Name: ________________________ Class: ___________________ Date: __________
ID: A
AP Practice Test - Calculator Part Multiple Choice Identify the choice that best completes the statement or answers the question. 4. A spherical tank contains 81.637 gallons of water at time t = 0 minutes. For the next 6 minutes, water flows out of the tank at the rate 9sin( t + 1 ) gallons per minute. How many gallons of water are in the tank at the end of the 6 minutes? a. 36.606 b. 45.031 c. 68.858 d. 77.355 e. 126.668
1. A particle moves along the x-axis so that at any time t ≥ 0 its velocity is given by v(t) = t 2 ln(t + 2). What is the acceleration of the particle at t = 6? a. 20.453 b. 29.453 c. 1.500 d. 133.084 e. 74.860
∫
2. If
∫
3
0
5
0
a. b. c. d. e.
f(x) dx = 4 and
∫
5
3
f(x) dx = 6 , then
(3 + 7f(x)) dx = 85 70 100 10 73
3. For t ≥ 0 hours, H is a differentiable function of t that gives the temperature, in degrees Celsius, at an Artic weather station. Which is the best interpretation of H ′(24)? a. The change in temperature during the first day b. The change in temperature during the 24th hour c. The average rate at which the temperature changed during the 24th hour d. The rate at which the temperature is changing during the first day e. The rate at which the temperature is changing at the end of the 24th hour
1
Name: ________________________
ID: A 7. If f is a continuous funtion on the closed interval [a,b], which of the following must be true? a. There is a number c in the open interval (a,b) such that f(c) = 0. b. There is a number c in the open interval (a,b) such that f(a) < f(c) < f(b). c. There is a number c in the closed interval [a,b] such that f(c) ≥ f(x) for all x in [a,b]. d. There is a number c in the open interval (a,b) such that f ′(c) = 0. e. There is a number c in the open interval f(b) − f(a) . (a,b) such that f ′(c) = b−a
5.
8.
A left Riemann sum, a right Reimann sum, and a trapezoidal sum are used to approximate the value of
∫
x 2.5 2.8 3.0 3.1 f(x) 31.25 39.20 45 48.05 The function f is differentiable and has values as shown in the table above. Both f and f ′ are strictly increasing on the interval 0 ≤ x ≤ 5. Which of the following could be the value of f ′(3)? a. 20 b. 27.5 c. 29 d. 30 e. 30.5
1
0
f(x) dx , each using the same
number of intervals. The graph of the function f is shown in the figure above. Which of the sums give an underestimate of the value of
∫
1
0
f(x) dx ?
I. Left sum II. Right sum III. Trapezoidal Sum a. I only b. II only c. III only d. I and III only e. II and III only 6. The first derivative of the function f is given by −sin(2x) f ′(x) = x − 4e . How many points of inflection does the graph of f have on the interval 0 < x < 2π ? a. Three b. Four c. Five d. Six e. Seven
2
Name: ________________________
9.
ID: A
10. The base of a loudspeaker is determined by x2 x2 the two curves y = and y = − for 1 ≤ x ≤ 4 10 10 as shown in the figure above. For this loudspeaker, the cross sections perpendicular to the x-axis are squares. what is the volume of the loudspeaker, in cubic units? a. 2.046 b. 4.092 c. 4.200 d. 8.184 e. 25.711
The graph of f ′, the derivate of the function f , is shown above. On which of the following intervals is f decreasing? a. [2, 4] only b. [3, 5] only c. [0, 1] and [3, 5] d. [2, 4] and [6, 7] e. [0, 2] and [4, 6]
11.
x 3 4 5 6 7 f(x) 20 17 12 16 20 The function f is continuous and differentiable on the closed interval [3, 7]. The table above gives selected values of f on this interval. Which of the following must be true? I. The minimum value of f on [3, 7] is 12. II. There exists a c, for 3 < x < 7, such that f ′(c) = 0 III. f ′(x) > 0 for 5 < x < 7 a. I only b. II only c. III only d. I and III only e. I, II, III
3
Name: ________________________
12.
