AQA FP1 2007 January — Question 1 10 marks

Exam BoardAQA
ModuleFP1 (Further Pure Mathematics 1)
Year2007
SessionJanuary
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicComplex Numbers Arithmetic
TypeVerifying roots satisfy equations
DifficultyEasy -1.2 This is a routine Further Maths question testing basic complex number operations: solving quadratic equations with complex roots (standard formula application), binomial expansion, and direct substitution to verify a root. All parts are straightforward recall and mechanical computation with no problem-solving or insight required, making it easier than average even for Further Maths.
Spec4.02e Arithmetic of complex numbers: add, subtract, multiply, divide4.02i Quadratic equations: with complex roots

1
  1. Solve the following equations, giving each root in the form \(a + b \mathrm { i }\) :
    1. \(x ^ { 2 } + 16 = 0\);
    2. \(x ^ { 2 } - 2 x + 17 = 0\).
    1. Expand \(( 1 + x ) ^ { 3 }\).
    2. Express \(( 1 + \mathrm { i } ) ^ { 3 }\) in the form \(a + b \mathrm { i }\).
    3. Hence, or otherwise, verify that \(x = 1 + \mathrm { i }\) satisfies the equation $$x ^ { 3 } + 2 x - 4 \mathrm { i } = 0$$

Part (a)(i)
AnswerMarks Guidance
Roots are \(\pm 4i\)M1A1 2 marks
Part (a)(ii)
AnswerMarks Guidance
Roots are \(1 \pm 4i\)M1A1 2 marks
Part (b)(i)
AnswerMarks Guidance
\((1 + x)^3 = 1 + 3x + 3x^2 + x^3\)M1A1 2 marks
Part (b)(ii)
AnswerMarks Guidance
\((1 + i)^3 = 1 + 3i - 3 - i = -2 + 2i\)M1A1 2 marks
Part (b)(iii)
\((1 + i)^3 + 2(1 + i) - 4i\)
AnswerMarks Guidance
\(\ldots = (-2 + 2i) + (2 + 2i) - 0\)M1, A1 2 marks
Total for Question 1: 10 marks
### Part (a)(i)
Roots are $\pm 4i$ | M1A1 | 2 marks | M1 for one correct root or two correct factors

### Part (a)(ii)
Roots are $1 \pm 4i$ | M1A1 | 2 marks | M1 for correct method

### Part (b)(i)
$(1 + x)^3 = 1 + 3x + 3x^2 + x^3$ | M1A1 | 2 marks | M1A0 if one small error

### Part (b)(ii)
$(1 + i)^3 = 1 + 3i - 3 - i = -2 + 2i$ | M1A1 | 2 marks | M1 if $i^2 = -1$ used

### Part (b)(iii)
$(1 + i)^3 + 2(1 + i) - 4i$
$\ldots = (-2 + 2i) + (2 + 2i) - 0$ | M1, A1 | 2 marks | with attempt to evaluate; convincingly shown (AG)

### **Total for Question 1: 10 marks**

---
1
\begin{enumerate}[label=(\alph*)]
\item Solve the following equations, giving each root in the form $a + b \mathrm { i }$ :
\begin{enumerate}[label=(\roman*)]
\item $x ^ { 2 } + 16 = 0$;
\item $x ^ { 2 } - 2 x + 17 = 0$.
\end{enumerate}\item \begin{enumerate}[label=(\roman*)]
\item Expand $( 1 + x ) ^ { 3 }$.
\item Express $( 1 + \mathrm { i } ) ^ { 3 }$ in the form $a + b \mathrm { i }$.
\item Hence, or otherwise, verify that $x = 1 + \mathrm { i }$ satisfies the equation

$$x ^ { 3 } + 2 x - 4 \mathrm { i } = 0$$
\end{enumerate}\end{enumerate}

\hfill \mbox{\textit{AQA FP1 2007 Q1 [10]}}