\begin{align*} \text{Let price of item} & = P \\ \\ \text{Initially, } P & = 300 \\ \\ \text{After 1 month, } P & = 300 \times 0.95 \phantom{00000} [100\% - 5\% = 95\% = {95 \over 100} = 0.95] \\ P & = 300(0.95) \\ \\ \text{After 2 months, } P & = 300(0.95) \times 0.95 \\ P & = 300(0.95)^2 \\ \\ \text{After 3 months, } P & = 300(0.95)^2 \times 0.95 \\ P & = 300(0.95)^3 \\ \\ \implies \text{After } n \text{ months, } P & = 300(0.95)^n \\ \\ \text{Let } & P = 120, \\ 120 & = 300 (0.95)^n \\ {120 \over 300} & = (0.95)^n \\ 0.4 & = (0.95)^n \\ \lg 0.4 & = \lg (0.95)^n \\ \lg 0.4 & = n \lg 0.95 \phantom{000000} [\text{Power law (logarithms)}] \\ \\ n & = {\lg 0.4 \over \lg 0.95} \\ n & = 17.86 \\ \\ \text{When } & n = 17, \\ P & = 300(0.95)^{17} \approx \$ 125.44 \\ \\ \text{When } & n =18, \\ P & = 300(0.95)^{18} \approx \$ 119.16 < \$ 120 \\ \\ \therefore 18 \text{ complete months} & \text{ have elapsed before item is sold.} \end{align*}

(a)(i)

\begin{align*} a^3 - b^3 & = (a - b)(a^2 + ab + b^2) \phantom{000000} [\text{Need to memorise}] \end{align*}

(a)(ii)

\begin{align*} 347^3 - 286^3 & = (347 - 286)[(347)^2 + (347)(286) + (286)^2] \\ & = (61)(301 \phantom{.} 447) \\ \\ \text{Factor} & = 61 \text{ or } 301 \phantom{.} 447 \end{align*}

(b)

\begin{align*} {(x + n)^3 \over 2} - {(x - n)^3 \over 2} & = { (x + n)^3 - (x - n)^3 \over 2 } \\ & = { [ (x + n) - (x - n) ] [ (x + n)^2 + (x + n)(x - n) + (x - n)^2 ] \over 2 } \phantom{000000} [a = x + n, b = x - n] \\ & = { ( x + n - x + n ) [ (x + n)^2 + (x + n)(x - n) + (x - n)^2 ] \over 2 } \\ & = { 2n [(x + n)^2 + (x + n)(x - n) + (x - n)^2] \over 2 } \\ & = n [(x + n)^2 + (x + n)(x - n) + (x - n)^2] \\ \\ \therefore \text{Since expression is a } & \text{multiple of } n, \text{ it is divisible by } n \end{align*}

(a)

\begin{align*} u & = x^{n + 1} &&& v &= \ln x \\ {du \over dx} & =(n + 1)x^n &&& {dv \over dx} & = {1 \over x} \end{align*} \begin{align*} {d \over dx} (x^{n + 1} \ln x) & = (x^{n + 1})\left(1 \over x\right) + (\ln x)(n + 1)(x^n) \phantom{000900} [\text{Product rule}] \\ & = (x^{n + 1}) (x^{-1}) + (\ln x)(n + 1)(x^n) \\ & = x^{n + 1 + (-1)} + (\ln x)(n + 1)(x^n) \phantom{000000000} [a^m \times a^n = a^{m + n} ] \\ & = x^n + (\ln x)(n + 1)(x^n) \end{align*}

(b)

\begin{align*} {d \over dx} (x^{n + 1} \ln x) & = x^n + (\ln x)(n + 1)(x^n) \\ \\ \implies \int x^n \phantom{.} dx + \int (\ln x)(n + 1)(x^n) \phantom{.} dx & = x^{n + 1} \ln x \\ \int (\ln x)(n + 1)(x^n) \phantom{.} dx & = x^{n + 1} \ln x - \int x^n \phantom{.} dx \\ \underbrace{ (n + 1) \int (\ln x)(x^n) \phantom{.} dx }_{\text{Can factorise } (n + 1) \text{ since it is a constant}} & = x^{n + 1} \ln x - {x^{n + 1} \over n + 1} \\ \int (\ln x)(x^n) \phantom{.} dx & = {1 \over n + 1} \left( {x^{n + 1} \ln x \over 1} - {x^{n + 1} \over n + 1} \right) \\ \int (\ln x)(x^n) \phantom{.} dx & = { x^{n + 1} \ln x \over n + 1} - {x^{n + 1} \over (n + 1)^2} + C \end{align*}