ID: A 15. The functions f and g are differentiable. For all x, f(g(x)) = x and g(f(x)) = x . If f(3) = 8 and f ′(3) = 9, what are the values of g ′(8) ? 1 1 a. g(8) = and g ′(8) = − 3 9 1 1 b. g(8) = and g ′(8) = 3 9 c. g(8) = 3 and g ′(8) = −9 1 d. g(8) = 3 and g ′(8) = − 9 1 e. g(8) = 3 and g ′(8) = 9
The figure above shows the graph of f ′, the derivate of f , on the open interval −7 < x < 7. If f ′ has four zeros on −7 < x < 7, how many maxima does f have on −7 < x < 7 ? a. One b. Two c. Three d. Four e. Five
16. A particle moves along the x-axis so that its velocity at any time t ≥ 0 is given by v(t) = 5te −t − 1. At t = 0, the particle is at position x = 1. What is the total distance traveled by the particle from t = 0 to t = 4 ? a. 0.366 b. 0.542 c. 1.542 d. 1.821 e. 2.821
13. The rate at which water is sprayed on a field of vegtables is given by R(t) = 2 1 + 5t 3 , where t is in minutes and R(t) is in gallons per minute. During the time interval 0 ≤ t ≤ 4, what is the average rate of water flow, in gallons per minute? a. 8.453 b. 13.395 c. 14.691 d. 18.916 e. 35.833 14.
17. Let f be the function with first derivative defined by f ′(x) = sin(x 3 ) for 0 ≤ x ≤ 2 . At what value of x does f attain its maximum value on the closed interval 0 ≤ x ≤ 2 ? a. 0 b. 1.162 c. 1.465 d. 1.845 e. 2
f ′(x) g(x) g ′(x) 1 3 -2 -3 4 The table above gives values for the differentiable functions f and g and their derivatives at x = 1. If h(x) = (2f(x) + 3)(1 + g(x)), then h ′(1) = ? a. -28 b. -16 c. 40 d. 44 e. 47
x
f(x)
4
ID: A
AP Practice Test - Calculator Part Answer Section MULTIPLE CHOICE 1. ANS: B
a(t) = v ′(t) = 2t ln(t + 2) + t 2 a(6) = 2(6) ln(6 + 2) +
1 t+2
62 = 29.453 6+2
PTS: 1 2. ANS: A
∫
5
0
(3 + 7f(x)) dx =
∫
5
0
∫
3dx + 7
5
0
ÊÁ Á È ˘5 f(x) dx = ÍÍÎ (3x) ˙˙˚ + 7 ÁÁÁÁ 0 ÁË
∫
3
0
f(x) dx +
∫
5
3
ˆ˜ ˜ f(x) dx ˜˜˜˜ ˜¯
= 5(3) − 0(3) + 7(4 + 6) = 85 PTS: 1 3. ANS: E The only interpretation of H ′(24) is the rate at which the temperature changes at the end of the 24th hour. PTS: 1 4. ANS: A
81.637 −
∫
6
0
9sin( t + 1)dx = 36.606
PTS: 1
1
ID: A 5. ANS: D
I. A left sum is an underestimate. II. A right sum is an overestimate. III. A trapezoidal sum is an underestimate. PTS: 1 6. ANS: B
The second derivative has 4 places where it changes sign on the interval 0 < x < 2π PTS: 1
2
ID: A 7. ANS: C Feedback A B C D E
