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, angles in radians, C, A and B are constants, and exact answers unless stated. This set is for both Calculus AB and Calculus BC students.
Question 1 (multiple choice · foundation)
Which function is a solution of dy/dx = 3y/x for x > 0?
- (A) y = 3x
- (B) y = x³
- (C) y = e^(3x)
- (D) y = x³ + 1
Answer and explanation
Answer: (B). For y = x³: left side dy/dx = 3x². Right side 3y/x = 3x³/x = 3x². Equal for all x > 0.
- (A) Left side 3; right side 3(3x)/x = 9. Never equal.
- (C) Left side 3e^(3x); right side 3e^(3x)/x. These agree only at x = 1, so a check at x = 1 alone would wrongly accept it.
- (D) Left side 3x²; right side 3(x³ + 1)/x = 3x² + 3/x. The extra 3/x is never 0, so the sides never agree.
Question 2 (multiple choice · core)
For which values of k is y = e^(kx) a solution of y″ + y′ − 6y = 0?
- (A) k = −2 and k = 3
- (B) k = 2 only
- (C) k = 2 and k = −3
- (D) Every real number k
Answer and explanation
Answer: (C). y′ = ke^(kx) and y″ = k²e^(kx). Substituting: e^(kx)(k² + k − 6) = 0. Since e^(kx) is never 0, k² + k − 6 = (k − 2)(k + 3) = 0, so k = 2 or k = −3.
- (A) solves k² − k − 6 = 0. It comes from a sign slip on the y′ term.
- (B) finds one root, perhaps by trial, and misses the other.
- (D) confuses “infinitely many solutions” with “every exponential works”. Only two values of k make the bracket zero.
Question 3 (multiple choice · core)
The function y = 5 − 2e^(−3t) is a solution of which differential equation?
- (A) dy/dt = 3(y − 5)
- (B) dy/dt = −3y
- (C) dy/dt = 5 − y
- (D) dy/dt = 3(5 − y)
Answer and explanation
Answer: (D). dy/dt = −2 × (−3)e^(−3t) = 6e^(−3t). And 5 − y = 2e^(−3t), so 3(5 − y) = 6e^(−3t). The sides match for all t.
- (A) gives 3(−2e^(−3t)) = −6e^(−3t): the right size but the wrong sign.
- (B) gives −15 + 6e^(−3t). The constant −15 does not match.
- (C) gives 2e^(−3t), which is missing the factor 3 from the chain rule.
Question 4 (multiple choice · core)
Which of these is not a solution of dy/dx = 2y?
- (A) y = 0
- (B) y = −4e^(2x)
- (C) y = e^(2x) + 1
- (D) y = e^(2x + 1)
Answer and explanation
Answer: (C). For y = e^(2x) + 1: left side 2e^(2x); right side 2e^(2x) + 2. They differ by 2 for every x.
- (A) Left side 0, right side 2 × 0 = 0. A constant solution.
- (B) Left side −8e^(2x), right side 2(−4e^(2x)) = −8e^(2x). This is Ce^(2x) with C = −4.
- (D) e^(2x + 1) = e × e^(2x), which is Ce^(2x) with C = e. Left side 2e^(2x + 1) equals the right side.
Question 5 (constructed response · core)
Let y = 2sin(3x) − cos(3x).
(a) Find y′ and y″. (b) Show that y is a solution of y″ + 9y = 0. (c) Explain why y = 2sin(3x) − cos(3x) + 1 is not a solution of the same equation. (d) Give two other solutions of y″ + 9y = 0 and explain how you know there are infinitely many.
Worked solution and suggested Marlbridge rubric
(a) y′ = 6cos(3x) + 3sin(3x). y″ = −18sin(3x) + 9cos(3x).
(b) Left side: y″ + 9y = (−18sin(3x) + 9cos(3x)) + (18sin(3x) − 9cos(3x)) = 0. Right side: 0. Equal for all x, so y is a solution.
(c) Adding 1 does not change y″, but it adds 9 × 1 = 9 to the left side. The left side becomes 9, not 0, for every x.