(a)

\begin{align*} y & = Ax^b \\ \ln y & = \ln (Ax^b) \\ \ln y & = \ln A + \ln x^b \phantom{000000} [\text{Product law (logarithms)}] \\ \ln y & = \ln A + b \ln x \phantom{00000.} [\text{Power law (logarithms)}] \\ \ln y & = b \ln x + \ln A \phantom{00000.}[Y = mX + c] \\ \\ \text{Plot } & \ln x \text{ on the } X\text{-axis} \end{align*}

(b)

\begin{align*} \ln y & = b \ln x + \ln A \phantom{00000.}[Y = mX + c] \\ \\ b & = \text{Gradient} \\ & = {Y_2 - Y_1 \over X_2 - X_1} \\ & = {12.1 - 6.3 \over 4.1 - 1.2} \\ & = 2 \\ \\ \ln y & = 2 \ln x + \ln A \\ \\ \text{Using } & (1.2, 6.3), \\ 6.3 & = 2(1.2) + \ln A \\ 6.3 & = 2.4 + \ln A \\ 3.9 & = \ln A \\ 3.9 & = \log_e A \\ e^{3.9} & = A \\ \\ A & \approx 49.4, b = 2 \end{align*}

(a)

\begin{align*} y & = {1 \over 2}x^2 - 3 x^{1 \over 2} + 1 \\ \\ {dy \over dx} & = {1 \over 2}(2)(x) - 3 \left(1 \over 2\right) x^{-{1 \over 2}} \\ & = x - {3 \over 2} \left(1 \over \sqrt{x}\right) \\ & = x - {3 \over 2 \sqrt{x}} \ \\ \text{Let } & {dy \over dx} = 0, \phantom{000000} [\text{At stationary point, } {dy \over dx} = 0 ] \\ 0 & = x - {3 \over 2 \sqrt{x}} \\ {3 \over 2 \sqrt{x}} & = x \\ 3 & = (2 \sqrt{x})(x) \\ 3 & = (2 x^{1 \over 2})(x) \\ 3 & = 2 x^{3 \over 2} \\ {3 \over 2} & = x^{3 \over 2} \\ \left(3 \over 2\right)^2 & = (x^{3 \over 2})^2 \\ \left(3 \over 2\right)^2 & = x^3 \\ \sqrt[3]{\left(3 \over 2\right)^2} & = x \\ \\ x & = \left[ \left(3 \over 2\right)^2 \right]^{1 \over 3} \\ x & = \left(3 \over 2\right)^{2 \over 3} \phantom{0} \text{ (Shown)} \end{align*}

(b)

\begin{align*} {dy \over dx} & = x - {3 \over 2 \sqrt{x}} \\ & = x - {3 \over 2} \left(1 \over \sqrt{x}\right) \\ & = x - {3 \over 2} x^{-{1 \over 2}} \\ \\ {d^2 y \over dx^2} & = 1 - {3 \over 2} \left(-{1 \over 2}\right) x^{-{3 \over 2}} \\ & = 1 + {3 \over 4} \left(1 \over x^{3 \over 2}\right) \\ & = 1 + {3 \over 4 x^{3 \over 2}} \\ \\ \text{When } & x = \left(3 \over 2\right)^{2 \over 3}, \\ {d^2 y \over dx^2 } & = 1 + {3 \over 4 \left[ \left(3 \over 2\right)^{2 \over 3}\right] ^{3 \over 2}} \\ & = 1.5 > 0 \\ \\ \therefore \text{The stationary } & \text{point at } x = \left(3 \over 2\right)^{2 \over 3} \text{ is a minimum point} \end{align*}

(a)

\begin{align*} \left({1 \over 2} - ax\right)^5 & = \left(1 \over 2\right)^5 + {5 \choose 1} \left(1 \over 2\right)^4 (-ax) + {5 \choose 2} \left(1 \over 2\right)^3 (-ax)^2 + {5 \choose 3} \left(1 \over 2\right)^2 (-ax)^3 + ... \\ & = {1 \over 32} + (5)\left(1 \over 16\right)(-ax) + (10)\left(1 \over 8\right)(a^2 x^2) + (10) \left(1 \over 4\right)(-a^3 x^3) + ... \\ & = {1 \over 32} - {5 \over 16} ax + {5 \over 4} a^2 x^2 - {5 \over 2} a^3 x^3 + ... \end{align*}

