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, and exact answers unless stated. Notation: sec²x means (sec x)², and fractions are written on one line. The context in Question 6 is invented. This set is for both Calculus AB and Calculus BC students.
Question 1 (multiple choice · foundation)
If f(x) = 3 tan x − 2 sec x, what is f′(x)?
- (A) 3 sec x − 2 sec x tan x
- (B) 3 sec²x + 2 sec x tan x
- (C) 3 sec²x − 2 tan x
- (D) 3 sec²x − 2 sec x tan x
Answer and explanation
Answer: (D). Use the constant multiple and difference rules with d/dx tan x = sec²x and d/dx sec x = sec x tan x: f′(x) = 3 sec²x − 2 sec x tan x.
- (A) drops the square: the derivative of tan x is sec²x, not sec x.
- (B) flips the sign of the second term. Only the “co” functions (cot, csc) bring in a new minus sign; sec x does not.
- (C) uses d/dx sec x = tan x, missing the factor sec x.
Question 2 (multiple choice · core)
What is the slope of the graph of y = sec x at x = π/3?
- (A) √3
- (B) 2√3/3
- (C) 2√3
- (D) 4
Answer and explanation
Answer: (C). dy/dx = sec x tan x. At π/3: sec(π/3) = 1/cos(π/3) = 1/(1/2) = 2 and tan(π/3) = √3. Slope = 2 × √3 = 2√3.
- (A) uses d/dx sec x = tan x, so it gives tan(π/3) only.
- (B) uses tan(π/3) = 1/√3 = √3/3, which is the value of tan(π/6).
- (D) uses sec²x, which is the derivative of tan x, not of sec x.
Question 3 (multiple choice · core)
Let h(x) = x cot x for 0 < x < π. What is h′(π/2)?
- (A) −π/2
- (B) 0
- (C) π/2
- (D) −1
Answer and explanation
Answer: (A). Product rule: h′(x) = (1)(cot x) + x(−csc²x) = cot x − x csc²x. At π/2: cot(π/2) = cos(π/2)/sin(π/2) = 0 and csc(π/2) = 1. So h′(π/2) = 0 − (π/2)(1) = −π/2.
- (B) keeps only the first product-rule term, cot x, which is 0 at π/2.
- (C) uses d/dx cot x = +csc²x, losing the minus sign.
- (D) multiplies the derivatives, (1)(−csc²x), instead of using the product rule.
Question 4 (multiple choice · core)
What is d/dx (sec²x − tan²x)?
- (A) 2 sec²x tan x
- (B) 0
- (C) 1
- (D) 4 sec²x tan x
Answer and explanation
Answer: (B). The identity 1 + tan²x = sec²x gives sec²x − tan²x = 1 wherever both are defined. The derivative of a constant is 0. You can also check with the product rule: d/dx sec²x = 2 sec x · sec x tan x = 2 sec²x tan x, and d/dx tan²x = 2 tan x sec²x. These are equal, so the difference is 0.
- (A) differentiates sec²x but forgets to subtract the derivative of tan²x.
- (C) gives the value of the expression, not its derivative.
- (D) adds the two derivatives instead of subtracting them.
Question 5 (constructed response · core)
(a) Write csc x in terms of sin x, and use the quotient rule to show that d/dx csc x = −csc x cot x. (b) For which values of x is the result in (a) valid? (c) Find the equation of the line tangent to y = csc x at x = π/6.
Worked solution
(a) csc x = 1/sin x. Top: 1, derivative 0. Bottom: sin x, derivative cos x.
d/dx csc x = [sin x · 0 − 1 · cos x]/sin²x = −cos x/sin²x = −(1/sin x)(cos x/sin x) = −csc x cot x.
(b) Wherever sin x ≠ 0, that is x ≠ kπ for any integer k.
(c) csc(π/6) = 1/(1/2) = 2. cot(π/6) = (√3/2)/(1/2) = √3. Slope = −(2)(√3) = −2√3. Tangent line: y = 2 − 2√3(x − π/6).
Suggested mark points (4): 1 for rewriting as 1/sin x and a correct quotient-rule set-up; 1 for splitting −cos x/sin²x into −csc x cot x; 1 for the domain x ≠ kπ; 1 for the tangent line with point (π/6, 2) and slope −2√3.
Question 6 (constructed response · core)
In an invented scenario, a camera on level ground is 50 metres from the point directly below a balloon. The balloon rises straight up. When the angle of elevation from the camera to the balloon is θ radians, the balloon’s height is
H(θ) = 50 tan θ metres, for 0 ≤ θ < π/2.
(a) Find H′(θ). (b) Find H′(π/6) and H′(π/3). Give units, and explain what the two values tell you. (c) Explain why H′(θ) ≥ 50 for every θ in the domain. (d) Use an identity to show that H′(θ) = 50 + [H(θ)]²/50.
