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Grade 12 Review Quiz — Easy Version 2
Grade 12 · MathSheets Name: ____________________ Date: __________
Answer every question. The lesson name before each question tells you which skill it checks — open that lesson if you need a reminder.
How to solve
This quiz mixes questions from different lessons, like a real test. Read each question carefully: the lesson name in front tells you what kind of problem it is. Check your answers, then practise the lessons where you made mistakes.
Example
Law of Sines & Law of Cosines: In a triangle, side a = 5, side b = 8, and the included angle C = 60°. Find side c (Law of Cosines).
Answer: 7
Not quite. The answer is
💡 Law of Cosines: c2 = a2 + b2 − 2ab cos C = 82 + 152 − 2(8)(15)cos 60°. cos 60° = 12, so 2(8)(15)(12) = 120 and c2 = 64 + 225 − 120 = 169. c = √169 = 13.
✓ Correct!
Not quite. The answer is
💡 Law of Cosines: c2 = a2 + b2 − 2ab cos C = 82 + 152 − 2(8)(15)cos 60°. cos 60° = 12, so 2(8)(15)(12) = 120 and c2 = 64 + 225 − 120 = 169. c = √169 = 13.
💡 How to solve: Law of Sines & Law of Cosines
When two sides and the included angle are given, use the Law of Cosines: c² = a² + b² - 2ab·cos C. Remember cos 60° = 12 and cos 120° = -12, then take the square root to find c.
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In a triangle, side a = 5, side b = 8, and the included angle C = 60°. Find side c (Law of Cosines).
✓ You've got it! Press Next to keep going. Not yet — the answer is . Read the steps above once more.
Not quite. The answer is
💡 Write y = 2x − 1. Swap x and y: x = 2y − 1. Add 1 to both sides: x + 1 = 2y. Divide both sides by 2: y = x + 12. So f⁻¹(x) = x + 12.
✓ Correct!
Not quite. The answer is
💡 Write y = 2x − 1. Swap x and y: x = 2y − 1. Add 1 to both sides: x + 1 = 2y. Divide both sides by 2: y = x + 12. So f⁻¹(x) = x + 12.
💡 How to solve: Inverse Functions
To find f⁻¹(k), set f(x) = k and solve for x. To find the inverse function, write y = mx + b, swap x and y, and solve for y, giving f⁻¹(x) = x - bm.
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If f(x) = x + 3, find f⁻¹(0).
✓ You've got it! Press Next to keep going. Not yet — the answer is . Read the steps above once more.
Not quite. The answer is
💡 −9 < −4, so use the first piece, f(x) = x − 3. Substitute: f(−9) = (−9) − 3 = −9 − 3 = −12. So f(−9) = −12.
✓ Correct!
Not quite. The answer is
💡 −9 < −4, so use the first piece, f(x) = x − 3. Substitute: f(−9) = (−9) − 3 = −9 − 3 = −12. So f(−9) = −12.
💡 How to solve: Evaluating Piecewise Functions
Compare the given x-value with each condition to choose the correct piece. Substitute x into only that piece and simplify.
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f(x) = x + 4 if x < 3; f(x) = -2x if x ≥ 3. Find f(7).
✓ You've got it! Press Next to keep going. Not yet — the answer is . Read the steps above once more.
Not quite. The answer is
💡 Use the power rule d/dx(axn) = (a × n)xn−1 on each term: d/dx(3x³) = 3 × 3x² = 9x² and d/dx(-6x) = -6. So f'(x) = 9x² - 6.
✓ Correct!
Not quite. The answer is
💡 Use the power rule d/dx(axn) = (a × n)xn−1 on each term: d/dx(3x³) = 3 × 3x² = 9x² and d/dx(-6x) = -6. So f'(x) = 9x² - 6.
💡 How to solve: Derivatives: The Power Rule
Power rule: the derivative of axn is (a·n)xn-1. Differentiate term by term; a constant term has derivative 0. To find f'(a), differentiate first, then substitute x = a.
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Find f'(x) for f(x) = 5x³ - 2.
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| 0 | 2 | ||
| -3 | 5 |
Not quite. The answer is
💡 For
| a | b | ||
| c | d |
✓ Correct!
Not quite. The answer is
💡 For
| a | b | ||
| c | d |
💡 How to solve: Matrices
To add matrices, add the numbers in the same positions. For the determinant of
| a | b | ||
| c | d |
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Add:
| -4 | 2 | ||
| 0 | -3 |
| -2 | 2 | ||
| 3 | -2 |
✓ You've got it! Press Next to keep going. Not yet — the answer is . Read the steps above once more.
Not quite. The answer is
💡 Work from the inside out. First f(4) = 3(4) − 2 = 12 − 2 = 10. Then f(10) = 3(10) − 2 = 30 − 2 = 28. So f(f(4)) = 28.
✓ Correct!
Not quite. The answer is
💡 Work from the inside out. First f(4) = 3(4) − 2 = 12 − 2 = 10. Then f(10) = 3(10) − 2 = 30 − 2 = 28. So f(f(4)) = 28.