(b)

\begin{align*} \left(4 + {5 \over ax}\right)\left({1 \over 2} - ax\right)^5 & = \left(4 + {5 \over ax}\right) \left( {1 \over 32} - {5 \over 16} ax + {5 \over 4} a^2 x^2 - {5 \over 2} a^3 x^3 + ... \right) \\ & = ... + (4) \left(-{5 \over 16}ax\right) + (4)\left({5 \over 4}a^2 x^2\right) + \left(5 \over ax\right)\left({5 \over 4} a^2 x^2\right) + \left({5 \over ax}\right)\left(- {5 \over 2} a^3 x^3\right) + ... \\ & = ... - {5 \over 4} ax + 5a^2 x^2 +{25 \over 4} ax - {25 \over 2} a^2 x^2 + ... \\ \\ \text{Coefficient of } x & = -{5 \over 4}a + {25 \over 4} a \\ & = 5a \\ \\ \text{Coefficient of } x^2 & = 5a^2 - {25 \over 2} a^2 \\ & = -{15 \over 2} a^2 \\ \\ \text{Coefficient of } x & = 2 \times \text{Coefficient of } x^2 \\ 5a & = 2 \times -{15 \over 2}a^2 \\ 5a & = -15a^2 \\ 5a + 15a^2 & = 0 \\ a + 3a^2 & = 0\\ a(1 + 3a) & = 0 \end{align*} \begin{align*} a & = 0 \text{ (Reject)} & \text{ or } && 1 + 3a & = 0 \\ & &&& 3a & = -1 \\ & &&& a & = -{1 \over 3} \end{align*}

(a)

\begin{align*} \text{Substitute } y = x + 1 \text{ into } { (x - a)^2 \over 2} & + (y - 1)^2 = 1, \\ { (x - a)^2 \over 2} + (x + 1 - 1)^2 & = 1 \\ { (x - a)^2 \over 2} + x^2 & = 1 \\ { (x)^2 - 2(x)(a) + (a)^2 \over 2} + {x^2 \over 1} & = 1 \phantom{00000000} [ (a - b)^2 = a^2 - 2ab + b^2 ] \\ { x^2 - 2ax + a^2 \over 2} + {2x^2 \over 2} & = 1 \\ { x^2 - 2ax + a^2 + 2x^2 \over 2 } & = 1 \\ { 3x^2 - 2ax + a^2 \over 2 } & = {1 \over 1} \\ 3x^2 - 2ax + a^2 & = 2 \\ 3x^2 - 2ax + a^2 - 2 & = 0 \\ \\ b^2 - 4ac & = (-2a)^2 - 4(3)(a^2 - 2) \\ & = 4a^2 - 12(a^2 - 2) \\ & = 4a^2 - 12a^2 + 24 \\ & = 24 - 8a^2 \\ \\ b^2 - 4ac & > 0 \phantom{00000000} [\text{Two real roots since line intersects curve twice}] \\ 24 - 8a^2 & > 0 \\ 3 - a^2 & > 0 \\ a^2 - 3 & < 0 \\ a^2 - (\sqrt{3})^2 & < 0 \\ (a + \sqrt{3})(a - \sqrt{3}) & < 0 \phantom{00000000} [ a^2 - b^2 = (a + b)(a - b)] \end{align*}

$$ \text{Set of values: } \{ a: a \in \mathbb{R}, -\sqrt{3} < a < \sqrt{3} \} $$

(b)

\begin{align*} \text{From (a), } b^2 - 4ac & = 24 - 8a^2 \\ \\ \text{If line is tangent to curve, } b^2 - 4ac & = 0 \\ 24 - 8a^2 & = 0 \\ -8a^2 & = -24 \\ a^2 & = {-24 \over -8} \\ a^2 & = 3 \\ a & = \pm \sqrt{3} \end{align*}

