Exam questions · Maths · Further Algebra
Algebraic Fractions and Proof
- 6 exam questions
- 17 marks
- 9 quick checks
-
1 Prove [3 marks]
Prove that the sum of any three consecutive odd numbers is a multiple of 3.
Show answerHide answer
Model answer
Let the numbers be \(2n + 1\), \(2n + 3\) and \(2n + 5\). Their sum is \(6n + 9 = 3(2n + 3)\), which is a multiple of 3.
Mark scheme
- \(2n + 1\), \(2n + 3\), \(2n + 5\) — M1
- \(6n + 9\) — M1
- \(3(2n + 3)\) with a conclusion — A1
-
2 Simplify [3 marks]
Simplify \(\dfrac{3x^2 + 6x}{x^2 - 4}\).
Show answerHide answer
Model answer
\(\dfrac{3x(x + 2)}{(x - 2)(x + 2)} = \dfrac{3x}{x - 2}\).
Mark scheme
- \(3x(x + 2)\) — M1
- \((x - 2)(x + 2)\) — M1
- \(\dfrac{3x}{x - 2}\) — A1
-
3 Solve [3 marks]
Solve \(\dfrac{x + 2}{5} + \dfrac{x - 3}{2} = 1\).
Show answerHide answer
Model answer
Multiply every term by 10: \(2(x + 2) + 5(x - 3) = 10\). Then \(7x - 11 = 10\), so \(x = 3\).
Mark scheme
- \(2(x + 2) + 5(x - 3) = 10\) — M1
- \(7x - 11 = 10\) — M1
- \(x = 3\) — A1
-
4 Show that [2 marks]
Show that the statement “\(n^2 + n + 11\) is always prime” is not true.
Show answerHide answer
Model answer
When \(n = 11\), \(121 + 11 + 11 = 143 = 11 \times 13\), which is not prime. So the statement is not true.
Mark scheme
- A counter-example, such as \(n = 11\), substituted — M1
- \(143 = 11 \times 13\) with a conclusion — A1
-
5 Prove [3 marks]
Prove that \((n + 2)^2 - n^2\) is a multiple of 4 for every integer \(n\).
Show answerHide answer
Model answer
\((n + 2)^2 - n^2 = n^2 + 4n + 4 - n^2 = 4n + 4 = 4(n + 1)\), which is a multiple of 4.
Mark scheme
- \(n^2 + 4n + 4\) — M1
- \(4n + 4\) — M1
- \(4(n + 1)\) with a conclusion — A1
-
6 Simplify [3 marks]
Simplify \(\dfrac{x^2 - 3x - 10}{x^2 - 4}\).
Show answerHide answer
Model answer
\(\dfrac{(x - 5)(x + 2)}{(x - 2)(x + 2)} = \dfrac{x - 5}{x - 2}\).
Mark scheme
- \((x - 5)(x + 2)\) or \((x - 2)(x + 2)\) — M1
- Both factorised — M1
- \(\dfrac{x - 5}{x - 2}\) — A1
Quick check
-
1
What may be cancelled in an algebraic fraction?
Show answerHide answer
B: Factors that multiply the whole top and the whole bottom
Terms that are added or subtracted cannot be cancelled.
-
2
Simplify \(\dfrac{x^2 - 9}{x + 3}\).
Show answerHide answer
A: \(x - 3\)
\(\dfrac{(x - 3)(x + 3)}{x + 3} = x - 3\).
-
3
Which statement about \(\dfrac{x + 3}{3}\) is correct?
Show answerHide answer
D: The 3s cannot be cancelled because the 3 on top is added
Only factors can be cancelled, not terms.
-
4
Solve \(\dfrac{x - 1}{3} + \dfrac{x + 2}{6} = 2\).
Show answerHide answer
C: \(x = 4\)
Multiply by 6: \(2(x - 1) + (x + 2) = 12\), so \(3x = 12\).
-
5
Which expression is an odd number for any whole number \(n\)?
Show answerHide answer
B: \(2n + 1\)
\(2n\) is even, so adding 1 makes it odd.
-
6
What is the sum of three consecutive whole numbers \(n\), \(n + 1\) and \(n + 2\)?
Show answerHide answer
A: \(3n + 3\)
\(n + n + 1 + n + 2 = 3n + 3 = 3(n + 1)\).
-
7
Which value of \(n\) is a counter-example to “\(n^2 + n + 1\) is always prime”?
Show answerHide answer
D: \(n = 4\)
\(16 + 4 + 1 = 21 = 3 \times 7\), which is not prime.
-
8
Simplify \(\dfrac{x^2 + 5x + 6}{x^2 + 3x + 2}\).
Show answerHide answer
C: \(\dfrac{x + 3}{x + 1}\)
\(\dfrac{(x + 2)(x + 3)}{(x + 1)(x + 2)} = \dfrac{x + 3}{x + 1}\).
-
9
Expand and simplify \((n + 1)^2 - (n - 1)^2\).
Show answerHide answer
B: \(4n\)
\(n^2 + 2n + 1 - n^2 + 2n - 1 = 4n\).