5 Steps to a 5 AP Calculus AB 2019 - William Ma

MA 3972-MA-Book April 11, 2018 16:32

AP Calculus AB Practice Exam 1 361

Solutions for AP Calculus AB Practice Exam 1 Section I

Section I Part A

1. The correct answer is (C).
   Note that limx→π(3 cosx+3)=3 cosπ+ 3 =0,
   and limx→π(x−π)=0. Therefore, the

xlim→π

3 cosx+ 3

x−π

is an indeterminate form of the

type

0

0

. ApplyingL’Hôpital’sRule, you have

xlim→π

−3 sinx

1

=

−3 sinπ

1

=0.

2. The correct answer is (D).
   Atx=−4, the graph of fhas a horizontal
   tangent, which implies thatfis differentiable
   and f′(−4)=0.
   Atx=−2,f is discontinuous and thus not
   differentiable. At each ofx=0 andx=2, the
   one-sided derivatives are different, sof is not
   differentiable. (Note that the graphs atx= 0
   andx=2 are sometimes referred to as “corners”.)


3. The correct answer is (A).
   Using theu-substitution method, let
   u=x^2 +1, and obtaindu= 2 xdx,or
   du
   2
   =xdx. Therefore,
   ∫
   x
   (x^2 +1)^2
   dx=

1

2

∫

du

u^2

=

1

2

∫

u−^2 du

=

1

2

(

u−^1

− 1

)

+c=−

1

2 u

+c=−

1

2 (x^2 + 1 )

+c.

4. The correct answer is (C).
   Examining the graph, note that
   lim
   x→ 2 +
   f(x)=lim
   x→ 2 −
   f(x)=3. Since the two
   one-sided limits are equal, limx→ 2 f(x) exists.
   Statement I is true. Also, note that
   xlim→ 0 −f(x)=1 and limx→ 0 + f(x)=2. Therefore,
   statement II is true, but statement III is false

because the two one-sided limits are not the

same.

5. The correct answer is (C).

Recall the formula

d

dx

(logau)=

1

(lna)(u)

·

du

dx

.

In this case,y=10 log 2 (x^4 +1), and you have

dy

dx

=(10)

1

ln(2)·(x^4 +1)

· 4 x^3 =

40 x^3

ln(2)·(x^4 +1)

.

6. The correct answer is (A).
   Letu=x^2 −4. Differentiating, you have
   du
   dx
   = 2 x,or
   du
   2
   =xdx. Usingu-substitution,

rewrite

∫

x(3x^2 −^4 )dxas

1

2

∫

3 udu. Recall the

formula

∫

axdx=

ax

lna

+c. Therefore,

1

2

∫

3 udu=

3 u

2ln3

+c,or

3 x^2 −^4

2ln3

+c.

7. The correct answer is (C).

If a function is increasing, then

dy

dx

>0, and if

a function is decreasing,

dy

dx

<0. Also, if a

function is concave up,

d^2 y

dx^2

>0, and if a

function is concave down,

d^2 y

dx^2

<0.

Note the following:

Point

dy

dx

d^2 y

dx^2

A Decreasing− Concave up+

B Increasing+ Concave up+

C Increasing+ Concave down−

D Decreasing− Concave down−

Thus, at pointC,

dy

dx

>0 and

d^2 y

dx^2

<0.