(a)

\begin{align*} \text{L.H.S} & = {1 + \sin 2 \theta \over \sin \theta} - {1 + \cos 2 \theta \over \cos \theta} \\ & = { \cos \theta (1 + \sin 2 \theta) \over \sin \theta \cos \theta} - {\sin \theta (1 + \cos 2 \theta) \over \sin \theta \cos \theta} \\ & = { \cos \theta (1 + \sin 2 \theta) - \sin \theta (1 + \cos 2 \theta) \over \sin \theta \cos \theta } \\ & = { \cos \theta (1 + 2 \sin \theta \cos \theta) - \sin \theta (1 + 2 \cos^2 \theta - 1) \over \sin \theta \cos \theta} \\ & = { \cos \theta + 2 \sin \theta \cos^2 \theta - \sin \theta (2 \cos^2 \theta) \over \sin \theta \cos \theta} \\ & = { \cos \theta ( 1 + 2 \sin \theta \cos \theta - 2 \sin \theta \cos \theta) \over \sin \theta \cos \theta} \\ & = { 1 + 2 \sin \theta \cos \theta - 2 \sin \theta \cos \theta \over \sin \theta} \\ & = {1 \over \sin \theta} \\ & = \text{cosec } \theta \\ & = \text{R.H.S} \end{align*}

(b)

\begin{align*} \underbrace{{1 + \sin 2 \theta \over \sin \theta} - {1 + \cos 2 \theta \over \cos \theta}}_\text{LHS of part a} & = 2 \sin \theta \\ \text{cosec } \theta & = 2 \sin \theta \\ {1 \over \sin \theta} & = {2 \sin \theta \over 1} \\ 1 & = 2 \sin^2 \theta \\ {1 \over 2} & = \sin^2 \theta \\ \\ \sin \theta & = \pm \sqrt{1 \over 2} \phantom{000000} [\text{All 4 quadrants}] \\ \\ \text{Basic angle, } \alpha & = \sin^{-1} \sqrt{1 \over 2} \\ & = 45^\circ \\ & = {\pi \over 4} \end{align*}

\begin{align*} \theta & = {\pi \over 4}, \pi - {\pi \over 4}, \pi + {\pi \over 4}, 2\pi - {\pi \over 4} \\ & = {\pi \over 4} \text{ (NA), } {3 \over 4} \pi \text{ (NA), } {5 \over 4} \pi, {7 \over 4} \pi \text{ (NA)} \\ \\ \therefore \theta & = {5 \over 4} \pi \end{align*}

(a)

\begin{align*} y & = 2 \sin ax - 3 \cos 2x \\ \\ \text{When } & x = 0, \\ y & = 2 \sin 0 - 3 \cos 0 \\ y & = 2(0) - 3(1) \\ y & = -3 \\ \\ \therefore & \phantom{.} P(0, - 3) \\ \\ {dy \over dx} & = 2(a)\cos ax - 3 (2) (- \sin 2x) \phantom{0000} [ {d \over dx} [\sin f(x)] = f'(x) . \cos f(x)] ] \\ & = 2a \cos ax + 6 \sin 2x \phantom{00000000000} [{d \over dx} [\cos f(x)] = f'(x) . - \sin f(x) ] \\ \\ \text{When } & x = 0, \\ {dy \over dx} & = 2a \cos 0 + 6 \sin 0 \\ & = 2a(1) + 6(0) \\ & = 2a \\ \\ \text{Gradient of normal at } P & = {-1 \over 2a} \phantom{000000000000000000000000} [m_1 \times m_2 = -1 \implies m_2 = {-1 \over m_1} ] \\ & = -{1 \over 2a} \\ \\ y & = mx + c \\ y & = -{1 \over 2a} x + c \\ \\ \text{Using } & P(0, -3), \\ -3 & = -{1 \over 2a}(0) + c \\ -3 & = c \\ \\ \text{Eqn of normal: } & y = -{1 \over 2a}x - 3 \end{align*}

(b)

\begin{align*} \text{Eqn of normal: } & y = -{1 \over 2a}x - 3 \\ \\ \text{Let } & y = 0, \\ 0 & = -{1 \over 2a}x - 3 \\ {1 \over 2a}x & = -3 \\ x & = (2a)(-3) \\ x & = -6a \\ \\ \therefore & \phantom{.} Q(-6a, 0) \end{align*}

\begin{align*} \text{Area of triangle } OPQ & = {1 \over 2} \times 6a \times 3 \\ & = 9a \\ \\ 9a & = 27 \\ a & = {27 \over 9} \\ a & = 3 \end{align*}

