Final Exam Model Paper 1

A function from A to B is a relation in which every element of A is related to exactly one element of B.

f(−2) = 5, f(−1) = 2, f(1) = 2, f(2) = 5, so
$$f = \{(-2, 5), (-1, 2), (1, 2), (2, 5)\}, \quad f^{-1} = \{(5, -2), (2, -1), (2, 1), (5, 2)\}.$$
She is not right. In $f^{-1}$ the element 2 is related to both −1 and 1 (and 5 to both −2 and 2), so it does not give one image to each element: $f^{-1}$ is a relation but not a function. Run backwards, the machine can't tell which number went in.

$2x - 3 > 1 \Rightarrow 2x > 4 \Rightarrow x > 2$: an open dot at 2 (2 itself is not included) and the line going right.

$a_{21}$ is in the 2nd row and 1st column: $a_{21} = 15$. The first stall sold 15 cups of tea on Sunday.

No. A symmetric matrix has $a_{ij} = a_{ji}$, but here $a_{12} = 8$ and $a_{21} = 15$.

$$A + B = \begin{pmatrix} 12 + 9 & 8 + 11 \\ 15 + 14 & 10 + 6 \end{pmatrix} = \begin{pmatrix} 21 & 19 \\ 29 & 16 \end{pmatrix}$$
It shows the two stalls' sales together: 21 teas and 19 coffees on Saturday, 29 teas and 16 coffees on Sunday.

The highest power of x is 3, so the degree is 3 (a cubic polynomial).

Divide by x − 3, so use 3 at the left; the coefficients are 1, −2, −5, 6.

Quotient $x^2 + x - 2$, remainder 0. Since the remainder is 0, p(3) = 0: 3 is a root of p(x) = 0. Check: 27 − 18 − 15 + 6 = 0 ✓.

$$\cos 150^\circ = \cos(180^\circ - 30^\circ) = -\cos 30^\circ = -\frac{\sqrt{3}}{2}$$

$\sin 300^\circ = \sin(360^\circ - 60^\circ) = -\sin 60^\circ = -\frac{\sqrt{3}}{2}$. Between 180∘° and 270∘° (the third quadrant) the sine is negative, and $\sin(180^\circ + 60^\circ) = -\sin 60^\circ$. So the angle is 240∘°.

$\sec^2\theta - \tan^2\theta = 1$ (that is, $1 + \tan^2\theta = \sec^2\theta$).

$\operatorname{cosec}\theta = \frac{1}{\sin\theta}$

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

The foot of the ladder is $\sqrt{17^2 - 8^2} = \sqrt{289 - 64} = \sqrt{225} = 15$ m from the wall. So
$$\cos\theta = \frac{\text{base}}{\text{hypotenuse}} = \frac{15}{17}, \qquad \cot\theta = \frac{\text{base}}{\text{perpendicular}} = \frac{15}{8}.$$

$$17 \times \frac{8}{17} + 8 \times \frac{15}{8} = 8 + 15 = 23$$

x cos α + y sin α = p, where p is the length of the perpendicular from the origin to the line and α the angle it makes with the X-axis.

Here p = 4 and $\alpha = 30^\circ$:
$$\begin{aligned} x\cos 30^\circ + y\sin 30^\circ &= 4 \\ \frac{\sqrt{3}}{2}x + \frac{1}{2}y &= 4 \\ \sqrt{3}x + y &= 8 \end{aligned}$$
The road is $\sqrt{3}x + y - 8 = 0$.

Let P(x, y) be such a point. Then PT = 5:
$$\begin{aligned} \sqrt{(x - 2)^2 + (y + 1)^2} &= 5 \\ (x - 2)^2 + (y + 1)^2 &= 25 \\ x^2 - 4x + 4 + y^2 + 2y + 1 &= 25 \\ x^2 + y^2 - 4x + 2y - 20 &= 0 \end{aligned}$$
The locus is a circle with center T and radius 5 km.

m₁ = 3, m₂ = 2, (x₁, y₁) = (−3, 4), (x₂, y₂) = (7, −1):
$$\begin{aligned} x &= \frac{3 \times 7 + 2 \times (-3)}{3 + 2} = \frac{21 - 6}{5} = 3 \\ y &= \frac{3 \times (-1) + 2 \times 4}{3 + 2} = \frac{-3 + 8}{5} = 1 \end{aligned}$$
P(3, 1). It lies between A and B, as it should.

P(x, y) → P′(−y, −x)

P(1, 1) → P′(−1, −1), Q(4, 2) → Q′(−2, −4), R(2, 4) → R′(−4, −2).

Each image point is as far from the mirror line as its object, on the other side.

$$\overrightarrow{AB} = \begin{pmatrix} 5 - 2 \\ 3 - (-1) \end{pmatrix} = \begin{pmatrix} 3 \\ 4 \end{pmatrix}$$

$|\overrightarrow{AB}| = \sqrt{3^2 + 4^2} = \sqrt{25} = 5$ units, so the drone flew 50 m.

$\overrightarrow{CD} = \begin{pmatrix} 6 \\ 8 \end{pmatrix} = 2\overrightarrow{AB}$: the same direction, but |CD| = 10 while |AB| = 5. She is not right: equal vectors need the same magnitude as well as the same direction. These are like vectors, not equal ones.

$Q_3 = \left(\frac{3(N + 1)}{4}\right)^{\text{th}}$ item, with the data in ascending order.

In order: 7, 8, 9, 10, 11, 12, 14, 15, 18, 20, 25. $\frac{3(11 + 1)}{4} = 9$, so Q₃ is the 9th item: Q₃ = 18 minutes.