💡 How to solve: Advanced Function Evaluation
To find f(k), substitute k for every x and simplify, squaring before multiplying. To find f(f(k)), first find f(k), then put that result back into f.
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If f(x) = 2x² + 2, find f(2).
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Not quite. The answer is
💡 Multiply both sides by (x − 2)(x − 4): 4x − 14 = A(x − 4) + B(x − 2). Put x = 2 (it makes x − 2 zero): the left side is −6 and the right side is A(−2), so −2A = −6 and A = 3. Put x = 4 (it makes x − 4 zero): the left side is 2 and the right side is B(2), so 2B = 2 and B = 1.
✓ Correct!
Not quite. The answer is
💡 Multiply both sides by (x − 2)(x − 4): 4x − 14 = A(x − 4) + B(x − 2). Put x = 2 (it makes x − 2 zero): the left side is −6 and the right side is A(−2), so −2A = −6 and A = 3. Put x = 4 (it makes x − 4 zero): the left side is 2 and the right side is B(2), so 2B = 2 and B = 1.
💡 How to solve: Partial Fraction Decomposition of Rational Functions
Factor the denominator and write one fraction with an unknown constant for each factor, e.g. 5x - 1x2 - x - 2 = A/(x + 1) + B/(x - 2). Multiply both sides by the whole denominator: 5x - 1 = A(x - 2) + B(x + 1). Then substitute the x-value that makes each factor 0: x = 2 gives 9 = 3B, so B = 3, and x = -1 gives -6 = -3A, so A = 2. (You can also expand and match the coefficients of x and the constants.)
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8x + 4x² + x = Ax + Bx + 1. Find A and B.
✓ You've got it! Press Next to keep going. Not yet — the answer is . Read the steps above once more.
Not quite. The answer is
💡 z = (x − μ) / σ = 650 − 59030 = 6030 = 2. So the z-score is 2.
✓ Correct!
Not quite. The answer is
💡 z = (x − μ) / σ = 650 − 59030 = 6030 = 2. So the z-score is 2.
💡 How to solve: Normal Distribution & z-Scores
The z-score z = (x - μ) / σ counts standard deviations from the mean. The 68-95-99.7 rule: about 68% of data lies within 1σ of the mean, 95% within 2σ, and 99.7% within 3σ, so each band holds 34%, 13.5%, 2.35%, and 0.15% on each side.
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The battery lifetimes (hours) are normally distributed with mean 590 and standard deviation 40. Find the z-score of a lifetime of 510.
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Not quite. The answer is
💡 Since 1 − sin2 x = cos2 x, sin2 x − 1 = −(1 − sin2 x) = -cos2 x.
✓ Correct!
Not quite. The answer is
💡 Since 1 − sin2 x = cos2 x, sin2 x − 1 = −(1 − sin2 x) = -cos2 x.
💡 How to solve: Trigonometric Identities
Rewrite everything in sin and cos: tan x = sin x / cos x, sec x = 1 / cos x, csc x = 1 / sin x, cot x = cos x / sin x. Then use sin2 x + cos2 x = 1 (and 1 + tan2 x = sec2 x, 1 + cot2 x = csc2 x) and cancel.
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Simplify: csc x · cos x
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Not quite. The answer is
💡 The top and the bottom both have degree 2. Divide every term by x²: every term except the leading ones goes to 0 as x → ∞, so the limit is the ratio of the leading coefficients: 8-4 = -2. The limit is -2.
✓ Correct!
Not quite. The answer is
💡 The top and the bottom both have degree 2. Divide every term by x²: every term except the leading ones goes to 0 as x → ∞, so the limit is the ratio of the leading coefficients: 8-4 = -2. The limit is -2.
💡 How to solve: Limits: Factoring & Limits at Infinity
If substituting gives 00, factor and cancel the common factor (or multiply by the conjugate for square roots), then substitute. As x → ∞, compare degrees: a bigger denominator degree gives 0, equal degrees give the ratio of the leading coefficients.
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Find the limit: lim(x→-8) of x² - 64x + 8
✓ You've got it! Press Next to keep going. Not yet — the answer is . Read the steps above once more.
| 1 | -3 | ||
| 0 | 1 |
Not quite. The answer is
💡 First the determinant: det = ad − bc = 1 × 1 − (-3) × 0 = 1 − 0 = 1. Swap a and d and change the signs of b and c:
| 1 | 3 | ||
| 0 | 1 |
| 1 | 3 | ||
| 0 | 1 |
✓ Correct!
Not quite. The answer is
💡 First the determinant: det = ad − bc = 1 × 1 − (-3) × 0 = 1 − 0 = 1. Swap a and d and change the signs of b and c:
| 1 | 3 | ||
| 0 | 1 |
| 1 | 3 | ||
| 0 | 1 |
💡 How to solve: Inverse Matrices & Cramer's Rule
For A =
| a | b | ||
| c | d |
| d | -b | ||
| -c | a |
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Find the inverse of
| -3 | -2 | ||
| -2 | -1 |
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Not quite. The answer is
💡 The coin and the die don't affect each other, so multiply: P(heads) × P(1) = 12 × 16 = 112.