(a)

\begin{align*} \text{Length of } AB, y & = \sqrt{ 3^2 + x^2 } \phantom{000000000000} [\text{Pythagoras theorem}] \\ y & = \sqrt{9 + x^2} \\ y & = (9 + x^2)^{1 \over 2} \\ \\ {dy \over dx} & = {1 \over 2} (9 + x^2)^{-{1 \over 2}} (2x) \phantom{000000} [\text{Chain rule}] \\ & = x (9 + x^2)^{-{1 \over 2}} \\ & = {x \over 1} \left(1 \over \sqrt{9 + x^2}\right) \\ & = {x \over \sqrt{9 + x^2}} \\ \\ {dy \over dt} & = {dy \over dx} \times {dx \over dt} \\ & = {x \over \sqrt{9 + x^2}} \times 2 \\ & = {x \over \sqrt{9 + x^2}} \times {2 \over 1} \\ & = {2x \over \sqrt{9 + x^2}} \\ \\ \text{When } & x = 2, \\ {dy \over dt} & = {2(4) \over \sqrt{9 + 4^2}} \\ & = 1.6 \\ \\ \text{Rate of change of length } AB & = 1.6 \text{ units per second} \end{align*}

(b)

\begin{align*} \text{Area of square, } z & = AB \times AB \\ & = \sqrt{9 + x^2} \times \sqrt{9 + x^2} \\ & = 9 + x^2 \\ \\ {dz \over dx} & = 2x \\ \\ {dz \over dt} & = {dz \over dx} \times {dx \over dt} \\ & = 2x \times 2 \\ & = 4x \\ \\ {dz \over dt} & \le 80 \\ 4x & \le 80 \\ x & \le {80 \over 4} \\ x & \le 20 \\ \\ \text{Max. value of } x & = 20 \end{align*}

(a)(i)

\begin{align*} y & = a \cos kx \\ \\ \text{Period} & = {2\pi \over k} \\ \\ x \text{-coordinate of } C & = {2\pi \over k} \div 2 \\ & = {2\pi \over k} \times {1 \over 2} \\ & = {\pi \over k} \\ \\ x \text{-coordinate of } A & = {\pi \over k} \div 2 \\ & = {\pi \over k} \times {1 \over 2} \\ & = {\pi \over 2k} \\ \\ x \text{-coordinate of } B & = {\pi \over 2k} \times 3 \\ & = {\pi \over 2k} \times {3 \over 1} \\ & = {3\pi \over 2k} \end{align*}

(a)(ii)

\begin{align*} y & = a \cos kx \\ \\ \text{Amplitude} & = a \\ \\ \text{Coordinates of } A: & \left( {\pi \over 2k}, 0\right) \\ \text{Coordinates of } C: & \left( {\pi \over k}, -a\right) \\ \\ \text{Length of } AB & = {1 \over 2} \times \text{Period} \\ & = {1 \over 2} \times {2\pi \over k} \\ & = {\pi \over k} \\ \\ \text{Length of } AC & = \sqrt{ \left( {\pi \over k} - {\pi \over 2k} \right)^2 + (-a - 0)^2 } \\ & = \sqrt{ \left( {2\pi \over 2k} - {\pi \over 2k} \right)^2 + (-a )^2 } \\ & = \sqrt{ \left(\pi \over 2k\right)^2 + a^2 } \\ \\ \text{Length of } AB & = \text{Length of } AC \\ {\pi \over k} & = \sqrt{ \left(\pi \over 2k\right)^2 + a^2 } \\ \left(\pi \over k\right)^2 & = \left(\pi \over 2k\right)^2 + a^2 \\ {\pi^2 \over k^2} & = {pi^2 \over 4k^2} + a^2 \\ \\ a^2 & = {\pi^2 \over k^2} - {pi^2 \over 4k^2} \\ a^2 & = {4\pi^2 \over 4k^2} - {\pi^2 \over 4k^2} \\ a^2 & = {3\pi^2 \over 4k^2} \\ a & = \pm \sqrt{ 3 \pi^2 \over 4k^2 } \\ a & = \pm { \sqrt{3} \sqrt{\pi^2} \over \sqrt{4k^2} }\\ a & = { \pi \sqrt{3} \over 2k } \text{ or } - { \pi \sqrt{3} \over 2k } \text{ (Reject, since } a > 0) \end{align*}

