Exam questions · Maths · Further Algebra
Algebraic Fractions and Proof
- 6 exam questions
- 17 marks
- 9 quick checks
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1 Show that [3 marks]
Show that the sum of three consecutive even numbers is a multiple of 6.
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Model answer
Let the numbers be \(2n\), \(2n + 2\) and \(2n + 4\). Their sum is \(6n + 6 = 6(n + 1)\), which is a multiple of 6.
Mark scheme
- \(2n\), \(2n + 2\), \(2n + 4\) — M1
- \(6n + 6\) — M1
- \(6(n + 1)\) with a conclusion — Q1
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2 Simplify [3 marks]
Simplify \(\dfrac{x^2 + 5x}{x^2 - 25}\).
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Model answer
\(\dfrac{x(x + 5)}{(x - 5)(x + 5)} = \dfrac{x}{x - 5}\).
Mark scheme
- \(x(x + 5)\) — M1
- \((x - 5)(x + 5)\) — M1
- \(\dfrac{x}{x - 5}\) — A1
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3 Solve [3 marks]
Solve \(\dfrac{x - 2}{3} + \dfrac{x + 1}{2} = 4\).
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Model answer
Multiply every term by 6: \(2(x - 2) + 3(x + 1) = 24\). Then \(5x - 1 = 24\), so \(x = 5\).
Mark scheme
- \(2(x - 2) + 3(x + 1) = 24\) — M1
- \(5x - 1 = 24\) — M1
- \(x = 5\) — A1
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4 Show that [2 marks]
Edith says, “\(n^2 > n\) for every number \(n\).” Show that Edith is wrong.
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Model answer
When \(n = 1\), \(n^2 = 1\) and \(n = 1\), so \(n^2\) is not greater than \(n\). This counter-example shows she is wrong. (\(n = 0.5\) also works.)
Mark scheme
- A valid counter-example, such as \(n = 1\) or \(n = 0.5\) — M1
- Substitutes and shows the statement is false — Q1
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5 Prove [3 marks]
Prove that \((2n + 1)^2 - (2n - 1)^2\) is a multiple of 8 for every positive integer \(n\).
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Model answer
\((2n + 1)^2 = 4n^2 + 4n + 1\) and \((2n - 1)^2 = 4n^2 - 4n + 1\). The difference is \(8n\), which is a multiple of 8.
Mark scheme
- \(4n^2 + 4n + 1\) or \(4n^2 - 4n + 1\) — M1
- \(8n\) — M1
- States that \(8n\) is a multiple of 8 — Q1
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6 Write [3 marks]
Write \(\dfrac{2}{x + 1} - \dfrac{1}{x - 2}\) as a single fraction, in its simplest form.
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Model answer
The common denominator is \((x + 1)(x - 2)\), so the fraction is \(\dfrac{2(x - 2) - (x + 1)}{(x + 1)(x - 2)} = \dfrac{x - 5}{(x + 1)(x - 2)}\).
Mark scheme
- Common denominator \((x + 1)(x - 2)\) — M1
- \(2(x - 2) - (x + 1)\) — M1
- \(\dfrac{x - 5}{(x + 1)(x - 2)}\) — A1
Quick check
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1
What may be cancelled in an algebraic fraction?
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B: Factors that multiply the whole top and the whole bottom
Terms that are added or subtracted cannot be cancelled.
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2
Simplify \(\dfrac{x^2 - 9}{x + 3}\).
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A: \(x - 3\)
\(\dfrac{(x - 3)(x + 3)}{x + 3} = x - 3\).
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3
Which statement about \(\dfrac{x + 3}{3}\) is correct?
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D: The 3s cannot be cancelled because the 3 on top is added
Only factors can be cancelled, not terms.
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4
Solve \(\dfrac{x - 1}{3} + \dfrac{x + 2}{6} = 2\).
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C: \(x = 4\)
Multiply by 6: \(2(x - 1) + (x + 2) = 12\), so \(3x = 12\).
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5
Which expression is an odd number for any whole number \(n\)?
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B: \(2n + 1\)
\(2n\) is even, so adding 1 makes it odd.
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6
What is the sum of three consecutive whole numbers \(n\), \(n + 1\) and \(n + 2\)?
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A: \(3n + 3\)
\(n + n + 1 + n + 2 = 3n + 3 = 3(n + 1)\).
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7
Which value of \(n\) is a counter-example to “\(n^2 + n + 1\) is always prime”?
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D: \(n = 4\)
\(16 + 4 + 1 = 21 = 3 \times 7\), which is not prime.
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8
Simplify \(\dfrac{x^2 + 5x + 6}{x^2 + 3x + 2}\).
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C: \(\dfrac{x + 3}{x + 1}\)
\(\dfrac{(x + 2)(x + 3)}{(x + 1)(x + 2)} = \dfrac{x + 3}{x + 1}\).
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9
Expand and simplify \((n + 1)^2 - (n - 1)^2\).
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B: \(4n\)
\(n^2 + 2n + 1 - n^2 + 2n - 1 = 4n\).