1.1 Consider the arithmetic sequence: 10; 7; 4; 1; ... 1.1.1 To find the general term $T_n$ of the sequence, use the formula for an arithmetic sequence: $$T_n = a + (n-1)d$$ where $a$ is the first term and $d$ is the common difference. Calculate $d$: $$d = 7 - 10 = -3$$ So, $$T_n = 10 + (n-1)(-3) = 10 - 3(n-1) = 10 - 3n + 3 = 13 - 3n$$ 1.1.2 To find which term equals $-113$, set $T_n = -113$: $$13 - 3n = -113$$ Subtract 13 from both sides: $$-3n = -113 - 13 = -126$$ Divide both sides by $-3$: $$n = \frac{\cancel{-126}}{\cancel{-3}} = 42$$ So, the 42nd term is $-113$. 1.2 Consider the sequence: 1; 6; 15; 28; ... 1.2.1 This is a quadratic sequence. To find the formula for the $n$th term, find the first and second differences: First differences: $6-1=5$, $15-6=9$, $28-15=13$ Second differences: $9-5=4$, $13-9=4$ Since second differences are constant and equal to 4, the sequence is quadratic with general term: $$T_n = an^2 + bn + c$$ The second difference equals $2a = 4$, so $a=2$. Use the first three terms to find $b$ and $c$: For $n=1$: $T_1 = 2(1)^2 + b(1) + c = 2 + b + c = 1$ For $n=2$: $T_2 = 2(2)^2 + 2b + c = 8 + 2b + c = 6$ For $n=3$: $T_3 = 2(3)^2 + 3b + c = 18 + 3b + c = 15$ From $n=1$: $$2 + b + c = 1 \Rightarrow b + c = -1$$ From $n=2$: $$8 + 2b + c = 6 \Rightarrow 2b + c = -2$$ Subtract the first from the second: $$(2b + c) - (b + c) = -2 - (-1) \Rightarrow b = -1$$ Then from $b + c = -1$: $$-1 + c = -1 \Rightarrow c = 0$$ So the formula is: $$T_n = 2n^2 - n$$ 1.2.2 To find the 100th term: $$T_{100} = 2(100)^2 - 100 = 2(10000) - 100 = 20000 - 100 = 19900$$ 1.3 Given quadratic sequence: 1; 5; $x$; 19; ... Use the equation: $$-4 \times 5 = 19 - x$$ Calculate left side: $$-20 = 19 - x$$ Add $x$ to both sides and add 20 to both sides: $$x = 19 + 20 = 39$$ So, $x = 39$. 2.1.1 Graph of $g(x) = -2x + 5$ is a straight line with slope $-2$ and y-intercept $5$. 2.1.2 Graph of $h(x) = 3^x$ is an exponential curve with base 3. 2.2 Given $k(x) = -2x^2 + 2$: 2.2.1 The graph is a downward opening parabola with vertex at $(0, 2)$. 2.2.2 The graph is increasing where the derivative $k'(x) > 0$. Calculate derivative: $$k'(x) = -4x$$ Set $k'(x) > 0$: $$-4x > 0 \Rightarrow x < 0$$ So, $k$ is increasing for $x < 0$. 2.2.3 The range is all $y$ values less than or equal to the vertex's y-coordinate: $$\text{Range} = (-\infty, 2]$$ 2.3 Given $f(x) = \frac{3}{x-5} - 4$: 2.3.1 Vertical asymptote where denominator is zero: $$x - 5 = 0 \Rightarrow x = 5$$ Horizontal asymptote is the limit as $x \to \pm \infty$: $$y = -4$$ 2.3.2 Domain is all real numbers except $x=5$: $$\text{Domain} = (-\infty, 5) \cup (5, \infty)$$ 2.4 Point $P(2,5)$: 2.4.1 Reflection along $y = x$ swaps coordinates: $$P' = (5, 2)$$ 2.4.2 Shift 2 units right adds 2 to x-coordinate: $$P' = (2 + 2, 5) = (4, 5)$$ 3.1 Given $5 \cos A = 3$, $A \in (90^\circ, 360^\circ)$: $$\cos A = \frac{3}{5}$$ Since $A$ is in the third or fourth quadrant, cosine is positive only in the fourth quadrant, so $A$ is in the fourth quadrant. Calculate $\sin A$ using Pythagorean identity: $$\sin^2 A = 1 - \cos^2 A = 1 - \left(\frac{3}{5}\right)^2 = 1 - \frac{9}{25} = \frac{16}{25}$$ Since $A$ is in the fourth quadrant, $\sin A < 0$: $$\sin A = -\frac{4}{5}$$ Calculate $\tan A$: $$\tan A = \frac{\sin A}{\cos A} = \frac{-\frac{4}{5}}{\frac{3}{5}} = -\frac{4}{3}$$ 3.2 Given $\sin 31^\circ = m$, find $\cos 149^\circ$: Note $149^\circ = 180^\circ - 31^\circ$, so $$\cos 149^\circ = -\cos 31^\circ$$ Using identity $\sin^2 \theta + \cos^2 \theta = 1$: $$\cos 31^\circ = \sqrt{1 - \sin^2 31^\circ} = \sqrt{1 - m^2}$$ Therefore, $$\cos 149^\circ = -\sqrt{1 - m^2}$$ 3.3 Simplify: $$\frac{\tan(-A) \cos(90^\circ - A)}{\sin(360^\circ - A)}$$ Use identities: $$\tan(-A) = -\tan A$$ $$\cos(90^\circ - A) = \sin A$$ $$\sin(360^\circ - A) = -\sin A$$ Substitute: $$\frac{-\tan A \sin A}{-\sin A} = \frac{-\tan A \sin A}{-\sin A} = \tan A$$ 3.4 Prove: $$\frac{\tan x \sin x \cos x}{\cos^2(180^\circ - x)} = \tan^2 x$$ Note: $$\cos(180^\circ - x) = -\cos x$$ So, $$\cos^2(180^\circ - x) = \cos^2 x$$ Rewrite left side: $$\frac{\tan x \sin x \cos x}{\cos^2 x} = \tan x \sin x \frac{\cos x}{\cos^2 x} = \tan x \sin x \frac{1}{\cos x}$$ Since $\tan x = \frac{\sin x}{\cos x}$: $$= \frac{\sin x}{\cos x} \sin x \frac{1}{\cos x} = \frac{\sin^2 x}{\cos^2 x} = \tan^2 x$$ 3.5 Solve general solution of: $$1 - 2 \cos x = 0$$ Rearranged: $$2 \cos x = 1 \Rightarrow \cos x = \frac{1}{2}$$ General solution for $\cos x = \frac{1}{2}$ is: $$x = 2k\pi \pm \frac{\pi}{3}, \quad k \in \mathbb{Z}$$ 4.1 Given quadrilateral with $AB=20$, $BC=12$, $CD=7$, $\angle B = 110^\circ$, $\angle CDA = 71^\circ$, find length $AC$. Use Law of Cosines in $\triangle ABC$: $$AC^2 = AB^2 + BC^2 - 2 \times AB \times BC \times \cos B$$ Calculate: $$AC^2 = 20^2 + 12^2 - 2 \times 20 \times 12 \times \cos 110^\circ = 400 + 144 - 480 \times \cos 110^\circ$$ Calculate $\cos 110^\circ \approx -0.3420$: $$AC^2 = 544 - 480 \times (-0.3420) = 544 + 164.16 = 708.16$$ So, $$AC = \sqrt{708.16} \approx 26.6$$ 4.2 Find $\angle DAC$ in $\triangle ADC$. Use Law of Cosines: $$AC^2 = AD^2 + DC^2 - 2 \times AD \times DC \times \cos DAC$$ We know $AC \approx 26.6$, $DC=7$, but $AD$ is unknown. We need $AD$ or another approach. Since $\angle CDA = 71^\circ$, use Law of Cosines in $\triangle ADC$ to find $AD$ first. Use Law of Cosines: $$AD^2 = AC^2 + DC^2 - 2 \times AC \times DC \times \cos 71^\circ$$ Calculate $\cos 71^\circ \approx 0.3256$: $$AD^2 = 26.6^2 + 7^2 - 2 \times 26.6 \times 7 \times 0.3256 = 707.56 + 49 - 121.2 = 635.36$$ So, $$AD = \sqrt{635.36} \approx 25.2$$ Now use Law of Cosines to find $\angle DAC$: $$AC^2 = AD^2 + DC^2 - 2 \times AD \times DC \times \cos DAC$$ Rearranged: $$\cos DAC = \frac{AD^2 + DC^2 - AC^2}{2 \times AD \times DC} = \frac{25.2^2 + 7^2 - 26.6^2}{2 \times 25.2 \times 7}$$ Calculate numerator: $$25.2^2 = 635.04, \quad 7^2 = 49, \quad 26.6^2 = 707.56$$ $$635.04 + 49 - 707.56 = -23.52$$ Calculate denominator: $$2 \times 25.2 \times 7 = 352.8$$ So, $$\cos DAC = \frac{-23.52}{352.8} \approx -0.0667$$ Therefore, $$\angle DAC = \cos^{-1}(-0.0667) \approx 94.0^\circ$$ 4.3 Find area of $\triangle ABC$ using formula: $$\text{Area} = \frac{1}{2} AB \times BC \times \sin B$$ Calculate $\sin 110^\circ \approx 0.9397$: $$\text{Area} = \frac{1}{2} \times 20 \times 12 \times 0.9397 = 10 \times 12 \times 0.9397 = 112.76$$ Final answers: - $T_n = 13 - 3n$ - Term equal to $-113$ is $n=42$ - $T_n = 2n^2 - n$ - $T_{100} = 19900$ - $x = 39$ - $g(x) = -2x + 5$ - $h(x) = 3^x$ - $k(x) = -2x^2 + 2$, increasing for $x<0$, range $(-\infty, 2]$ - $f(x) = \frac{3}{x-5} - 4$, vertical asymptote $x=5$, horizontal asymptote $y=-4$, domain all real except $x=5$ - Reflected $P = (5, 2)$, shifted $P = (4, 5)$ - $\tan A = -\frac{4}{3}$ - $\cos 149^\circ = -\sqrt{1 - m^2}$ - Simplified expression equals $\tan A$ - Identity proven - General solution $x = 2k\pi \pm \frac{\pi}{3}$ - $AC \approx 26.6$ - $\angle DAC \approx 94.0^\circ$ - Area $\triangle ABC \approx 112.76$