Time: designed to take 90 minutes. The real exam gives you 3 hours for this section plus 3 written-response questions, and you may take up to 6 hours.
You will need a graphing calculator and the Math 30-1 formula sheet. Keep all decimals through a question and round only the final answer, to the accuracy the question asks for.
Multiple choice: choose the best of the four answers. Numerical response: write your answer in the boxes, first digit on the left, unused boxes blank, and a 0 before the decimal point for answers between 0 and 1.
Use the following information to answer question 1.
The graph of y = f(x) is transformed to the graph of y = −2f(x − 5) + 3.
1. Which of the following lists the transformations that are applied to the graph of y = f(x)?
- A. A vertical stretch about the x-axis by a factor of 2, a reflection in the x-axis, then a translation of 5 units right and 3 units up
- B. A vertical stretch about the x-axis by a factor of 2, a reflection in the y-axis, then a translation of 5 units left and 3 units up
- C. A translation of 5 units right and 3 units up, then a vertical stretch about the x-axis by a factor of 2 and a reflection in the x-axis
- D. A vertical stretch about the x-axis by a factor of 1/2, a reflection in the x-axis, then a translation of 5 units right and 3 units up
Use the following information to answer question 2.
The point (−6, 4) lies on the graph of y = f(x).
2. The corresponding point on the graph of y = ½f(3x) − 1 is
- A. (−18, 1)
- B. (−2, 1)
- C. (−2, 5)
- D. (−18, 7)
Use the following information to answer question 3.
The domain of y = f(x) is −4 ≤ x ≤ 8.
3. The domain of y = f(2(x + 1)) is
- A. −5 ≤ x ≤ 3
- B. −9 ≤ x ≤ 15
- C. −3 ≤ x ≤ 3
- D. −2 ≤ x ≤ 5
Use the following information to answer numerical-response question 1.
Four functions are listed below.
- 1 y = f(x − 3)
- 2 y = f(x) − 3
- 3 y = f(x + 3)
- 4 y = f(x) + 3
Numerical Response
1. The graph of y = f(x) translated 3 units right is function ___, translated 3 units left is function ___, translated 3 units up is function ___, and translated 3 units down is function ___.
(Record all four digits of your answer in the numerical-response section on the answer sheet.)
4. If f(x) = 2x³ − 5, then f⁻¹(x) =
- A. 1/(2x³ − 5)
- B. (∛x + 5)/2
- C. ∛((x − 5)/2)
- D. ∛((x + 5)/2)
5. The solution to the equation 4^(x + 1) = 8^(x − 2) is
- A. x = 3
- B. x = 4
- C. x = 8
- D. x = 10
Use the following information to answer question 6.
log_b 2 = m and log_b 5 = n
6. Written in terms of m and n, log_b 50 =
- A. m + 2n
- B. 2m + n
- C. mn²
- D. m + n²
Numerical Response
2. The solution to the equation 3^(2x − 1) = 81 is x = ______.
(Record your answer to the nearest tenth in the numerical-response section on the answer sheet.)
7. The expression 2 log x − ½ log y + log 3, where x > 0 and y > 0, written as a single logarithm, is
- A. log(6x²/y)
- B. log(3x²/√y)
- C. log(x²√y/3)
- D. log(2x − ½y + 3)
Use the following information to answer question 8.
The graph of y = log₂ x is transformed to the graph of y = log₂(x + 4) − 3.
8. Which statement about the transformed graph is true?
- A. Its horizontal asymptote is y = −3 and its domain is x > 0
- B. Its vertical asymptote is x = −4 and its domain is x > 4
- C. Its vertical asymptote is x = 4 and its domain is x > 4
- D. Its vertical asymptote is x = −4 and its domain is x > −4
9. When P(x) = 2x³ − 5x² + x − 4 is divided by x + 2, the remainder is
- A. −42
- B. −22
- C. −6
- D. 38
Use the following information to answer numerical-response question 3.
The polynomial function P(x) = x³ + kx² − 4x − 12 has x − 2 as a factor.
Numerical Response
3. The value of k is ______.
(Record your answer in the numerical-response section on the answer sheet.)
Use the following information to answer question 10.
A polynomial function has zeros of −2 with multiplicity 2 and 3 with multiplicity 1, and its graph passes through the point (0, −12).
10. An equation of this function is
- A. y = 2(x + 2)²(x − 3)
- B. y = (x + 2)²(x − 3)
- C. y = −(x + 2)²(x − 3)
- D. y = (x − 2)²(x + 3)
11. Which of the following best describes the end behaviour of the graph of y = −2x⁴ + 3x² − 7?
- A. The graph extends up in quadrant 2 and down in quadrant 4
- B. The graph extends up in quadrant 2 and up in quadrant 1
- C. The graph extends down in quadrant 3 and down in quadrant 4
- D. The graph extends down in quadrant 3 and up in quadrant 1
Use the following information to answer numerical-response question 4.
