f x f x− + →→ →

= ∞ = −∞ So, ( )1

lim is false.x

f x→

= ∞

B: Because ( ) ( ) ( )3 3 3 3

lim 3 and lim 2, lim lim .x x x x

f x f x f x− + − +→ → → →

= = > So, ( ) ( )3 3

lim lim is false.x x

f x f x− +→ →

<

C: Because ( ) ( ) ( )33 3

lim 3 and lim 2, lim does not exist.xx x

f x f x f x− + →→ →

= = So, ( )3

lim 1 is false.x

f x→

=

D: Because ( ) ( ) ( ) ( )0 3 0 3

lim 2 and lim 2, lim lim .x x x x

f x f x f x f x+ + + +→ → → →

= = = So, ( ) ( )0 3

lim lim is true.x x

f x f x+ +→ →

=


2 AP® Exam Practice Questions for Chapter 1


9. The function ( ) 4

10f x

x= increases without bound as x

approaches 0 from the left and as x approaches 0 from the right. So, the limit is nonexistent.


10. Because 1

1lim 2

1x

x

x−→

− =−

and 1

1lim 2,

1x

x

x+→

− =−

1

1lim 2.

1x

x

x→

− =−

So, the answer is C.

11. (a) ( )( )1 393.1

2 378.4

s

s

=

=

Because s is continuous on [ ]1, 2 and

( ) ( )2 382 1 ,s s< < by the Intermediate Value Theorem

there exists a time t in the open interval ( )1, 2 such that

( ) 382 meters.s t =

(b) ( ) 2

2

4.9 398

0 4.9 398

9.012

s t t

t

t

= − +

= − +≈

The negative solution does not make sense in the context of the problem. So, the object hits the ground after approximately 9.012 seconds.

(c) ( ) ( ) ( ) ( )( )

( )

( )

( )( )

( )( )

22

3 3

2

3

2

3

3

3

4.9 398 4.9 3 3983lim lim

3 3

4.9 4.9 9lim

3

4.9 9lim

3

4.9 3 3lim

3

lim 4.9 3

4.9 3 3

29.4 m sec

t t

t

t

t

t

ts t s

t t

t

t

t

t

t t

t

t

→ →

→

→

→

→

− + − − +−=

− −− +

=−

− −=

−− − +

=−

= − +

= − +

= −

Note: Merely saying “because s is differentiable” or “because s is decreasing” would not earn the justification points.

1 pt: answer with units



12. (a) ( ) ( )1 1 1

lim 4 lim lim 4

2 4

6

x x xf x f x

→ → → + = +

= +=

(b) ( )3

5 5lim 5

1x g x−→= =

(c) ( ) ( ) ( )( )2

lim 2 0 0x

f x g x→

= =

(d) ( )

( )( )( )

( )( )

3 3

3

3

2 6lim lim

1 2 1

2 6lim

3

2 3lim

3

2

x x

x

x

f x x

g x x

x

x

x

x

→ →

→

→

− +=

− − −

− +=−

− −=

−

= −

13. (a) Because ( ) ( )2 2

lim 1 and lim 1,x x

f x f x− +→ →

= = and

the limits are the same, ( )2

lim 1.x

f x→

=

(b) Because ( ) ( )2

2 3, lim 1,x

f f x→

= = and

( ) ( )2

2 lim , x

f f x f→

≠ is not continuous at 2.x =

(c) ( )( ) ( )2 2

lim sin sin lim sin 1x x

f x f x→ →

= =

2 pts: finds limit

1 pt: answer

1 pt: answer

Note: The equations for f and g must both be correct to be eligible for the last three points (i.e., the limit must yield the indeterminant form



14. (a) ( ) ( )( )( )( )

2

2

2 35 6

2 7 3 2 1 3

2, 3

2 1

x xx xf x

x x x x

xx

x

+ ++ += =+ + + +

+= ≠ −+

( )f x has discontinuities at 1

2x = −

and 3.x = −

(b) ( )

( )( )( )( )

( )

2

23 3

3

3

5 6lim lim

2 7 3

2 3lim

2 1 3

2lim

2 13 2

2 3 1

1

5

x x

x

x

x xf x

x x

x x

x x

x

x

→ − → −

→ −

→ −

+ +=+ +

+ +=

+ +

+=+

− +=− +

=

(c) ( )f x has a vertical asymptote at 1

.2

x = −

4 pts: answers with justification (justify where denominator equals zero)

2 pts: answer

Note: Including in the answer (a removable

discontinuity) would lose one of the answer points.



15. (a) Because ( )T x is continuous on [ )0, 10 ,

( ) ( )4

lim 4 172.x

T x T→

= =

(b) ( ) ( )8 3 164 174

8 3 8 310

52

T T− −=− −

−=

= −

The average rate of change is −2°F per minute.

(c) ( )T x is continuous and when 6,x =

( ) 166.5T x > ° and when 8,x =

( ) 166.5 .T x < °

So, the shortest interval is ( )6, 8 .

(d) Because ( )T x is continuous, the average

rate of change for 6 9x≤ ≤ is

( ) ( )9 6 162 168

9 6 9 66

32.

T T− −=− −

−=

= −

So, the tangent line at 8x = has a slope

of about 2.−

1 pt: answer2 pts:

1 pt: justification (appeals to the continuity of )T

1 pt: justification (evidence of a difference quotient)2 pts:

1 pt: answer with units

( ) ( )( )

1 pt: shows 8 166.5 6

places 166.5 in this interval

1 pt: answer (an open or closed interval would 3 pts:

generally be accepted here)

1 pt: justification (appeals to the continuity of or the Intermed

T T

T

< <

iate Value Theorem)

Note: Merely stating “because T is differentiable” or “because T is decreasing” would not earn the justification point.

([ ])

1 pt: justification evidence of a difference quotient

on 6, 92 pts:

1 pt: answer



16. (a) ( )( ) ( ) ( )

2

22 2 2

4 2

f x ax x b

f a b

a b

= + −

= + −

= + −

( )( ) ( )2 2

2

f x ax b

f a b

a b

= +

= +

= +

f is continuous at 2x = when 4 2 2 .a b a b+ − = +

( )( ) ( )5 5

5

f x ax b

f a b

a b

= +

= +

= +

( )( ) ( )

2 7

5 2 5 7

10 7

f x ax

f a

a

= −

= −

= −

f is continuous at 5x = when 5 10 7.a b a+ = −

4 2 2 2 2 2

5 10 7 5 7

a b a b a b

a b a a b

+ − = + − = −+ = − − + = −

Multiply both sides of the second equation by 2.

2 2 2

10 2 14

8 16

2

a b

a b

a

a

− = −− + = −− = −

=

When ( )2, 5 2 7 10 7 3.a b= = − = − =

So, f is continuous when 2 and 3.a b= =

(b) ( ) ( ) ( )3 3

lim lim 2 3 2 3 3 9x x

f x x→ →

= + = + =

(c) ( ) ( )

( )( )

( )( )

1 1

2

1

1

1

lim lim1

2 3lim

1

2 3 1lim

1

lim 2 3

2 1 3

5

x x

x

x

x

f xg x

x

x x

x

x x

x

x

→ →

→

→

→

=−

+ −=−

+ −=

−= +

= +

=

2 pts: answer from using values of a and b from part (a) and the correct piece of f

Note: Incorrect values of a and b may be imported from part (a) as long as a is nonzero (i.e., an answer consistent with values for part (a) can generally earn these two points).

Note: To be eligible for these points, the limit must

yield the indeterminant form using the values of

a and b from part (a) (i.e., incorrect values of a and b generally cannot be imported here).

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