Final Exam Model Paper 2

A relation R in a set A is reflexive if every element of A is related to itself: (x, x) ∈ R for every x ∈ A.

Reflexive: (1, 1), (2, 2) and (3, 3) are all in R, so it is reflexive ✓.

Symmetric: (1, 2) ∈ R but (2, 1) ∉ R (2 is not less than or equal to 1), so it is not symmetric.

She is only half right: R is reflexive but not symmetric.

Swap each pair: $R^{-1} = \{(1, 1), (2, 1), (3, 1), (2, 2), (3, 2), (3, 3)\}$. It is the relation “is greater than or equal to”.

Yes. It is square, and the elements on either side of the main diagonal are equal: $m_{12} = m_{21} = 4$. So M is its own mirror image in the main diagonal, whatever a is.

Equal matrices have equal corresponding elements, so every element of N − M is 0: N − M is the zero (null) matrix $\begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}$.

Corresponding elements: a + 1 = 5, so a = 4; b − 2 = 4, so b = 6.

$\sqrt{5} + \sqrt{3}$

$$\begin{aligned} \frac{4}{\sqrt{5} - \sqrt{3}} &= \frac{4(\sqrt{5} + \sqrt{3})}{(\sqrt{5} - \sqrt{3})(\sqrt{5} + \sqrt{3})} \\ &= \frac{4(\sqrt{5} + \sqrt{3})}{5 - 3} \\ &= 2\sqrt{5} + 2\sqrt{3} \end{aligned}$$
With $\sqrt{5} \approx 2.236$ and $\sqrt{3} \approx 1.732$: about 4.47 + 3.46 = 7.94 m.

$$\tan 135^\circ = \tan(180^\circ - 45^\circ) = -\tan 45^\circ = -1$$

tan (−θ) = − tan θ, so $\tan(-45^\circ) = -\tan 45^\circ = -1 = \tan 135^\circ$. The angle is −45∘°.

$\cot\theta = \frac{\cos\theta}{\sin\theta}$

$\operatorname{cosec}^2\theta - \cot^2\theta = 1$

$$\begin{aligned} \text{LHS} &= \frac{\sec^2\theta}{\operatorname{cosec}^2\theta} = \frac{\frac{1}{\cos^2\theta}}{\frac{1}{\sin^2\theta}} \\ &= \frac{\sin^2\theta}{\cos^2\theta} = \tan^2\theta = \text{RHS} \end{aligned}$$
Proved.

In a right-angled triangle with perpendicular 7 and base 24, the hypotenuse is $\sqrt{7^2 + 24^2} = \sqrt{49 + 576} = \sqrt{625} = 25$. So
$$\sin\theta = \frac{7}{25}, \qquad \cos\theta = \frac{24}{25}.$$

$$25\left(\frac{7}{25} + \frac{24}{25}\right) = 7 + 24 = 31$$

$\frac{x}{a} + \frac{y}{b} = 1$, where a is the x-intercept and b the y-intercept.

The intercepts are equal, b = a: $\frac{x}{a} + \frac{y}{a} = 1$, that is, x + y = a. It passes through (3, 4): 3 + 4 = a, so a = 7.
The canal is x + y = 7 (x + y − 7 = 0).

Let P(x, y). Its distance from the X-axis is |y|, so PA = |y|:
$$\begin{aligned} \sqrt{x^2 + (y - 3)^2} &= |y| \\ x^2 + y^2 - 6y + 9 &= y^2 \\ x^2 - 6y + 9 &= 0 \end{aligned}$$

Let it divide CD in the ratio m₁:m₂. On the X-axis y = 0:
$$\frac{m_1 \times 6 + m_2 \times (-3)}{m_1 + m_2} = 0 \Rightarrow 6m_1 = 3m_2 \Rightarrow m_1 : m_2 = 1 : 2.$$
Then $x = \frac{1 \times 5 + 2 \times 2}{1 + 2} = \frac{9}{3} = 3$. The ratio is 1 : 2, at the point (3, 0).

P(x, y) → P′(x, 2b − y)

Here b = 2, so (x, y) → (x, 4 − y): A′(1, 5), B′(4, 4), C′(2, 3).

Each vertex and its image are the same distance from y = 2, on opposite sides.

It goes from (−1, −2) to (3, 1): $\vec{m} = \begin{pmatrix} 3 - (-1) \\ 1 - (-2) \end{pmatrix} = \begin{pmatrix} 4 \\ 3 \end{pmatrix}$.

$|\vec{m}| = \sqrt{4^2 + 3^2} = 5$, so $\hat{m} = \frac{1}{5}\begin{pmatrix} 4 \\ 3 \end{pmatrix} = \begin{pmatrix} \frac{4}{5} \\ \frac{3}{5} \end{pmatrix}$.

No. $-\vec{m} = \begin{pmatrix} -4 \\ -3 \end{pmatrix}$ has the same magnitude, 5, but points the opposite way: it is the negative vector of m, from (3, 1) back to (−1, −2).

$Q_1 = \left(\frac{N + 1}{4}\right)^{\text{th}}$ item, where $N = \Sigma f$.

$\frac{30 + 1}{4} = 7.75$. The c.f. just greater than 7.75 is 11, for x = 1: Q₁ = 1 child.

No. Q₃ is the $\frac{3(30 + 1)}{4} = 23.25$th item; the c.f. just greater than that is 24, for x = 2, so Q₃ = 2. Quartiles divide the households into four equal groups by count, so they depend on the frequencies, not on where the values sit between the smallest and the largest.

x takes values closer and closer to 3 from the right, greater than 3 (3.1, 3.01, 3.001, …), never equal to 3.

