AP Calculus AB Practice Exam 1
Name: ______________________ Date: __________
Part A: multiple choice: 29 questions, 62 minutes, no calculator. Part B: multiple choice: 13 questions, 38 minutes, graphing calculator. Part A: free response: 2 questions, 30 minutes, graphing calculator. Part B: free response: 4 questions, 60 minutes, no calculator. steplyclass.com/ap-calculus-ab/practice-exam-1/
Section I, part A: no calculator
- What is the limit of x2 − 9x − 3 as x approaches 3?
- A) 6
- B) 3
- C) 9
- D) 0
- What is the limit of 5x3 − 2x2x3 + 7 as x increases without bound?
- A) 0
- B) infinity
- C) 57
- D) 52
- What is the limit of √x − 2x − 4 as x approaches 4?
- A) 0
- B) 4
- C) 14
- D) 12
- The function f is defined by f(x) = x2 for x < 2 and f(x) = 3x − 2 for x ≥ 2. What is the limit of f(x) as x approaches 2?
- A) 2
- B) 4
- C) 8
- D) 6
- What is the limit of 3x2 + 1x3 − 4 as x increases without bound?
- A) 0
- B) 3
- C) −14
- D) infinity
- If f(x) = 2x2 − 3x, what is the limit of (f(x + h) − f(x))/h as h approaches 0?
- A) 2x − 3
- B) 4x
- C) 0
- D) 4x − 3
- If f(x) = 6x2 − 4x + 9, what is f′(x)?
- A) 6x − 4
- B) 12x + 5
- C) 12x − 4
- D) 2x3 − 2x2 + 9x
- If f(x) = (x2 + 1)(3x − 2), what is f′(x)?
- A) 6x
- B) 9x2 − 4x + 3
- C) 6x2 − 4x + 3
- D) 9x2 + 3
- If f(x) = 3x − 1x + 2, what is f′(x)?
- A) 3
- B) −7/(x + 2)2
- C) 5/(x + 2)2
- D) 7/(x + 2)2
- If f(x) = 2sin x − 4cos x, what is f′(x)?
- A) 2cos x + 4sin x
- B) 2cos x − 4sin x
- C) −2cos x + 4sin x
- D) −2cos x − 4sin x
- If f(x) = (2x + 1)5, what is f′(x)?
- A) 5(2x + 1)4
- B) 10(2x + 1)4
- C) 10(2x + 1)5
- D) 2(2x + 1)4
- For the curve x2 + y2 = 25, what is dy/dx at the point (3, 4)?
- A) 34
- B) −43
- C) −34
- D) 43
- If f(x) = ln(x2 + 1), what is f′(x)?
- A) 1/(x2 + 1)
- B) 2x/(x2 + 1)
- C) 2x
- D) 1/(2x)
- For the curve x2 − y2 = 9, what is the slope of the tangent line at the point (5, 4)?
- A) 45
- B) −54
- C) 54
- D) 1
- The tangent line to f(x) = √x at x = 9 is used to estimate √9.2. What is the estimate?
- A) 3.2
- B) 3.1
- C) 3
- D) 3.033
- What is the limit of sin(3x)/x as x approaches 0?
- A) 3
- B) 0
- C) 1
- D) 13
- The function f(x) = x3 − x satisfies the Mean Value Theorem on [0, 2]. What value of c in (0, 2) does the theorem give?
- A) 1
- B) 43
- C) 2
- D) 2/√3
- At what point does f(x) = −x2 + 4x + 1 have a relative maximum?
- A) (−2, −11)
- B) (2, 5)
- C) (2, −3)
- D) (4, 1)
- At what point does the graph of f(x) = x3 − 9x2 + 24x have a point of inflection?
- A) (3, 18)
- B) (2, 20)
- C) (4, 16)
- D) (0, 0)
- Two positive numbers have a product of 49. What is the smallest possible value of their sum?
- A) 7
- B) 50
- C) 14
- D) 49
- Which is ∫ (6x2 − 4x + 1) dx?
- A) 12x − 4 + C
- B) 6x3 − 4x2 + x + C
- C) 2x3 − 4x2 + x + C
- D) 2x3 − 2x2 + x + C
- What is the value of ∫02(3x2 + 1) dx?
- A) 8
- B) 10
- C) 13
- D) 12
- Which is ∫ 2x(x2 + 1)3 dx?
- A) (x2 + 1)4 + C
- B) x2(x2 + 1)4/4 + C
- C) (x2 + 1)4/4 + C
- D) 3(x2 + 1)2 + C
- What is the derivative with respect to x of ∫1x(t3 + 1) dt?
- A) x3 + 1
- B) x44 + x
- C) 3x2
- D) x3
- The area under y = 9 − x2 from x = 0 to x = 3 is estimated with 3 rectangles of equal width, using right endpoints. What is the estimate?
