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
- 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).
- 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.
- 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.
- 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.