These are original Marlbridge practice questions, not past exam questions. The rubrics are a suggested Marlbridge rubric to help you check your work; they are not an official scoring guideline. Assumptions for every question: no calculator, exact answers, and every function called f or r is continuous. Notation: ∫ (a to x) f(t) dt means the definite integral of f(t) from t = a to t = x. The context in Question 6 and all its data are invented. This set is for both Calculus AB and Calculus BC students.
Question 1 (multiple choice · foundation)
A function f is continuous on [−4, 6]. You are told that f(t) > 0 for −4 < t < −1, f(t) < 0 for −1 < t < 3 and f(t) > 0 for 3 < t < 6. Also, f is decreasing on (−4, 1) and increasing on (1, 6). Let g(x) = ∫ (0 to x) f(t) dt. At which value of x in the open interval (−4, 6) does g have a relative maximum?
- (A) x = −1
- (B) x = 1
- (C) x = 3
- (D) both x = −1 and x = 3
Answer and explanation
Answer: (A). g′ = f. At x = −1, f changes from positive to negative, so g changes from increasing to decreasing: a relative maximum.
- (B) is where f has its minimum, so g has a point of inflection there, not an extremum.
- (C) is a relative minimum: f changes from negative to positive at 3.
- (D) treats every zero of f as a maximum. The direction of the sign change decides the type.
Question 2 (multiple choice · core)
Let g(x) = ∫ (0 to x) (t² − 6t + 8) dt. On which interval is g both decreasing and concave down?
- (A) (2, 3)
- (B) (3, 4)
- (C) (2, 4)
- (D) (−∞, 3)
Answer and explanation
Answer: (A). g′(x) = x² − 6x + 8 = (x − 2)(x − 4), which is negative on (2, 4), so g is decreasing there. g″(x) = 2x − 6, which is negative for x < 3, so g is concave down there. Both hold on (2, 3).
- (B) is where g is decreasing but concave up (g″ > 0 for x > 3).
- (C) is where g is decreasing; it ignores the concavity condition.
- (D) is where g is concave down; it ignores the decreasing condition (g is increasing for x < 2).
Question 3 (multiple choice · core)
Let g(x) = ∫ (0 to x) (t − 1)²(t + 2) dt. How many relative extrema does g have?
- (A) 0
- (B) 1
- (C) 2
- (D) 4
Answer and explanation
Answer: (B). g′(x) = (x − 1)²(x + 2). The factor (x − 1)² is never negative, so the sign of g′ is the sign of x + 2, except that g′(1) = 0. g′ changes from negative to positive at x = −2 (a relative minimum). At x = 1, g′ is positive on both sides, so there is no extremum there.
- (A) misses the sign change at x = −2.
- (C) counts both zeros of g′ without checking for a sign change.
- (D) also counts the zeros of g″ = 3(x − 1)(x + 1). Those are points of inflection, not extrema.
Question 4 (multiple choice · core)
A function f is continuous on [0, 10]. Its graph crosses the t-axis only at t = 3 and t = 7. The graph and the t-axis enclose three regions: from 0 to 3 the region lies above the axis and has area 5; from 3 to 7 it lies below and has area 8; from 7 to 10 it lies above and has area 6. Let g(x) = ∫ (0 to x) f(t) dt. What is the absolute maximum value of g on [0, 10]?
- (A) 3
- (B) 5
- (C) 11
- (D) 19
Answer and explanation
Answer: (B). Candidates: endpoints 0 and 10 and the sign changes at 3 and 7. g(0) = 0, g(3) = 5, g(7) = 5 − 8 = −3, g(10) = −3 + 6 = 3. The largest is 5, at x = 3.
- (A) is g(10). The endpoint is not the maximum, because g lost 8 units between 3 and 7.
- (C) adds the two positive areas, 5 + 6, and ignores the negative region in between.
- (D) adds all three areas as positive: the total area, not a value of g.
Question 5 (graph · core)
The graph of the continuous function f on −4 ≤ t ≤ 4 is shown above. Let g(x) = ∫ (0 to x) f(t) dt.
(a) Find g(−4), g(−2) and g(4). (b) Find the intervals on which g is increasing, and the x-value of each relative extremum of g in (−4, 4). Justify. (c) Find the x-values of the points of inflection of the graph of g. Give a reason. (d) Find the absolute maximum and absolute minimum values of g on [−4, 4]. Justify.
Worked solution and suggested Marlbridge rubric
(a) The semicircle has area (1/2)π(2²) = 2π and lies below the axis, so ∫ (−4 to 0) f(t) dt = −2π. Then g(−4) = ∫ (0 to −4) f = −(−2π) = 2π. Half the semicircle lies on [−2, 0], so g(−2) = −(−π) = π. From 0 to 3: a triangle with base 3 and height 2, area 3. From 3 to 4: a triangle below the axis with area 1. So g(4) = 3 − 1 = 2.
(b) f > 0 on (0, 3), so g is increasing on [0, 3]. f < 0 on (−4, 0) and (3, 4), so g is decreasing there. g′ = f changes from negative to positive at x = 0, so g has a relative minimum at x = 0. g′ = f changes from positive to negative at x = 3, so g has a relative maximum at x = 3.
(c) g″ = f′. f is decreasing on (−4, −2), increasing on (−2, 2) and decreasing on (2, 4). f′ changes sign at t = −2 and t = 2, so g has points of inflection at x = −2 and x = 2.