(d) For example y = sin(3x) and y = 5cos(3x). In general, y = A sin(3x) + B cos(3x) gives y″ = −9A sin(3x) − 9B cos(3x) = −9y, so y″ + 9y = 0 for every pair of constants A and B. There are infinitely many choices of A and B, so infinitely many solutions.
| Point | What earns it |
|---|---|
| 1 | Correct y′ and y″, using the chain rule (factor 3, then 9) |
| 1 | Substitutes into y″ + 9y and simplifies to 0, concluding “for all x” |
| 1 | Explains that the constant 1 adds 9 to the left side |
| 1 | Two correct further solutions and a general argument for infinitely many |
A check at one value of x, such as x = 0, earns no mark for (b).
Question 6 (constructed response · core)
Let y = 1/(4 − x).
(a) Show that y is a solution of dy/dx = y² on the interval x < 4. (b) Show that y = 1/(C − x) is a solution of dy/dx = y² for every constant C. (c) Explain why y = 1/(4 − x) cannot be a solution on an interval that contains x = 4. (d) Show that y = 0 is a solution of dy/dx = y² and explain why it is not in the family in (b).
Worked solution and suggested Marlbridge rubric
(a) Write y = (4 − x)^(−1). By the chain rule, dy/dx = −(4 − x)^(−2) × (−1) = 1/(4 − x)². Right side: y² = 1/(4 − x)². Equal for all x < 4.
(b) dy/dx = −(C − x)^(−2) × (−1) = 1/(C − x)², and y² = 1/(C − x)². Equal wherever y is defined, for any C.
(c) At x = 4 the function is undefined (division by zero), so it is not differentiable there. A solution must be differentiable at every point of its interval.
(d) For y = 0: left side 0, right side 0² = 0. Equal for all x. But 1/(C − x) is never 0 for any C, because a fraction with numerator 1 cannot equal 0. So y = 0 is an extra solution outside the family.
| Point | What earns it |
|---|---|
| 1 | Correct derivative 1/(4 − x)², with the chain-rule factor −1 handled |
| 1 | Shows both sides equal for the general C |
| 1 | States that the function is undefined, so not differentiable, at x = 4 |
| 1 | Verifies y = 0 and explains why no C gives it |
Question 7 (constructed response · stretch)
Consider the differential equation dy/dx = 2y − 4x + 1.
(a) Find the constants a and b so that y = ax + b is a solution. (b) Show that y = 2x + 1/2 + Ce^(2x) is a solution for every constant C. (c) A student tests y = x³ + 2x + 1/2 at x = 0, finds both sides equal 2, and concludes it is a solution. Show that the student is wrong.
Worked solution and suggested Marlbridge rubric
(a) For y = ax + b, the left side is a. The right side is 2(ax + b) − 4x + 1 = (2a − 4)x + (2b + 1). For these to be equal for all x, the x-coefficients and the constants must match:
- x terms: 0 = 2a − 4, so a = 2.
- constants: a = 2b + 1, so 2 = 2b + 1 and b = 1/2.
So y = 2x + 1/2.
(b) Left side: dy/dx = 2 + 2Ce^(2x). Right side: 2(2x + 1/2 + Ce^(2x)) − 4x + 1 = 4x + 1 + 2Ce^(2x) − 4x + 1 = 2 + 2Ce^(2x). Equal for all x and every C.
(c) Left side: dy/dx = 3x² + 2. Right side: 2(x³ + 2x + 1/2) − 4x + 1 = 2x³ + 2. At x = 0 both are 2, but at x = 1 the left side is 5 and the right side is 4. The sides are not identical, so the function is not a solution. (They agree only at x = 0 and x = 3/2.)
| Point | What earns it |
|---|---|
| 1 | Substitutes y = ax + b and matches the x-coefficients to get a = 2 |
| 1 | Matches constants to get b = 1/2 |
| 1 | Shows both sides equal 2 + 2Ce^(2x) |
| 1 | Finds both sides of (c) as expressions in x and gives an x where they differ |
| 1 | States that a solution must make the sides equal for all x, so one point is not enough |
How did you do?
- Q1, Q3 or Q4 wrong: revisit “The verification method” in the study guide and set out left side and right side separately.
- Q2 or Q5 wrong: redo Worked example 2 and the “Second-order equations” section.
- Q6 wrong: see Worked example 3 on intervals and extra solutions.
- Q7 wrong: reread “Why one point is not enough”.
Then tick off the topic checklist.