The graph of y = (2x² − 5x − 3)/(x − 3) has a point of discontinuity at (3, b).
Numerical Response
4. The value of b is ______.
(Record your answer in the numerical-response section on the answer sheet.)
12. The domain of the function y = √(5 − 2x) + 3 is
- A. x ≥ −3
- B. x ≤ 5
- C. x ≥ 5/2
- D. x ≤ 5/2
13. The solution to the equation √(3x + 4) = x is
- A. x = 4 only
- B. x = −1 and x = 4
- C. There is no solution
- D. x = −1 only
14. The exact value of cos(5π/6) is
- A. √3/2
- B. −√3/2
- C. −1/2
- D. 1/2
Use the following information to answer question 15.
The terminal arm of an angle θ in standard position passes through the point P(−8, 15).
15. The value of sin θ is
- A. −15/8
- B. 8/17
- C. 15/17
- D. −8/17
Numerical Response
5. The number of solutions to the equation sin 2θ = −½ on the domain 0 ≤ θ < 2π is ______.
(Record your answer in the numerical-response section on the answer sheet.)
16. A central angle of 2.5 radians in a circle of radius 12 cm subtends an arc of length
- A. 4.8 cm
- B. 30 cm
- C. 60 cm
- D. 94.2 cm
17. The general solution to the equation 2 cos θ + √3 = 0, where n ∈ I, is
- A. θ = π/3 + 2πn and θ = 2π/3 + 2πn
- B. θ = 5π/6 + πn
- C. θ = π/6 + 2πn and θ = 11π/6 + 2πn
- D. θ = 5π/6 + 2πn and θ = 7π/6 + 2πn
18. The expression (1 − cos²θ)/(sin θ cos θ), where sin θ ≠ 0 and cos θ ≠ 0, simplifies to
- A. tan θ
- B. csc θ
- C. cot θ
- D. sec θ
Use the following information to answer question 19.
sin θ = 3/5, where θ terminates in quadrant II.
19. The value of sin 2θ is
- A. 6/5
- B. 24/25
- C. −24/25
- D. −7/25
Use the following information to answer question 20.
In the expansion of (2x − 3)⁸, written in descending powers of x, the general term is t_(k+1) = ₈C_k (2x)^(8−k) (−3)^k.
20. The coefficient of the term containing x⁵ is
- A. −48 384
- B. −6 048
- C. 1 512
- D. 48 384
Use the following information to answer question 21.
y = −4 cos[2(x − π/3)] + 7
21. Which statement about the graph of the function above is true?
- A. The range is −4 ≤ y ≤ 4 and the period is π
- B. The range is 3 ≤ y ≤ 11 and the period is 4π
- C. The amplitude is −4 and the period is 2π
- D. The range is 3 ≤ y ≤ 11 and the period is π
Use the following information to answer question 22.
The depth of water in a harbour varies sinusoidally. The maximum depth is 8.2 m, the minimum depth is 2.4 m, and one complete cycle takes 12.4 h. The depth is modelled by d(t) = a cos[b(t − c)] + d.
22. The values of a and d, respectively, are
- A. 5.8 and 5.3
- B. 2.9 and 5.3
- C. 2.9 and 10.6
- D. 5.3 and 2.9
Use the following information to answer numerical-response question 6.
A 100 mg sample of a radioactive isotope decays according to A(t) = 100(½)^(t/5.7), where A is the mass remaining, in milligrams, and t is the time, in days.
Numerical Response
6. The time it takes for the sample to decay to 12 mg is ______ days.
(Record your answer to the nearest tenth in the numerical-response section on the answer sheet.)
23. In how many different ways can 7 students be seated in a row of 7 chairs if 2 particular students must sit beside each other?
- A. 720
- B. 1 440
- C. 2 520
- D. 5 040
Use the following information to answer question 24.
A committee of 4 people is chosen from a group of 6 boys and 5 girls.
24. The number of different committees that contain at least 3 girls is
- A. 60
- B. 65
- C. 165
- D. 330
Use the following information to answer numerical-response question 7.
Four expressions are listed below.
Numerical Response
7. The three expressions above that have a value of 120 are numbered ______, ______, and ______.
(Record all three digits of your answer in any order in the numerical-response section on the answer sheet.)
Numerical Response
8. The number of different arrangements of all the letters of the word BANANA is ______.
(Record your answer in the numerical-response section on the answer sheet.)
