This mock exam mirrors the College Board CLEP Precalculus exam: 48 questions, 90 minutes, computer-delivered. On the real exam the questions are split into two sections with a calculator quirk:
Some real items are numeric-entry (you type the answer). For clean self-grading, all 48 questions here are presented as five-choice multiple choice (A–E), exactly one correct. Practice Section 2 discipline by working the no-calculator topics (identities, exact values, factoring, exponent rules) entirely by hand.
Scoring: scores are reported on a 20–80 scaled score. The American Council on Education (ACE) recommends granting credit at a scaled score of 50, which corresponds to roughly half the questions correct. Use the estimator at the bottom to convert your raw score.
| Official content area | Weight | Questions |
|---|---|---|
| Algebraic Expressions, Equations, and Inequalities | 20% | 10 |
| Functions: Concept, Properties, and Operations | 15% | 7 |
| Representations of Functions (Symbolic, Graphical, Tabular) | 30% | 14 |
| Analytic Geometry | 10% | 5 |
| Trigonometry and its Applications | 15% | 7 |
| Functions as Models | 10% | 5 |
| Total | 100% | 48 |
Bloom target: Apply + Analyze + Evaluate ≥ 60% of items (this exam runs ~88%). Answer-distribution target: each letter 15–25% of the key. Topics are interleaved so the sequence feels like a real mixed CLEP form.
1. A) 4x⁴. Distribute the outer exponent to every factor: 8^(2/3) = (∛8)² = 2² = 4, and (x⁶)^(2/3) = x^(6·2/3) = x⁴, giving 4x⁴. Distractors: B) 8x⁴ leaves the 8 untouched, but the exponent applies to it too. C) 16x⁴ computes 8·2 = 16 instead of (∛8)². D) 4x⁹ inverts the exponent on x, dividing 6 by 2/3 instead of multiplying. E) 6x⁴ mis-evaluates 8^(2/3) as 6. Fix: on a power, take the root (denominator) first, then the power (numerator), and apply the outer exponent to each inside factor. [Apply]
2. C) 2a² + a. Substitute: 2(a+1)² − 3(a+1) + 1 = 2(a²+2a+1) − 3a − 3 + 1 = 2a² + 4a + 2 − 3a − 2 = 2a² + a. Distractors: A) 2a² − 3a + 2 computes f(a) + 1, treating f(a+1) as f(a) + 1. B) 2a² + 4a + 2 expands only the squared term and drops −3(a+1) + 1. D) 2a² − 3a uses (a+1)² = a² + 1, losing the middle term 2a. E) 2a² + 7a + 6 flips the sign of the −3(a+1) term to +3(a+1). Fix: replace every x with the whole input (a+1), then expand (a+1)² = a² + 2a + 1 in full. [Apply]
| x | 0 | 1 | 2 | 3 |
|---|---|---|---|---|
| f(x) | 3 | 6 | 12 | 24 |
Which of the following defines f?
3. E) f(x) = 3·2^x. Outputs multiply by 2 at each step (3→6→12→24), a constant ratio, so the function is exponential with initial value 3 and base 2: f(x) = 3·2^x, which reproduces 3, 6, 12, 24. Distractors: A) 3x + 3 is linear, matching only the first gap (gives 3, 6, 9, 12). B) 2·3^x swaps the initial value and the growth factor (gives 2, 6, 18, 54). C) 3x² + 3 forces a quadratic (gives 3, 6, 15, 30). D) 3 + 2^x gives 4, 5, 7, 11 and fails even at x = 0. Fix: constant differences signal linear; constant ratios signal exponential y = (initial)·(ratio)^x. [Analyze]
4. B) −√3/2. The angle 5π/6 = 150° lies in Quadrant II with reference angle π/6; cosine is negative there, so cos(5π/6) = −cos(π/6) = −√3/2. Distractors: A) √3/2 drops the negative sign required in Quadrant II. C) −1/2 is cos(2π/3), not cos(5π/6). D) 1/2 confuses it with sin(5π/6). E) −√2/2 belongs to 135° (3π/4), a different reference angle. Fix: find the reference angle, take the exact value there, then attach the sign from the quadrant. [Remember]
5. D) center (3, −2), radius 5. Complete the square: (x² − 6x) + (y² + 4y) = 12 → (x − 3)² − 9 + (y + 2)² − 4 = 12 → (x − 3)² + (y + 2)² = 25, so center (3, −2), radius √25 = 5. Distractors: A) (−3, 2), r = 5 flips both center signs. B) (3, −2), r = 25 reports r² as the radius. C) combines both errors. E) (6, −4), √12 reads coefficients directly without completing the square. Fix: center coordinates are the opposite of the numbers inside the squared terms, and the radius is the square root of the right-hand side. [Apply]
6. A) f(x − 3) − 2. A shift right by 3 replaces x with (x − 3) inside the function; a shift down by 2 subtracts 2 outside: g(x) = f(x − 3) − 2. Distractors: B) f(x + 3) − 2 shifts left (inside sign reversed). C) f(x − 3) + 2 shifts up instead of down. D) f(x − 2) − 3 swaps the horizontal and vertical amounts. E) f(x + 3) + 2 reverses both directions. Fix: inside changes act horizontally and opposite to the sign; outside changes act vertically and with the sign. [Understand]
7. C) P(t) = 500·2^(t/4). Doubling means base 2; "every 4 hours" makes the exponent t/4 (the number of 4-hour periods), so P(t) = 500·2^(t/4). Check t = 4: 500·2¹ = 1000, one doubling. Distractors: A) 500·2^(4t) doubles every 1/4 hour. B) 500·2^t doubles every hour. D) 500·4^t quadruples every hour. E) 500 + 2^(t/4) adds instead of scaling, breaking the initial value. Fix: exponent = (elapsed time) ÷ (time per one doubling); the growth factor multiplies, never adds. [Apply]
8. B) x = −1/2 and x = 3. Factor: 2x² − 5x − 3 = (2x + 1)(x − 3) = 0, so x = −1/2 or x = 3. Distractors: A) x = 1/2, −3 flips both signs. C) x = −1/2, −3 gets the sign of the 3-root wrong. D) x = 1/2, 3 mishandles the 2x + 1 factor. E) x = 2, −3 factors the constant without the leading 2. Fix: set each factor to zero; from (2x + 1) the root is −1/2, not −2. [Apply]
