AP Calculus BC Practice Exam 1 391
- The correct answer is (B).
∫ 20
0
v(t)dt=
1
2
(40)(5)+
1
2
(10)(−20)
+
1
2
(5)(20)= 50
- The correct answer is (D).
Letu= 5 x;du= 5 dxor
du
5
=dx
∫
k(5x)dx=
1
5
∫
k(u)du=
1
5
h(u)+C
=
1
5
h(5x)+C
∫ 1
− 1
k(5x)dx=
1
5
h(5x)
] 1
− 1
=
1
5
h(5)−
1
5
h(−5).
- The correct answer is (D).
v(t)=s′(t)=
t^2
2
−t+1 anda(t)=t−1.
a(t)
V(t)
[ [
0 14
0
decreasing increasing
rel. min.
–++++–––
Seta(t)= 0 ⇒t=1. Thus,v(t) has a relative
minimum att=1 andv(1)=
1
2
.Since it is the
only relative extremum, it is an absolute
minimum. And, sincev(t) is continuous on
the closed interval [0, 4],v(t) has an absolute
maximum at the endpoints
v(0)=1 andv(4)= 8 − 4 + 1 =5.
Therefore, the maximum velocity of the
particle on [1, 4] is 5.
- The correct answer is (D).
x= 3 t^2 − 2 ⇒
dx
dt
= 6 tand
y= 2 t^3 + 2 ⇒
dy
dt
= 6 t^2 , and since
dy
dx
=
(
dy
dt
)/(
dx
dt
)
dy
dx
∣∣
∣∣
t= 1
=
6 t^2
6 t
∣∣
∣∣
t= 1
=1 is the slope of the tangent
line. Att=1, x=1, y=4, so the point of
tangency is (1, 4). Equation of tangent:
y− 4 =1(x−1)⇒y=x+3.
- The correct answer is (A).
A possible graph of f
x
y
(1,–2)
(–1,2)
–1 0 1
–1
1
2
–2
Ifb=0, then 0 is a root and thus,r=0; but
r<0. Ifb=1 or 2, then the graph of fmust
cross thex-axis, which implies there is another
rootr, and thatr>0. Thus,b=−1.