(b)

\begin{align*} \text{Area of triangle } ABC & = {1 \over 2} \times b \times h \\ & = {1 \over 2} \times AB \times a \\ & = {1 \over 2} \times {\pi \over k} \times a \\ & = {\pi \over 2k} \times {a \over 1} \\ & = {a \pi \over 2k} \\ \\ 1 & = {a\pi \over 2k} \\ 2k & = a \pi \\ {2k \over \pi} & = a \\ \\ \text{From (a)(ii), } & a = {\pi \sqrt{3} \over 2k}, \\ {2k \over \pi} & = {\pi \sqrt{3} \over 2k} \\ (2k)(2k) & = \pi (\pi \sqrt{3}) \\ 4k^2 & = \pi^2 \sqrt{3} \\ k^2 & = {\pi^2 \sqrt{3} \over 4} \\ k^2 & = {\pi^2 3^{1 \over 2} \over 4} \\ k & = \pm \sqrt{\pi^2 3^{1 \over 2} \over 4} \\ k & = \pm \left( {\pi^2 3^{1 \over 2} \over 4} \right)^{1 \over 2} \\ k & = \pm { \pi^{ 2 \times {1 \over 2} } (3)^{ {1 \over 2} \times {1 \over 2} } \over 4^{1 \over 2} } \\ k & = \pm { \pi (3)^{1 \over 4} \over 2 } \\ k & = { \pi (3)^{1 \over 4} \over 2} \text{ or } - { \pi (3)^{1 \over 4} \over 2} \text{ (Reject since } k > 0) \\ \\ \text{Period} & = {2\pi \over k} \\ & = {2\pi \over { \pi (3)^{1 \over 4} \over 2}} \\ & = 2\pi \div { \pi (3)^{1 \over 4} \over 2} \\ & = {2\pi \over 1} \times {2 \over \pi (3)^{1 \over 4}} \\ & = {4 \pi \over \pi (3)^{1 \over 4}} \\ & = {4 \over (3)^{1 \over 4}} \\ \\ \text{Amplitude} & = a \\ & = {2k \over \pi} \\ & = { 2 \left[ {\pi (3)^{1 \over 4} \over 2}\right] \over \pi} \\ & = { \pi (3)^{1 \over 4} \over \pi} \\ & = (3)^{1 \over 4} \end{align*}

(a)

\begin{align*} \text{Eqn of chord: } & 2y - 3x =1 7 \\ 2y & = 3x + 17 \\ y & = {3 \over 2}x + {17 \over 2} \phantom{000000} [y = mx + c] \\ \\ \text{Gradient of chord} & = {3 \over 2} \\ \text{Gradient of diameter // } AB & = {3 \over 2} \\ \\ y & = mx + c \\ y & = {3 \over 2}x + c \\ \\ \text{Using } & D\left(-{1 \over 2}, {1 \over 4}\right), \\ {1 \over 4} & = {3 \over 2}\left(-{1 \over 2}\right) + c \\ {1 \over 4} & = -{3 \over 4} + c \\ 1 & = c \\ \\ \text{Eqn of diameter // } AB: & y = {3 \over 2}x + 1 \\ \\ \text{Let } & y = -2, \phantom{000000} [y \text{-coordinate of centre} = -2] \\ -2 & = {3 \over 2}x + 1 \\ -{3 \over 2}x & = 3 \\ x & = 3 \div -{3 \over 2} \\ x & = -2 \\ \\ \text{Centre, }C & (-2, -2) \\ \\ \text{Radius} & = AC \\ & = \sqrt{ [-2 -(-1)]^2 + (-2 - 7)^2 } \\ & = \sqrt{82} \\ \\ \text{Eqn of circle: } & [x - (-2)]^2 + [y - (-2)]^2 = (\sqrt{82})^2 \\ & \phantom{000-.} (x + 2)^2 + (y + 2)^2 = 82 \end{align*}

