Final Exam Model Paper 3

A × B is the set of all ordered pairs whose first element is from A and second element from B: A × B = {(x, y):x ∈ A, y ∈ B}.

R₁ is a function: each of 1, 2 and 3 has exactly one image (1 and 2 may share the image a).
R₂ is not: 1 has two images, a and b, and 3 has none.

(−1, 3] = {x: − 1 < x ≤ 3, x ∈ R}: an open dot at −1, a filled dot at 3, and the line between them.

A diagonal matrix: it is square, and every element off the main diagonal is 0.

No. A scalar matrix is a diagonal matrix whose diagonal elements are all equal, but here they are 2, 5 and −1.

$$2A + I = \begin{pmatrix} 4 & 0 & 0 \\ 0 & 10 & 0 \\ 0 & 0 & -2 \end{pmatrix} + \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} 5 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & -1 \end{pmatrix}$$

The roots are where the graph cuts the X-axis: $x = 2 + \sqrt{3}$ and $x = 2 - \sqrt{3}$.

$$\begin{aligned} (2 + \sqrt{3}) + (2 - \sqrt{3}) &= 4 \\ (2 + \sqrt{3})(2 - \sqrt{3}) &= 2^2 - (\sqrt{3})^2 = 4 - 3 = 1 \end{aligned}$$
Sum 4, product 1: both rational, though each root is irrational. (So $2 - \sqrt{3} = \frac{1}{2 + \sqrt{3}}$: the two roots are reciprocals.)

$\cos(-60^\circ) = \cos 60^\circ = \frac{1}{2}$

$\sin 330^\circ = \sin(360^\circ - 30^\circ) = -\sin 30^\circ = -\frac{1}{2}$, a negative number. Between 90∘° and 180∘° (the second quadrant) every sine is positive. So there is no such angle. (The angles with sine $-\frac{1}{2}$ up to 360∘° are 210∘° and 330∘°.)

$\sin^2\theta = 1 - \cos^2\theta$

$\sec\theta = \frac{1}{\cos\theta}$

$$\begin{aligned} \text{LHS} &= (\sin^2\theta + 2\sin\theta\cos\theta + \cos^2\theta) + (\sin^2\theta - 2\sin\theta\cos\theta + \cos^2\theta) \\ &= 2\sin^2\theta + 2\cos^2\theta \\ &= 2(\sin^2\theta + \cos^2\theta) = 2 \times 1 = 2 = \text{RHS} \end{aligned}$$
Proved.

The kite's height is $\sqrt{50^2 - 40^2} = \sqrt{2500 - 1600} = \sqrt{900} = 30$ m. So $\sin\theta = \frac{30}{50} = \frac{3}{5}$, and
$$\tan\theta = \frac{30}{40} = \frac{3}{4}, \qquad \operatorname{cosec}\theta = \frac{50}{30} = \frac{5}{3}.$$

$$\frac{\frac{3}{5} - \frac{4}{5}}{\frac{3}{5} + \frac{4}{5}} = \frac{-\frac{1}{5}}{\frac{7}{5}} = -\frac{1}{7}$$

$m = -\frac{a}{b}$ (the coefficient of x over the coefficient of y, with the sign changed).

$$\begin{aligned} 3x - 4y &= -20 \\ \frac{3x}{-20} - \frac{4y}{-20} &= 1 \\ \frac{x}{-\frac{20}{3}} + \frac{y}{5} &= 1 \end{aligned}$$
x-intercept $-\frac{20}{3}$, y-intercept 5. Check: y = 0 gives 3x = −20; x = 0 gives −4y = −20, y = 5 ✓.

Let P(x, y).
$$\begin{aligned} (x - 1)^2 + y^2 + (x + 1)^2 + y^2 &= 20 \\ x^2 - 2x + 1 + x^2 + 2x + 1 + 2y^2 &= 20 \\ 2x^2 + 2y^2 &= 18 \\ x^2 + y^2 &= 9 \end{aligned}$$
The locus is a circle with center O and radius 3.

Externally, m₂ is taken as −m₂:
$$\begin{aligned} x &= \frac{2 \times 4 - 1 \times 1}{2 - 1} = 7 \\ y &= \frac{2 \times 7 - 1 \times (-2)}{2 - 1} = 16 \end{aligned}$$
The point is (7, 16), beyond D. Check: it is twice as far from C as from D ($6\sqrt{10}$ from C, $3\sqrt{10}$ from D).

P(x, y) → P′(2a − x, 2b − y)

With (a, b) = (1, 2), (x, y) → (2 − x, 4 − y): A′(0, 0), B′(−2, 1), C′(−1, −2).

The center (1, 2) is the midpoint of each vertex and its image.

If two vectors are the two sides of a triangle taken in order, their sum is the third side taken the opposite way: in △ABC, AB + BC = AC.

$$\overrightarrow{AC} = \begin{pmatrix} 3 + 5 \\ 4 - 2 \end{pmatrix} = \begin{pmatrix} 8 \\ 2 \end{pmatrix}, \qquad |\overrightarrow{AC}| = \sqrt{64 + 4} = \sqrt{68} = 2\sqrt{17} \approx 8.25$$

No. He walked $|\overrightarrow{AB}| + |\overrightarrow{BC}| = 5 + \sqrt{29} \approx 10.39$ units, but |AC| is the straight distance from A to C, about 8.25. Vectors add head to tail, not by adding their lengths; the two sides of a triangle together are longer than the third.

The lowest value, the lower quartile Q₁, the median, the upper quartile Q₃ and the highest value.