Answer key · Sample diploma exam
Math 30-1 · maddyhelps.com
Multiple choice
1. A · 2. B · 3. C · 4. D · 5. C · 6. A · 7. B · 8. D · 9. A · 10. B · 11. C · 12. D · 13. A · 14. B · 15. C · 16. B · 17. D · 18. A · 19. C · 20. A · 21. D · 22. B · 23. B · 24. B
Numerical response
1. 1342 · 2. 2.5 · 3. 3 · 4. 7 · 5. 4 · 6. 17.4 · 7. 234 · 8. 60
Question 1 — A. A vertical stretch about the x-axis by a factor of 2, a reflection in the x-axis, then a translation of 5 units right and 3 units up
In y = af[b(x − h)] + k, a = −2 gives a vertical stretch by a factor of |−2| = 2 together with a reflection in the x-axis; h = 5 is 5 units right and k = 3 is 3 units up. Stretches and reflections are applied before translations, so the choice that translates first is out.
Question 2 — B. (−2, 1)
The mapping is (x, y) → (x/3, ½y − 1). The x-coordinate is divided by 3, because b = 3 compresses horizontally: −6 ÷ 3 = −2. The y-coordinate is ½(4) − 1 = 1, so the image is (−2, 1). Multiplying by 3 instead of dividing gives the distractor (−18, 1).
Question 3 — C. −3 ≤ x ≤ 3
Whatever goes inside f must still land between −4 and 8, so −4 ≤ 2(x + 1) ≤ 8. Divide by 2: −2 ≤ x + 1 ≤ 4. Subtract 1: −3 ≤ x ≤ 3. Undo the parameters in the reverse order you apply them.
Numerical Response 1 — 1342
Inside the brackets moves the graph horizontally and does the opposite of what it looks like: f(x − 3) is 3 right, f(x + 3) is 3 left. Outside the brackets moves it vertically and does what it looks like: +3 is up, −3 is down. Right, left, up, down = 1, 3, 4, 2.
Question 4 — D. ∛((x + 5)/2)
Swap x and y: x = 2y³ − 5. Solve for y: x + 5 = 2y³, so y³ = (x + 5)/2 and y = ∛((x + 5)/2). The inverse is not the reciprocal, which is where 1/(2x³ − 5) comes from.
Question 5 — C. x = 8
Write both sides base 2: 2^(2(x + 1)) = 2^(3(x − 2)). Equate exponents: 2x + 2 = 3x − 6, so x = 8. Check: 4⁹ = 262 144 and 8⁶ = 262 144.
Question 6 — A. m + 2n
50 = 2 × 5², so log_b 50 = log_b 2 + log_b 5² = log_b 2 + 2 log_b 5 = m + 2n. The product law adds; the power law brings the exponent out front as a multiplier, not as an exponent on n.
Numerical Response 2 — 2.5
81 = 3⁴, so 2x − 1 = 4, giving 2x = 5 and x = 2.5. A numerical-response answer between 0 and 1 would need the 0 recorded before the decimal point; here the answer is 2.5.
Question 7 — B. log(3x²/√y)
Power law first: 2 log x = log x² and ½ log y = log √y. Then the product and quotient laws: log x² − log √y + log 3 = log(3x²/√y). Logarithms of different terms cannot be combined inside one log the way log(2x − ½y + 3) suggests.
Question 8 — D. Its vertical asymptote is x = −4 and its domain is x > −4
x + 4 replaces x, which translates the graph 4 units left, so the asymptote moves from x = 0 to x = −4 and the domain becomes x > −4. The −3 shifts the graph down and does not affect the domain; a logarithmic function has no horizontal asymptote.
Question 9 — A. −42
By the remainder theorem the remainder is P(−2), because x + 2 = x − (−2). P(−2) = 2(−8) − 5(4) + (−2) − 4 = −16 − 20 − 2 − 4 = −42. Using P(2) instead is the classic slip.
Numerical Response 3 — 3
By the factor theorem, P(2) = 0. So 8 + 4k − 8 − 12 = 0, which gives 4k = 12 and k = 3. Check: P(x) = x³ + 3x² − 4x − 12 = (x − 2)(x + 2)(x + 3).
Question 10 — B. y = (x + 2)²(x − 3)
The zeros give y = a(x + 2)²(x − 3). Substituting (0, −12): −12 = a(4)(−3) = −12a, so a = 1 and y = (x + 2)²(x − 3). The sign inside each factor is the opposite of the zero.
Question 11 — C. The graph extends down in quadrant 3 and down in quadrant 4
The degree is even and the leading coefficient is negative, so both ends point down: the graph falls in quadrant 3 on the left and in quadrant 4 on the right. Only the leading term matters for end behaviour.
Numerical Response 4 — 7
Factor the numerator: 2x² − 5x − 3 = (2x + 1)(x − 3). The factor x − 3 cancels, leaving y = 2x + 1 with a hole where x = 3. Substituting: b = 2(3) + 1 = 7. A cancelled factor gives a hole, not a vertical asymptote.
