2025 IJMB mathematics paper I

**SECTION A

1. Find the first three terms in the expansion of $\left(1+\frac{1}{x}\right)^5(1-x)^2$ in descending power of x. [4 marks]

2. Obtain the equation whose roots are the reciprocals of those of the equation $ax^2 - x + c = 0$. [4 marks]

3. Express $\left(\dfrac{\sqrt{5}-i}{\sqrt{5}+i}\right)^2$ in the form $a+ib$. [4 marks]

4. Given that $\tan 30° = \dfrac{1}{\sqrt{3}}$, obtain in surd form $\tan 15°$. [4 marks]

5. Show that the point (3,0) lies on the line $x=3$ and the circle $x^2+y^2=9$. Find the point of intersection. What conclusion can be drawn? [4 marks]

SECTION B: ALGEBRA — Attempt at least ONE

6. (a) Given that $A=\begin{pmatrix}1&0&1\\-1&2&0\\1&2&3\end{pmatrix}$, $B=\begin{pmatrix}3&0&-1\\2&4&0\\-1&0&6\end{pmatrix}$, show that $(AB)^T \neq A^TB^T$. [08 marks]

(b) Using mathematical induction, show that $\displaystyle\sum_{r=1}^{n}\frac{1}{r(r+1)}=\frac{n}{n+1}$. [08 marks]

© Find the value of $b$ for which $bx^2+bx+1=0$ has equal roots. [04 marks]

SECTION C: TRIGONOMETRY AND COMPLEX NUMBERS — Attempt at least ONE

7. (a) Find the set of values of $y$ for which $9^{y+5}=5^{y+9}$. [10 marks]

(b) The expression $ax^2+bx+c$ is divisible by $(x-1)$, has remainder 2 when divided by $(x+1)$, and remainder 8 when divided by $(x-2)$. Find $a$, $b$ and $c$, and factorize completely. [10 marks]

8. (a) Resolve $\dfrac{x+1}{(x-1)(x-2)(x+3)}$ into partial fractions. Hence obtain its binomial expansion up to the term in $x^2$. [08 marks]

(b) If $\alpha$ and $\beta$ are roots of $ax^2+bx+1=0$, form the equation whose roots are $\dfrac{1}{\alpha+2\beta}$ and $\dfrac{1}{\beta+2\alpha}$ in terms of $a$ and $b$. [08 marks]

© If one root of $x^2+px+q=0$ is twice the other, prove that $2p^2=9q$. [04 marks]

9. (a) Find the roots of $z^5+1=0$, where $z=x+iy$. [10 marks]

(b) Given $z=r(\cos\theta+i\sin\theta)$, show that $z^n=r^n(\cos n\theta+i\sin n\theta)$. [10 marks]

10. (a)(i) If $z_1=\dfrac{1}{1+i}$, $z_2=\dfrac{1}{1-i}$, find $z_1^2-z_2^2$ in the form $x+iy$. [12 marks]

(ii) Simplify $x^2-\dfrac{1}{x^2}$, given $x=\dfrac{1}{1+i\sqrt{3}}$.

(b) Given $\sin^{-1}x+\cos^{-1}(x\sqrt{3})=\dfrac{\pi}{2}$, find $x$. [08 marks]

11. (a) Using expressions for $\cos(A+B)$ and $\sin(A+B)$, obtain $\tan(A+B)$. Hence find $A+B$ if $\tan A=5$ and $\tan B=1$. [10 marks]

(b) Express $12\sin x-5\cos x$ in the form $R\sin(x-\alpha)$ and solve $12\sin x-5\cos x=13$ for $0°\leq x\leq360°$. [10 marks]

SOLUTIONS

SECTION A

Q1. Expand $\left(1+\frac{1}{x}\right)^5(1-x)^2$ — first three terms in descending powers of $x$