- A) 22
- B) 8
- C) 18
- D) 13
- Which function y solves dy/dx = 2xy with y(0) = 3?
- A) ex2 + 2
- B) 3ex2
- C) 3e2x
- D) 3x2
- Which is the general solution of dy/dx = 3x2 − 2?
- A) 6x + C
- B) x3 − 2 + C
- C) x3 − 2x + C
- D) 3x3 − 2x + C
- What is the average value of f(x) = x2 on the interval [0, 3]?
- A) 3
- B) 4.5
- C) 9
- D) 27
- What is the area of the region between y = x and y = x2?
- A) 12
- B) 16
- C) 56
- D) 13
Section I, part B: calculator
- The radius of a circle grows at 2 cm per second. How fast is the area growing when the radius is 5 cm, in square cm per second, to the nearest hundredth?
- A) 31.42
- B) 62.83
- C) 78.54
- D) 12.57
- A 10-foot ladder leans against a wall. The top slides down the wall at 2 feet per second. How fast, in feet per second, is the bottom moving away from the wall when the top is 8 feet above the floor?
- A) 2
- B) 1.5
- C) 1.6
- D) 2.67
- Air flows into a balloon at a rate of r(t) = 2t + 10 litres per second. How many litres flow in from t = 2 to t = 6 seconds?
- A) 22
- B) 40
- C) 72
- D) 76
- The function f(x) = −x2 + 3x satisfies the Mean Value Theorem on [−1, 2]. What value of c does the theorem give?
- A) 12
- B) 32
- C) 0
- D) −12
- A rectangular pen with an area of 72 square feet is fenced on three sides; a barn wall is the fourth side. What is the least length of fence needed, in feet?
- A) 30
- B) 36
- C) 24
- D) 18
- At what point does the graph of f(x) = x3 + 3x2 − 2 have a point of inflection?
- A) (0, −2)
- B) (−2, 2)
- C) (1, 2)
- D) (−1, 0)
- What is the value of ∫01(x2 + 1) dx?
- A) 43
- B) 13
- C) 1
- D) 2
- The area under y = 2x + 1 from x = 0 to x = 4 is estimated with 4 rectangles of width 1, using left endpoints. What is the estimate?
- A) 24
- B) 16
- C) 20
- D) 12
- A population P grows by dP/dt = 0.05P, with P(0) = 200. What is P(10), to the nearest hundredth?
- A) 300
- B) 210
- C) 329.74
- D) 543.66
- What is the average value of f(x) = 3x2 − 2x on the interval [1, 3]?
- A) 18
- B) 9
- C) 6
- D) 11
- The region between y = x and y = x2 is turned around the x-axis. What is the volume of the solid, to the nearest hundredth?
- A) 0.42
- B) 0.10
- C) 1.05
- D) 0.13
- What is the area of the region between y = x2 + 1 and y = 2x + 4?
- A) 9
- B) 24
- C) −323
- D) 323
- What is the area of the region between y = x2 − 1 and the x-axis from x = 0 to x = 3?
- A) 6
- B) 203
- C) 223
- D) 8
Section II, part A: calculator
Question 1: Water in a tank (9 points)
Water is pumped into a tank at the rate r(t) = 3t2 + 2t litres per minute, for 0 ≤ t ≤ 3. At the same time water drains out of the tank at a constant rate of 10 litres per minute. At t = 0 the tank holds 50 litres.
(a) How many litres of water are pumped into the tank from t = 0 to t = 3? Show the integral you use.
(b) Is the amount of water in the tank increasing or decreasing at t = 1? Give a reason.
(c) Write an expression for A(t), the litres of water in the tank at time t. Use it to find A(3).
(d) At what time t, for 0 < t < 3, is the amount of water in the tank least? Justify your answer.
Question 2: A particle on a line (9 points)
A particle moves along the x-axis. Its velocity is v(t) = t2 − 4t + 3 feet per second, for 0 ≤ t ≤ 5. At t = 0 the particle is at x = 2.
(a) At what times in 0 < t < 5 is the particle at rest? Show how you find them.
(b) Is the particle speeding up or slowing down at t = 4? Give a reason.
(c) Find the displacement of the particle from t = 0 to t = 5. Show the integral you use.
(d) Find the position of the particle at t = 5 and the total distance it travels from t = 0 to t = 5.
Section II, part B: no calculator
Question 3: Analysing a cubic (9 points)
Let f(x) = x3 − 3x2 − 9x + 5.
(a) Find f′(x).
(b) Find the x-values where f has a relative maximum and a relative minimum. Justify your answers.
(c) On what interval is the graph of f concave up? Give a reason.
(d) Find the absolute minimum value of f on the closed interval [0, 4]. Justify your answer.
Question 4: A differential equation (9 points)
Consider the differential equation dy/dx = 6x2 y.
