These questions practise the five skills in advanced non-calculator reasoning: exact values, choosing algebra, structure, proof and checking. They run from easier to harder and are original.
Use no calculator, write the structure you notice before you begin, and open each answer only after a full attempt. The non-calculator working trainer can give a second opinion on exact arithmetic afterwards.
The questions
Q1. Find the exact value of sin² 45° + cos² 60°.
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sin 45° = √2/2, so sin² 45° = 2/4 = 1/2. cos 60° = 1/2, so cos² 60° = 1/4.
Sum: 1/2 + 1/4 = 3/4.
Q2. Rationalise the denominator of 4/(√7 − √3).
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Multiply top and bottom by √7 + √3. The bottom is 7 − 3 = 4. The numerator is 4(√7 + √3).
So the result is 4(√7 + √3)/4 = √7 + √3.
Check: 4/(2.646 − 1.732) = 4/0.914 ≈ 4.38, and √7 + √3 ≈ 4.38.
Q3. Write (1 + √3)/(2 − √3) in the form a + b√3.
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Multiply top and bottom by 2 + √3. The bottom is 4 − 3 = 1.
The numerator is (1 + √3)(2 + √3) = 2 + √3 + 2√3 + 3 = 5 + 3√3.
So the answer is 5 + 3√3. Check: 2.732/0.268 ≈ 10.2, and 5 + 5.196 = 10.196.
Q4. Given that a + b = 6 and ab = 4, find (a) a² + b², (b) (a − b)².
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(a) a² + b² = (a + b)² − 2ab = 36 − 8 = 28.
(b) (a − b)² = a² + b² − 2ab = 28 − 8 = 20.
Check: the numbers with sum 6 and product 4 are 3 ± √5. Then a² + b² = 2(9 + 5) = 28, and a − b = 2√5, so (a − b)² = 20.
Q5. Given that x + 1/x = 6, find (a) x² + 1/x², (b) x³ + 1/x³.
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(a) Square the given: x² + 2 + 1/x² = 36, so x² + 1/x² = 34.
(b) Cube the given: (x + 1/x)³ = x³ + 3x + 3/x + 1/x³ = x³ + 1/x³ + 3(x + 1/x). So 216 = x³ + 1/x³ + 18, and x³ + 1/x³ = 198.
Check: x = 3 + 2√2 gives 1/x = 3 − 2√2. Then x² = 17 + 12√2 and 1/x² = 17 − 12√2, sum 34. Also x³ = 99 + 70√2, and with 1/x³ = 99 − 70√2 the sum is 198.
Q6. Simplify (x² − 25)/(x² + 2x − 15).
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x² − 25 = (x − 5)(x + 5), and x² + 2x − 15 = (x + 5)(x − 3).
Cancel (x + 5): (x − 5)/(x − 3), for x ≠ −5 and x ≠ 3.
Check x = 6: the original is 11/33 = 1/3 and the answer is 1/3.
Q7. Simplify (4ⁿ⁺¹ − 4ⁿ)/(3 × 4ⁿ⁻¹).
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Numerator: 4ⁿ⁺¹ − 4ⁿ = 4ⁿ(4 − 1) = 3 × 4ⁿ. Bottom: 3 × 4ⁿ⁻¹.
Divide: 4ⁿ/4ⁿ⁻¹ = 4, so the answer is 4.
Check n = 1: (16 − 4)/3 = 4. Check n = 2: (64 − 16)/12 = 4.
Q8. Prove that the product of two consecutive even integers is a multiple of 8.
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Let the integers be 2n and 2n + 2, where n is an integer. The product is 2n(2n + 2) = 4n² + 4n = 4n(n + 1).
n and n + 1 are consecutive integers, so one is even and n(n + 1) = 2k for some integer k.
So the product is 4 × 2k = 8k, a multiple of 8.
Check: 2 × 4 = 8, 4 × 6 = 24 and 6 × 8 = 48.
Q9. Solve √(2x + 3) = x, checking each answer.
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Square both sides: 2x + 3 = x², so x² − 2x − 3 = 0 and (x − 3)(x + 1) = 0.
Test x = 3: √9 = 3. ✓
Test x = −1: √1 = 1, but the right side is −1. ✗
So x = 3 is the only solution. The squaring step created the extra answer.
Q10. Show that √(11 + 6√2) = 3 + √2.
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Square the right side: (3 + √2)² = 9 + 6√2 + 2 = 11 + 6√2.
Since 3 + √2 is positive, it is the positive square root of 11 + 6√2, so the statement is true.
Check: 11 + 8.485 = 19.485, and its square root is 4.414. Also 3 + 1.414 = 4.414.
Q11. Given that x = √5 + 2, find the exact values of (a) x + 1/x, (b) x² + 1/x².
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(a) 1/x = 1/(√5 + 2). Multiply top and bottom by √5 − 2. The bottom is 5 − 4 = 1, so 1/x = √5 − 2.
Then x + 1/x = (√5 + 2) + (√5 − 2) = 2√5.
(b) x² + 1/x² = (x + 1/x)² − 2 = (2√5)² − 2 = 20 − 2 = 18.
Check: x² = 9 + 4√5 and 1/x² = 9 − 4√5, and the sum is 18.
If you got these wrong
- Q1 to Q3: the exact value, a rationalising step or a missing cross term. Revisit exact surd and trigonometric values.
- Q4 and Q5: an identity with a dropped middle term. Revisit choosing algebra before numerical substitution.
- Q6 and Q7: cancelling across a sum or missing a common power. Revisit simplifying by recognising structure.
- Q8: the reason in the proof was missing. Revisit explaining a proof step.
- Q9 and Q10: no independent check, or an extra root accepted. Revisit checking with a second method.
- Q11: a mix of rationalising and structure. Revisit L01 and L02 together.
Record each slip in the mistake log, and test a quadratic idea with the quadratic structure explorer. When a whole group keeps going wrong, Additional Mathematics tuition can work through that group with you one-to-one.