MA 3972-MA-Book April 11, 2018 16:20
392 STEP 5. Build Your Test-Taking Confidence
- The correct answer is (A).
01 3 5
(3, 3)
(3, 1)
(3, –1)
(5, –1)
(1, –2)
y
x
A possible
graph of f
Ifb=0, thenr=3, butrcannot be 3.
Ifb=1, or 3, fwould have more than one
root. Thus, of all the choices, the only possible
value forbis−1.
- The correct answer is (A).
V=
1
3
πr^2 (5)−
1
3
πr^2 (r)
=
5
3
πr^2 −
1
3
πr^3
dV
dr
=
10
3
πr−πr^2
dV
dr
∣∣
∣∣
r= 5
=
10
3
π( 5 )−π( 25 )=
− 25 π
3
- The correct answer is (D).
x^3 −y^2 =1; 3x^2 − 2 y
dy
dx
= 0 ⇒
dy
dx
=
3 x^2
2 y
Atx=1, 1^3 −y^2 = 1 ⇒y= 0 ⇒(1, 0)
dy
dx
∣∣
∣
∣x= 1 =
3
(
12
)
2 ( 0 )
undefined.
- The correct answer is (C).
I. f is differentiable on (a,b) since the
graph is a smooth curve.
II. There exists a horizontal tangent to the
graph on (a,b); thus,f′(k)=0 for somek
on (a,b).
III. The graph is concave downward; thus,
f′′<0.
- The correct answer is (D).
v(t)=t^2 − 3 t−10; setv(t)= 0
⇒(t−5)(t +2)= 0
⇒t=5ort=− 2
a(t)= 2 t−3; seta(t)= 0
⇒ 2 t− 3 =0ort=
3
2
.
V(t)
a(t)
- – – – – – – – – – – – – – – – – – – 0 + + +
- – – – – 0 + + + + + + + + + + + + + + + + +
t
3
2
05
Sincev(t) anda(t) are both negative on
(0, 3/ 2 ), and are both positive on (5,∞), the
particle’s speed is increasing on these intervals.
- The correct answer is (C).
∫ 2
− 2
f(x)dx=
∫ 0
− 2
f(x)dx+
∫ 2
0
f(x)dx
=
1
2
( 2 )(− 1 )+
(
−
(
1
2
)
π( 1 )^2
)
=− 1 −
π
2
- The correct answer is (C).
Average value=
1
π
2
−
(
−
π
2
)
∫π/ 2
−π/ 2
3 cos( 2 x)dx
=
1
π
[
3 sin( 2 x)
2
]π/ 2
−π/ 2
=
3
2 π
[sinπ−(sin[−π])]= 0.
