What this is for. Use this diagnostic to find which topics of Unit 5, Analytical Applications of Differentiation, to revisit. There is one question per topic. These are original Marlbridge practice questions, not past exam questions. All contexts and data are invented. The questions are not calibrated, and your result is not a predicted score.
Rules. No calculator; about 30 minutes. Answer everything before opening any answer, and give a reason for every conclusion. The unit is shared by Calculus AB and Calculus BC, and there are no BC-only questions: every question is for both courses.
Question 1 (multiple choice · 5.1)
A function f is differentiable for all real x, with f(2) = 7 and f(6) = −1. Which statement must be true?
- (A) f′(c) = 0 for some c in (2, 6).
- (B) f′(4) = −2.
- (C) f′(c) = −2 for some c in (2, 6).
- (D) f′(c) = 3 for some c in (2, 6).
Answer and explanation
Answer: (C). f is differentiable, so both conditions hold on [2, 6]. The average rate is (−1 − 7)/(6 − 2) = −2, so by the Mean Value Theorem f′(c) = −2 for some c in (2, 6).
- (A) would need f(2) = f(6).
- (B) assumes c is the midpoint.
- (D) averages the two outputs instead of finding the secant slope.
If you missed this: Topic 5.1 study guide.
Question 2 (multiple choice · 5.2)
Which is the complete list of critical points of f(x) = x² − 8 ln x?
- (A) x = 2 only
- (B) x = −2 and x = 2
- (C) x = 0 and x = 2
- (D) x = −2, x = 0 and x = 2
Answer and explanation
Answer: (A). f′(x) = 2x − 8/x = 2(x − 2)(x + 2)/x. A critical point must be in the domain of f, which is x > 0. So only x = 2 counts.
- (B) keeps x = −2, where ln x is undefined.
- (C) and (D) count x = 0, where f itself does not exist.
If you missed this: Topic 5.2 study guide.
Question 3 (multiple choice · 5.3)
Let f(x) = ln(x² + 1) − x. Which statement is true?
- (A) f is increasing on (−∞, 1) and decreasing on (1, ∞).
- (B) f is decreasing on (−∞, 1) and increasing on (1, ∞).
- (C) f is increasing on (0, ∞).
- (D) f is decreasing on (−∞, ∞).
Answer and explanation
Answer: (D). f′(x) = 2x/(x² + 1) − 1 = −(x − 1)²/(x² + 1), which is negative for every x except x = 1. A single zero with the same sign on both sides does not end an interval of decrease.
- (A) and (B) assume f′ changes sign at x = 1. A squared factor does not.
- (C) forgets the − 1: 2x/(x² + 1) alone is positive for x > 0.
If you missed this: Topic 5.3 study guide.
Question 4 (short answer · 5.4)
Let f(x) = x − 4 arctan x.
(a) Show that f′(x) = (x² − 3)/(x² + 1) and find the critical points of f. (b) Classify each critical point using the First Derivative Test. Justify. (c) Find the exact relative maximum value of f.
Worked answer
(a) f′(x) = 1 − 4/(1 + x²) = (x² − 3)/(x² + 1). The bottom is never 0, so the critical points are where x² = 3: x = −√3 and x = √3.
(b) f′(−2) = 1/5 > 0, f′(0) = −3 < 0, f′(2) = 1/5 > 0. f′ changes from positive to negative at x = −√3: relative maximum. f′ changes from negative to positive at x = √3: relative minimum.
(c) f(−√3) = −√3 − 4(−π/3) = 4π/3 − √3 (about 2.46). Substitute into f, not f′.
If you missed this: Topic 5.4 study guide.
Question 5 (multiple choice · 5.5)
What are the absolute maximum and absolute minimum values of f(x) = x − 2√x on [0, 9]?
- (A) Maximum 3, at x = 9; minimum 0, at x = 0
- (B) Maximum 3, at x = 9; minimum −1, at x = 1
- (C) Maximum 0, at x = 0; minimum −1, at x = 1
- (D) There is no absolute maximum, because f′(0) does not exist.