Question 12 — D. x ≤ 5/2
A square root needs a radicand that is not negative: 5 − 2x ≥ 0, so 5 ≥ 2x and x ≤ 5/2. The + 3 shifts the graph up and affects the range, not the domain.
Question 13 — A. x = 4 only
Square both sides: 3x + 4 = x², so x² − 3x − 4 = 0 and (x − 4)(x + 1) = 0, giving x = 4 or x = −1. Check both: √16 = 4 works, but √1 = 1 ≠ −1, so x = −1 is extraneous. Squaring can create roots that do not satisfy the original equation.
Question 14 — B. −√3/2
5π/6 is in quadrant II with reference angle π/6. Cosine is negative in quadrant II and cos(π/6) = √3/2, so cos(5π/6) = −√3/2. The value −1/2 comes from using π/3 as the reference angle instead of π/6.
Question 15 — C. 15/17
r = √((−8)² + 15²) = √289 = 17. Then sin θ = y/r = 15/17. It is positive because the point is in quadrant II, where sine is positive. The value −8/17 is cos θ and −15/8 is tan θ.
Numerical Response 5 — 4
Let u = 2θ. As θ runs over 0 ≤ θ < 2π, u runs over 0 ≤ u < 4π — two full turns. sin u = −½ has two solutions per turn, so there are 4 solutions in total. The b value tells you how many cycles fit in the domain.
Question 16 — B. 30 cm
a = θr with θ in radians, so a = 2.5 × 12 = 30 cm. Converting to degrees first is unnecessary, and dividing 12 by 2.5 gives the 4.8 cm distractor.
Question 17 — D. θ = 5π/6 + 2πn and θ = 7π/6 + 2πn
cos θ = −√3/2, so the reference angle is π/6 and cosine is negative in quadrants II and III: θ = π − π/6 = 5π/6 and θ = π + π/6 = 7π/6, each plus 2πn. The pair π/6 and 11π/6 is where cosine is +√3/2, not −√3/2.
Question 18 — A. tan θ
By the Pythagorean identity, 1 − cos²θ = sin²θ. So the expression is sin²θ/(sin θ cos θ) = sin θ/cos θ = tan θ. Cancelling one sine from the top and bottom is the whole question.
Question 19 — C. −24/25
Using x² + y² = r²: x = −4 since θ is in quadrant II, so cos θ = −4/5. Then sin 2θ = 2 sin θ cos θ = 2(3/5)(−4/5) = −24/25. Forgetting that cosine is negative in quadrant II gives 24/25; −7/25 is cos 2θ.
Question 20 — A. −48 384
The power of x is 8 − k, so 8 − k = 5 and k = 3, which makes it the 4th term. Its coefficient is ₈C₃(2)⁵(−3)³ = 56 × 32 × (−27) = −48 384. Dropping the 2⁵ from (2x)⁵ gives −1 512, and losing the sign on (−3)³ gives 48 384.
Question 21 — D. The range is 3 ≤ y ≤ 11 and the period is π
The amplitude is |a| = 4 and the midline is y = 7, so the range is 7 − 4 ≤ y ≤ 7 + 4, or 3 ≤ y ≤ 11. The period is 2π/|b| = 2π/2 = π. Amplitude is never negative — the minus sign is a reflection.
Question 22 — B. 2.9 and 5.3
The amplitude is half the difference: a = (8.2 − 2.4)/2 = 2.9. The midline is the average: d = (8.2 + 2.4)/2 = 5.3. Using the whole difference instead of half gives 5.8.
Numerical Response 6 — 17.4
12 = 100(½)^(t/5.7), so (½)^(t/5.7) = 0.12. Taking logs: (t/5.7) log 0.5 = log 0.12, so t = 5.7 × log 0.12/log 0.5 = 17.435… ≈ 17.4 days. Keep every decimal until the final rounding.
Question 23 — B. 1 440
Treat the pair as one block: 6 objects arrange in 6! = 720 ways, and the pair can sit in 2! = 2 orders inside the block. 720 × 2 = 1 440. Forgetting to swap the pair gives 720; 5 040 is 7! with no restriction.
Question 24 — B. 65
"At least 3" splits into cases. Exactly 3 girls: ₅C₃ × ₆C₁ = 10 × 6 = 60. Exactly 4 girls: ₅C₄ = 5. Total 60 + 5 = 65. Committees are unordered, so combinations, and the cases add.
Numerical Response 7 — 234
₆C₂ = 15, ₆P₃ = 6 × 5 × 4 = 120, 5! = 120, and ₁₀C₃ = (10 × 9 × 8)/(3 × 2 × 1) = 120. So expressions 2, 3 and 4. Any order of those three digits is scored correct.
Numerical Response 8 — 60
There are 6 letters with 3 A's and 2 N's repeated, so the count is 6!/(3!2!) = 720/12 = 60. Divide by the factorial of each repeat, or you count identical arrangements more than once.