5 Steps to a 5 AP Calculus BC 2019

AP-Calculus-BC 2727-MA-Book May 11, 2018 14:23

396 STEP 5. Build Your Test-Taking Confidence

Solutions to BC Practice Exam 1---Section II

86. The correct answer is (C).

[−1.5π,1.5π] by [−1,2]

Using the [Inflection] function on your

calculator, you obtainx=− 2 .08 andx= 2 .08.

Thus, there are two points of inflection on

(−π,π).

87. The correct answer is (A).

f(x)=x^2 exUsing your calculator, you obtain

f(1)≈ 2 .7183 and f′(1)≈ 8 .15485.

Equation of tangent line atx=1:

y− 2. 7183 = 8 .15485(x−1)

y= 8 .15485(x−1)+ 2. 7183

f(1.1)≈ 8 .15485(1. 1 −1)+ 2. 7183

≈ 3. 534.

88. The correct answer is (A).
    The area bounded byy=− 3 x^2 +kx−1 and
    thex-axis, the linesx=1 andx=2is
    A=

∫ 2

1

(

− 3 x^2 +kx− 1

)

dx=−x^3 +

kx^2

2

−x

∣∣

∣∣

2

1

=

(

− 23 +

k

2

22 − 2

)

−

(

− 13 +

k

2

12 − 1

)

=(− 10 + 2 k)−

(

− 2 +

k

2

)

.

Since the area is known to be 5.5,

setA=− 8 +

3

2

k= 5 .5 and solve:

3

2

k= 5. 5 + 8 = 13. 5

⇒

3

2

k= 13. 5 ⇒k=9. Alternatively, you

could also use a TI-89 graphing calculator and

solve the equation

∫ 2

1

(

− 3 x^2 +kx− 1

)

dx= 5. 5

and obtaink=9.

89. The correct answer is (C).
    y=x^2 ;
    dy
    dx

= 2 x

y=−

√

x=−x^1 /^2 ;

dy

dx

=−

1

2

x−^1 /^2 =−

1

2

√

x

Perpendicular tangent lines⇒slopes are

negative reciprocals.

Thus, (2x)

(

−

1

2

√

x

)

=− 1

−

√

x=− 1 ⇒

√

x=1orx= 1.