Which statement best describes the sequence 3, 6, 12, 24, …?
AArithmetic with common difference 2
BArithmetic with common difference 3
CGeometric with common ratio 3
DGeometric with common ratio 2
Explanation
Each term is found by multiplying the previous term by the same factor: 6/3 = 12/6 = 24/12 = 2. A constant ratio (not a constant difference) makes the sequence geometric with common ratio 2.
Question 2 of 10
MCQU1Topic 1.6Easy No calc
Consider the polynomial function f(x) = -3x⁴ + 5x³ - x + 2. Which statement best describes the end behavior of the graph of f?
AAs x → -∞, f(x) → -∞; as x → +∞, f(x) → +∞.
BAs x → -∞, f(x) → -∞; as x → +∞, f(x) → -∞.
CAs x → -∞, f(x) → +∞; as x → +∞, f(x) → +∞.
DAs x → -∞, f(x) → +∞; as x → +∞, f(x) → -∞.
Explanation
End behavior is governed by the leading term, -3x⁴. Because the degree 4 is even, both ends of the graph go the same direction; because the leading coefficient -3 is negative, both ends fall. So f(x) → -∞ as x → -∞ and as x → +∞. Lower-degree terms do not affect end behavior.
Question 3 of 10
MCQU3Topic 3.14Medium No calc
What is the graph of the polar equation r = 5?
AA circle of radius 5 passing through the origin
BA circle of radius 5 centered at the origin
CA vertical line
DA spiral
Explanation
r = 5 fixes the distance from the origin at 5 for every angle θ, tracing a circle of radius 5 centered at the origin. A circle that merely passes through the origin would have a varying r.
Question 4 of 10
MCQU3Topic 3.9Medium No calc
What is the range of y = arcsin x (the inverse sine function)?
A[−π/2, π/2]
B[0, π]
C[0, 2π]
D(−π/2, π/2)
Explanation
To be invertible, sine is restricted to [−π/2, π/2], so its inverse outputs values in that closed interval. The interval [0, π] is the range of arccos, not arcsin, and the endpoints are included.
Question 5 of 10
MCQU1Topic 1.11Medium No calc
When the polynomial p(x) = 2x³ − 3x² + 4x − 5 is divided by (x − 2), what is the remainder?
A−5
B11
C7
D0
Explanation
By the Remainder Theorem the remainder equals p(2) = 2(8) − 3(4) + 4(2) − 5 = 16 − 12 + 8 − 5 = 7. A remainder of 0 would mean (x − 2) is a factor, and −5 is just the constant term.
Question 6 of 10
MCQU2Topic 2.3Easy No calc
For f(x) = 7(2)ˣ, what is f(0), and what does it represent in the model?
A7, the initial value
B0, the starting point
C14, the initial value
D2, the growth factor
Explanation
Any nonzero base raised to the 0 power is 1, so f(0) = 7·2⁰ = 7·1 = 7. In an exponential model a·bˣ, the coefficient a is the initial value (the output when x = 0).
Question 7 of 10
MCQU3Topic 3.5Medium No calc
A sinusoidal function oscillates between a minimum of y = 2 and a maximum of y = 8. What are its amplitude and midline?
AAmplitude 6, midline y = 4
BAmplitude 3, midline y = 5
CAmplitude 5, midline y = 3
DAmplitude 3, midline y = 4
Explanation
The amplitude is half the peak-to-trough distance: (8 − 2)/2 = 3. The midline is the average of the max and min: (8 + 2)/2 = 5. Using the full distance 6 for amplitude is a common error.
Question 8 of 10
MCQU2Topic 2.9Medium No calc
What is the value of log₂(32)?
A4
B5
C16
D6
Explanation
log₂(32) asks for the exponent on 2 that gives 32. Since 2⁵ = 32, the value is 5. The answer 16 is 32/2, and 4 would give only 2⁴ = 16.
Question 9 of 10
MCQU1Topic 1.10Medium No calc
What are the coordinates of the hole in the graph of r(x) = (x² − x − 6) / (x − 3)?
AThere is no hole; x = 3 is a vertical asymptote
B(−2, 5)
C(3, 0)
D(3, 5)
Explanation
Factor the numerator: x² − x − 6 = (x − 3)(x + 2). The common factor (x − 3) cancels, creating a hole at x = 3. Substituting x = 3 into the simplified form (x + 2) gives 5, so the hole is at (3, 5).
Question 10 of 10
MCQU3Topic 3.3Easy No calc
What is the exact value of cos(π/3)?
A1
B√3/2
C√2/2
D1/2
Explanation
On the unit circle, the angle π/3 (60°) has cosine 1/2. The value √3/2 is the sine of π/3, so the two are easy to swap.
Free Response 1 · Section II
FRQModeling a Periodic ContextU3 No calc
The figure shows the graph of h, the height (in meters, relative to a fixed reference) of a marked point on a slowly rotating wheel, as a function of time t in seconds. No calculator is allowed. (a) From the graph, state the amplitude, the midline, and the period of h. (b) Write an equation for h(t) using a cosine function. (c) Find all values of t in 0 ≤ t ≤ 8 at which h(t) equals the midline value. (d) Explain what the amplitude and the midline represent physically for the marked point.
Free response is self-scored — work it out, then reveal the model answer and scoring checklist to compare.
Model answer
(a) Reading the graph, the curve rises to a maximum of 8 meters and falls to a minimum of -2 meters. Amplitude = (maximum - minimum)/2 = (8 - (-2))/2 = 10/2 = 5 meters. Midline: h = (maximum + minimum)/2 = (8 + (-2))/2 = 6/2 = 3 meters, so the midline is h = 3 (matching the dashed line). Period: one full cycle (maximum back to maximum) takes 4 seconds, so the period is 4 seconds.
(b) A cosine model has the form h(t) = A cos(b·t) + D. Since the graph begins at its maximum at t = 0, an unshifted cosine (with positive A) fits directly. A = amplitude = 5, D = midline = 3, and b = 2π/period = 2π/4 = π/2. h(t) = 5 cos((π/2)t) + 3, with h in meters and t in seconds.
(c) h(t) equals the midline value when h(t) = 3: 5 cos((π/2)t) + 3 = 3 5 cos((π/2)t) = 0 cos((π/2)t) = 0. Cosine is 0 at odd multiples of π/2: (π/2)t = π/2 + kπ for integer k. Dividing by π/2: t = 1 + 2k. For 0 ≤ t ≤ 8: t = 1, 3, 5, 7 seconds.
(d) The midline, h = 3 meters, is the height of the center (axle) of the wheel above the fixed reference; it is the constant height about which the marked point oscillates. The amplitude, 5 meters, is the radius of the wheel — the greatest distance the marked point rises above or falls below that center as the wheel turns. So the point moves between 3 + 5 = 8 meters at the top and 3 - 5 = -2 meters at the bottom.
Scoring · 6 points
1. (a) Correct amplitude = 5 m and midline h = 3 m read from the graph (1 pt)
2. (a) Correct period = 4 seconds identified from one full cycle (1 pt)
3. (b) Correct cosine equation h(t) = 5 cos((π/2)t) + 3 with b = π/2 (1 pt)
4. (c) Sets h(t) = 3 and reduces to cos((π/2)t) = 0 correctly (1 pt)
5. (c) Finds all four solutions t = 1, 3, 5, 7 seconds on 0 ≤ t ≤ 8 (1 pt)
6. (d) Explains midline = axle/center height and amplitude = wheel radius (max distance from center) (1 pt)