9. D) y = −(x + 1)² + 4. Vertex (−1, 4) gives y = a(x + 1)² + 4; using (0, 3): a(1) + 4 = 3, so a = −1. Then y = −(x + 1)² + 4, and (1, 0) checks: −(2²) + 4 = 0. Distractors: A) opens upward (a = +1), contradicting the downward shape. B) y = −(x − 1)² + 4 puts the vertex at x = +1. C) has vertex (−1, −4), below the x-axis. E) y = −(x + 1)² + 3 lowers the vertex to 3 and fails (0, 3). Fix: read the vertex into a(x − h)² + k (inside sign reversed), then solve for a with a second point. [Analyze]
10. E) 10. Work inside out: g(−2) = (−2)² = 4, then f(4) = 3(4) − 2 = 10. Distractors: A) 64 reverses the order, computing g(f(−2)) = g(−8). B) −14 uses g(−2) = −4 (sign error squaring). C) 16 stops at g of an intermediate value. D) −8 reports only f(−2). Fix: (f ∘ g)(x) means f(g(x)) — evaluate the inner function g first. [Apply]
11. A) 12/13. For an acute angle, cos²θ = 1 − sin²θ = 1 − 25/169 = 144/169, so cos θ = 12/13 (positive in Quadrant I). This is the 5–12–13 right triangle. Distractors: B) 13/12 is sec θ, the reciprocal. C) 5/12 is tan θ. D) 12/5 is cot θ. E) 8/13 uses a wrong leg length (8 is not the missing side). Fix: with sin = opp/hyp = 5/13, the adjacent leg is √(13² − 5²) = 12, so cos = 12/13. [Apply]
12. C) x ≥ 3, x ≠ 5. The radicand needs 2x − 6 ≥ 0, i.e. x ≥ 3, and the denominator forbids x = 5; combine to x ≥ 3 with x ≠ 5. Distractors: A) x ≥ 3 ignores the denominator restriction. B) x > 3 wrongly excludes the endpoint 3, where the numerator is 0 but the function is defined. D) drops the square-root requirement entirely. E) 3 ≤ x ≤ 5 invents an upper bound. Fix: intersect two conditions — radicand ≥ 0 (endpoint included) AND denominator ≠ 0. [Analyze]
13. B) −2 < x < 3. |2x − 1| < 5 means −5 < 2x − 1 < 5; add 1: −4 < 2x < 6; divide by 2: −2 < x < 3. Distractors: A) x < 3 keeps only one side. C) x > −2 keeps only the other side. D) x < −2 or x > 3 is the solution to the greater-than inequality |2x − 1| > 5. E) −3 < x < 2 mishandles the arithmetic after isolating x. Fix: |A| < c becomes the compound −c < A < c (an "and"); |A| > c becomes an "or." [Apply]
14. D) 10. Distance = √((4 − (−2))² + (−5 − 3)²) = √(6² + (−8)²) = √(36 + 64) = √100 = 10. Distractors: A) √14 adds the coordinate differences (6 + 8 = 14) before rooting. B) 14 adds |6| + |8| with no rooting. C) √28 mishandles the squares. E) √52 squares only part of the differences (e.g., 6² + 4²). Fix: distance = √((Δx)² + (Δy)²); square the differences, add, then take one square root. [Apply]
15. E) f(x) = 4·(1/2)^x. The curve decays with initial value 4 and halves each step (4→2→1), so base 1/2: f(x) = 4·(1/2)^x, matching all three points. Distractors: A) 4·2^x grows instead of decaying. B) 4 + (1/2)^x has y-intercept 5 and asymptote y = 4, not y = 0. C) 4/x is undefined at x = 0. D) 2·(1/2)^x has y-intercept 2, not 4. Fix: decay → base between 0 and 1; the y-intercept fixes the initial value, the ratio fixes the base. [Analyze]
16. A) $40. Cost = 25 + 0.15(100) = 25 + 15 = $40. Distractors: B) $15 drops the base fee. C) $125 ignores the per-minute rate and adds the minutes as dollars (25 + 100). D) $375 multiplies base × rate × minutes. E) $2,500 multiplies base × minutes, ignoring the rate. Fix: a linear model is (fixed base) + (rate × quantity); add the flat fee once, scale only the variable part. [Apply]
17. B) f(x) = x³ − x. A function is odd when f(−x) = −f(x). Here f(−x) = (−x)³ − (−x) = −x³ + x = −(x³ − x) = −f(x), so it is odd. Distractors: A) x² + 1 is even, f(−x) = f(x). C) |x| is even. D) x³ + 1 gives f(−x) = −x³ + 1, neither odd nor even. E) x² − x mixes an even and an odd term, so it is neither. Fix: odd means every term has an odd power and no constant; test f(−x) = −f(x). [Analyze]
18. C) 11. Match each input to its condition: −3 < 0 → f(−3) = −3 + 4 = 1; 0 ≤ 2 < 3 → f(2) = 2² = 4; 3 ≥ 3 → f(3) = 2(3) = 6. Sum = 1 + 4 + 6 = 11. Distractors: A) 9 uses the wrong branch for one input. B) 13 squares −3 by using the middle branch. D) 17 uses 2x for f(2) as well. E) 7 evaluates only f(3). Fix: for each input, first decide which condition it satisfies, then use only that branch. [Apply]
19. D) x = 3. Square both sides: x + 6 = x², so x² − x − 6 = 0 = (x − 3)(x + 2), giving x = 3 or x = −2. Check in the original: √9 = 3 ✓, but √4 = 2 ≠ −2, so x = −2 is extraneous. Only x = 3 works. Distractors: A) keeps both roots without checking. B) keeps the extraneous root and discards the valid one. C) declares both valid. E) claims no solution, but x = 3 checks. Fix: squaring can create extraneous roots — always substitute back into the original radical equation. [Evaluate]
20. E) x = π/6 and x = 5π/6. Solve sin x = 1/2 on [0, 2π): the reference angle is π/6, and sine is positive in Quadrants I and II, giving x = π/6 and x = π − π/6 = 5π/6. Distractors: A) π/6 only omits the second-quadrant solution. B) π/3, 2π/3 solves sin x = √3/2. C) π/6, 7π/6 uses Quadrant III (where sine is negative). D) π/6, 11π/6 uses Quadrant IV. Fix: after the reference angle, place solutions in the quadrants where the function has the required sign — sine positive means Q I and Q II. [Apply]
21. A) 32. log₂ x = 5 means x = 2⁵ = 32. Distractors: B) 25 computes 5². C) 10 computes 5·2. D) 7 computes 5 + 2. E) 64 computes 2⁶. Fix: log_b x = y converts to x = b^y — the base raised to the log. [Apply]