Expand $\left(1+\frac{1}{x}\right)^5$ using binomial theorem:

$$= 1 + \frac{5}{x} + \frac{10}{x^2} + \frac{10}{x^3}+\cdots$$

Expand $(1-x)^2$:

$$= 1 - 2x + x^2$$

Multiply, tracking descending powers from $x^2$:

| Power | Sources | Value |

||||

| $x^2$ | $x^2 \times 1$ | $x^2$ |

| $x^1$ | $(-2x)\times 1$ | $-2x$ |

| $x^0$ | $(1)(1)+(-2x)\left(\frac{5}{x}\right)+(x^2)\left(\frac{10}{x^2}\right)$ | $1-10+10=1$ |

$$\boxed{\text{First three terms: } x^2 - 2x + 1}$$

Q2. Equation with reciprocal roots of $ax^2-x+c=0$

For original roots $\alpha, \beta$:

$$\alpha+\beta = \frac{1}{a}, \qquad \alpha\beta = \frac{c}{a}$$

New roots $\frac{1}{\alpha}, \frac{1}{\beta}$:

$$\text{Sum} = \frac{1}{\alpha}+\frac{1}{\beta} = \frac{\alpha+\beta}{\alpha\beta} = \frac{1/a}{c/a} = \frac{1}{c}$$

$$\text{Product} = \frac{1}{\alpha\beta} = \frac{a}{c}$$

New equation: $x^2 - \frac{1}{c}x + \frac{a}{c} = 0$

Multiply through by $c$:

$$\boxed{cx^2 - x + a = 0}$$

Q3. Express $\left(\dfrac{\sqrt{5}-i}{\sqrt{5}+i}\right)^2$ in form $a+ib$

Step 1: Simplify the fraction by multiplying by conjugate:

$$\frac{\sqrt{5}-i}{\sqrt{5}+i} \times \frac{\sqrt{5}-i}{\sqrt{5}-i} = \frac{(\sqrt{5}-i)^2}{5+1} = \frac{5-2\sqrt{5}i+i^2}{6} = \frac{4-2\sqrt{5}i}{6} = \frac{2-\sqrt{5}i}{3}$$

Step 2: Square the result:

$$\left(\frac{2-\sqrt{5}i}{3}\right)^2 = \frac{4-4\sqrt{5}i+5i^2}{9} = \frac{4-4\sqrt{5}i-5}{9} = \frac{-1-4\sqrt{5}i}{9}$$

$$\boxed{a+ib = -\frac{1}{9} - \frac{4\sqrt{5}}{9}i}$$

Q4. Find $\tan 15°$ given $\tan 30° = \frac{1}{\sqrt{3}}$

Use the double angle formula: $\tan 30° = \dfrac{2\tan 15°}{1-\tan^2 15°}$

Let $t = \tan 15°$:

$$\frac{1}{\sqrt{3}} = \frac{2t}{1-t^2}$$

$$1-t^2 = 2\sqrt{3}t$$

$$t^2+2\sqrt{3}t-1=0$$

Using quadratic formula:

$$t = \frac{-2\sqrt{3}\pm\sqrt{12+4}}{2} = \frac{-2\sqrt{3}\pm 4}{2} = -\sqrt{3}\pm 2$$

Since $15°$ is in the first quadrant, $\tan 15° > 0$:

$$\boxed{\tan 15° = 2-\sqrt{3}}$$

Q5. Point (3,0) on line $x=3$ and circle $x^2+y^2=9$

On the line: $x=3$ → substituting gives $3=3$ ✓

On the circle: $3^2+0^2=9$ ✓

Intersection of line and circle:

Substitute $x=3$ into $x^2+y^2=9$:

$$9+y^2=9 \Rightarrow y^2=0 \Rightarrow y=0$$

Only intersection point is (3, 0)

Conclusion: The line $x=3$ is a tangent to the circle $x^2+y^2=9$ at the point (3, 0) — it meets the circle at exactly one point.

SECTION B: ALGEBRA

Q6(a) Show $(AB)^T \neq A^TB^T$

Compute AB:

$$AB = \begin{pmatrix}1&0&1\\-1&2&0\\1&2&3\end{pmatrix}\begin{pmatrix}3&0&-1\\2&4&0\\-1&0&6\end{pmatrix}$$

• Row 1: $(3+0-1,\ 0+0+0,\ -1+0+6) = (2,\ 0,\ 5)$

• Row 2: $(-3+4+0,\ 0+8+0,\ 1+0+0) = (1,\ 8,\ 1)$

• Row 3: $(3+4-3,\ 0+8+0,\ -1+0+18) = (4,\ 8,\ 17)$

$$AB = \begin{pmatrix}2&0&5\\1&8&1\\4&8&17\end{pmatrix}$$

$$(AB)^T = \begin{pmatrix}2&1&4\\0&8&8\\5&1&17\end{pmatrix}$$

Compute $A^TB^T$:

$$A^T = \begin{pmatrix}1&-1&1\\0&2&2\\1&0&3\end{pmatrix}, \quad B^T = \begin{pmatrix}3&2&-1\\0&4&0\\-1&0&6\end{pmatrix}$$

