5 Steps to a 5 AP Calculus BC 2019

426 STEP 5. Build Your Test-Taking Confidence

84. The correct answer is (D).

y

x

0 0.5 1.0

(0.5, 1.5)

y = f(x)

(1, 2.324)

1.0

Sincey(0)=1, the graph ofypasses through

the point (0, 1). The slope of the tangent line

at (0, 1) is

dy

dx

∣∣

∣∣

x= 0

=e^0 =1. The equation of the

tangent isy− 1 =1(x−0) ory=x+1. At

x= 0 .5,y= 0. 5 + 1 = 1 .5. Thus, the point

(0.5, 1.5) is your next starting point, and

dy

dx

∣∣

∣

∣x= 0. 5 =e

0. (^5) =√e. The equation of the next
   line isy− 1. 5 =

√

e(x− 0 .5) or

y=

√

e(x− 0 .5)+ 1 .5. Atx=1,

y=

√

e(1− 0 .5)+ 1. 5 ≈ 2 .324.

85. The correct answer is (B).

The property

∫ 0

−k

f(x)dx=−

∫k

0

f(x)dx

implies that the regions bounded by the graph

offand thex-axis are such that one region is

above thex-axis and the other region is below.

The property also implies thatf is an odd

function, which means thatf(−x)=−f(x)or

that the graph off is symmetrical with respect

to the origin. The only graph that satisfies

those conditions is choice (B).

86. The correct answer is (C).

If 2x

dy

dx

− 7 = 1 ⇒

dy

dx

=

8

2 x

=

4

x

.

Integratedy=

4 dx

x

to get

y=4ln|x|+C. Sincey= 4 .5 whenx=3,

4. 5 =4ln3+C⇒ 0. 10555 =C. Thus,

y=4ln|x|+ 0 .10555. Atx= 3 .1,y= 4 .631.

87. The correct answer is (A)
    Step 1. Begin by finding the first non-negative
    value oftsuch thatv(t)=0. To accomplish
    this, use your graphing calculator, set
    y 1 =sin(x^2 +1), and graph.

MAIN RAD EXACT FUNC

F1-

Tools

F2-

Zoom

F3

Trace

F4

Regraph

F5-

Math

F6-

Draw

F7-

Pen

Use the [Zero] function and find the first

non-negative value ofxsuch thaty 1 =0. Note

thatx= 1 .46342.

Step 2. The position function of the particle is

s(t)=

∫

v(t)dt.Sinces(1)=5, we have

∫ 1. 46342

1

v(t)=s(1.46342)−s(1).Using your

calculator, evaluate

∫ 1. 46342

1

sin (x^2 +1)dxand

obtain 0.250325. Therefore,

. 250325 =s(1.46342)−s(1) or
. 250325 =s(1.46342)− 5 .Thus,s(1.46342)
= 5. 250325 ≈ 5 .250.