Answer and explanation
Answer: (B). f is continuous on [0, 9], so the Candidates Test applies. f′(x) = 1 − 1/√x = 0 at x = 1. Candidates: f(0) = 0, f(1) = −1, f(9) = 3.
- (A) misses x = 1.
- (C) misses the right endpoint.
- (D) The Extreme Value Theorem needs continuity, not differentiability.
If you missed this: Topic 5.5 study guide.
Question 6 (multiple choice · 5.6)
Let f(x) = xeˣ. Which statement about the graph of f is true?
- (A) Concave up on (−1, ∞); point of inflection at x = −1
- (B) Concave up for all x, because eˣ > 0
- (C) Concave up on (−2, ∞); point of inflection at x = −2
- (D) Concave down on (−2, ∞); point of inflection at x = −2
Answer and explanation
Answer: (C). f′(x) = (x + 1)eˣ and f″(x) = (x + 2)eˣ, which has the sign of x + 2. f″ changes from negative to positive at x = −2.
- (A) uses the zero of f′, a critical point.
- (B) ignores the factor x + 2.
- (D) reverses the sign of f″.
If you missed this: Topic 5.6 study guide.
Question 7 (multiple choice · 5.7)
Let f(x) = eˣ − 2x for all real x. Which statement is true?
- (A) f has a relative maximum at x = ln 2, because f′(ln 2) = 0.
- (B) f has a relative minimum at x = ln 2, but with no closed interval there is no absolute minimum.
- (C) The absolute minimum value of f is e² − 4, at x = 2.
- (D) The absolute minimum value of f is 2 − 2 ln 2, at x = ln 2.
Answer and explanation
Answer: (D). f′(x) = eˣ − 2 = 0 only at x = ln 2, and f″(ln 2) = 2 > 0: a relative minimum. It is the only critical point of a continuous function, so it is the absolute minimum.
- (A) f′ = 0 only marks a candidate; the sign of f″ decides.
- (B) The one-critical-point rule works on open intervals too.
- (C) solves eˣ = 2 as x = 2.
If you missed this: Topic 5.7 study guide.
Question 8 (multiple choice · 5.8)
You are sketching f(x) = x⁴ − 6x². Which gives the x-coordinates of every relative extremum and point of inflection?
- (A) Minimums at ±√3; maximum at 0; inflection at 0 only
- (B) Minimums at ±√3; maximum at 0; inflection at ±1
- (C) Minimums at ±1; maximum at 0; inflection at ±√3
- (D) Minimums at ±√3; no maximum; inflection at ±1
Answer and explanation
Answer: (B). f′(x) = 4x(x² − 3) has signs −, +, −, + across −√3, 0, √3: minimums at ±√3 (value −9), maximum at 0 (value 0). f″(x) = 12(x² − 1) changes sign at ±1 (value −5).
- (A) guesses the centre of symmetry; f″(0) = −12.
- (C) swaps the zeros of f′ and f″.
- (D) misses the sign change of f′ at 0.
If you missed this: Topic 5.8 study guide.
Question 9 (multiple choice · 5.9)
f is twice differentiable on (0, 6). The graph of f′ rises on (0, 2), falls on (2, 6), and crosses the x-axis only at x = 5, from above to below. Which statement must be true?
- (A) f has a point of inflection at x = 2 and a relative maximum at x = 5.
- (B) f has a relative maximum at x = 2 and a point of inflection at x = 5.
- (C) f is decreasing on (2, 6).
- (D) f has a relative minimum at x = 5.
Answer and explanation
Answer: (A). f′ changes from increasing to decreasing at x = 2, so f changes from concave up to concave down. f′ changes from positive to negative at x = 5: a relative maximum.
- (B) swaps the roles. A peak of f′ is about concavity of f.
- (C) confuses “f′ decreasing” with “f′ negative”. f′ > 0 on (2, 5).
- (D) reverses the First Derivative Test.
If you missed this: Topic 5.9 study guide.
Question 10 (multiple choice · 5.10)
At a ticket price of p dollars, a fictional concert hall sells N(p) = 900 − 30p tickets, for 10 ≤ p ≤ 30. Which price gives the greatest revenue?
