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

MA 3972-MA-Book April 11, 2018 16:20

390 STEP 5. Build Your Test-Taking Confidence

Solutions to AB Practice Exam 2 Section I

Section I Part A

1. The correct answer is (B).
   ∫ 8

0

x^2 /^3 dx=

x^5 /^3

5 / 3

] 8

0

=

3 x^5 /^3

5

] 8

0

=

3(8)^5 /^3

5

− 0 =

3(32)

5

=

96

5

2. The correct answer is (B).
   The lim
   x→ 0

2 −2 cosx= 2 −2 cos(0)=0, and

lim

x→ 0

x=0. Therefore, lim

x→ 0

2 −2 cosx

x

is an

indeterminate form of

0

0

. ApplyingL'Hôpital's

Rule, you have lim

x→ 0

2 sinx

1

=

2 sin 0

1

=0.

3. The correct answer is (C).
   lim
   x→− 2 +

∣

∣x− 1

∣

∣=

∣

∣− 2 − 1

∣

∣= 3

lim

x→− 2 −

(2x+7)=2(−2)+ 7 = 3

Thus, lim

x→− 2

f(x)= 3.

4. The correct answer is (D).

decr. decr.

decr.

incr.

incr.

ab

0

f′ –

f′

f

f′′

f

+ –

Concave

upward

Concave

downward

+ –

5. The correct answer is (D).∫
   x
   π/ 2

2 costdt=2 sint

]x

π/ 2 =2 sinx−2(1)

=2 sinx− 2

6. The correct answer is (A).
   y= 3 e−^2 x;
   dy
   dx
   = 3 e−^2 x(−2)=− 6 e−^2 x

dy

dx

∣∣

∣

∣x=ln 2 =−^6 e

−2ln2=− 6 (eln 2)−^2

=−6(2)−^2

=− 6

(

1

4

)

=−

3

2

Slope of normal line atx=ln 2 is

2

3

.

Atx=ln 2,y= 3 e−2ln2=

3

4

; point

(

ln 2,

3

4

)

.

Equation of normal line:

y−

3

4

=

2

3

(x −ln 2) ory=

2

3

(x −ln 2)+

3

4

.

7. The correct answer is (B).
   f′(x)=lim
   h→ 0

f(x+h)−f(x)

h

Thus, lim

h→ 0

csc

(

π

4

+h

)

−csc

(

π

4

)

h

=

d(cscx)

dx

∣

∣∣

∣

x=π 4

=−csc

(

π

4

)

cot

(

π

4

)

=−

√

2(1)=−

√

2.

8. The correct answer is (D).
   Letu=x^3 +1,du= 3 x^2 dxor
   du
   3
   =x^2 dx.

f(x)=

∫

x^2

√

x^3 + 1 dx=

∫√

u

du

3

=

1

3

∫

u^1 /^2 du