(a) Write an equation of the tangent line to the solution curve through the point (1, 2). Use it to estimate y(1.1).
(b) Find d²y/dx² at the point (1, 2). Is the estimate in part (a) too big or too small? Give a reason.
(c) Find the particular solution y = f(x) with f(0) = 2. Show your separation of variables.
Question 5: Area and volume (9 points)
Let R be the region between the graph of y = 4x − x2 and the x-axis.
(a) Find the area of R.
(b) R is turned around the x-axis. Find the volume of the solid.
(c) The vertical line x = k divides R into two regions of equal area. Write, but do not solve, an equation for k.
Question 6: An accumulation function (9 points)
Let g(x) = ∫1x(t2 − 4) dt.
(a) Find g′(x).
(b) For 0 < x < 5, at what x-value does g have a relative minimum? Justify your answer.
(c) Find g(3).
(d) Find the limit of g(x)/(x − 1) as x approaches 1.
Answer key
Section I, part A: no calculator
- A
- D
- C
- B
- A
- D
- C
- B
- D
- A
- B
- C
- B
- C
- D
- A
- D
- B
- A
- C
- D
- B
- C
- A
- D
- B
- C
- A
- B
Section I, part B: calculator
- B
- D
- C
- A
- C
- D
- A
- B
- C
- B
- A
- D
- C
Section II
Question 1: Water in a tank (9 points)
(a) How many litres of water are pumped into the tank from t = 0 to t = 3? Show the integral you use.
Scoring guide (2 points):
- 1 point: working: one mark for each right line (Check my steps)
- 1 point: the right answer, finished
(b) Is the amount of water in the tank increasing or decreasing at t = 1? Give a reason.
At t = 1 water comes in at r(1) = 3 + 2 = 5 litres per minute.
It drains out at 10 litres per minute, which is more than 5.
So the amount of water is decreasing at t = 1.
Scoring guide (2 points):
- 1 point: finds r(1) = 5
- 1 point: compares 5 with the drain rate 10 and says decreasing
(c) Write an expression for A(t), the litres of water in the tank at time t. Use it to find A(3).
A(t) = 50 + ∫0t(r(s) − 10) ds.
At t = 3: ∫03r(s) ds = 36 (part (a)), and 10 × 3 = 30 litres drain out.
A(3) = 50 + 36 − 30 = 56 litres.
Scoring guide (3 points):
- 1 point: the integral of r(s) − 10 (in minus out)
- 1 point: the starting 50 litres added
- 1 point: A(3) = 56 litres
(d) At what time t, for 0 < t < 3, is the amount of water in the tank least? Justify your answer.
A′(t) = r(t) − 10 = 3t2 + 2t − 10. A′(t) = 0 when t = (−2 + √124)/6 ≈ 1.523 (the other answer is negative).
A′(t) < 0 before 1.523 and A′(t) > 0 after it: the water goes down, then up.
So the amount is least at t ≈ 1.523 minutes.
Scoring guide (2 points):
- 1 point: sets A′(t) = r(t) − 10 = 0 and finds t ≈ 1.523
- 1 point: justifies: A′ changes from negative to positive there
Question 2: A particle on a line (9 points)
(a) At what times in 0 < t < 5 is the particle at rest? Show how you find them.
At rest means v(t) = 0: t2 − 4t + 3 = 0.
(t − 1)(t − 3) = 0, so t = 1 and t = 3 seconds.
Scoring guide (2 points):
- 1 point: sets v(t) = 0
- 1 point: t = 1 and t = 3
(b) Is the particle speeding up or slowing down at t = 4? Give a reason.
v(4) = 16 − 16 + 3 = 3, which is positive.
a(t) = v′(t) = 2t − 4, so a(4) = 4, which is positive.
The velocity and the acceleration have the same sign, so the particle is speeding up.
Scoring guide (2 points):
- 1 point: finds v(4) = 3 and a(4) = 4
- 1 point: speeding up, because v and a have the same sign
(c) Find the displacement of the particle from t = 0 to t = 5. Show the integral you use.
Scoring guide (3 points):
- 1 point: the integral of v(t) from 0 to 5
- 1 point: the antiderivative t33 − 2t2 + 3t
- 1 point: displacement 203 feet
(d) Find the position of the particle at t = 5 and the total distance it travels from t = 0 to t = 5.
Position: x(5) = 2 + 203 = 263 feet.
Distance: the particle turns at t = 1 and t = 3. ∫01v dt = 43, ∫13v dt = −43, ∫35v dt = 20/3.
Total distance = 43 + 43 + 203 = 283 feet.
Scoring guide (2 points):
- 1 point: x(5) = 263
- 1 point: total distance 283 (the part from 1 to 3 counted as positive)
Question 3: Analysing a cubic (9 points)
(a) Find f′(x).