- (A) $30
- (B) $10
- (C) $15
- (D) $6,750
Answer and explanation
Answer: (C). R(p) = p(900 − 30p), so R′(p) = 900 − 60p = 0 at p = 15. Candidates: R(10) = 6,000, R(15) = 6,750, R(30) = 0.
- (A) solves N(p) = 0 instead of R′(p) = 0.
- (B) gives the most tickets, not the most revenue.
- (D) is the greatest revenue, not the price.
If you missed this: Topic 5.10 study guide.
Question 11 (short answer · 5.11)
A fictional print studio can make between 1 and 200 figurines a day. Making x figurines costs C(x) = 0.02x² + 3x + 288 dollars, so the average cost per figurine is A(x) = C(x)/x = 0.02x + 3 + 288/x.
(a) Find the number of figurines that minimises the average cost on [1, 200]. Justify that it gives the absolute minimum. (b) Find the minimum average cost and interpret it in context. (c) A manager says making 200 a day is cheapest per figurine, because the fixed $288 is spread more thinly. Respond.
Worked answer
(a) A′(x) = 0.02 − 288/x² = 0 gives x = 120 in the interval. A is continuous on [1, 200]; candidates: A(1) = 291.02, A(120) = 7.80, A(200) = 8.44. The smallest is at x = 120.
(b) The lowest possible average cost is $7.80 per figurine, when the studio makes 120 figurines a day.
(c) Wrong. A′(x) > 0 on (120, 200), so past 120 the average cost rises: it is $8.44 at 200.
If you missed this: Topic 5.11 study guide.
Question 12 (multiple choice · 5.12)
The curve eʸ + y = x² − 2x + 2 passes through (1, 0) and defines y as a differentiable function of x. Which statement is true at (1, 0)?
- (A) dy/dx = 0 and d²y/dx² = −1, so y has a relative maximum at x = 1.
- (B) dy/dx = 0 and d²y/dx² = 1, so y has a relative minimum at x = 1.
- (C) dy/dx = 0 and d²y/dx² = 2, so y has a relative minimum at x = 1.
- (D) dy/dx does not exist, so (1, 0) is not a critical point.
Answer and explanation
Answer: (B). (eʸ + 1)y′ = 2x − 2, so y′ = 0 at (1, 0). Differentiate again: eʸ(y′)² + (eʸ + 1)y″ = 2. At (1, 0): 2y″ = 2, so y″ = 1 > 0.
- (A) gets the sign of y″ wrong, which reverses the conclusion.
- (C) forgets to divide by eʸ + 1 = 2.
- (D) eʸ + 1 is never 0, so dy/dx always exists.
If you missed this: Topic 5.12 study guide.
Your next step
| Topic | Question(s) | If you missed it, read |
|---|---|---|
| 5.1 Mean Value Theorem | 1 | Guide 5.1 |
| 5.2 Critical points | 2 | Guide 5.2 |
| 5.3 Increasing and decreasing | 3 | Guide 5.3 |
| 5.4 First Derivative Test | 4 | Guide 5.4 |
| 5.5 Candidates Test | 5 | Guide 5.5 |
| 5.6 Concavity | 6 | Guide 5.6 |
| 5.7 Second derivative test | 7 | Guide 5.7 |
| 5.8 Sketching graphs | 8 | Guide 5.8 |
| 5.9 Connecting f, f′ and f″ | 9 | Guide 5.9 |
| 5.10 Optimization set-up | 10 | Guide 5.10 |
| 5.11 Optimization in context | 11 | Guide 5.11 |
| 5.12 Implicit relations | 12 | Guide 5.12 |
How to use your result
- Mark each topic secure, shaky (unsure or a slip) or gap (wrong).
- Fix gaps in Topics 5.2 and 5.3 first. Critical points and sign charts are the first step of every later topic in the unit.
- Check your written answers to Questions 4 and 11: did you give a reason, not just a number?
- For a gap, read the guide, then do its practice set.
- Then try the Unit 5 mixed review, where each question combines topics.