22. B) y = −(3/2)x + 5. The perpendicular slope is the negative reciprocal of 2/3, which is −3/2. Point-slope through (4, −1): y + 1 = −(3/2)(x − 4) = −(3/2)x + 6, so y = −(3/2)x + 5. Distractors: A) reuses the original slope 2/3. C) y = −(3/2)x − 7 uses the right slope but a sign error placing the point. D) y = (3/2)x − 7 forgets the negative in the reciprocal. E) uses the negative of the original slope (−2/3) rather than its reciprocal. Fix: perpendicular slope = negative reciprocal (flip and change sign), then anchor with the given point. [Apply]
23. C) (2x + 1)/(1 − x). Set y = (x − 1)/(x + 2) and solve for x: y(x + 2) = x − 1 → yx + 2y = x − 1 → x(y − 1) = −1 − 2y → x = (2y + 1)/(1 − y). Swapping variable names, f⁻¹(x) = (2x + 1)/(1 − x). Check x = 0: f(0) = −1/2 and f⁻¹(−1/2) = 0 ✓. Distractors: A) (x + 2)/(x − 1) merely flips the fraction. B) (2x + 1)/(x − 1) has the wrong denominator sign. D) (1 − x)/(2x + 1) inverts the correct answer. E) is the original function unchanged. Fix: swap x and y, then solve for y — collect all y-terms on one side and factor. [Apply]
24. D) y = −f(x) + 3. Reflecting across the x-axis negates outputs: −f(x); shifting up 3 adds 3 outside: −f(x) + 3. Distractors: A) f(−x) + 3 reflects across the y-axis instead. B) −f(x + 3) shifts horizontally, not vertically. C) −f(x) − 3 shifts down. E) f(−x) − 3 reflects the wrong axis and shifts down. Fix: x-axis reflection negates the whole function; vertical shift adds outside. [Understand]
25. E) (3, 1). Add the equations to eliminate y: (2x + y) + (x − y) = 7 + 2 → 3x = 9 → x = 3; then y = 7 − 2(3) = 1. Check: 3 − 1 = 2 ✓. Distractors: A) (1, 3) swaps the coordinates. B) (2, 3) fails both equations. C) (3, −1) has a sign error in y. D) (5, −3) satisfies neither equation. Fix: with opposite y-coefficients, add the equations to eliminate y in one step, then back-substitute. [Apply]
26. A) 64 ft. The vertex of h(t) = −16t² + 32t + 48 is at t = −b/(2a) = −32/(2·−16) = 1 s; h(1) = −16 + 32 + 48 = 64 ft. Distractors: B) 48 ft reports h(0), the initial height, not the max. C) 32 ft misreads the linear coefficient as the height. D) 80 ft adds 32 + 48 without evaluating at the vertex. E) 96 ft doubles the initial height. Fix: maximum of a downward parabola occurs at t = −b/(2a); substitute that t back to get the height. [Apply]
| x | 1 | 3 | 6 |
|---|---|---|---|
| f(x) | 4 | 10 | 25 |
The average rate of change of f from x = 1 to x = 6 is
27. B) 21/5. Average rate of change = (f(6) − f(1))/(6 − 1) = (25 − 4)/5 = 21/5. Distractors: A) 21 forgets to divide by the change in x. C) 7/2 uses the wrong interval or averages differently. D) 5/21 inverts the ratio. E) 3 divides 21 by 7 instead of 5. Fix: average rate of change is Δy/Δx between the two endpoints — the slope of the secant line. [Apply]
28. C) sin θ. Using 1 − cos²θ = sin²θ, the expression is sin²θ / sin θ = sin θ. Distractors: A) cos θ misapplies the Pythagorean identity. B) csc θ inverts the result. D) tan θ comes from dividing by cos θ. E) 1 cancels incorrectly, treating sin²θ/sin θ as 1. Fix: replace 1 − cos²θ with sin²θ first, then cancel one factor of sin θ. [Apply]
29. D) no real solutions. The discriminant is b² − 4ac = (−4)² − 4(1)(7) = 16 − 28 = −12 < 0, so there are no real solutions (two complex conjugates). Distractors: A) and B) require a positive discriminant. C) requires discriminant = 0. E) infinitely many never occurs for a genuine quadratic. Fix: the sign of b² − 4ac decides: positive → two real, zero → one repeated, negative → none real. [Analyze]
30. A) It touches the x-axis at x = −2 and crosses at x = 3. An even-multiplicity zero (x = −2, mult 2) makes the graph touch and turn; an odd-multiplicity zero (x = 3, mult 1) makes it cross. Distractors: B) reverses which zero touches and which crosses. C) treats both as odd. D) treats both as even. E) invents an extra zero at x = 0 that the factored form does not provide. Fix: even multiplicity → bounce (touch); odd multiplicity → pass through (cross). [Analyze]
31. E) x ≥ 4. (f ∘ g)(x) = √(x − 4) requires x − 4 ≥ 0, so x ≥ 4. Distractors: A) all reals ignores the radical. B) x ≥ 0 uses the domain of √x without shifting. C) x ≤ 4 reverses the inequality. D) x > 4 wrongly excludes the endpoint 4, where √0 = 0 is defined. Fix: for √(inner), require inner ≥ 0 (endpoint included) and solve. [Apply]
32. B) vertical, length 10. In x²/9 + y²/25 = 1 the larger denominator (25) sits under y², so the major axis is vertical with semi-axis a = √25 = 5 and full length 2a = 10. Distractors: A) uses the smaller denominator and wrong orientation. C) right length but horizontal orientation. D) length 25 forgets to take the square root and double. E) uses √9 = 3 for the semi-axis. Fix: the larger denominator marks the major axis; its length is 2√(that denominator). [Understand]
33. C) horizontal compression by a factor of 1/2. Replacing x with 2x inside f speeds the input, compressing the graph horizontally toward the y-axis by a factor of 1/2. Distractors: A) a stretch by 2 comes from f(x/2), not f(2x). B) and D) describe vertical changes, which come from a coefficient outside f. E) a horizontal shift comes from f(x − 2), not f(2x). Fix: f(bx) with b > 1 compresses horizontally by 1/b; the effect is the reciprocal of the coefficient. [Understand]
34. D) x = 6. Isolate: 1/x = 1/2 − 1/3 = 3/6 − 2/6 = 1/6, so x = 6. Check: 1/6 + 1/3 = 1/6 + 2/6 = 3/6 = 1/2 ✓. Distractors: A) 1/6 reports 1/x instead of x. B) 5 comes from subtracting denominators. C) 1 mishandles the fraction subtraction. E) −6 sign error. Fix: combine the constant fractions with a common denominator first, then reciprocate to solve for x. [Apply]