$A^TB^T$:

• Row 1: $(3+0-1,\ 2-4+0,\ -1+0+6)=(2,\ -2,\ 5)$

• Row 2: $(0+0-2,\ 0+8+0,\ 0+0+12)=(-2,\ 8,\ 12)$

• Row 3: $(3+0-3,\ 2+0+0,\ -1+0+18)=(0,\ 2,\ 17)$

$$A^TB^T = \begin{pmatrix}2&-2&5\\-2&8&12\\0&2&17\end{pmatrix}$$

Comparing Row 1, Col 2: $(AB)^T$ gives 1, $A^TB^T$ gives −2

$$\boxed{(AB)^T \neq A^TB^T \checkmark}$$

(Note: The correct identity is $(AB)^T = B^TA^T$)

Q6(b) Prove by induction: $\displaystyle\sum_{r=1}^{n}\frac{1}{r(r+1)}=\frac{n}{n+1}$

Base case $n=1$:

$$\text{LHS} = \frac{1}{1\times2} = \frac{1}{2}, \qquad \text{RHS} = \frac{1}{1+1} = \frac{1}{2} \checkmark$$

Inductive hypothesis: Assume true for $n=k$:

$$\sum_{r=1}^{k}\frac{1}{r(r+1)} = \frac{k}{k+1}$$

Inductive step — show true for $n=k+1$:

$$\sum_{r=1}^{k+1}\frac{1}{r(r+1)} = \frac{k}{k+1}+\frac{1}{(k+1)(k+2)}$$

$$= \frac{k(k+2)+1}{(k+1)(k+2)} = \frac{k^2+2k+1}{(k+1)(k+2)} = \frac{(k+1)^2}{(k+1)(k+2)} = \frac{k+1}{k+2}$$

This equals $\dfrac{k+1}{(k+1)+1}$, which is the formula for $n=k+1$.

Hence proved by mathematical induction ✓

Q6© Value of $b$ for equal roots of $bx^2+bx+1=0$

Equal roots require discriminant $= 0$:

$$\Delta = b^2-4(b)(1)=0$$

$$b^2-4b=0 \Rightarrow b(b-4)=0$$

$$\boxed{b=0 \text{ or } b=4}$$

($b=0$ reduces it to a linear equation, so the meaningful answer giving equal roots is $b=4$)

Verification: $4x^2+4x+1=0 \Rightarrow (2x+1)^2=0 \Rightarrow x=-\frac{1}{2}$ (repeated) ✓

SECTION C

Q7(a) Solve $9^{y+5}=5^{y+9}$

Take natural log of both sides:

$$(y+5)\ln 9 = (y+9)\ln 5$$

$$y\ln 9 + 5\ln 9 = y\ln 5 + 9\ln 5$$

$$y(\ln 9 - \ln 5) = 9\ln 5 - 5\ln 9$$

$$y = \frac{9\ln 5-5\ln 9}{\ln 9-\ln 5}$$

Since $\ln 9 = 2\ln 3$:

$$y = \frac{9\ln 5-10\ln 3}{2\ln 3-\ln 5}$$

Numerical evaluation: $\ln 5\approx1.6094$, $\ln 3\approx1.0986$

$$y = \frac{9(1.6094)-10(1.0986)}{2(1.0986)-1.6094} = \frac{14.485-10.986}{2.197-1.609} = \frac{3.499}{0.588}$$

$$\boxed{y \approx 5.95}$$

Q7(b) Find $a$, $b$, $c$ and factorize $f(x)=ax^2+bx+c$

By the Remainder Theorem:

• $f(1)=0$ (divisible by $x-1$): $\quad a+b+c=0$ …(i)

• $f(-1)=2$: $\quad a-b+c=2$ …(ii)

• $f(2)=8$: $\quad 4a+2b+c=8$ …(iii)

(i) − (ii): $2b=-2 \Rightarrow \boxed{b=-1}$

(i): $a+c=1$ …(iv)

