This set has ten original questions covering algebra, logarithms, trigonometry and calculus. They are ordered from easier to harder and carry no mark labels. Check your own syllabus year on the Cambridge page before treating any topic as in scope.
Use paper, write every step, and open the answer only when you have tried. The original mixed-practice builder can assemble longer sessions, and the non-calculator working trainer helps with exact arithmetic.
The questions
1. Solve 3x² − 12x = 0.
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Factorise: 3x(x − 4) = 0. So x = 0 or x = 4. Check x = 4: 3(16) − 48 = 0.
A common slip is to divide both sides by x and lose the root x = 0.
2. Write x² + 6x + 2 in the form (x + p)² + q, and state the minimum value of the expression.
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Half of 6 is 3, so (x + 3)² = x² + 6x + 9. Then x² + 6x + 2 = (x + 3)² − 9 + 2 = (x + 3)² − 7.
The square is never negative, so the minimum value is −7, reached when x = −3.
3. Find the values of k for which x² + kx + 9 = 0 has equal roots.
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Equal roots need b² − 4ac = 0. Here k² − 4(1)(9) = k² − 36 = 0, so k = 6 or k = −6.
Check k = 6: x² + 6x + 9 = (x + 3)², one repeated root.
4. Find the remainder when f(x) = x³ − 2x² + x − 5 is divided by (x − 2).
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By the remainder theorem, the remainder is f(2) = 8 − 8 + 2 − 5 = −3.
5. Solve 3x+1 = 20, giving x to 3 significant figures.
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Take logarithms: (x + 1) lg 3 = lg 20, so x + 1 = 1.30103 / 0.47712 = 2.7268.
Then x = 1.7268, which is x = 1.73. Check: 32.727 is about 20.
6. Find the stationary points of y = x³ − 3x² + 2 and decide whether each is a maximum or a minimum.
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dy/dx = 3x² − 6x = 3x(x − 2) = 0, so x = 0 or x = 2.
When x = 0, y = 2. When x = 2, y = 8 − 12 + 2 = −2.
The second derivative is 6x − 6. At x = 0 it is −6, so (0, 2) is a maximum. At x = 2 it is 6, so (2, −2) is a minimum.
7. Evaluate the definite integral of (2x + 1) from x = 1 to x = 3.
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An antiderivative is x² + x. At x = 3 it equals 12, and at x = 1 it equals 2. So the value is 12 − 2 = 10.
Check: the region is a trapezium with parallel sides 3 and 7 and width 2, area (3 + 7)/2 × 2 = 10.
8. Find the coefficient of x² in the expansion of (1 + 2x)⁵.
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The x² term is C(5, 2) × (2x)² = 10 × 4x² = 40x². The coefficient is 40.
9. Solve tan x = 2 for 0° ≤ x ≤ 360°, giving answers to 1 decimal place.
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The principal value is tan⁻¹ 2 = 63.4°. Tangent repeats every 180°, so the second solution is 63.4° + 180° = 243.4°.
x = 63.4° or x = 243.4°
10. A curve has gradient function dy/dx = 4x − 6 and passes through (2, 5). Find the equation of the curve, then the equation of the tangent at x = 2.
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Integrate: y = 2x² − 6x + c. Substitute (2, 5): 5 = 8 − 12 + c, so c = 9. The curve is y = 2x² − 6x + 9.
At x = 2 the gradient is 4(2) − 6 = 2, and y = 5. The tangent is y − 5 = 2(x − 2), so y = 2x + 1.
If you got these wrong
| What went wrong | Go to |
|---|---|
| Lost a root, or factorising errors (Q1) | algebraic equations and inequalities |
| Completing the square or discriminant (Q2, Q3) | quadratic structure and discriminants |
| Remainder theorem (Q4) | polynomial factors and remainders |
| Logarithms and exponents (Q5) | exponential and logarithmic reasoning |
| Stationary points, or choosing the test (Q6) | stationary points and optimisation |
| Integration and finding c (Q7, Q10) | integration methods |
| Binomial coefficients (Q8) | binomial expansion |
| Missing the second solution (Q9) | trigonometric equations and graphs |
| Tangent equation (Q10) | tangents, normals and rates |
For a fuller picture of how to choose methods, read the help pages on forming an optimisation model and keeping every solution in a trigonometric interval. Log your errors using the revision page.
Using the set well
Mark yourself on method first and the final value second. A correct answer reached by a lucky guess should be treated as a gap.
If one topic keeps appearing in your wrong answers, return to the learning guide. Where unfamiliar questions stop you before the first line, online one-to-one Additional Mathematics tuition is the place to practise starting them with a teacher beside you.