35. A) π. The period of sin(Bx + C) is 2π/|B|; here B = 2, so the period is 2π/2 = π. The amplitude 3 and phase shift −π do not affect the period. Distractors: B) 2π ignores the factor 2. C) π/2 divides by B². D) 4π multiplies by 2 instead of dividing. E) 3π uses the amplitude 3. Fix: period depends only on the coefficient of x: 2π/|B|. [Apply]
36. E) 11 − 10i. FOIL: (3 + 2i)(1 − 4i) = 3 − 12i + 2i − 8i² = 3 − 10i − 8(−1) = 3 − 10i + 8 = 11 − 10i. Distractors: A) 3 − 8i drops the outer/inner terms. B) 11 + 10i has the imaginary sign wrong. C) −5 − 10i uses i² = +1. D) 3 − 10i forgets to convert −8i². Fix: multiply as binomials, then replace i² with −1 and combine real and imaginary parts. [Apply]
37. B) 10 g. In 18 years there are 18/6 = 3 half-lives, so the amount is 80·(1/2)³ = 80/8 = 10 g. Distractors: A) 20 g stops after 2 half-lives. C) 5 g runs 4 half-lives. D) 40 g runs only 1 half-life. E) 26.7 g divides by 3 linearly instead of halving repeatedly. Fix: number of half-lives = elapsed time ÷ half-life; multiply the start by (1/2) that many times. [Apply]
38. C) 2x + h. Expand: [f(x+h) − f(x)]/h = [(x + h)² − x²]/h = [x² + 2xh + h² − x²]/h = (2xh + h²)/h = 2x + h. Distractors: A) 2x drops the h before it can be canceled properly (that is the limit, not the quotient). B) 2x + h² fails to divide h² by h. D) x + h forgets the factor of 2 from the middle term. E) 2xh cancels incorrectly. Fix: expand (x + h)², subtract x², then divide every remaining term by h. [Apply]
39. D) y = 3 sin x − 1. Amplitude 3 (half of 2 − (−4) = 6) and midline y = −1 give y = 3 sin x − 1; it passes through (0, −1) rising to a max of 2 at x = π/2, matching sine. Distractors: A) sin x + 3 has amplitude 1 and the wrong midline. B) 3 sin x + 1 uses midline +1. C) −3 sin x − 1 is reflected, so it falls from (0, −1). E) 3 cos x − 1 starts at its maximum at x = 0, not on the midline. Fix: amplitude = (max − min)/2, midline = (max + min)/2; sine starts on the midline rising, cosine starts at a peak. [Analyze]
40. A) x < −2 or x > 3. Factor: x² − x − 6 = (x − 3)(x + 2) > 0. The product is positive outside the roots −2 and 3, so x < −2 or x > 3. Distractors: B) −2 < x < 3 is where the product is negative (< 0). C) and D) use wrong roots (−3 and 2). E) x > 3 only omits the left branch. Fix: find the roots, then a "> 0" parabola opening up is positive outside the roots, negative between them. [Analyze]
41. B) 40√3/3 ft. The height is opposite the 30° angle and 40 ft is adjacent, so tan 30° = h/40, giving h = 40 tan 30° = 40·(1/√3) = 40√3/3 ≈ 23.1 ft. Distractors: A) 20 ft uses sin 30° = 1/2 as the ratio. C) 40√3 ft uses tan 60° (or multiplies by √3). D) 40 ft ignores the trig. E) 80 ft doubles. Fix: pick the ratio linking the known side to the unknown: opposite over adjacent is tangent. [Apply]
42. E) (−1, 2). A function decreases where its graph falls; between the local maximum at x = −1 and the local minimum at x = 2 the graph descends, so it decreases on (−1, 2). Distractors: A) (−∞, −1) and B) (2, ∞) are where it rises. C) (−1, ∞) and D) (−∞, 2) each mix an increasing part with the decreasing part. Fix: decreasing = falling from left to right, which happens between a max (on the left) and the next min (on the right). [Analyze]
43. C) (x + 1)² + (y − 4)² = 25. The radius is the distance from center (−1, 4) to (2, 0): √((2 + 1)² + (0 − 4)²) = √(9 + 16) = 5, so r² = 25 and the equation is (x + 1)² + (y − 4)² = 25. Distractors: A) flips the center signs. B) uses r = 5 as r² (should be 25). D) flips signs and uses r for r². E) 100 squares the radius twice. Fix: standard form uses (x − h)² + (y − k)² = r²; find r as a distance, then square it. [Apply]
44. D) x = 3. log(x − 3) is defined only when x − 3 > 0; as x → 3⁺ the argument → 0⁺ and the log → −∞, giving a vertical asymptote at x = 3. Distractors: A) y = 3 and C) y = 0 are horizontal lines; logs have vertical asymptotes. B) x = 0 is the asymptote of log x before the shift. E) x = −3 has the shift direction backward. Fix: a log's vertical asymptote sits where its argument equals 0; solve x − 3 = 0. [Understand]
45. B) x = 3 and x = −5/3. |3x − 2| = 7 splits into 3x − 2 = 7 → x = 3, and 3x − 2 = −7 → 3x = −5 → x = −5/3. Distractors: A) x = 3 only keeps one case. C) x = 3, 5/3 gets the sign of the second root wrong. D) x = −3, 5/3 mishandles both. E) x = 9, −5/3 adds 7 to 2 before dividing. Fix: |A| = c (with c > 0) yields two equations, A = c and A = −c; solve each. [Apply]
46. A) $14,450. Depreciating 15% per year multiplies by 0.85 annually: 20,000·(0.85)² = 20,000·0.7225 = $14,450. Distractors: B) $14,000 subtracts a flat 30% (15% × 2), treating it as linear. C) $17,000 applies 15% just once. D) $13,000 subtracts 35%. E) $15,000 rounds arbitrarily. Fix: a constant percent change compounds — multiply by (1 − rate) once per year, not subtract a flat total. [Apply]
47. E) f(x) = x³ + 1. A one-to-one function passes the horizontal-line test. The cubic x³ + 1 is strictly increasing, so each output comes from exactly one input. Distractors: A) x², B) |x|, and C) x² − 4 each hit the same output for ±x (fail the horizontal-line test). D) the constant 4 sends every input to one output — as far from one-to-one as possible. Fix: one-to-one means no horizontal line meets the graph twice; strictly increasing (or decreasing) functions qualify. [Analyze]