No. They agree for every x except 3, but $f(3) = \frac{0}{0}$ has no value while g(3) = 6. A function is fixed by its domain as well as its rule, and 3 is not in the domain of f: its graph is the line y = x + 3 with a hole at (3, 6).

From both sides f(x) heads for 6: $\lim_{x \to 3} f(x) = 6$.

Putting x = 3 gives $\frac{9 - 9}{3 - 3} = \frac{0}{0}$, an indeterminate form, so f(3) has no value. The limit asks what f(x) heads for near 3, which is why it is found from values on both sides.

By the section formula with m₁ = k, m₂ = 1:
$$x = \frac{k \times 4 + 1 \times (-2)}{k + 1} = \frac{4k - 2}{k + 1}, \qquad y = \frac{k \times 0 + 1 \times 6}{k + 1} = \frac{6}{k + 1}.$$

$\frac{4k - 2}{k + 1} = \frac{4 - \frac{2}{k}}{1 + \frac{1}{k}} \to \frac{4}{1} = 4$ and $\frac{6}{k + 1} \to 0$. So P heads for (4, 0), the point B: the larger the first part of the ratio, the closer P is to B. (For k = 100, P is about (3.94, 0.06).)

In double-intercept form: the line through B(4, 0) with slope $\frac{0 - 6}{4 - (-2)} = -1$ meets the Y-axis at y = 4, so $\frac{x}{4} + \frac{y}{4} = 1$, that is, x + y = 4. Check with A: −2 + 6 = 4 ✓.

Multiply top and bottom by $\sqrt{2} - 1$, since $(\sqrt{2} + 1)(\sqrt{2} - 1) = 2 - 1 = 1$:
$$\begin{aligned} x &= \frac{4\sqrt{2} - 2}{\sqrt{2} + 1} = (4\sqrt{2} - 2)(\sqrt{2} - 1) = 8 - 4\sqrt{2} - 2\sqrt{2} + 2 = 10 - 6\sqrt{2} \\ y &= \frac{6}{\sqrt{2} + 1} = 6(\sqrt{2} - 1) = 6\sqrt{2} - 6 \end{aligned}$$
Then $x + y = 10 - 6\sqrt{2} + 6\sqrt{2} - 6 = 4$ ✓, so P lies on AB. (P is about (1.51, 2.49).)

$$f = \{(-2, 4), (-1, 1), (0, 0), (1, 1), (2, 4)\}, \qquad \text{range} = \{0, 1, 4\}.$$

No, it is into. Its range {0, 1, 4} is not the whole co-domain B: 2 and 3 have no pre-image (no whole number in A squares to 2 or 3).

$\overline{X} = \frac{4 + 1 + 0 + 1 + 4}{5} = \frac{10}{5} = 2$.

$$\sigma = \sqrt{\frac{\Sigma d^2}{N}} = \sqrt{\frac{14}{5}} = \sqrt{2.8} \approx 1.67$$

$$150^{\mathrm{g}} = \left(150 \times \frac{9}{10}\right)^\circ = 135^\circ, \qquad 150^{\mathrm{g}} = \left(150 \times \frac{\pi}{200}\right)^{\mathrm{c}} = \left(\frac{3\pi}{4}\right)^{\mathrm{c}}$$

$$l = r\theta = 21 \times \frac{3}{4} \times \frac{22}{7} = 49.5 \text{ cm}$$
For 66 cm: $\theta = \frac{l}{r} = \frac{66}{21} = \frac{22}{7}$ radians, which is π radians, 180∘°: half a turn.

No. $150^{\mathrm{g}} = 135^\circ$, so $\sin 150^{\mathrm{g}} = \sin 135^\circ = \sin(180^\circ - 45^\circ) = \frac{1}{\sqrt{2}} \approx 0.71$, but $\sin 150^\circ = \sin(180^\circ - 30^\circ) = \frac{1}{2}$. They are different angles, measured in different units.

No. The distance is 49.5t cm, which grows past any number we name as t grows without bound: it tends to infinity, and infinity is not a number, so there is no limit.

$\overrightarrow{AB} = \begin{pmatrix} 2 \\ 0 \end{pmatrix}$ and $\overrightarrow{A'B'} = \begin{pmatrix} 0 \\ 2 \end{pmatrix}$: both have magnitude 2, but one points right and the other up. He is right: the vectors are not equal, though their magnitudes are.

$$\overrightarrow{BC} = \begin{pmatrix} 3 - 3 \\ 5 - 2 \end{pmatrix} = \begin{pmatrix} 0 \\ 3 \end{pmatrix}, \qquad \overrightarrow{B'C'} = \begin{pmatrix} -3 - 0 \\ 3 - 3 \end{pmatrix} = \begin{pmatrix} -3 \\ 0 \end{pmatrix}$$
|BC| = 3 and |B′C′| = 3.

Not about the origin: +90∘° about O would send A(1, 2) to (−2, 1), not (0, 1).

The center is as far from each point as from its image, so it lies on the locus of points equidistant from A and A', and from B and B'. For P(x, y):
$$\begin{aligned} PA = PA' &: (x - 1)^2 + (y - 2)^2 = x^2 + (y - 1)^2 \Rightarrow x + y = 2 \\ PB = PB' &: (x - 3)^2 + (y - 2)^2 = x^2 + (y - 3)^2 \Rightarrow 3x - y = 2 \end{aligned}$$
Adding: 4x = 4, x = 1, y = 1. The center is (1, 1).

The angle: from the center, A is 1 unit up, (0, 1), and A' is 1 unit left, (−1, 0): a quarter turn anticlockwise. So the rotation is +90∘° (anticlockwise) about (1, 1). Check with the rule (x, y) → (−(y − 1) + 1, (x − 1) + 1) = (2 − y, x): C(3, 5) → (−3, 3) = C′ ✓.