(iii) − (i): $3a+b=8 \Rightarrow 3a=9 \Rightarrow \boxed{a=3}$

From (iv): $\boxed{c=-2}$

$$f(x) = 3x^2-x-2$$

Factorize (looking for factors of form $(3x+p)(x+q)$ where $pq=-2$, $p+3q=-1$):

$$3x^2-x-2 = (3x+2)(x-1)$$

Verify: $(3x+2)(x-1)=3x^2-3x+2x-2=3x^2-x-2$ ✓

$$\boxed{f(x)=(3x+2)(x-1)}$$

Q8(a) Partial fractions of $\dfrac{x+1}{(x-1)(x-2)(x+3)}$

Let:

$$\frac{x+1}{(x-1)(x-2)(x+3)} = \frac{A}{x-1}+\frac{B}{x-2}+\frac{C}{x+3}$$

Find A (multiply both sides by $(x-1)$, set $x=1$):

$$A = \frac{1+1}{(1-2)(1+3)} = \frac{2}{(-1)(4)} = -\frac{1}{2}$$

Find B ($x=2$):

$$B = \frac{2+1}{(2-1)(2+3)} = \frac{3}{(1)(5)} = \frac{3}{5}$$

Find C ($x=-3$):

$$C = \frac{-3+1}{(-3-1)(-3-2)} = \frac{-2}{(-4)(-5)} = \frac{-2}{20} = -\frac{1}{10}$$

$$\boxed{\frac{x+1}{(x-1)(x-2)(x+3)} = \frac{-1/2}{x-1}+\frac{3/5}{x-2}-\frac{1/10}{x+3}}$$

Binomial Expansion to $x^2$ (valid for $|x|<1$):

Term 1:

$$\frac{-1/2}{x-1} = \frac{1/2}{1-x} = \frac{1}{2}(1+x+x^2+\cdots)$$

Term 2:

$$\frac{3/5}{x-2} = \frac{-3/10}{1-x/2} = -\frac{3}{10}\left(1+\frac{x}{2}+\frac{x^2}{4}+\cdots\right)$$

Term 3:

$$\frac{-1/10}{x+3} = \frac{-1/30}{1+x/3} = -\frac{1}{30}\left(1-\frac{x}{3}+\frac{x^2}{9}-\cdots\right)$$

Constant terms:

$$\frac{1}{2}-\frac{3}{10}-\frac{1}{30} = \frac{15-9-1}{30} = \frac{5}{30} = \frac{1}{6}$$

Coefficient of $x$:

$$\frac{1}{2}-\frac{3}{20}+\frac{1}{90} = \frac{90-27+2}{180} = \frac{65}{180} = \frac{13}{36}$$

Coefficient of $x^2$:

$$\frac{1}{2}-\frac{3}{40}-\frac{1}{270} = \frac{540-81-2}{1080} = \frac{457}{1080}$$

$$\boxed{\frac{x+1}{(x-1)(x-2)(x+3)} \approx \frac{1}{6}+\frac{13}{36}x+\frac{457}{1080}x^2+\cdots}$$

Q8(b) New equation with roots $\frac{1}{\alpha+2\beta}$ and $\frac{1}{\beta+2\alpha}$

For $ax^2+bx+1=0$: $\quad \alpha+\beta=-\dfrac{b}{a}$, $\quad \alpha\beta=\dfrac{1}{a}$

New roots sum $S$:

$$S = \frac{1}{\alpha+2\beta}+\frac{1}{\beta+2\alpha} = \frac{(\beta+2\alpha)+(\alpha+2\beta)}{(\alpha+2\beta)(\beta+2\alpha)} = \frac{3(\alpha+\beta)}{(\alpha+2\beta)(\beta+2\alpha)}$$

Expand denominator:

$$(\alpha+2\beta)(\beta+2\alpha) = \alpha\beta+2\alpha^2+2\beta^2+4\alpha\beta = 2(\alpha^2+\beta^2)+5\alpha\beta$$

$$\alpha^2+\beta^2 = (\alpha+\beta)^2-2\alpha\beta = \frac{b^2}{a^2}-\frac{2}{a}$$

$$\text{Denominator} = 2\left(\frac{b^2}{a^2}-\frac{2}{a}\right)+\frac{5}{a} = \frac{2b^2}{a^2}+\frac{1}{a} = \frac{2b^2+a}{a^2}$$

$$S = \frac{3(-b/a)}{(2b^2+a)/a^2} = \frac{-3b/a \times a^2}{2b^2+a} = \frac{-3ab}{2b^2+a}$$