48. D) 3/2. tan(π/4) = 1 and sin(π/6) = 1/2, so the sum is 1 + 1/2 = 3/2. Distractors: A) 1 uses only the tangent term. B) 1/2 uses only the sine term. C) √3/2 + 1 uses sin(π/3) instead of sin(π/6). E) √2/2 + 1/2 uses sin(π/4) for the first term. Fix: recall the exact values separately — tan(π/4) = 1, sin(π/6) = 1/2 — then add. [Apply]
1. A) 4x⁴. Distribute the outer exponent to every factor: 8^(2/3) = (∛8)² = 2² = 4, and (x⁶)^(2/3) = x^(6·2/3) = x⁴, giving 4x⁴. Distractors: B) 8x⁴ leaves the 8 untouched, but the exponent applies to it too. C) 16x⁴ computes 8·2 = 16 instead of (∛8)². D) 4x⁹ inverts the exponent on x, dividing 6 by 2/3 instead of multiplying. E) 6x⁴ mis-evaluates 8^(2/3) as 6. Fix: on a power, take the root (denominator) first, then the power (numerator), and apply the outer exponent to each inside factor. [Apply]
2. C) 2a² + a. Substitute: 2(a+1)² − 3(a+1) + 1 = 2(a²+2a+1) − 3a − 3 + 1 = 2a² + 4a + 2 − 3a − 2 = 2a² + a. Distractors: A) 2a² − 3a + 2 computes f(a) + 1, treating f(a+1) as f(a) + 1. B) 2a² + 4a + 2 expands only the squared term and drops −3(a+1) + 1. D) 2a² − 3a uses (a+1)² = a² + 1, losing the middle term 2a. E) 2a² + 7a + 6 flips the sign of the −3(a+1) term to +3(a+1). Fix: replace every x with the whole input (a+1), then expand (a+1)² = a² + 2a + 1 in full. [Apply]
3. E) f(x) = 3·2^x. Outputs multiply by 2 at each step (3→6→12→24), a constant ratio, so the function is exponential with initial value 3 and base 2: f(x) = 3·2^x, which reproduces 3, 6, 12, 24. Distractors: A) 3x + 3 is linear, matching only the first gap (gives 3, 6, 9, 12). B) 2·3^x swaps the initial value and the growth factor (gives 2, 6, 18, 54). C) 3x² + 3 forces a quadratic (gives 3, 6, 15, 30). D) 3 + 2^x gives 4, 5, 7, 11 and fails even at x = 0. Fix: constant differences signal linear; constant ratios signal exponential y = (initial)·(ratio)^x. [Analyze]
4. B) −√3/2. The angle 5π/6 = 150° lies in Quadrant II with reference angle π/6; cosine is negative there, so cos(5π/6) = −cos(π/6) = −√3/2. Distractors: A) √3/2 drops the negative sign required in Quadrant II. C) −1/2 is cos(2π/3), not cos(5π/6). D) 1/2 confuses it with sin(5π/6). E) −√2/2 belongs to 135° (3π/4), a different reference angle. Fix: find the reference angle, take the exact value there, then attach the sign from the quadrant. [Remember]
5. D) center (3, −2), radius 5. Complete the square: (x² − 6x) + (y² + 4y) = 12 → (x − 3)² − 9 + (y + 2)² − 4 = 12 → (x − 3)² + (y + 2)² = 25, so center (3, −2), radius √25 = 5. Distractors: A) (−3, 2), r = 5 flips both center signs. B) (3, −2), r = 25 reports r² as the radius. C) combines both errors. E) (6, −4), √12 reads coefficients directly without completing the square. Fix: center coordinates are the opposite of the numbers inside the squared terms, and the radius is the square root of the right-hand side. [Apply]
6. A) f(x − 3) − 2. A shift right by 3 replaces x with (x − 3) inside the function; a shift down by 2 subtracts 2 outside: g(x) = f(x − 3) − 2. Distractors: B) f(x + 3) − 2 shifts left (inside sign reversed). C) f(x − 3) + 2 shifts up instead of down. D) f(x − 2) − 3 swaps the horizontal and vertical amounts. E) f(x + 3) + 2 reverses both directions. Fix: inside changes act horizontally and opposite to the sign; outside changes act vertically and with the sign. [Understand]
7. C) P(t) = 500·2^(t/4). Doubling means base 2; "every 4 hours" makes the exponent t/4 (the number of 4-hour periods), so P(t) = 500·2^(t/4). Check t = 4: 500·2¹ = 1000, one doubling. Distractors: A) 500·2^(4t) doubles every 1/4 hour. B) 500·2^t doubles every hour. D) 500·4^t quadruples every hour. E) 500 + 2^(t/4) adds instead of scaling, breaking the initial value. Fix: exponent = (elapsed time) ÷ (time per one doubling); the growth factor multiplies, never adds. [Apply]
8. B) x = −1/2 and x = 3. Factor: 2x² − 5x − 3 = (2x + 1)(x − 3) = 0, so x = −1/2 or x = 3. Distractors: A) x = 1/2, −3 flips both signs. C) x = −1/2, −3 gets the sign of the 3-root wrong. D) x = 1/2, 3 mishandles the 2x + 1 factor. E) x = 2, −3 factors the constant without the leading 2. Fix: set each factor to zero; from (2x + 1) the root is −1/2, not −2. [Apply]
9. D) y = −(x + 1)² + 4. Vertex (−1, 4) gives y = a(x + 1)² + 4; using (0, 3): a(1) + 4 = 3, so a = −1. Then y = −(x + 1)² + 4, and (1, 0) checks: −(2²) + 4 = 0. Distractors: A) opens upward (a = +1), contradicting the downward shape. B) y = −(x − 1)² + 4 puts the vertex at x = +1. C) has vertex (−1, −4), below the x-axis. E) y = −(x + 1)² + 3 lowers the vertex to 3 and fails (0, 3). Fix: read the vertex into a(x − h)² + k (inside sign reversed), then solve for a with a second point. [Analyze]
10. E) 10. Work inside out: g(−2) = (−2)² = 4, then f(4) = 3(4) − 2 = 10. Distractors: A) 64 reverses the order, computing g(f(−2)) = g(−8). B) −14 uses g(−2) = −4 (sign error squaring). C) 16 stops at g of an intermediate value. D) −8 reports only f(−2). Fix: (f ∘ g)(x) means f(g(x)) — evaluate the inner function g first. [Apply]
11. A) 12/13. For an acute angle, cos²θ = 1 − sin²θ = 1 − 25/169 = 144/169, so cos θ = 12/13 (positive in Quadrant I). This is the 5–12–13 right triangle. Distractors: B) 13/12 is sec θ, the reciprocal. C) 5/12 is tan θ. D) 12/5 is cot θ. E) 8/13 uses a wrong leg length (8 is not the missing side). Fix: with sin = opp/hyp = 5/13, the adjacent leg is √(13² − 5²) = 12, so cos = 12/13. [Apply]