Worked solution and suggested Marlbridge rubric
(a) H′(θ) = 50 sec²θ.
(b) sec(π/6) = 1/(√3/2) = 2/√3, so sec²(π/6) = 4/3 and H′(π/6) = 50 × 4/3 = 200/3 metres per radian. sec(π/3) = 2, so sec²(π/3) = 4 and H′(π/3) = 200 metres per radian. The height changes three times as fast, per unit of angle, at θ = π/3 as at θ = π/6. As the balloon gets higher, a small increase in the angle corresponds to a much larger increase in height.
(c) sec²θ = 1/cos²θ. For 0 ≤ θ < π/2, 0 < cos θ ≤ 1, so 0 < cos²θ ≤ 1 and 1/cos²θ ≥ 1. So H′(θ) = 50 sec²θ ≥ 50. Equality holds only at θ = 0.
(d) sec²θ = 1 + tan²θ, so H′(θ) = 50 + 50 tan²θ. Since tan θ = H/50, 50 tan²θ = 50 × H²/2500 = H²/50. So H′(θ) = 50 + H²/50. Check at θ = π/3: H = 50√3, H² = 7500, so 50 + 150 = 200, as in (b).
| Point | What earns it |
|---|---|
| 1 | H′(θ) = 50 sec²θ |
| 1 | Both values, 200/3 and 200, with units of metres per radian |
| 1 | A correct interpretation comparing the two rates |
| 1 | Argument that sec²θ ≥ 1 because 0 < cos²θ ≤ 1 on the domain |
| 1 | Uses 1 + tan²θ = sec²θ and tan θ = H/50 to reach 50 + H²/50 |
A statement such as “the balloon is 200 metres high” in (b) earns no interpretation mark: H′ is a rate, not a height.
Question 7 (constructed response · stretch)
Let k(x) = tan x/(1 + sec x), for 0 < x < π/2.
(a) Use the quotient rule and an identity to show that k′(x) = 1/(1 + cos x). (b) Show that k(x) = sin x/(1 + cos x) on this interval, and differentiate this form to confirm your answer to (a). (c) Find the equation of the line tangent to the graph of k at x = π/3.
Worked solution and suggested Marlbridge rubric
(a) Top: tan x, derivative sec²x. Bottom: 1 + sec x, derivative sec x tan x.
k′(x) = [(1 + sec x)sec²x − tan x · sec x tan x]/(1 + sec x)².
Factor sec x from the top: sec x[(1 + sec x)sec x − tan²x] = sec x[sec x + sec²x − tan²x]. Since sec²x − tan²x = 1, the bracket is sec x + 1. So
k′(x) = sec x(1 + sec x)/(1 + sec x)² = sec x/(1 + sec x).
Multiply top and bottom by cos x: k′(x) = 1/(cos x + 1).
(b) Multiply top and bottom of k by cos x: tan x · cos x = sin x and (1 + sec x)cos x = cos x + 1. So k(x) = sin x/(1 + cos x).
Quotient rule: [(1 + cos x)cos x − sin x(−sin x)]/(1 + cos x)² = (cos x + cos²x + sin²x)/(1 + cos x)² = (1 + cos x)/(1 + cos x)² = 1/(1 + cos x). The two methods agree.
(c) k(π/3) = sin(π/3)/(1 + cos(π/3)) = (√3/2)/(3/2) = √3/3. k′(π/3) = 1/(1 + 1/2) = 2/3. Tangent line: y = √3/3 + (2/3)(x − π/3).
| Point | What earns it |
|---|---|
| 1 | Correct quotient-rule set-up with sec²x and sec x tan x |
| 1 | Uses sec²x − tan²x = 1 to simplify the numerator |
| 1 | Reaches 1/(1 + cos x) |
| 1 | Rewrites k as sin x/(1 + cos x) and differentiates it correctly |
| 1 | Tangent line with point (π/3, √3/3) and slope 2/3 |
Acceptable alternative for (a): rewrite first, as in (b), then differentiate. Full credit for (a) needs the identity step or an equivalent simplification.
How did you do?
- Q1 or Q2 wrong: relearn the table of four results in the study guide, and practise deriving sec x from 1/cos x.
- Q3 wrong: redo Worked example 2 (product with tan x) and check the sign pattern for “co” functions.
- Q4 or Q7 wrong: see Worked example 3, where 1 + tan²x = sec²x simplifies a quotient.
- Q5 wrong: reread “Deriving d/dx sec x” and Worked example 1; the csc x derivation follows the same steps.
- Q6 wrong: see “Reading the derivative on a graph” and Figure 1.
Then tick off the topic checklist.