We would have to know if f(x) changes sign on the interval for this to be true We would have to know that f(b) > f(a) for this to be true Yes We would have to know that the derivative exsists on (a,b) and that f(a) = f(b) for this to be true We would have to know that the derivative exists on (a,b) for this to be true
PTS: 1 8. ANS: D By the mean value theorem:
f ′(c) =
f(3.0) − f(2.8) 45 − 39.2 = = 29 3.0 − 2.8 .2
where 2.8 < c < 3.0 And f ′(c) =
f(3.1) − f(3.0) 48.05 − 45 = = 30.5 .1 3.1 − 3.0
where 3.0 < c < 3.1 So 29 < f ′(3) < 30.5 PTS: 1 9. ANS: E Function is increasing when f ′(x) < 0 this occurs [0,2] and [4,6] PTS: 1 10. ANS: D ÊÁ 2 ÊÁ 2 ˆ˜ ˆ˜ 2 4 Á Á x ˜ ˜˜ ÁÁ x Á − ÁÁÁÁ ˜˜˜˜ ˜˜˜ dx = Á 1 Á Á 10 ÁË 10 ˜¯ ˜˜ Ë ¯
∫
∫
4
1
È ˘4 ÊÁ 2 ˆ˜ 2 4 Ê ÁÁ x 4 ˆ˜˜ ÍÍÍ 1 x 5 ˙˙˙ ÁÁ x ˜˜ ÁÁ ˜˜ dx = ∫ ÁÁÁ ˜˜˜ = ÍÍÍ ˙˙˙ = 1024 − 1 = 1023 == 8.184 ÁÁ 5 ˜˜ Á ˜ Í ˙˙ 25 5 25 125 125 125 1 Á ˜ Í ˙˚ Ë ¯ Ë ¯ ÍÎ 1
PTS: 1 11. ANS: B I. We do not know this. We would have to know that f ′(5) = 0. II. By the mean value theorem we know this must be true. III. We do not know this. This would be true only if f was increasing. PTS: 1
3
ID: A 12. ANS: A There is only one place where f ′ goes from + to - through 0 so there is only one maxima. PTS: 1 13. ANS: C
∫
4
0
R(t)dt =
∫
4
0
2 1 + 5t 3 dt ≈ 14.691
PTS: 1 14. ANS: D Using the product property and the chain rule d (2f(x) + 3) = 2f ′(x) dx
d (1 + g(x)) = g ′(x) dx and d(uv) du dv = (v) + (u) dx dx dx h ′(x) = (2f ′(x))(1 + g(x)) + (2f(x) + 3)(g' (x)) h' (1) = (2f ′(1))(1 + g(1)) + (2f(1) + 3)(g ′(1)) = (2(−2))(1 + (−3)) + (2(3) + 3)(4) = (−4)(−2) + (6 + 3)(4) = 8 + 9(4) = 8 + 36 = 44 PTS: 1 15. ANS: E g(8) = g(f(3)) = 3
Since g(f(x)) = x
d (f(g(x)) = f ′(g(x))g ′(x) dx Since f(g(x)) = x then
Using the chain rule
d d (f(g(x))) = (x) = 1 dx dx
So d (f(g(x))) = 1 = f ′(g(x))g ′(x) dx 1 = f ′(g(8))g' (8) = f ′(3)g ′(8) = 9g ′(8) So g' (8) =
1 9
PTS: 1
4
ID: A 16. ANS: D b 4 distance=∫ || v(t) ||dt = ∫ || 5te −t − 1 || dt = 1.821 a 0
PTS: 1 17. ANS: C f ′(x) = cos(x 3 )
f ′(x) = sin(x 3 ) = 0 when sin(x 3 ) = 0 2x 2 = 0 when x = 0, sin(x 3 ) = 0 when x 3 = 0, π f ″(x) = 3x 2 cos(x 3 ) At x = 0 f ″(x) = 0, for x < 0 f ′(x) < 0, and for x > 0 f ′(x) > 0 so x = 0 is a minimum 2/3 At x 3 = π , f ″(x) = 3(π ) cos (π ) < 0 so x = is a maximum
x = (π ) ≈ 1.464591887 3
PTS: 1
5