12. C) x ≥ 3, x ≠ 5. The radicand needs 2x − 6 ≥ 0, i.e. x ≥ 3, and the denominator forbids x = 5; combine to x ≥ 3 with x ≠ 5. Distractors: A) x ≥ 3 ignores the denominator restriction. B) x > 3 wrongly excludes the endpoint 3, where the numerator is 0 but the function is defined. D) drops the square-root requirement entirely. E) 3 ≤ x ≤ 5 invents an upper bound. Fix: intersect two conditions — radicand ≥ 0 (endpoint included) AND denominator ≠ 0. [Analyze]
13. B) −2 < x < 3. |2x − 1| < 5 means −5 < 2x − 1 < 5; add 1: −4 < 2x < 6; divide by 2: −2 < x < 3. Distractors: A) x < 3 keeps only one side. C) x > −2 keeps only the other side. D) x < −2 or x > 3 is the solution to the greater-than inequality |2x − 1| > 5. E) −3 < x < 2 mishandles the arithmetic after isolating x. Fix: |A| < c becomes the compound −c < A < c (an "and"); |A| > c becomes an "or." [Apply]
14. D) 10. Distance = √((4 − (−2))² + (−5 − 3)²) = √(6² + (−8)²) = √(36 + 64) = √100 = 10. Distractors: A) √14 adds the coordinate differences (6 + 8 = 14) before rooting. B) 14 adds |6| + |8| with no rooting. C) √28 mishandles the squares. E) √52 squares only part of the differences (e.g., 6² + 4²). Fix: distance = √((Δx)² + (Δy)²); square the differences, add, then take one square root. [Apply]
15. E) f(x) = 4·(1/2)^x. The curve decays with initial value 4 and halves each step (4→2→1), so base 1/2: f(x) = 4·(1/2)^x, matching all three points. Distractors: A) 4·2^x grows instead of decaying. B) 4 + (1/2)^x has y-intercept 5 and asymptote y = 4, not y = 0. C) 4/x is undefined at x = 0. D) 2·(1/2)^x has y-intercept 2, not 4. Fix: decay → base between 0 and 1; the y-intercept fixes the initial value, the ratio fixes the base. [Analyze]
16. A) $40. Cost = 25 + 0.15(100) = 25 + 15 = $40. Distractors: B) $15 drops the base fee. C) $125 ignores the per-minute rate and adds the minutes as dollars (25 + 100). D) $375 multiplies base × rate × minutes. E) $2,500 multiplies base × minutes, ignoring the rate. Fix: a linear model is (fixed base) + (rate × quantity); add the flat fee once, scale only the variable part. [Apply]
17. B) f(x) = x³ − x. A function is odd when f(−x) = −f(x). Here f(−x) = (−x)³ − (−x) = −x³ + x = −(x³ − x) = −f(x), so it is odd. Distractors: A) x² + 1 is even, f(−x) = f(x). C) |x| is even. D) x³ + 1 gives f(−x) = −x³ + 1, neither odd nor even. E) x² − x mixes an even and an odd term, so it is neither. Fix: odd means every term has an odd power and no constant; test f(−x) = −f(x). [Analyze]
18. C) 11. Match each input to its condition: −3 < 0 → f(−3) = −3 + 4 = 1; 0 ≤ 2 < 3 → f(2) = 2² = 4; 3 ≥ 3 → f(3) = 2(3) = 6. Sum = 1 + 4 + 6 = 11. Distractors: A) 9 uses the wrong branch for one input. B) 13 squares −3 by using the middle branch. D) 17 uses 2x for f(2) as well. E) 7 evaluates only f(3). Fix: for each input, first decide which condition it satisfies, then use only that branch. [Apply]
19. D) x = 3. Square both sides: x + 6 = x², so x² − x − 6 = 0 = (x − 3)(x + 2), giving x = 3 or x = −2. Check in the original: √9 = 3 ✓, but √4 = 2 ≠ −2, so x = −2 is extraneous. Only x = 3 works. Distractors: A) keeps both roots without checking. B) keeps the extraneous root and discards the valid one. C) declares both valid. E) claims no solution, but x = 3 checks. Fix: squaring can create extraneous roots — always substitute back into the original radical equation. [Evaluate]
20. E) x = π/6 and x = 5π/6. Solve sin x = 1/2 on [0, 2π): the reference angle is π/6, and sine is positive in Quadrants I and II, giving x = π/6 and x = π − π/6 = 5π/6. Distractors: A) π/6 only omits the second-quadrant solution. B) π/3, 2π/3 solves sin x = √3/2. C) π/6, 7π/6 uses Quadrant III (where sine is negative). D) π/6, 11π/6 uses Quadrant IV. Fix: after the reference angle, place solutions in the quadrants where the function has the required sign — sine positive means Q I and Q II. [Apply]
21. A) 32. log₂ x = 5 means x = 2⁵ = 32. Distractors: B) 25 computes 5². C) 10 computes 5·2. D) 7 computes 5 + 2. E) 64 computes 2⁶. Fix: log_b x = y converts to x = b^y — the base raised to the log. [Apply]
22. B) y = −(3/2)x + 5. The perpendicular slope is the negative reciprocal of 2/3, which is −3/2. Point-slope through (4, −1): y + 1 = −(3/2)(x − 4) = −(3/2)x + 6, so y = −(3/2)x + 5. Distractors: A) reuses the original slope 2/3. C) y = −(3/2)x − 7 uses the right slope but a sign error placing the point. D) y = (3/2)x − 7 forgets the negative in the reciprocal. E) uses the negative of the original slope (−2/3) rather than its reciprocal. Fix: perpendicular slope = negative reciprocal (flip and change sign), then anchor with the given point. [Apply]
23. C) (2x + 1)/(1 − x). Set y = (x − 1)/(x + 2) and solve for x: y(x + 2) = x − 1 → yx + 2y = x − 1 → x(y − 1) = −1 − 2y → x = (2y + 1)/(1 − y). Swapping variable names, f⁻¹(x) = (2x + 1)/(1 − x). Check x = 0: f(0) = −1/2 and f⁻¹(−1/2) = 0 ✓. Distractors: A) (x + 2)/(x − 1) merely flips the fraction. B) (2x + 1)/(x − 1) has the wrong denominator sign. D) (1 − x)/(2x + 1) inverts the correct answer. E) is the original function unchanged. Fix: swap x and y, then solve for y — collect all y-terms on one side and factor. [Apply]
24. D) y = −f(x) + 3. Reflecting across the x-axis negates outputs: −f(x); shifting up 3 adds 3 outside: −f(x) + 3. Distractors: A) f(−x) + 3 reflects across the y-axis instead. B) −f(x + 3) shifts horizontally, not vertically. C) −f(x) − 3 shifts down. E) f(−x) − 3 reflects the wrong axis and shifts down. Fix: x-axis reflection negates the whole function; vertical shift adds outside. [Understand]