New roots product $P$:

$$P = \frac{1}{(\alpha+2\beta)(\beta+2\alpha)} = \frac{a^2}{2b^2+a}$$

New equation $x^2 - Sx + P = 0$, multiply through by $(2b^2+a)$:

$$\boxed{(2b^2+a)x^2+3abx+a^2=0}$$

Q8© Prove $2p^2=9q$ if one root is twice the other in $x^2+px+q=0$

Let roots be $\alpha$ and $2\alpha$:

Sum of roots:

$$\alpha+2\alpha = -p \Rightarrow 3\alpha=-p \Rightarrow \alpha=-\frac{p}{3}$$

Product of roots:

$$\alpha \cdot 2\alpha = q \Rightarrow 2\alpha^2=q$$

Substitute $\alpha$:

$$2\left(-\frac{p}{3}\right)^2=q \Rightarrow 2\cdot\frac{p^2}{9}=q \Rightarrow \frac{2p^2}{9}=q$$

$$\boxed{2p^2=9q \checkmark}$$

Q9(a) Roots of $z^5+1=0$

$$z^5=-1$$

Write $-1$ in polar form: $-1 = e^{i\pi}$

General form: $-1 = e^{i\pi(2k+1)}$ for $k=0,1,2,3,4$

$$z_k = e^{i\pi(2k+1)/5} = \cos\frac{(2k+1)\pi}{5}+i\sin\frac{(2k+1)\pi}{5}$$

| $k$ | Angle | Root |

||||

| 0 | $36°$ | $\cos36°+i\sin36°$ |

| 1 | $108°$ | $\cos108°+i\sin108°$ |

| 2 | $180°$ | $-1$ |

| 3 | $252°$ | $\cos252°+i\sin252°$ |

| 4 | $324°$ | $\cos324°+i\sin324°$ |

The five roots are the fifth roots of $-1$, equally spaced at $72°$ apart on the unit circle.

Q9(b) Prove $z^n=r^n(\cos n\theta+i\sin n\theta)$ by induction

Base case $n=1$:

$$z^1=r(\cos\theta+i\sin\theta) \checkmark$$

Inductive hypothesis: Assume $z^k=r^k(\cos k\theta+i\sin k\theta)$

Inductive step:

$$z^{k+1}=z^k\cdot z=r^k(\cos k\theta+i\sin k\theta)\cdot r(\cos\theta+i\sin\theta)$$

$$=r^{k+1}[(\cos k\theta\cos\theta-\sin k\theta\sin\theta)+i(\sin k\theta\cos\theta+\cos k\theta\sin\theta)]$$

Using compound angle formulas:

$$=r^{k+1}[\cos(k+1)\theta+i\sin(k+1)\theta]$$

This is the formula for $n=k+1$.

Hence by induction, $z^n=r^n(\cos n\theta+i\sin n\theta)$ for all $n\in\mathbb{Z}^+$ ✓

Q10(a)(i) Find $z_1^2-z_2^2$ where $z_1=\frac{1}{1+i}$, $z_2=\frac{1}{1-i}$

Rationalize:

$$z_1=\frac{1}{1+i}\times\frac{1-i}{1-i}=\frac{1-i}{2}$$

$$z_2=\frac{1}{1-i}\times\frac{1+i}{1+i}=\frac{1+i}{2}$$

Square each:

$$z_1^2=\frac{(1-i)^2}{4}=\frac{1-2i+i^2}{4}=\frac{1-2i-1}{4}=\frac{-2i}{4}=\frac{-i}{2}$$

$$z_2^2=\frac{(1+i)^2}{4}=\frac{1+2i+i^2}{4}=\frac{2i}{4}=\frac{i}{2}$$

Subtract:

$$z_1^2-z_2^2=\frac{-i}{2}-\frac{i}{2}=-i$$

$$\boxed{z_1^2-z_2^2=0-i, \text{ i.e., } x=0,\ y=-1}$$

Q10(a)(ii) Simplify $x^2-\frac{1}{x^2}$ given $x=\frac{1}{1+i\sqrt{3}}$

Rationalize $x$:

$$x=\frac{1}{1+i\sqrt{3}}\times\frac{1-i\sqrt{3}}{1-i\sqrt{3}}=\frac{1-i\sqrt{3}}{1+3}=\frac{1-i\sqrt{3}}{4}$$

Find $\frac{1}{x}$:

$$\frac{1}{x}=1+i\sqrt{3}$$

Find $x^2$:

$$x^2=\left(\frac{1-i\sqrt{3}}{4}\right)^2=\frac{1-2i\sqrt{3}-3}{16}=\frac{-2-2i\sqrt{3}}{16}=\frac{-1-i\sqrt{3}}{8}$$

