Let $z_{1} = 2 + i$ and $z_{2} = 1 – 3i$. Find $z_{1}\overline{z_{2}}$ in the form $a + bi$, and find $\left|z_{1}\overline{z_{2}}\right|$.
Question HKC78-AIHL-CPX-01
[Maximum mark: 6]
Let $z_{1} = 2 + i$ and $z_{2} = 1 – 3i$. Find $z_{1}\overline{z_{2}}$ in the form $a + bi$, and find $\left|z_{1}\overline{z_{2}}\right|$.
Mark Scheme
$overline{z_{2}} = 1 + 3i$. [1]
$z_{1}overline{z_{2}} = (2 + i)(1 + 3i) = 2 + 6i + i + 3i^{2} = -1 + 7i$. [3]
$left|-1 + 7iright| = sqrt{1 + 49} = sqrt{50} = 5sqrt{2}$. [2]
Total: 6 marks
Let $A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$. Find $\det A$ and $A^{-1}$.
Question HKC78-AIHL-MAT-01
[Maximum mark: 5]
Let $A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$. Find $\det A$ and $A^{-1}$.
Mark Scheme
$det A = 4 – 6 = -2$. [1]
$A^{-1} = -dfrac{1}{2}begin{pmatrix} 4 & -2 \ -3 & 1 end{pmatrix} = begin{pmatrix} -2 & 1 \ dfrac{3}{2} & -dfrac{1}{2} end{pmatrix}$. [3]
Check the top row against the first column of $A$: $-2cdot 1 + 1cdot 3 = 1$ and $-2cdot 2 + 1cdot 4 = 0$. [1]
Total: 5 marks
An arithmetic sequence has a first term of 5 and a common difference of 3. Find the 10th term of the sequence.
Question Q1
An arithmetic sequence has a first term of 5 and a common difference of 3. Find the 10th term of the sequence.
Mark Scheme
Step 1: Identify given information.
$u_1 = 5$, $d = 3$, $n = 10$.
Step 2: Use formula $u_n = u_1 + (n-1)d$.
$u_{10} = 5 + (10-1)3$
Step 3: Calculate.
$u_{10} = 5 + (9)3 = 5 + 27 = 32$.
Answer: 32
An arithmetic sequence has $u_{3} = 11$ and $u_{8} = 26$. Find the first term and the sum of the first $15$ terms.
Question HKC78-AIHL-SEQ-01
[Maximum mark: 5]
An arithmetic sequence has $u_{3} = 11$ and $u_{8} = 26$. Find the first term and the sum of the first $15$ terms.
Mark Scheme
$u_{3} = a + 2d = 11$ and $u_{8} = a + 7d = 26$. Subtract: $5d = 15$, so $d = 3$ and $a = 5$. [3]
$S_{15} = dfrac{15}{2}big(2cdot 5 + 14cdot 3big) = dfrac{15}{2} cdot 52 = 390$. [2]
Total: 5 marks
$\$5000$ is invested at a nominal annual rate of $6\%$, compounded monthly. Find the value after $2$ years, to the nearest cent.
Question HKC78-AIHL-FIN-01
[Maximum mark: 4]
$\$5000$ is invested at a nominal annual rate of $6\%$, compounded monthly. Find the value after $2$ years, to the nearest cent.
Mark Scheme
The monthly rate is $dfrac{0.06}{12} = 0.005$, and there are $24$ months. [1]
Value $= 5000(1.005)^{24}$. [1]
$(1.005)^{24} = 1.1271598ldots$, so the value is $5635.798ldots$. [1]
To the nearest cent, the value is $$5635.80$. [1]
Total: 4 marks
$\$400$ is deposited at the end of each month for $3$ years. The account pays a nominal annual rate of $4.8\%$, compounded monthly. Find the amount in the account immediately after the last deposit, to the nearest dollar.
Question HKC78-AIHL-FIN-02
[Maximum mark: 5]
$\$400$ is deposited at the end of each month for $3$ years. The account pays a nominal annual rate of $4.8\%$, compounded monthly. Find the amount in the account immediately after the last deposit, to the nearest dollar.
Mark Scheme
Monthly rate $i = dfrac{0.048}{12} = 0.004$, and $n = 36$ deposits. [1]
Future value of an ordinary annuity: $400 cdot dfrac{(1.004)^{36} – 1}{0.004}$. [2]
$(1.004)^{36} approx 1.154552$, so the value is $15455.24ldots$. [1]
To the nearest dollar, the amount is $$15455$. [1]
Total: 5 marks
Express $z = -1 + i$ in modulus-argument form, with the argument in $(-\pi, \pi]$.
Question HKC78-AIHL-CPX-02
[Maximum mark: 4]
Express $z = -1 + i$ in modulus-argument form, with the argument in $(-\pi, \pi]$.
Mark Scheme
$|z| = sqrt{(-1)^{2} + 1^{2}} = sqrt{2}$. [1]
The point $(-1, 1)$ is in the second quadrant. The reference angle is $dfrac{pi}{4}$, so $arg z = pi – dfrac{pi}{4} = dfrac{3pi}{4}$. [2]
$z = sqrt{2}operatorname{cis}dfrac{3pi}{4}$, or $sqrt{2}left(cosdfrac{3pi}{4} + isindfrac{3pi}{4}right)$. [1]
Total: 4 marks
Use an inverse matrix to solve $\begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 7 \\ 4 \end{pmatrix}$.
Question HKC78-AIHL-MAT-02
[Maximum mark: 5]
Use an inverse matrix to solve $\begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 7 \\ 4 \end{pmatrix}$.
Mark Scheme
Let $M = begin{pmatrix} 2 & 1 \ 1 & 1 end{pmatrix}$. Then $det M = 1$, so $M^{-1} = begin{pmatrix} 1 & -1 \ -1 & 2 end{pmatrix}$. [2]
$begin{pmatrix} x \ y end{pmatrix} = begin{pmatrix} 1 & -1 \ -1 & 2 end{pmatrix}begin{pmatrix} 7 \ 4 end{pmatrix} = begin{pmatrix} 3 \ 1 end{pmatrix}$. [2]
Check: $2(3) + 1 = 7$ and $3 + 1 = 4$. So $x = 3$ and $y = 1$. [1]
Total: 5 marks
Four notebooks and six pens cost $\$38$. Three notebooks and two pens cost $\$20$. Find the price of one notebook and the price of one pen.
Question HKC78-AIHL-SYS-02
[Maximum mark: 5]
Four notebooks and six pens cost $\$38$. Three notebooks and two pens cost $\$20$. Find the price of one notebook and the price of one pen.
Mark Scheme
Let $n$ be the notebook price and $p$ the pen price. Then $4n + 6p = 38$ and $3n + 2p = 20$. Divide the first equation by $2$: $2n + 3p = 19$. [1]
Multiply $2n + 3p = 19$ by $3$ and the second equation by $2$: $6n + 9p = 57$ and $6n + 4p = 40$. [1]
Subtract: $5p = 17$, so $p = dfrac{17}{5} = 3.4$. [2]
$3n + 2(3.4) = 20$, so $3n + 6.8 = 20$, $3n = 13.2$ and $n = 4.4$. A notebook costs $$4.40$ and a pen costs $$3.40$. [1]
Total: 5 marks
