# CSEC Math Jan 2012 P2 Q11

CSEC Math Jan 2012 Paper 2 Question 11

Updated on April 20th, 2012: Added some video explanations

In part (a), $$\overset{\longrightarrow}{OA} = \begin{pmatrix}-1 \\ 3\end{pmatrix}$$, $$\overset{\longrightarrow}{OB} = \begin{pmatrix}5 \\ 1\end{pmatrix}$$ and $$\overset{\longrightarrow}{BA} = \overset{\longrightarrow}{BO} + \overset{\longrightarrow}{OA} = \begin{pmatrix}-5 \\ -1\end{pmatrix} + \begin{pmatrix}-1 \\ 3\end{pmatrix} = \begin{pmatrix}-6 \\ 2\end{pmatrix}$$.

Given that $$G$$ is the mid-point of the line $$AB$$, $$\overset{\longrightarrow}{BG} = \frac{1}{2} \overset{\longrightarrow}{BA} = \begin{pmatrix}-3 \\ 1\end{pmatrix}$$ and $$\overset{\longrightarrow}{OG} = \overset{\longrightarrow}{OB} + \overset{\longrightarrow}{BG} = \begin{pmatrix}5 \\ 1\end{pmatrix} + \begin{pmatrix}-3 \\ 1\end{pmatrix} = \begin{pmatrix}2 \\ 2\end{pmatrix}$$.

In part (b), $$L = \begin{pmatrix}3 & 2 \\ 1 & 4\end{pmatrix}$$ and $$M = \begin{pmatrix}-1 & 3 \\ 0 & 2\end{pmatrix}$$.

$$\begin{eqnarray} L + 2M &=& \begin{pmatrix}3 & 2 \\ 1 & 4\end{pmatrix} + 2 \begin{pmatrix}-1 & 3 \\ 0 & 2\end{pmatrix} \\ &=& \begin{pmatrix}3-2 & 2+6 \\ 1+0 & 4+4\end{pmatrix} \\ &=& \begin{pmatrix}1 & 8 \\ 1 & 8\end{pmatrix}. \end{eqnarray}$$

$$\begin{eqnarray} LM &=& \begin{pmatrix}3 & 2 \\ 1 & 4\end{pmatrix} \begin{pmatrix}-1 & 3 \\ 0 & 2\end{pmatrix} \\ &=& \begin{pmatrix}3 \times -1 + 2 \times 0 & 3 \times 3 + 2 \times 2 \\ 1 \times -1 + 4 \times 0 & 1 \times 3 + 4 \times 2\end{pmatrix} \\ &=& \begin{pmatrix}-3 & 13 \\ -1 & 11\end{pmatrix}. \end{eqnarray}$$

Finally, in part (c) we are given a matrix $$Q$$ such that $$Q = \begin{pmatrix}4 & 2 \\ 1 & 1\end{pmatrix}$$.

$$\begin{eqnarray} Q^{-1} &=& \frac{1}{|Q|} \begin{pmatrix}1 & -2 \\ -1 & 4\end{pmatrix} \\ &=& \frac{1}{4 – 2} \begin{pmatrix}1 & -2 \\ -1 & 4\end{pmatrix} \\ &=& \frac{1}{2} \begin{pmatrix}1 & -2 \\ -1 & 4\end{pmatrix}. \end{eqnarray}$$

Using a matrix method, we need to find the values of $$x$$ and $$y$$ in the equation $$\begin{pmatrix}4 & 2 \\ 1 & 1\end{pmatrix} \begin{pmatrix}x \\ y\end{pmatrix} = \begin{pmatrix}8 \\ 3\end{pmatrix}$$.

$$\begin{eqnarray} \begin{pmatrix}4 & 2 \\ 1 & 1\end{pmatrix} \begin{pmatrix}x \\ y\end{pmatrix} &=& \begin{pmatrix}8 \\ 3\end{pmatrix} \\ QX &=& B \\ X &=& Q^{-1}B \\ &=& \frac{1}{2} \begin{pmatrix}1 & -2 \\ -1 & 4\end{pmatrix} \begin{pmatrix}8 \\ 3\end{pmatrix} \\ &=& \frac{1}{2} \begin{pmatrix}2 \\ 4\end{pmatrix} \\ &=& \begin{pmatrix}1 \\ 2\end{pmatrix}. \end{eqnarray}$$

Therefore, $$x = 1$$ and $$y = 2$$.

# CSEC Math Jan 2012 P2 Q10

CSEC Math Jan 2012 Paper 2 Question 10

Updated on April 20th, 2012: Added some video explanations

In part (a), we have a regular hexagon, not drawn to scale, with centre, $$O$$, and $$AO = 5\text{ cm}$$. The size of angle $$AOB$$ is $$\frac{360^\circ}{6} = 60^\circ$$.

A regular hexagon

Since the hexagon is regular it means that the 6 triangles are congruent. So, by finding the area of one of the triangles and multiplying the result by 6 we would get the area of the hexagon.

$$\begin{eqnarray} \text{Area of triangle } AOB &=& \left ( \frac{1}{2} \times OA \times OB \times \sin \angle AOB \right ) \text{ cm}^2 \\ &=& \left ( \frac{1}{2} \times 5 \times 5 \times \sin 60^\circ \right ) \text{ cm}^2 \\ &=& \left ( \frac{25}{2} \times \frac{\sqrt{3}}{2} \right ) \text{ cm}^2 \\ &=& \frac{25 \sqrt{3}}{4} \text{ cm}^2. \end{eqnarray}$$

Therefore, the area of the hexagon is 6 times the area of triangle $$AOB$$.

$$\begin{eqnarray} \text{Area of hexagon} &=& 6 \times \frac{25 \sqrt{3}}{4} \text{ cm}^2 \\ &=& \frac{75 \sqrt{3}}{2} \text{ cm}^2 \\ &\approx& 65 \text{ cm}^2. \end{eqnarray}$$

For part (b), the diagram below, not drawn to scale, shows a vertical pole, $$PL$$, standing on a horizontal plane, $$KLM$$. The angle of elevation of $$P$$ from $$K$$ is $$28^\circ$$, $$KL = 15\text{ m}$$, $$LM = 19\text{ m}$$ and $$\angle KLM = 115^\circ$$.

