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, axes drawn with equal scales, and y a differentiable function of x (or C of t) unless stated. Contexts and models are invented. This set is for both Calculus AB and Calculus BC students.
Question 1 (multiple choice · foundation)
Which of the following is a constant solution of dy/dx = (y + 1)(x − 2)?
- (A) y = −1
- (B) y = 1
- (C) x = 2
- (D) y = 2
Answer and explanation
Answer: (A). If y = −1, then dy/dx = 0 because y is constant, and the equation gives (−1 + 1)(x − 2) = 0 for every x. Both sides agree, so y = −1 is a solution.
- (B) solves y + 1 = 0 with the wrong sign. At x = 0, y = 1 gives a slope of (2)(−2) = −4, not 0.
- (C) is where the segments are flat, but x = 2 is a vertical line. It is not a function of x, so it cannot be a solution y(x).
- (D) uses the number from the other factor. At x = 0, y = 2 gives a slope of (3)(−2) = −6.
Question 2 (multiple choice · core)
Let y = f(x) be the solution of dy/dx = (5 − y)/3 with f(0) = 1. What is the limit of f(x) as x → ∞?
- (A) 1
- (B) 3
- (C) 5
- (D) The limit is infinite.
Answer and explanation
Answer: (C). The constant solution is y = 5. For y < 5 the slope (5 − y)/3 is positive, so f increases from 1. As y gets closer to 5 the slope gets closer to 0, so the curve levels off and approaches the line y = 5 without crossing it.
- (A) is the starting value. f does not stay at 1, because the slope there is 4/3, not 0.
- (B) takes the denominator 3 as the level where the curve settles. The slope at y = 3 is 2/3, not 0.
- (D) notices that f is increasing but ignores that the slopes shrink to 0 as y approaches 5.
Question 3 (multiple choice · core)
Let y = g(x) be the solution of dy/dx = x² + y through the point (1, −4). At x = 1, which statement is true?
- (A) g is increasing and concave up.
- (B) g is increasing and concave down.
- (C) g is decreasing and concave up.
- (D) g is decreasing and concave down.
Answer and explanation
Answer: (D). dy/dx = 1² + (−4) = −3 < 0, so g is decreasing. Differentiate implicitly: d²y/dx² = 2x + dy/dx = 2x + x² + y. At (1, −4): 2 + 1 − 4 = −1 < 0, so g is concave down.
- (A) and (B) get the sign of dy/dx wrong, for example by computing x² − y = 5.
- (C) differentiates y as if it were a constant, giving d²y/dx² = 2x = 2 > 0. Because y depends on x, its derivative dy/dx must be included.
Question 4 (multiple choice · core)
A slope field has identical segments in every vertical column. The segments are horizontal along the y-axis, rise for x > 0 and fall for x < 0, and are steeper further from the y-axis. Which could be the general solution of the differential equation it shows?
- (A) y = x²/2 + C
- (B) y = Ceˣ
- (C) y = (x + C)²/2
- (D) y = −x²/2 + C
Answer and explanation
Answer: (A). Each function in (A) has derivative x. That depends on x only (identical columns), is 0 on the y-axis, positive for x > 0, negative for x < 0 and larger in size further out. The members are vertical shifts of one parabola, as expected when dy/dx depends on x only.
- (B) has derivative Ceˣ = y, which depends on y. The rows, not the columns, would be identical.
- (C) has derivative x + C, so different members have different slopes at the same x (and at x = 0 the slope is C, not 0). The members are horizontal shifts, which matches a field that depends on y.
- (D) has derivative −x: falling for x > 0 and rising for x < 0, the opposite of the description.
Question 5 (sketch · core)
The slope field for dy/dx = −x(y − 2)/2 is shown below.
(a) On a copy of the field, sketch the solution curves through (0, 3) and through (0, 1). (b) Let y = h(x) be the solution through (0, 3). Find the limit of h(x) as x → ∞, and explain how the field supports your answer. (c) Show that y = 2 is a solution of the differential equation. (d) Use d²y/dx² to show that h has a relative maximum at x = 0.
Worked solution and suggested Marlbridge rubric
(a)
(b) The limit is 2. Above y = 2, the segments to the right of the y-axis fall, so h decreases for x > 0. As h gets closer to 2 the factor y − 2 shrinks, and the segments near y = 2 are almost flat, so the curve levels off towards the constant solution y = 2 rather than crossing it. (Check, for interest: h(x) = 2 + e^(−x²/4), which tends to 2.)
(c) If y = 2, then dy/dx = 0. The equation gives −x(2 − 2)/2 = 0 for every x. Both sides agree, so y = 2 is a solution.
(d) At (0, 3), dy/dx = −0 × (1)/2 = 0. Differentiate using the product rule, with y a function of x:
d²y/dx² = −(y − 2)/2 − (x/2) · dy/dx.
