AP Calculus BC Practice Exam 2 425
- The correct answer is (C) as shown below.
y
x
0 0.1
(0, 1) (0.1, 1.2)
f
l
Sincef′′(0)=3, the graph of fis concave
upward atx=0. Sincef′(0)=2 the slope of
linelis 2. The equation of linelusing
y−y 1 =m(x−x 1 )isy− 1 =2(x−0) or
y= 2 x+1. Atx= 0 .1,y=2(0.1)+ 1 = 1 .2.
Thus,f(0.1)≈ 1 .2.
- The correct answer is (D).
The average value off′(x) on [1, 3] is
f′average=
1
3 − 1
∫ 3
1
f′(x)dx
=
1
2
(
f(3)−f(1)
)
.
- The correct answer is (C).
The velocity of the particle isx′(t)=−8 sin(4t)
andy′(t)=t. The speed of the particle is
|v(t)|=
√(
−8 sin (4t)
) 2
+(t)^2 , and att= 1
the speed of the particle is
√(
−8 sin (4(1))
) 2
+(1)^2 ≈ 6. 136.
- The correct answer is (C).
Sincef′>0 on the interval (−∞,−1),f
is increasing on (−∞,−1). Thus, statement I is
false. Alsof′>0on(−∞, 2) and f′<0on
(2,∞), which implies thatf is increasing on
(−∞, 2) and decreasing on (2,∞), respectively.
Therefore,fhas an absolute maximum atx=2.
Statement II is true. Finally,f′is decreasing
on (−∞,−1) and increasing on (−1, 0), which
meansf′′<0on(−∞,−1) andf′′>0on
(−1, 0). Thus, fis concave down on (−∞,−1)
and concave up on (−1, 0) producing a point
of inflection atx=−1. Statement III is true.
- The correct answer is (A).
∑∞
k= 0
(− 1 )k
π(^2 k)
( 2 k)!
= 1 −
π^2
2!
+
π^4
4!
−
π^6
6!
+···
Note that cosx= 1 −
x^2
2!
+
x^4
4!
−
x^6
6!
+···.
Thus,
∑∞
k= 0
(− 1 )k
π(^2 k)
( 2 k)!
=cosπ =−1.
- The correct answer is (A).
The slope of thetangentline toy=e−^2 x
is
dy
dx
=− 2 e−^2 x. The slope of thenormal
is the negative reciprocal,m=
1
2 e−^2 x
=
e^2 x
2
.
Whenx= 1 .158,m=
e2(1.158)
2
≈ 5 .068.
The slope of the normal line is approximately
5.068.
