Exam Board WJEC Module Further Unit 1 (Further Unit 1) Session Specimen Marks 7 Paper Download PDF ↗ Mark scheme Download PDF ↗ Topic Proof by induction Type Prove divisibility Difficulty Standard +0.8 This is a standard proof by induction question requiring students to verify the base case, assume for n=k, and prove for n=k+1. While induction is a Further Maths topic making it inherently more challenging than Core content, this particular question follows a routine template with straightforward algebra (factoring out 4^k and showing divisibility by 6). It's moderately above average difficulty due to the proof requirement and algebraic manipulation needed, but doesn't require novel insight. Spec 4.01a Mathematical induction: construct proofs
Use mathematical induction to prove that \(4^n + 2\) is divisible by 6 for all positive integers \(n\). [7]
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Answer Marks
Guidance
When \(n = 1\), \(4^6 + 2 = 6\) which is divisible by 6 so the proposition is true for \(n = 1\). B1
AO2
Assume the proposition to be true for \(n = k\) so that \(4^k + 2\) is divisible by 6 and equals \(6N\). M1
AO2
Consider (for \(n = k + 1\)):
Answer Marks
Guidance
\[4^{k+1} + 2 = 4 \times 4^k + 2 = 4 \times (6N - 2) + 2 = 24N - 6\] M1, A1, A1, A1
AO2
This is divisible by 6 so true for \(k \Rightarrow\) true for \(k + 1\). Since true for \(n = 1\), the result is proved by induction. A1
AO2
Total: [7]
Question 2(a)
Answer Marks
Guidance
\[\frac{-1 + 7i}{2 + i} = \frac{(-1 + 7i)(2 - i)}{(2 + i)(2 - i)} = \frac{5 + 15i}{5} = 1 + 3i\] M1, A1, A1
AO3
\(2(x + iy) + i(x - iy) = 1 + 3i\) M1
AO3
\(2x + y = 1\) A1
AO3
\(x + 2y = 3\) A1
AO3
\[\left( x = -\frac{1}{3}; y = \frac{5}{3} \right)\] A1
AO1
\[z = \frac{-1 + 5i}{3}\] A1
AO1
Question 2(b)
Answer Marks
Guidance
\[ z
= \frac{\sqrt{26}}{3} = 1.70 \text{ (1.699673171)}\]
\[\tan^{-1}(-5) = -1.3734...\] B1
AO1
\[\arg(z) = -1.3734 + \pi = 1.77 \text{ (1.768191887)}\] B1
AO1
\[z = 1.70(\cos 1.77 + i \sin 1.77)\] B1
AO1
Total: [11]
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When $n = 1$, $4^6 + 2 = 6$ which is divisible by 6 so the proposition is true for $n = 1$. | B1 | AO2
Assume the proposition to be true for $n = k$ so that $4^k + 2$ is divisible by 6 and equals $6N$. | M1 | AO2
Consider (for $n = k + 1$):
$$4^{k+1} + 2 = 4 \times 4^k + 2 = 4 \times (6N - 2) + 2 = 24N - 6$$ | M1, A1, A1, A1 | AO2
This is divisible by 6 so true for $k \Rightarrow$ true for $k + 1$. Since true for $n = 1$, the result is proved by induction. | A1 | AO2
| **Total: [7]** |
# Question 2(a)
$$\frac{-1 + 7i}{2 + i} = \frac{(-1 + 7i)(2 - i)}{(2 + i)(2 - i)} = \frac{5 + 15i}{5} = 1 + 3i$$ | M1, A1, A1 | AO3
$2(x + iy) + i(x - iy) = 1 + 3i$ | M1 | AO3
$2x + y = 1$ | A1 | AO3
$x + 2y = 3$ | A1 | AO3
$$\left( x = -\frac{1}{3}; y = \frac{5}{3} \right)$$ | A1 | AO1
$$z = \frac{-1 + 5i}{3}$$ | A1 | AO1
# Question 2(b)
$$|z| = \frac{\sqrt{26}}{3} = 1.70 \text{ (1.699673171)}$$ | B1 | AO1
$$\tan^{-1}(-5) = -1.3734...$$ | B1 | AO1
$$\arg(z) = -1.3734 + \pi = 1.77 \text{ (1.768191887)}$$ | B1 | AO1
$$z = 1.70(\cos 1.77 + i \sin 1.77)$$ | B1 | AO1
| **Total: [11]** |
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Use mathematical induction to prove that $4^n + 2$ is divisible by 6 for all positive integers $n$. [7]
\hfill \mbox{\textit{WJEC Further Unit 1 Q1 [7]}}