✓ Correct!
Not quite. The answer is
💡 The coin and the die don't affect each other, so multiply: P(heads) × P(1) = 12 × 16 = 112.
💡 How to solve: Statistics & Probability
To count ways to choose r items from n, use C(n, r) = n! ÷ (r! × (n - r)!). For independent events, P(A and B) = P(A) × P(B); write the probability as a fraction in simplest form.
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How many ways can you choose 1 item from a set of 4?
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| -1 | 1 | ||
| 0 | -1 |
Not quite. The answer is
💡 Multiply every entry by 3: 3 × (-1) = -3, 3 × 1 = 3, 3 × 0 = 0, 3 × (-1) = -3. The answer is
| -3 | 3 | ||
| 0 | -3 |
✓ Correct!
Not quite. The answer is
💡 Multiply every entry by 3: 3 × (-1) = -3, 3 × 1 = 3, 3 × 0 = 0, 3 × (-1) = -3. The answer is
| -3 | 3 | ||
| 0 | -3 |
💡 How to solve: Matrix Multiplication
To multiply two matrices, each entry is a row of the first times a column of the second:
| a | b | ||
| c | d |
| e | f | ||
| g | h |
| ae + bg | af + bh | ||
| ce + dg | cf + dh |
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Multiply:
| -1 | 2 | ||
| -2 | -2 |
| 3 | -2 | ||
| -3 | 1 |
✓ You've got it! Press Next to keep going. Not yet — the answer is . Read the steps above once more.
Not quite. The answer is
💡 A polynomial is continuous, so the limit is just its value at x = -4. Substitute x = -4: 4(-4) − 3 = -16 − 3 = -19. The limit is -19.
✓ Correct!
Not quite. The answer is
💡 A polynomial is continuous, so the limit is just its value at x = -4. Substitute x = -4: 4(-4) − 3 = -16 − 3 = -19. The limit is -19.
💡 How to solve: Introduction to Limits
For a polynomial, the limit as x approaches a equals the function value at a. Use direct substitution: replace x with a and simplify.
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Find the limit: lim(x→-4) of x - 8
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Not quite. The answer is
💡 From the left (x < -2) use -3x - 3: -3(-2) − 3 = 6 − 3 = 3. From the right (x > -2) use 3x + 9: 3(-2) + 9 = -6 + 9 = 3. Both sides approach 3, so the limit is 3.
✓ Correct!
Not quite. The answer is
💡 From the left (x < -2) use -3x - 3: -3(-2) − 3 = 6 − 3 = 3. From the right (x > -2) use 3x + 9: 3(-2) + 9 = -6 + 9 = 3. Both sides approach 3, so the limit is 3.
💡 How to solve: One-Sided Limits & Continuity
For the limit from the left, use the piece defined for x < c and substitute c; for the limit from the right, use the piece for x > c. The two-sided limit exists only when both one-sided limits are equal; otherwise write DNE. The value f(c) itself does not affect the limit. f is continuous at c when the left limit, the right limit and f(c) are all the same number. If the limit exists but f(c) is missing or different, the discontinuity is removable (a hole); if the one-sided limits are different numbers it is a jump; if the function grows without bound (a vertical asymptote) it is infinite. To find k, set the left piece equal to the right piece at x = c and solve.
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f(x) = x + 1 if x < 1; f(x) = -2x - 5 if x ≥ 1. Find the limit of f(x) as x approaches 1 from the left.
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Not quite. The answer is
💡 Use sin2 θ + cos2 θ = 1: cos2 θ = 1 − (1517)2 = 1 − 225289 = 64289, so cos θ = ±817. θ is acute, so cos θ = 817. Double-angle formula: cos 2θ = cos2 θ − sin2 θ = (817)2 − (1517)2 = 64289 − 225289 = -161289. So cos 2θ = -161289.
✓ Correct!
Not quite. The answer is
💡 Use sin2 θ + cos2 θ = 1: cos2 θ = 1 − (1517)2 = 1 − 225289 = 64289, so cos θ = ±817. θ is acute, so cos θ = 817. Double-angle formula: cos 2θ = cos2 θ − sin2 θ = (817)2 − (1517)2 = 64289 − 225289 = -161289. So cos 2θ = -161289.
💡 How to solve: Double-Angle Formulas
sin 2θ = 2 sin θ cos θ, cos 2θ = cos2 θ - sin2 θ = 1 - 2sin2 θ = 2cos2 θ - 1, and tan 2θ = 2tan θ / (1 - tan2 θ). First find the missing sin or cos with a right triangle (watch the quadrant signs), then substitute.
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sin θ = 817 and θ is acute. Find the exact value of sin 2θ.
✓ You've got it! Press Next to keep going. Not yet — the answer is . Read the steps above once more.
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Grade 12 Review Quiz — Easy Version 2
Answer every question. The lesson name before each question tells you which skill it checks — open that lesson if you need a reminder.