25. E) (3, 1). Add the equations to eliminate y: (2x + y) + (x − y) = 7 + 2 → 3x = 9 → x = 3; then y = 7 − 2(3) = 1. Check: 3 − 1 = 2 ✓. Distractors: A) (1, 3) swaps the coordinates. B) (2, 3) fails both equations. C) (3, −1) has a sign error in y. D) (5, −3) satisfies neither equation. Fix: with opposite y-coefficients, add the equations to eliminate y in one step, then back-substitute. [Apply]
26. A) 64 ft. The vertex of h(t) = −16t² + 32t + 48 is at t = −b/(2a) = −32/(2·−16) = 1 s; h(1) = −16 + 32 + 48 = 64 ft. Distractors: B) 48 ft reports h(0), the initial height, not the max. C) 32 ft misreads the linear coefficient as the height. D) 80 ft adds 32 + 48 without evaluating at the vertex. E) 96 ft doubles the initial height. Fix: maximum of a downward parabola occurs at t = −b/(2a); substitute that t back to get the height. [Apply]
27. B) 21/5. Average rate of change = (f(6) − f(1))/(6 − 1) = (25 − 4)/5 = 21/5. Distractors: A) 21 forgets to divide by the change in x. C) 7/2 uses the wrong interval or averages differently. D) 5/21 inverts the ratio. E) 3 divides 21 by 7 instead of 5. Fix: average rate of change is Δy/Δx between the two endpoints — the slope of the secant line. [Apply]
28. C) sin θ. Using 1 − cos²θ = sin²θ, the expression is sin²θ / sin θ = sin θ. Distractors: A) cos θ misapplies the Pythagorean identity. B) csc θ inverts the result. D) tan θ comes from dividing by cos θ. E) 1 cancels incorrectly, treating sin²θ/sin θ as 1. Fix: replace 1 − cos²θ with sin²θ first, then cancel one factor of sin θ. [Apply]
29. D) no real solutions. The discriminant is b² − 4ac = (−4)² − 4(1)(7) = 16 − 28 = −12 < 0, so there are no real solutions (two complex conjugates). Distractors: A) and B) require a positive discriminant. C) requires discriminant = 0. E) infinitely many never occurs for a genuine quadratic. Fix: the sign of b² − 4ac decides: positive → two real, zero → one repeated, negative → none real. [Analyze]
30. A) It touches the x-axis at x = −2 and crosses at x = 3. An even-multiplicity zero (x = −2, mult 2) makes the graph touch and turn; an odd-multiplicity zero (x = 3, mult 1) makes it cross. Distractors: B) reverses which zero touches and which crosses. C) treats both as odd. D) treats both as even. E) invents an extra zero at x = 0 that the factored form does not provide. Fix: even multiplicity → bounce (touch); odd multiplicity → pass through (cross). [Analyze]
31. E) x ≥ 4. (f ∘ g)(x) = √(x − 4) requires x − 4 ≥ 0, so x ≥ 4. Distractors: A) all reals ignores the radical. B) x ≥ 0 uses the domain of √x without shifting. C) x ≤ 4 reverses the inequality. D) x > 4 wrongly excludes the endpoint 4, where √0 = 0 is defined. Fix: for √(inner), require inner ≥ 0 (endpoint included) and solve. [Apply]
32. B) vertical, length 10. In x²/9 + y²/25 = 1 the larger denominator (25) sits under y², so the major axis is vertical with semi-axis a = √25 = 5 and full length 2a = 10. Distractors: A) uses the smaller denominator and wrong orientation. C) right length but horizontal orientation. D) length 25 forgets to take the square root and double. E) uses √9 = 3 for the semi-axis. Fix: the larger denominator marks the major axis; its length is 2√(that denominator). [Understand]
33. C) horizontal compression by a factor of 1/2. Replacing x with 2x inside f speeds the input, compressing the graph horizontally toward the y-axis by a factor of 1/2. Distractors: A) a stretch by 2 comes from f(x/2), not f(2x). B) and D) describe vertical changes, which come from a coefficient outside f. E) a horizontal shift comes from f(x − 2), not f(2x). Fix: f(bx) with b > 1 compresses horizontally by 1/b; the effect is the reciprocal of the coefficient. [Understand]
34. D) x = 6. Isolate: 1/x = 1/2 − 1/3 = 3/6 − 2/6 = 1/6, so x = 6. Check: 1/6 + 1/3 = 1/6 + 2/6 = 3/6 = 1/2 ✓. Distractors: A) 1/6 reports 1/x instead of x. B) 5 comes from subtracting denominators. C) 1 mishandles the fraction subtraction. E) −6 sign error. Fix: combine the constant fractions with a common denominator first, then reciprocate to solve for x. [Apply]
35. A) π. The period of sin(Bx + C) is 2π/|B|; here B = 2, so the period is 2π/2 = π. The amplitude 3 and phase shift −π do not affect the period. Distractors: B) 2π ignores the factor 2. C) π/2 divides by B². D) 4π multiplies by 2 instead of dividing. E) 3π uses the amplitude 3. Fix: period depends only on the coefficient of x: 2π/|B|. [Apply]
36. E) 11 − 10i. FOIL: (3 + 2i)(1 − 4i) = 3 − 12i + 2i − 8i² = 3 − 10i − 8(−1) = 3 − 10i + 8 = 11 − 10i. Distractors: A) 3 − 8i drops the outer/inner terms. B) 11 + 10i has the imaginary sign wrong. C) −5 − 10i uses i² = +1. D) 3 − 10i forgets to convert −8i². Fix: multiply as binomials, then replace i² with −1 and combine real and imaginary parts. [Apply]
37. B) 10 g. In 18 years there are 18/6 = 3 half-lives, so the amount is 80·(1/2)³ = 80/8 = 10 g. Distractors: A) 20 g stops after 2 half-lives. C) 5 g runs 4 half-lives. D) 40 g runs only 1 half-life. E) 26.7 g divides by 3 linearly instead of halving repeatedly. Fix: number of half-lives = elapsed time ÷ half-life; multiply the start by (1/2) that many times. [Apply]
