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Grade 12 Quiz: Calculus — 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
Antiderivatives & Definite Integrals: Evaluate the definite integral ∫ from 2 to 3 of (8x³) dx.
Answer: 130
Not quite. The answer is
💡 Product rule: f'(x) = g'(x)h(x) + g(x)h'(x), with g(x) = x² - x + 5 and h(x) = x + 2. By the power rule, g'(x) = 2x - 1 and h'(x) = 1. g(-2) = (-2)2 − (-2) + 5 = 4 + 2 + 5 = 11, g'(-2) = 2(-2) − 1 = -4 − 1 = -5, h(-2) = -2 + 2 = 0 and h'(-2) = 1. f'(-2) = -5 × 0 + 11 × 1 = 0 + 11 = 11.
✓ Correct!
Not quite. The answer is
💡 Product rule: f'(x) = g'(x)h(x) + g(x)h'(x), with g(x) = x² - x + 5 and h(x) = x + 2. By the power rule, g'(x) = 2x - 1 and h'(x) = 1. g(-2) = (-2)2 − (-2) + 5 = 4 + 2 + 5 = 11, g'(-2) = 2(-2) − 1 = -4 − 1 = -5, h(-2) = -2 + 2 = 0 and h'(-2) = 1. f'(-2) = -5 × 0 + 11 × 1 = 0 + 11 = 11.
💡 How to solve: Product, Quotient & Chain Rules
Product: (gh)' = g'h + gh'. Quotient: (gh)' = g'h - gh'h2. Chain: the derivative of (g(x))n is n(g(x))n-1·g'(x), of √(g(x)) is g'(x) / (2√(g(x))), and of e^(g(x)) is e^(g(x))·g'(x). Find g(a), g'(a), h(a), h'(a) first, then substitute.
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f(x) = (2x² + 2)(-x - 6). Use the product rule to find f'(3).
✓ 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
💡 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: 54. The limit is 54.
✓ 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: 54. The limit is 54.
💡 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→-9) of x² - 81x + 9
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💡 A polynomial is continuous, so the limit is just its value at x = 6. Substitute x = 6: 6(6) + 1 = 36 + 1 = 37. The limit is 37.
✓ Correct!
Not quite. The answer is
💡 A polynomial is continuous, so the limit is just its value at x = 6. Substitute x = 6: 6(6) + 1 = 36 + 1 = 37. The limit is 37.
💡 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→6) of 2x - 5
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💡 From the left (x < -2) use -3x: -3(-2) = 6. From the right (x > -2) use -x + 4: -(-2) + 4 = 2 + 4 = 6. Both sides approach 6, so the limit is 6.
✓ Correct!
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💡 From the left (x < -2) use -3x: -3(-2) = 6. From the right (x > -2) use -x + 4: -(-2) + 4 = 2 + 4 = 6. Both sides approach 6, so the limit is 6.
💡 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 - 4 if x ≤ 4; f(x) = -x + 4 if x > 4. Find the limit of f(x) as x approaches 4 from the left.
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💡 Use the power rule d/dx(axn) = (a × n)xn−1 on each term: d/dx(4x³) = 4 × 3x² = 12x² and d/dx(6x²) = 6 × 2x = 12x. So f'(x) = 12x² + 12x.
✓ Correct!
Not quite. The answer is
💡 Use the power rule d/dx(axn) = (a × n)xn−1 on each term: d/dx(4x³) = 4 × 3x² = 12x² and d/dx(6x²) = 6 × 2x = 12x. So f'(x) = 12x² + 12x.
💡 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 + 2x.
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💡 Use the reverse power rule: add 1 to the exponent and divide by the new exponent. ∫6x² dx = 6x³/3 = 2x³ and ∫4x dx = 4x²/2 = 2x². Add the constant of integration: 2x³ + 2x² + C.
✓ Correct!
Not quite. The answer is
💡 Use the reverse power rule: add 1 to the exponent and divide by the new exponent. ∫6x² dx = 6x³/3 = 2x³ and ∫4x dx = 4x²/2 = 2x². Add the constant of integration: 2x³ + 2x² + C.