Q₁ = 35, Q₃ = 60: $\text{Q.D.} = \frac{60 - 35}{2} = 12.5$ marks.

Yes. The middle half lies inside the box. Section B's box is 56 − 42 = 14 marks wide; Section A's is 60 − 35 = 25 marks wide. So Section B's middle half is much more consistent (its Q.D. is 7 against A's 12.5).

The sum of an infinite series is the limit of the sequence of its partial sums (the sums of its first 1, 2, 3, … terms), when they head for a number.

No. Every partial sum, 0.3, 0.33, 0.333, …, is less than $\frac{1}{3}$, but 0.333… with 3s going on for ever is the sum of the whole infinite series, which means the limit of the partial sums, and that limit is exactly $\frac{1}{3}$.

S₁ = 0.3, S₂ = 0.33, S₃ = 0.333, S₄ = 0.3333. They get closer and closer to $0.333\ldots = \frac{1}{3}$: the sum is $\frac{1}{3}$.

No. Its partial sums 1, 3, 6, 10, 15, … grow without bound, so they don't head for any number.

The intercepts are −2 and 4, so in double-intercept form
$$\frac{x}{-2} + \frac{y}{4} = 1 \Rightarrow -2x + y = 4 \Rightarrow y = 2x + 4.$$
For P(k, 2k + 4): 2 × k + 4 = 2k + 4 ✓, so P is on the road whatever k is.

$\frac{2k + 4}{k} = 2 + \frac{4}{k}$. As k → ∞, $\frac{4}{k} \to 0$, so the slope heads for 2, the road's own slope: seen from far along the road, OP runs nearly along it. (For k = 10, 100, 1000: 2.4, 2.04, 2.004.)

P(−1, 2 × (−1) + 4) = P(−1, 2), and $OP = \sqrt{(-1)^2 + 2^2} = \sqrt{5} \approx 2.24$ km.

Put (0, 0) in 2x − y + 4: 4 > 0. So the origin's side is 2x − y + 4 > 0. For F: 2 − 3 + 4 = 3 > 0, so the farmhouse is on the same side as the origin.

$$f = \{(1, 2), (2, 5), (3, 10), (4, 17), (5, 26), (6, 37), (7, 50)\}, \qquad \text{range} = \{2, 5, 10, 17, 26, 37, 50\}.$$

Yes, on {1, …, 7}: the squares of different positive numbers are different, so no two patterns use the same number of tiles. With −1 added, no: f(−1) = 1 + 1 = 2 = f(1), two elements with the same image.

In order: 2, 5, 10, 17, 26, 37, 50 (N = 7).
$Q_1 = \left(\frac{7 + 1}{4}\right)^{\text{th}} = 2^{\text{nd}}$ item = 5; $Q_3 = \left(\frac{3(7 + 1)}{4}\right)^{\text{th}} = 6^{\text{th}}$ item = 37.
$$\text{Q.D.} = \frac{37 - 5}{2} = 16, \qquad \text{Coefficient of Q.D.} = \frac{37 - 5}{37 + 5} = \frac{32}{42} = \frac{16}{21} \approx 0.76$$

$$120^\circ = \left(120 \times \frac{10}{9}\right)^{\mathrm{g}} = \left(\frac{400}{3}\right)^{\mathrm{g}} = 133\tfrac{1}{3}^{\mathrm{g}}, \qquad 120^\circ = \left(120 \times \frac{\pi}{180}\right)^{\mathrm{c}} = \left(\frac{2\pi}{3}\right)^{\mathrm{c}}$$

$$l = r\theta = 42 \times \frac{2}{3} \times \frac{22}{7} = 88 \text{ cm}$$
In an hour it turns 360∘°, $2\pi$ radians: $l = 42 \times 2 \times \frac{22}{7} = 264$ cm (three times 88).

After turning 120∘° from the upright position, the tip's height above the center is $42\cos 120^\circ = 42 \times \left(-\frac{1}{2}\right) = -21$ cm, since $\cos 120^\circ = -\cos 60^\circ$. Yes, she is right: the tip is 21 cm below the center (at the 4 o'clock mark).

$\frac{l}{\theta} = \frac{r\theta}{\theta} = r = 42$ for every angle other than 0, so as θ → 0 it heads for 42 cm, the length of the hand, even though at θ = 0 itself the ratio would be $\frac{0}{0}$.

$\overrightarrow{AA'} = \begin{pmatrix} 3 \\ -3 \end{pmatrix}$ and $\overrightarrow{BB'} = \begin{pmatrix} -2 \\ 2 \end{pmatrix} = -\frac{2}{3}\overrightarrow{AA'}$. One is a number times the other, so yes, they are parallel (unlike vectors, pointing opposite ways).

$|\overrightarrow{AA'}| = \sqrt{9 + 9} = 3\sqrt{2}$, $|\overrightarrow{BB'}| = \sqrt{4 + 4} = 2\sqrt{2}$, and $\overrightarrow{CC'} = \begin{pmatrix} -2 \\ 2 \end{pmatrix}$, so $|\overrightarrow{CC'}| = 2\sqrt{2}$.

From A(1, 4) → A′(4, 1) the coordinates have swapped: try (x, y) → (y, x).
Check: B(3, 1) → (1, 3) = B′ ✓, C(5, 3) → (3, 5) = C′ ✓.
So it is the reflection in the line y = x. This fits part a: every point moves at right angles to the mirror line, along $\begin{pmatrix} 1 \\ -1 \end{pmatrix}$ or its opposite, which is why AA′ and BB′ are parallel.