38. C) 2x + h. Expand: [f(x+h) − f(x)]/h = [(x + h)² − x²]/h = [x² + 2xh + h² − x²]/h = (2xh + h²)/h = 2x + h. Distractors: A) 2x drops the h before it can be canceled properly (that is the limit, not the quotient). B) 2x + h² fails to divide h² by h. D) x + h forgets the factor of 2 from the middle term. E) 2xh cancels incorrectly. Fix: expand (x + h)², subtract x², then divide every remaining term by h. [Apply]
39. D) y = 3 sin x − 1. Amplitude 3 (half of 2 − (−4) = 6) and midline y = −1 give y = 3 sin x − 1; it passes through (0, −1) rising to a max of 2 at x = π/2, matching sine. Distractors: A) sin x + 3 has amplitude 1 and the wrong midline. B) 3 sin x + 1 uses midline +1. C) −3 sin x − 1 is reflected, so it falls from (0, −1). E) 3 cos x − 1 starts at its maximum at x = 0, not on the midline. Fix: amplitude = (max − min)/2, midline = (max + min)/2; sine starts on the midline rising, cosine starts at a peak. [Analyze]
40. A) x < −2 or x > 3. Factor: x² − x − 6 = (x − 3)(x + 2) > 0. The product is positive outside the roots −2 and 3, so x < −2 or x > 3. Distractors: B) −2 < x < 3 is where the product is negative (< 0). C) and D) use wrong roots (−3 and 2). E) x > 3 only omits the left branch. Fix: find the roots, then a "> 0" parabola opening up is positive outside the roots, negative between them. [Analyze]
41. B) 40√3/3 ft. The height is opposite the 30° angle and 40 ft is adjacent, so tan 30° = h/40, giving h = 40 tan 30° = 40·(1/√3) = 40√3/3 ≈ 23.1 ft. Distractors: A) 20 ft uses sin 30° = 1/2 as the ratio. C) 40√3 ft uses tan 60° (or multiplies by √3). D) 40 ft ignores the trig. E) 80 ft doubles. Fix: pick the ratio linking the known side to the unknown: opposite over adjacent is tangent. [Apply]
42. E) (−1, 2). A function decreases where its graph falls; between the local maximum at x = −1 and the local minimum at x = 2 the graph descends, so it decreases on (−1, 2). Distractors: A) (−∞, −1) and B) (2, ∞) are where it rises. C) (−1, ∞) and D) (−∞, 2) each mix an increasing part with the decreasing part. Fix: decreasing = falling from left to right, which happens between a max (on the left) and the next min (on the right). [Analyze]
43. C) (x + 1)² + (y − 4)² = 25. The radius is the distance from center (−1, 4) to (2, 0): √((2 + 1)² + (0 − 4)²) = √(9 + 16) = 5, so r² = 25 and the equation is (x + 1)² + (y − 4)² = 25. Distractors: A) flips the center signs. B) uses r = 5 as r² (should be 25). D) flips signs and uses r for r². E) 100 squares the radius twice. Fix: standard form uses (x − h)² + (y − k)² = r²; find r as a distance, then square it. [Apply]
44. D) x = 3. log(x − 3) is defined only when x − 3 > 0; as x → 3⁺ the argument → 0⁺ and the log → −∞, giving a vertical asymptote at x = 3. Distractors: A) y = 3 and C) y = 0 are horizontal lines; logs have vertical asymptotes. B) x = 0 is the asymptote of log x before the shift. E) x = −3 has the shift direction backward. Fix: a log's vertical asymptote sits where its argument equals 0; solve x − 3 = 0. [Understand]
45. B) x = 3 and x = −5/3. |3x − 2| = 7 splits into 3x − 2 = 7 → x = 3, and 3x − 2 = −7 → 3x = −5 → x = −5/3. Distractors: A) x = 3 only keeps one case. C) x = 3, 5/3 gets the sign of the second root wrong. D) x = −3, 5/3 mishandles both. E) x = 9, −5/3 adds 7 to 2 before dividing. Fix: |A| = c (with c > 0) yields two equations, A = c and A = −c; solve each. [Apply]
46. A) $14,450. Depreciating 15% per year multiplies by 0.85 annually: 20,000·(0.85)² = 20,000·0.7225 = $14,450. Distractors: B) $14,000 subtracts a flat 30% (15% × 2), treating it as linear. C) $17,000 applies 15% just once. D) $13,000 subtracts 35%. E) $15,000 rounds arbitrarily. Fix: a constant percent change compounds — multiply by (1 − rate) once per year, not subtract a flat total. [Apply]
47. E) f(x) = x³ + 1. A one-to-one function passes the horizontal-line test. The cubic x³ + 1 is strictly increasing, so each output comes from exactly one input. Distractors: A) x², B) |x|, and C) x² − 4 each hit the same output for ±x (fail the horizontal-line test). D) the constant 4 sends every input to one output — as far from one-to-one as possible. Fix: one-to-one means no horizontal line meets the graph twice; strictly increasing (or decreasing) functions qualify. [Analyze]
48. D) 3/2. tan(π/4) = 1 and sin(π/6) = 1/2, so the sum is 1 + 1/2 = 3/2. Distractors: A) 1 uses only the tangent term. B) 1/2 uses only the sine term. C) √3/2 + 1 uses sin(π/3) instead of sin(π/6). E) √2/2 + 1/2 uses sin(π/4) for the first term. Fix: recall the exact values separately — tan(π/4) = 1, sin(π/6) = 1/2 — then add. [Apply]
CLEP reports a 20–80 scaled score; ACE-recommended credit is granted at 50. This mock uses the approximation scaled ≈ 20 + 1.25 × (raw correct), which places the credit threshold at 24 of 48 correct (≈50%).
| Raw correct (of 48) | Approx. scaled score |
|---|---|
| 0 | 20 |
| 4 | 25 |
| 8 | 30 |
| 12 | 35 |
| 16 | 40 |
| 20 | 45 |
| 24 | 50 ← ACE credit threshold |
| 28 | 55 |
| 32 | 60 |
| 36 | 65 |
| 40 | 70 |
| 44 | 75 |
| 48 | 80 |
Intermediate raw scores interpolate (each additional correct answer ≈ +1.25 scaled points). Disclaimer: CLEP's exact raw-to-scaled conversion is proprietary and varies by form; this table is an approximation for self-assessment only, not an official predictor.
Your running multiple-choice score appears in the bar below. Self-score the free-response section with the rubrics in the answer key, then use the diagnostic table to target review.