💡 How to solve: Antiderivatives & Definite Integrals
Reverse power rule: the antiderivative of axn is (an + 1)xn+1 + C. For a definite integral from a to b, find an antiderivative F and compute F(b) - F(a). When f(x) ≥ 0, that number is the area under the curve.
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Find the antiderivative: ∫(3x² + 4x) dx
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💡 By the power rule, d/dx(x²) = 1 × 2x = 2x, d/dx(4x) = 4 and d/dx(-1) = 0, so f'(x) = 2x + 4. The slope at x = -2 is m = f'(-2) = 2(-2) + 4 = -4 + 4 = 0. The slope of the tangent line is 0.
✓ Correct!
Not quite. The answer is
💡 By the power rule, d/dx(x²) = 1 × 2x = 2x, d/dx(4x) = 4 and d/dx(-1) = 0, so f'(x) = 2x + 4. The slope at x = -2 is m = f'(-2) = 2(-2) + 4 = -4 + 4 = 0. The slope of the tangent line is 0.
💡 How to solve: Slope & Equation of a Tangent Line
The slope of the tangent line at x = a is m = f'(a). The tangent passes through (a, f(a)), so y - f(a) = m(x - a); solve for y to get y = mx + b. The tangent is horizontal where f'(x) = 0.
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Find the slope of the tangent line to f(x) = x² - x - 3 at x = 2.
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💡 Product rule: f'(x) = g'(x)h(x) + g(x)h'(x), with g(x) = -x² - x + 3 and h(x) = -x - 1. By the power rule, g'(x) = -2x - 1 and h'(x) = -1. g(1) = -(1)2 − (1) + 3 = -1 − 1 + 3 = 1, g'(1) = -2(1) − 1 = -2 − 1 = -3, h(1) = -(1) − 1 = -1 − 1 = -2 and h'(1) = -1. f'(1) = -3 × (-2) + 1 × (-1) = 6 − 1 = 5.
✓ Correct!
Not quite. The answer is
💡 Product rule: f'(x) = g'(x)h(x) + g(x)h'(x), with g(x) = -x² - x + 3 and h(x) = -x - 1. By the power rule, g'(x) = -2x - 1 and h'(x) = -1. g(1) = -(1)2 − (1) + 3 = -1 − 1 + 3 = 1, g'(1) = -2(1) − 1 = -2 − 1 = -3, h(1) = -(1) − 1 = -1 − 1 = -2 and h'(1) = -1. f'(1) = -3 × (-2) + 1 × (-1) = 6 − 1 = 5.
💡 How to solve: Product, Quotient & Chain Rules
Product: (gh)' = g'h + gh'. Quotient: (gh)' = g'h - gh'h2. Chain: the derivative of (g(x))n is n(g(x))n-1·g'(x), of √(g(x)) is g'(x) / (2√(g(x))), and of e^(g(x)) is e^(g(x))·g'(x). Find g(a), g'(a), h(a), h'(a) first, then substitute.
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f(x) = (3x² - x + 6)(x + 1). Use the product rule to find f'(-3).
✓ 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
💡 Substituting x = 4 gives 00, so factor the top: x² - 11x + 28 = (x - 4)(x - 7). Cancel (x - 4) to get x - 7, then substitute x = 4: 4 − 7 = -3. The limit is -3.
✓ Correct!
Not quite. The answer is
💡 Substituting x = 4 gives 00, so factor the top: x² - 11x + 28 = (x - 4)(x - 7). Cancel (x - 4) to get x - 7, then substitute x = 4: 4 − 7 = -3. The limit is -3.
💡 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.
Try a similar one
Find the limit: lim(x→3) of x² - 9x - 3
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💡 A polynomial is continuous, so the limit is just its value at x = -6. Substitute x = -6: 5(-6) + 4 = -30 + 4 = -26. The limit is -26.
✓ Correct!
Not quite. The answer is
💡 A polynomial is continuous, so the limit is just its value at x = -6. Substitute x = -6: 5(-6) + 4 = -30 + 4 = -26. The limit is -26.
💡 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.
Try a similar one
Find the limit: lim(x→1) of 5x + 2
✓ 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 Quiz: Calculus — 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.