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, ln means the natural logarithm, and exact answers unless stated. Useful facts: e ≈ 2.718, so 2 < e < 3. This set is for both Calculus AB and Calculus BC students.
Question 1 (multiple choice · foundation)
Let f(x) = 5 sin x − 2 cos x. What is f′(π/3)?
- (A) 5/2 − √3
- (B) 5/2 + √3
- (C) 5√3/2 − 1
- (D) −5/2 − √3
Answer and explanation
Answer: (B). f′(x) = 5 cos x − 2(−sin x) = 5 cos x + 2 sin x. At π/3: 5(1/2) + 2(√3/2) = 5/2 + √3.
- (A) uses d/dx cos x = sin x (no minus sign), giving f′(x) = 5 cos x − 2 sin x.
- (C) is f(π/3) = 5(√3/2) − 2(1/2), the value of the function, not its derivative.
- (D) swaps the signs of both rules, using d/dx sin x = −cos x and d/dx cos x = sin x.
Question 2 (multiple choice · core)
What is lim (h → 0) [ln(e + h) − 1]/h?
- (A) 0
- (B) 1/e
- (C) 1
- (D) e
Answer and explanation
Answer: (B). Since ln e = 1, the limit is [f(e + h) − f(e)]/h with f(x) = ln x and a = e. So it equals f′(e) = 1/e.
- (A) treats the 0/0 form that substitution gives as the number 0.
- (C) uses the slope of ln x at x = 1 instead of at x = e. The point is a = e because the input is e + h.
- (D) inverts the rule, using x instead of 1/x as the derivative of ln x.
Question 3 (multiple choice · foundation)
If y = 3eˣ − x³ + ln 7, what is dy/dx?
- (A) 3eˣ − 3x² + 1/7
- (B) 3eˣ − 3x²
- (C) 3x·e^(x − 1) − 3x²
- (D) 3eˣ − 3x² + ln 7
Answer and explanation
Answer: (B). Term by term: d/dx (3eˣ) = 3eˣ; d/dx (−x³) = −3x²; ln 7 is a constant, so its derivative is 0.
- (A) treats ln 7 as if it were ln x evaluated at 7. ln 7 is a fixed number, so its derivative is 0.
- (C) applies the power rule to eˣ. The power rule needs a variable base and a constant power; eˣ is the other way round.
- (D) keeps the constant instead of differentiating it.
Question 4 (multiple choice · core)
For 0 < x < π, at which value of x does the graph of y = sin x + cos x have a horizontal tangent line?
- (A) π/4
- (B) π/2
- (C) 3π/4
- (D) π
Answer and explanation
Answer: (A). dy/dx = cos x − sin x. A horizontal tangent needs dy/dx = 0, so cos x = sin x. In 0 < x < π this happens only at x = π/4.
- (B) is where sin x alone has a horizontal tangent. The cos x term also contributes to the slope.
- (C) solves sin x + cos x = 0. That is where y = 0, or what you get by using d/dx cos x = +sin x. Neither is the condition for a horizontal tangent.
- (D) is where cos x alone has a horizontal tangent (and it is not even inside the open interval).
Question 5 (constructed response · core)
Let f(x) = eˣ − 2 ln x for x > 0.
(a) Find f′(x). (b) Find the equation of the tangent line to the graph of f at x = 1. (c) Explain why there is at least one value c with 1/2 < c < 1 at which the tangent line to the graph of f is horizontal.
Worked solution and suggested Marlbridge rubric
(a) f′(x) = eˣ − 2/x.
(b) f(1) = e − 2 ln 1 = e − 0 = e. f′(1) = e − 2. The tangent line is
y = e + (e − 2)(x − 1)
Since 2 < e < 3, the slope e − 2 is positive (about 0.718). Expanding gives y = (e − 2)x + 2, so the line crosses the y-axis at 2.
(c) f′ is continuous on [1/2, 1], because eˣ and 2/x are both continuous for x > 0.
- f′(1/2) = e^(1/2) − 4. Since e < 4, √e < 2, so f′(1/2) < 2 − 4 < 0.