(d) Candidates: g(−4) = 2π ≈ 6.28, g(0) = 0, g(3) = 3, g(4) = 2. Absolute maximum 2π, at x = −4. Absolute minimum 0, at x = 0. The maximum is at an endpoint, because area below the axis left of 0 counts as positive when you move from 0 back to −4.
| Point | What earns it |
|---|---|
| 1 | g(−4) = 2π, g(−2) = π and g(4) = 2, with the sign change from reversed limits shown |
| 1 | g increasing on [0, 3]; relative min at 0 and relative max at 3, each justified by the sign change of g′ = f |
| 1 | Inflection points at x = −2 and x = 2, because f changes between increasing and decreasing there |
| 1 | Candidates test with all four values; absolute max 2π at x = −4 and absolute min 0 at x = 0 |
Question 6 (constructed response · core)
A museum opens at t = 0 and closes at t = 8, where t is in hours. The net rate at which people enter the museum (arrivals minus departures) is R(t) people per hour. R is continuous and strictly decreasing on [0, 8]. Some values are shown.
| t (hours) | 0 | 2 | 4 | 6 | 8 |
|---|---|---|---|---|---|
| R(t) (people per hour) | 90 | 50 | 10 | −40 | −80 |
At t = 0 there are 120 people in the museum. Let P(x) = 120 + ∫ (0 to x) R(t) dt for 0 ≤ x ≤ 8.
(a) Find P′(2) and interpret it in context. (b) At what time is the number of people in the museum greatest? Give an interval of length 2 hours that contains this time, and justify your answer. (c) Is the graph of P concave up or concave down on (0, 8)? Explain. Does P have a point of inflection? (d) Use a trapezoidal sum with the four intervals in the table to estimate P(8).
Worked solution and suggested Marlbridge rubric
(a) P′(x) = R(x), so P′(2) = 50 people per hour. Two hours after opening, the number of people in the museum is increasing at 50 people per hour.
(b) R is continuous with R(4) = 10 > 0 and R(6) = −40 < 0, so by the Intermediate Value Theorem R(c) = 0 for some c in (4, 6). Because R is strictly decreasing, this is the only zero, R > 0 before c and R < 0 after c. So P′ = R changes from positive to negative at c, and c is the only critical point. P increases up to c and decreases after it, so P is greatest at t = c, between 4 and 6 hours.
(c) P′ = R, and R is strictly decreasing, so the graph of P is concave down on (0, 8). P′ never switches to increasing, so P has no point of inflection.
(d) Each interval is 2 hours wide:
2 × (90 + 50)/2 + 2 × (50 + 10)/2 + 2 × (10 − 40)/2 + 2 × (−40 − 80)/2 = 140 + 60 − 30 − 120 = 50
So P(8) ≈ 120 + 50 = 170 people.
| Point | What earns it |
|---|---|
| 1 | P′(2) = 50 with a correct interpretation (rate of change of the number of people, with units and time) |
| 1 | Interval (4, 6) from the IVT, with the change of sign of R from positive to negative |
| 1 | Concave down because P′ = R is decreasing, and no inflection point |
| 1 | Trapezoidal sum 50 and P(8) ≈ 170 |
Question 7 (constructed response · stretch)
Let g(x) = ∫ (0 to x) t(t − 3)² dt for all real x.
(a) Find the critical points of g. (b) Classify each critical point. A student says “g′(3) = 0, so g has a relative maximum or minimum at x = 3”. Explain whether the student is right. (c) Find the x-values of the points of inflection of the graph of g. Justify. (d) Without finding a formula for g, explain why g(x) ≥ 0 for every real x.
Worked solution and suggested Marlbridge rubric
(a) The integrand is a polynomial, so it is continuous and g′(x) = x(x − 3)². This is 0 at x = 0 and x = 3.
(b) (x − 3)² ≥ 0, so the sign of g′ follows the sign of x except at x = 3. For example g′(−1) = −16 and g′(2) = 2. At x = 0, g′ changes from negative to positive: relative minimum. At x = 3, g′ is positive on both sides (g′(2) = 2, g′(4) = 4): no extremum. The student is wrong: g′(3) = 0 gives a horizontal tangent, but an extremum needs a sign change.
(c) g″(x) = (x − 3)² + 2x(x − 3) = (x − 3)(3x − 3) = 3(x − 1)(x − 3). g″ > 0 for x < 1, g″ < 0 for 1 < x < 3, g″ > 0 for x > 3. g″ changes sign at both, so the points of inflection are at x = 1 and x = 3.
(d) g′ ≤ 0 for x ≤ 0, so g is decreasing on (−∞, 0]. g′ ≥ 0 for x ≥ 0, so g is increasing on [0, ∞). So the smallest value of g anywhere is g(0) = ∫ (0 to 0) … = 0. Therefore g(x) ≥ 0 for every x.
| Point | What earns it |
|---|---|
| 1 | g′(x) = x(x − 3)² and critical points x = 0 and x = 3 |
| 1 | Relative minimum at 0, and no extremum at 3 because g′ does not change sign there |
| 1 | g″ = 3(x − 1)(x − 3), with sign changes giving inflection points at 1 and 3 |
| 1 | g decreasing then increasing about x = 0, with g(0) = 0 the minimum, so g ≥ 0 |
Acceptable alternative for (d): for x > 0 the integrand is ≥ 0, so the area is ≥ 0; for x < 0 the integrand is ≤ 0 and the limits run right to left, so the two negatives make the integral ≥ 0.
How did you do?
- Q1, Q3 or Q7(b) wrong: reread “How features of f show up in g” in the study guide. An extremum needs a sign change in f.
- Q2 or Q7(c) wrong: redo Worked example 2: concavity of g comes from f′.
- Q4 or Q5 wrong: redo Worked example 1 and its candidates table. Watch the sign of areas to the left of the starting point.
- Q6 wrong: redo Worked example 3 on tables and descriptions.
Then tick off the topic checklist.