Scoring guide (2 points):
- 1 point: working: one mark for each right line (Check my steps)
- 1 point: the right answer, finished
(b) Find the x-values where f has a relative maximum and a relative minimum. Justify your answers.
f′(x) = 3x2 − 6x − 9 = 3(x − 3)(x + 1) = 0, so x = −1 or x = 3.
f′ changes from positive to negative at x = −1: a relative maximum.
f′ changes from negative to positive at x = 3: a relative minimum.
Scoring guide (3 points):
- 1 point: x = −1 and x = 3 from f′(x) = 0
- 1 point: relative maximum at x = −1, with the sign change of f′
- 1 point: relative minimum at x = 3, with the sign change of f′
(c) On what interval is the graph of f concave up? Give a reason.
f″(x) = 6x − 6.
f″(x) > 0 when x > 1, so the graph is concave up for x > 1.
Scoring guide (2 points):
- 1 point: f″(x) = 6x − 6
- 1 point: concave up for x > 1, because f″ > 0 there
(d) Find the absolute minimum value of f on the closed interval [0, 4]. Justify your answer.
Check the ends and the critical point inside: f(0) = 5, f(3) = 27 − 27 − 27 + 5 = −22, f(4) = 64 − 48 − 36 + 5 = −15.
The smallest is −22, so the absolute minimum is −22, at x = 3.
Scoring guide (2 points):
- 1 point: compares f(0), f(3) and f(4)
- 1 point: absolute minimum −22
Question 4: A differential equation (9 points)
(a) Write an equation of the tangent line to the solution curve through the point (1, 2). Use it to estimate y(1.1).
At (1, 2): dy/dx = 6 × 12 × 2 = 12.
Tangent line: y − 2 = 12(x − 1).
At x = 1.1: y ≈ 2 + 12 × 0.1 = 3.2.
Scoring guide (2 points):
- 1 point: slope 12 and the tangent line
- 1 point: estimate 3.2
(b) Find d²y/dx² at the point (1, 2). Is the estimate in part (a) too big or too small? Give a reason.
d²y/dx² = 12xy + 6x2 (dy/dx) = 12xy + 6x2 (6x2 y).
At (1, 2): 12 × 2 + 6 × 12 = 24 + 72 = 96, which is positive.
The curve is concave up there, so the tangent line lies below it: the estimate is too small.
Scoring guide (2 points):
- 1 point: d²y/dx² = 96 at (1, 2)
- 1 point: too small, because the curve is concave up
(c) Find the particular solution y = f(x) with f(0) = 2. Show your separation of variables.
Scoring guide (5 points):
- 1 point: separates the variables: dy/y = 6x2 dx
- 1 point: ln|y| on the left
- 1 point: 2x3 + C on the right
- 1 point: uses y(0) = 2 to find the constant
- 1 point: y = 2e2x3
Question 5: Area and volume (9 points)
(a) Find the area of R.
Scoring guide (3 points):
- 1 point: the curve meets the axis at x = 0 and x = 4
- 1 point: the integral of 4x − x2 from 0 to 4
- 1 point: area 323
(b) R is turned around the x-axis. Find the volume of the solid.
Scoring guide (3 points):
- 2 points: working: one mark for each right line (Check my steps)
- 1 point: the right answer, finished
(c) The vertical line x = k divides R into two regions of equal area. Write, but do not solve, an equation for k.
The whole area is 323, so each part has area 16/3.
∫0k(4x − x2) dx = 16/3.
Scoring guide (3 points):
- 1 point: the integral of 4x − x2 from 0 to k (or from k to 4)
- 1 point: uses half the area, 163
- 1 point: a complete equation
Question 6: An accumulation function (9 points)
(a) Find g′(x).
Scoring guide (2 points):
- 1 point: working: one mark for each right line (Check my steps)
- 1 point: the right answer, finished
(b) For 0 < x < 5, at what x-value does g have a relative minimum? Justify your answer.
g′(x) = x2 − 4 = 0 when x = 2 (x = −2 is not in the interval).
g′ changes from negative to positive at x = 2, so g has a relative minimum at x = 2.
Scoring guide (2 points):
- 1 point: g′(x) = 0 at x = 2
- 1 point: the sign change of g′ from negative to positive
(c) Find g(3).
Scoring guide (2 points):
- 1 point: working: one mark for each right line (Check my steps)
- 1 point: the right answer, finished
(d) Find the limit of g(x)/(x − 1) as x approaches 1.
g(1) = ∫ from 1 to 1 = 0, so the limit has the form 0/0.
L'Hôpital's rule: the limit = g′(1)/1 = (1 − 4)/1.
The limit is −3.
Scoring guide (3 points):
- 1 point: notes the form 00 (g(1) = 0)
- 1 point: uses L'Hôpital's rule with g′(x)
- 1 point: the limit −3
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