Yes. The quartiles are the 3rd and 9th items in order (9 and 18 minutes); changing the largest value, the 11th, doesn't move them, so $\text{Q.D.} = \frac{18 - 9}{2} = 4.5$ minutes either way. The standard deviation uses every value's distance from the mean, so a 60 makes it much larger (from about 5.3 to about 14.2 minutes).

The left-hand limit is the number f(x) gets as close as we like to as x takes values closer and closer to a from the left, less than a ($x \to a^-$), never equal to a. It is written $\lim_{x \to a^-} f(x)$.

No. If $\frac{0}{0} = m$, then 0 = 0 × m, which is true for every number m, not just 1. So $\frac{0}{0}$ can't be given one value: it is an indeterminate form. The rule “a number divided by itself is 1” holds only for numbers other than 0.

From the left of 2 the graph is the line y = x + 1, heading for the open circle at height 3: $\lim_{x \to 2^-} f(x) = 3$.

From the right, the line y = 6 − x heads for 4: $\lim_{x \to 2^+} f(x) = 4$. Since 3 ≠ 4, the limit does not exist. (The filled dot, f(2) = 1, doesn't matter for the limit.)

(2, 5) is on the line: 5 = 2m + c, so c = 5 − 2m.
The x-intercept is where y = 0: 0 = mx + 5 − 2m, so mx = 2m − 5 and $x = \frac{2m - 5}{m}$.

$\frac{2m - 5}{m} = 2 - \frac{5}{m}$. As m → ∞, $\frac{5}{m} \to 0$, so the x-intercept heads for 2: the line becomes the vertical line x = 2. (For m = 10, 100, 1000 it is 1.5, 1.95, 1.995.)

c = 5 − 2 × 3 = −1, so y = 3x − 1, that is, 3x − y − 1 = 0.

Put (0, 0) in 3x − y − 1: 0 − 0 − 1 = −1 < 0. So the side containing the origin is 3x − y − 1 < 0, that is, y > 3x − 1.

f(1) = 2, f(2) = 7, f(3) = 12, f(4) = 17, f(5) = 22:
$$f = \{(1, 2), (2, 7), (3, 12), (4, 17), (5, 22)\}, \qquad \text{range} = \{2, 7, 12, 17, 22\}.$$

Yes, it is one-to-one: different inputs give different outputs (each output is 5 more than the one before). So every output has exactly one input, and the inverse {(2, 1), (7, 2), (12, 3), (17, 4), (22, 5)} gives each element one image: it is a function too.

$\overline{X} = \frac{2 + 7 + 12 + 17 + 22}{5} = \frac{60}{5} = 12$.

$$\begin{aligned} \text{M.D.} &= \frac{\Sigma\lvert X - \overline{X}\rvert}{N} = \frac{30}{5} = 6 \\ \text{Coefficient of M.D.} &= \frac{\text{M.D.}}{\overline{X}} = \frac{6}{12} = 0.5 \end{aligned}$$

$$30^\circ = \left(30 \times \frac{10}{9}\right)^{\mathrm{g}} = \left(\frac{100}{3}\right)^{\mathrm{g}} = 33\tfrac{1}{3}^{\mathrm{g}}, \qquad 30^\circ = \left(30 \times \frac{\pi}{180}\right)^{\mathrm{c}} = \left(\frac{\pi}{6}\right)^{\mathrm{c}}$$

$$l = r\theta = 50 \times \frac{\pi}{6} = 50 \times \frac{22}{7} \times \frac{1}{6} = \frac{550}{21} \approx 26.19 \text{ cm}$$
In a minute: $40 \times \frac{550}{21} = \frac{22000}{21} \approx 1047.6$ cm, about 10.48 m.

Only half right. The arc is rθ, so doubling θ doubles it (to about 52.38 cm). But $\sin 60^\circ = \frac{\sqrt{3}}{2} \approx 0.87$, while $2\sin 30^\circ = 2 \times \frac{1}{2} = 1$: the sine is not doubled.

$\theta = \frac{l}{r} = \frac{26.19}{r}$ radians. As r → ∞, $\frac{26.19}{r} \to 0$: the angle heads for 0. A very long pendulum barely turns to travel the same arc.

$\overrightarrow{PP'} = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$ and $\overrightarrow{QQ'} = \begin{pmatrix} 3 \\ 0 \end{pmatrix}$. They point the same way but their magnitudes are 1 and 3, so he is not right: they are like vectors, not equal ones. (In an enlargement, points farther from the center move farther.)

$$\overrightarrow{PR} = \begin{pmatrix} 4 - 2 \\ 3 - 1 \end{pmatrix} = \begin{pmatrix} 2 \\ 2 \end{pmatrix}, \qquad \overrightarrow{P'R'} = \begin{pmatrix} 7 - 3 \\ 5 - 1 \end{pmatrix} = \begin{pmatrix} 4 \\ 4 \end{pmatrix} = 2\overrightarrow{PR}$$
$|\overrightarrow{PR}| = \sqrt{4 + 4} = 2\sqrt{2}$ and $|\overrightarrow{P'R'}| = \sqrt{16 + 16} = 4\sqrt{2}$: the image side is in the same direction and twice as long.

Scale factor: P′Q′ = 7 − 3 = 4 and PQ = 4 − 2 = 2, so $k = \frac{4}{2} = 2$ (positive: the image is on the same side of the center and not turned over).

Center: with center (a, b) and k = 2, P(x, y) → P′(2(x − a) + a,  2(y − b) + b) = (2x − a,  2y − b). For P: 2 × 2 − a = 3 and 2 × 1 − b = 1, so a = 1, b = 1. The center is (1, 1).

Check: Q(4, 1) → (8 − 1, 2 − 1) = (7, 1) = Q′ ✓ and R(4, 3) → (8 − 1, 6 − 1) = (7, 5) = R′ ✓. The enlargement is E[(1, 1), 2].