Find $\frac{1}{x^2}$:

$$\frac{1}{x^2}=(1+i\sqrt{3})^2=1+2i\sqrt{3}-3=-2+2i\sqrt{3}$$

Subtract:

$$x^2-\frac{1}{x^2}=\frac{-1-i\sqrt{3}}{8}-(-2+2i\sqrt{3})$$

$$=\frac{-1-i\sqrt{3}+16-16i\sqrt{3}}{8}=\frac{15-17i\sqrt{3}}{8}$$

$$\boxed{x^2-\frac{1}{x^2}=\frac{15}{8}-\frac{17\sqrt{3}}{8}i}$$

Q10(b) Solve $\sin^{-1}x+\cos^{-1}(x\sqrt{3})=\frac{\pi}{2}$

Use identity: $\sin^{-1}x = \frac{\pi}{2}-\cos^{-1}x$

$$\frac{\pi}{2}-\cos^{-1}x+\cos^{-1}(x\sqrt{3})=\frac{\pi}{2}$$

$$\cos^{-1}(x\sqrt{3})=\cos^{-1}x$$

$$x\sqrt{3}=x$$

$$x\sqrt{3}-x=0 \Rightarrow x(\sqrt{3}-1)=0$$

$$\boxed{x=0}$$

Verify: $\sin^{-1}(0)+\cos^{-1}(0)=0+\frac{\pi}{2}=\frac{\pi}{2}$ ✓

Q11(a) Derive $\tan(A+B)$ and find $A+B$ when $\tan A=5$, $\tan B=1$

$$\tan(A+B)=\frac{\sin(A+B)}{\cos(A+B)}=\frac{\sin A\cos B+\cos A\sin B}{\cos A\cos B-\sin A\sin B}$$

Divide numerator and denominator by $\cos A\cos B$:

$$\boxed{\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}}$$

With $\tan A=5$, $\tan B=1$:

$$\tan(A+B)=\frac{5+1}{1-(5)(1)}=\frac{6}{-4}=-\frac{3}{2}$$

$$A+B=\tan^{-1}\left(-\frac{3}{2}\right)$$

Since $\tan A=5>0$ and $\tan B=1>0$, both $A$ and $B$ are in the first quadrant, so $A+B$ is in the second quadrant:

$$A+B=180°-56.31°\approx\boxed{123.69°}$$

Q11(b) Express $12\sin x-5\cos x$ as $R\sin(x-\alpha)$ and solve $=13$

Expand $R\sin(x-\alpha)$:

$$R\sin(x-\alpha)=R\cos\alpha\sin x-R\sin\alpha\cos x$$

Compare coefficients:

$$R\cos\alpha=12, \qquad R\sin\alpha=5$$

$$R=\sqrt{12^2+5^2}=\sqrt{144+25}=\sqrt{169}=13$$

$$\tan\alpha=\frac{5}{12} \Rightarrow \alpha=\tan^{-1}\left(\frac{5}{12}\right)\approx22.62°$$

$$\boxed{12\sin x-5\cos x=13\sin(x-22.62°)}$$

Solve $13\sin(x-22.62°)=13$:

$$\sin(x-22.62°)=1$$

$$x-22.62°=90°$$

$$x=112.62°$$

Check second solution: $\sin=1$ has only one solution in $[0°,360°]$

$$\boxed{x\approx112.62°}$$

SUMMARY TABLE

| Q | Answer |

|—||

| 1 | $x^2-2x+1$ |

| 2 | $cx^2-x+a=0$ |

| 3 | $-\frac{1}{9}-\frac{4\sqrt5}{9}i$ |

| 4 | $2-\sqrt3$ |

| 5 | Tangent at (3,0) |

| 6c | $b=4$ |

| 7a | $y\approx5.95$ |

| 7b | $a=3,b=-1,c=-2$; $(3x+2)(x-1)$ |

| 8c | $2p^2=9q$ proved |

| 9a | $z_k=\cos\frac{(2k+1)\pi}{5}+i\sin\frac{(2k+1)\pi}{5}$, $k=0..4$ |

| 10a(i) | $-i$ |

| 10a(ii) | $\frac{15}{8}-\frac{17\sqrt3}{8}i$ |

| 10b | $x=0$ |

| 11a | $A+B\approx123.69°$ |

| 11b | $x\approx112.62°$ |