At (0, 3): d²y/dx² = −(1)/2 − 0 = −1/2 < 0. Since dy/dx = 0 and d²y/dx² < 0 at x = 0, h has a relative maximum there, by the second derivative test.
| Point | What earns it |
|---|---|
| 1 | Both curves pass through their points, follow the segments, and are drawn on both sides of the y-axis |
| 1 | (b) Limit 2, with a reason from the field: falling towards y = 2 and levelling off |
| 1 | (c) Substitutes y = 2 into both sides and shows they agree for all x |
| 1 | (d) Correct d²y/dx² including the dy/dx term, value −1/2 at (0, 3), and the conclusion with dy/dx = 0 |
Acceptable alternative for (d): a first-derivative argument. For y > 2, dy/dx > 0 when x < 0 and dy/dx < 0 when x > 0, so dy/dx changes from positive to negative at x = 0.
Question 6 (constructed response · core)
A medicine is given through a drip. In an invented model, its concentration C (mg/L) in a patient’s blood t hours after the drip starts satisfies
dC/dt = 2 − 0.5C, with C(0) = 0.
(a) Find the constant solution of the differential equation and explain what it means in context. (b) Show that the solution is increasing and concave down while 0 ≤ C < 4. (c) Use the tangent line at t = 0 to estimate C(0.5). Is your estimate too high or too low? Explain. (d) Describe what happens to the concentration in the long run.
Worked solution and suggested Marlbridge rubric
(a) dC/dt = 0 when 2 − 0.5C = 0, so C = 4. If the concentration were 4 mg/L, it would stay at 4 mg/L: the drip adds medicine at the same rate as the body removes it.
(b) For 0 ≤ C < 4, 0.5C < 2, so dC/dt > 0: C is increasing. Differentiate with respect to t: d²C/dt² = −0.5 · dC/dt. Since dC/dt > 0, d²C/dt² < 0: the graph is concave down.
(c) At t = 0, dC/dt = 2 − 0 = 2 mg/L per hour. The tangent line is C ≈ 0 + 2t, so C(0.5) ≈ 1 mg/L. The solution is concave down, so it lies below its tangent line: the estimate is too high. (Check, for interest: the exact solution C = 4(1 − e^(−t/2)) gives C(0.5) ≈ 0.885 mg/L.)
(d) The concentration rises towards 4 mg/L. As C approaches 4, the rate 2 − 0.5C approaches 0, so the increase slows and C levels off just below 4 mg/L without exceeding it.
| Point | What earns it |
|---|---|
| 1 | (a) C = 4 with a correct interpretation in context |
| 1 | (b) Sign of dC/dt from 0 ≤ C < 4 and d²C/dt² = −0.5 dC/dt < 0 |
| 1 | (c) Tangent line C ≈ 2t and estimate 1 mg/L |
| 1 | (c) and (d) “Too high” justified by concave down, and long-run value 4 mg/L with units |
Question 7 (constructed response · stretch)
Consider dy/dx = y − 2x + 1.
(a) Show that y = 2x + 1 is a solution. (b) Let y = k(x) be the solution through (0, 0). Is k increasing or decreasing at x = 0? Is its graph concave up or down there? (c) A student says: “The line y = 2x + 1 is a solution, so every other solution gets closer to it as x increases.” Let D = y − (2x + 1), the vertical distance of a solution above the line. Show that dD/dx = D, and use this to explain why the student is wrong for the solution through (0, 0).
Worked solution and suggested Marlbridge rubric
(a) For y = 2x + 1, dy/dx = 2. The right side is (2x + 1) − 2x + 1 = 2. They agree for every x.
(b) At (0, 0): dy/dx = 0 − 0 + 1 = 1 > 0, so k is increasing. Differentiate: d²y/dx² = dy/dx − 2 = (y − 2x + 1) − 2. At (0, 0): 1 − 2 = −1 < 0, so the graph is concave down.
(c) dD/dx = dy/dx − 2 = (y − 2x + 1) − 2 = y − 2x − 1 = D. For the solution through (0, 0), D starts at 0 − 1 = −1, so the curve starts below the line. While D < 0, dD/dx = D < 0, so D keeps decreasing: it becomes more negative, and its size grows. The curve moves further below the line as x increases, so the student is wrong. The line attracts nothing here: solutions on either side move away from it. (Check, for interest: k(x) = 2x + 1 − eˣ, so D = −eˣ, which grows in size.)
| Point | What earns it |
|---|---|
| 1 | (a) Both sides equal 2 for all x |
| 1 | (b) Increasing, from dy/dx = 1 at (0, 0) |
| 1 | (b) Concave down, from d²y/dx² = dy/dx − 2 = −1 |
| 1 | (c) Shows dD/dx = D |
| 1 | (c) Uses D(0) = −1 < 0 and dD/dx = D < 0 to conclude the distance grows, so the student is wrong |
Acceptable alternative for (c): an argument from the field that points below the line have dy/dx < 2, so the solution rises more slowly than the line and falls further behind, provided it is linked to D < 0.
How did you do?
- Q1 or Q5(c) wrong: reread “Constant solutions” in the study guide: substitute y = c into both sides.
- Q2, Q5(b) or Q6(d) wrong: redo Worked example 1 (c): curves level off towards constant solutions.
- Q3, Q5(d) or Q7(b) wrong: practise implicit differentiation for d²y/dx², remembering that y depends on x.
- Q4 wrong: see “Solutions are functions, and there are many” and the table of vertical and horizontal shifts.
- Q6(c) or Q7(c) wrong: redo Worked example 2: tangent lines, concavity and how curves behave near a straight-line solution.
Then tick off the topic checklist.