A pole on a horizontal plane

The diagram below shows the $$28^\circ$$ angle of elevation and one right angle.

A pole on a horizontal plane showing the angle of elevation and one right angle

To calculate the length of $$PL$$ we notice that $$\tan \angle PKL = \frac{PL}{KL}$$, i.e. $$PL = KL \tan \angle PKL = (15 \tan 28^\circ) \text{ m} \approx 8.0 \text{ m}$$.

To calculate the length of $$KM$$ we can use the cosine rule on triangle $$KLM$$.

$$\begin{eqnarray} (KM)^2 &=& (LM)^2 + (KL)^2 – 2(LM)(KL)\cos \angle KLM \\ &=& 19^2 + 15^2 – 2(19)(15)\cos 115^\circ \\ KM &\approx& 29 \text{ m}. \end{eqnarray}$$

And finally, the angle of elevation of $$P$$ from $$M$$ can be calculated by noticing that $$\tan \angle PML = \frac{PL}{LM} = \frac{8}{19}$$. So, $$\angle PML = \tan^{-1} \frac{8}{19} \approx 23^\circ$$.

# CSEC Math Jan 2012 P2 Q9

CSEC Math Jan 2012 Paper 2 Question 9

Updated on April 19th, 2012: Added some video explanations

In part (a),

1. Making $$x$$ the subject of the formula, we get
$$\begin{eqnarray} y &=& \frac{2x + 3}{x – 4} \\ y(x – 4) &=& 2x + 3 \\ xy – 4y &=& 2x + 3 \\ xy – 2x &=& 4y + 3 \\ x(y – 2) &=& 4y + 3 \\ x &=& \frac{4y + 3}{y – 2} & \text{, } y \neq 2. \end{eqnarray}$$
2. Hence, the inverse of $$y = f(x) = \frac{2x + 3}{x – 4}$$, where $$x \neq 4$$ is
$$f^{-1}(y) = \frac{4y + 3}{y – 2} \text{, where } y \neq 2.$$
3. $$f(x) = 0$$ whenever $$2x + 3 = 0$$, i.e. $$x = -\frac{3}{2}$$.

For part (b),

1. The inequality associated with the line $$x = 6$$ is $$x >= 6$$. And the inequality associated with the line $$x + y = 40$$ is $$x + y <= 40$$.

We are told that the function $$p = 4x + 3y$$ satisfies the solution set represented by the closed triangular region.

1. Now, the maximum and minimum values lie on the corners of the triangular region. Hence, the three corners at $$(6, 2)$$, $$(6, 34)$$ and $$(30, 10)$$ are the three candidates for which $$p$$ will have a maximum or a minimum.
2. The point $$(30, 10)$$ makes $$p$$ a maximum since it has the largest value of all the corner points.

# CSEC Math Jan 2012 P2 Q8

CSEC Math Jan 2012 Paper 2 Question 8

Updated on April 18th, 2012: Added a video explanation

Sarah is making a pattern of squares using straws. She uses four straws for the sides and two longer straws for the diagonals. The first three figures in her sequence of shapes are shown in the image above.

For part (a), the fourth figure in the pattern is shown in the image below.

Figure 4

In part (b), the completed table for Figure 4 and Figure 10 is shown below.

Table for Figure 4 and Figure 10

Using some trial and error in part (c) we get that Figure 21 uses 106 straws since $$21(6) – 20 = 106$$.

Finally, in part (d) we obtain an expression in $$n$$ for the total number of staws used in the $$n^{\text{th}}$$ figure. Following the pattern, we calculate that Figure $$n$$ uses $$n(6) – (n-1)$$ straws; i.e. $$6n – n + 1 = 5n + 1$$ straws.

# CSEC Math Jan 2012 P2 Q7

CSEC Math Jan 2012 Paper 2 Question 7

Updated on April 17th, 2012: Added a video explanation

In part (a) we use the histogram to complete the frequency table as shown below.

A completed frequency table for the data in the sample

For part (b),

1. The modal class interval is the interval with the highest frequency. So, in this case, the modal class interval is $$11 – 20$$ since 25 seedlings (the most in any interval) had a height in that interval.
2. The number of seedlings in the sample is given by summing the frequencies. Hence, there are $$18 + 25 + 23 + 20 + 14 = 100$$ seedlings.
3. To answer this part of the question we make the assumption that the average height of the seedlings in any of the intervals is the mid-point height. Hence, the mean height of the seedlings is

$$[18(5.5) + 25(15.5) + 23(25.5) + 20(35.5) + 14(45.5)] / 100 = 24.2 \text{ cm.}$$

4. There are $$20 + 14 = 34$$ seedlings with a height that is greater than 30 cm. Since there are 100 seedlings in all, we get a probability of 0.34 that a seedling chosen at random has a height that is greater than 30 cm.