- f′(1) = e − 2 > 0.
f′ changes sign on the interval, so by the intermediate value theorem there is a c with 1/2 < c < 1 and f′(c) = 0. At that c the tangent line is horizontal. (A calculator check, not required, gives c ≈ 0.8526.)
| Point | What earns it |
|---|---|
| 1 | f′(x) = eˣ − 2/x |
| 1 | Correct point (1, e) and slope e − 2, with a correct tangent line equation |
| 1 | Shows f′(1/2) < 0 and f′(1) > 0 with a reason, not just decimals |
| 1 | States that f′ is continuous on [1/2, 1] and applies the intermediate value theorem to conclude f′(c) = 0 |
A conclusion with no mention of continuity does not earn the last point. The intermediate value theorem needs it.
Question 6 (constructed response · core)
Evaluate each limit. In each case, name the function f and the number a that you are using.
(a) lim (h → 0) [sin(π/6 + h) − 1/2]/h (b) lim (x → 0) (eˣ − 1)/x (c) lim (x → π/2) cos x/(x − π/2)
Worked solution and suggested Marlbridge rubric
(a) sin(π/6) = 1/2, so this is [f(a + h) − f(a)]/h with f(x) = sin x, a = π/6. The limit is f′(π/6) = cos(π/6) = √3/2.
(b) e⁰ = 1, so this is [f(x) − f(a)]/(x − a) with f(x) = eˣ, a = 0. The limit is f′(0) = e⁰ = 1.
(c) cos(π/2) = 0, so cos x = cos x − cos(π/2). This is [f(x) − f(a)]/(x − a) with f(x) = cos x, a = π/2. The limit is f′(π/2) = −sin(π/2) = −1.
| Point | What earns it |
|---|---|
| 1 | Recognises that each limit is a definition of the derivative (shown at least once, e.g. by writing f′(a) = lim [f(a + h) − f(a)]/h) |
| 1 | (a): f = sin x, a = π/6, answer √3/2 |
| 1 | (b): f = eˣ, a = 0, answer 1 |
| 1 | (c): f = cos x, a = π/2, answer −1, with the correct sign |
A correct value with no f and a named earns the value but not the first point. In (c), +1 means the sign of d/dx cos x was lost.
Question 7 (constructed response · stretch)
(a) Find, in terms of a, the equation of the tangent line to y = ln x at the point where x = a (a > 0). (b) Find the value of a for which this tangent line passes through the origin. Give the equation of that line. (c) Find the point on y = eˣ whose tangent line passes through the origin, and give the equation of that line. (d) Explain how your answers to (b) and (c) are related.
Worked solution and suggested Marlbridge rubric
(a) The point is (a, ln a) and the slope is 1/a. Tangent: y = ln a + (1/a)(x − a).
(b) Put x = 0, y = 0: 0 = ln a + (1/a)(−a) = ln a − 1. So ln a = 1 and a = e. The line is y = 1 + (1/e)(x − e) = x/e.
(c) At (b, eᵇ) the slope is eᵇ. Tangent: y = eᵇ + eᵇ(x − b). Put x = 0, y = 0: 0 = eᵇ(1 − b). Since eᵇ > 0, b = 1. The point is (1, e) and the line is y = e + e(x − 1) = ex.
(d) y = ln x and y = eˣ are reflections in y = x. The point (e, 1) on ln x is the mirror image of (1, e) on eˣ. Reflecting swaps x and y, so the line y = x/e reflects to x = y/e, which is y = ex. The slopes 1/e and e are reciprocals, as the reflection predicts.
| Point | What earns it |
|---|---|
| 1 | Correct tangent line in terms of a, using slope 1/a |
| 1 | Solves ln a − 1 = 0 to get a = e and the line y = x/e |
| 1 | Point (1, e) and line y = ex on eˣ, with a reason why eᵇ ≠ 0 |
| 1 | Explains the reflection in y = x and the reciprocal slopes |
How did you do?
- Q1, Q3 or Q4 wrong: revisit “The four new rules” and the misconceptions list in the study guide. Check the minus sign on d/dx cos x and that constants differentiate to 0.
- Q2 or Q6 wrong: redo Worked example 3, “a limit that is really a derivative”. Always name f and a.
- Q5 wrong: redo Worked example 1 (tangent lines), then review the intermediate value theorem from Topic 1.16.
- Q7 wrong: reread “Why ln x has derivative 1/x” and Figure 2.
Then tick off the topic checklist.