(b)

\begin{align*} \text{Eqn of circle: } & (x + 2)^2 + (y + 2)^2 = 82 \\ \\ \text{Let } & x = 7, \\ (7 + 2)^2 + (y + 2)^2 & = 82 \\ 81 + (y + 2)^2 & = 82 \\ (y + 2)^2 & = 1 \\ y + 2 & = \pm \sqrt{1} \\ y + 2 & = 1 \text{ or } - 1 \\ y & = -1 \text{ or } - 3 \end{align*}

\begin{align*} [\text{Any point on the } \text{blue line will have } & x \text{-coordinate }7 \text{ AND be inside the circle}] \\ \\ \text{Set of values: } \{ \alpha: & \alpha \in \mathbb{R}, -3 < \alpha < -1 \} \end{align*}

\begin{align*} \text{Area of trapezium} & = {1 \over 2} \times \text{Sum of parallel sides} \times \text{Height} \\ \\ \implies \text{Sum of parallel sides} & = { \text{Area of trapezium} \over {1 \over 2} \times \text{Height} }\\ & = {26 \over {1 \over 2} (4 - \sqrt{3}) } \times {2 \over 2} \\ & = {52 \over 4 - \sqrt{3} } \times {4 + \sqrt{3} \over 4 + \sqrt{3} } \phantom{000000} [\text{Rationalise denominator}] \\ & = { 52 (4 + \sqrt{3}) \over (4 - \sqrt{3})(4 + \sqrt{3}) } \\ & = { 208 + 52 \sqrt{3} \over (4)^2 - (\sqrt{3})^2 } \phantom{00000000000} [(a - b)(a + b) = a^2 - b^2 ] \\ & = { 208 + 52 \sqrt{3} \over 13 } \\ & = { 208 \over 13} + {52 \sqrt{3} \over 13} \\ & = 16 + 4 \sqrt{3} \\ \\ BC & = 16 + 4\sqrt{3} - (6 + \sqrt{3}) \\ & = 16 + 4 \sqrt{3} - 6 - \sqrt{3} \\ & = (10 + 3 \sqrt{3}) \text{ cm} \end{align*}

(a)

\begin{align*} T & = 4 + 2h - {1 \over 2} h^2 \\ T & = - {1 \over 2} h^2 + 2h + 4 \\ T & = -{1 \over 2} \left( h^2 - 4h \right) + 4 \\ T & = -{1 \over 2} \left[ h^2 - 4h + \left(4 \over 2\right)^2 - \left(4 \over 2\right)^2 \right] + 4 \\ T & = -{1 \over 2} [ (h - 2)^2 - 4] + 4 \\ T & = -{1 \over 2} (h - 2)^2 + 2 + 4 \\ T & = -{1 \over 2} (h - 2)^2 + 6 \\ T & = 6 - {1 \over 2} (h - 2)^2 \\ \\ \text{Maximum } & \text{temperature, } T = 6^\circ\text{C, when } h = 2 \text{ hours} \end{align*}

(b)

\begin{align*} \text{Following day's temperature, } F & = {1 \over 2} T \\ & = {1 \over 2} \left[ 6 - {1 \over 2} (h - 2)^2 \right] \\ & = 3 - {1 \over 4} (h - 2)^2 \\ \\ \text{Maximum } \text{temperature, } T & = 3^\circ\text{C, when } h = 2 \text{ hours} \end{align*}

\begin{align*} \text{Gradient, } {dy \over dx} & = e^x( 3 e^{2x} - e^{-4x} ) - \cos 2x \\ & = 3e^{2x + x} - e^{-4x + x} - \cos 2x \\ & = 3e^{3x} - e^{-3x} - \cos 2x \\ \\ y & = \int 3e^{3x} - e^{-3x} - \cos 2x \phantom{.} dx \\ y & = 3 \left(e^{3x} \over 3\right) - {e^{-3x} \over -3} - {\sin 2x \over 2} + c \phantom{000000} \left[ \int e^{f(x)} \phantom{.} dx = {e^{f(x)} \over f'(x)} \right] \\ y & = e^{3x} + {1 \over 3} e^{-3x} - {1 \over 2} \sin 2x + c \phantom{000000000} \left[ \int \cos f(x) \phantom{.} dx = {\sin f(x) \over f'(x)} \right] \\ \\ \text{Using } & O(0, 0), \\ 0 & = e^0 + {1 \over 3} e^0 - {1 \over 2} \sin 0 + c \\ 0 & = 1 + {1 \over 3}(1) - {1 \over 2} (0) + c \\ 0 & = {4 \over 3} + c \\ -{4 \over 3} & = c \\ \\ y & = e^{3x} + {1 \over 3}e^{-3x} - {1 \over 2} \sin 2x - {4 \over 3} \end{align*}

(a)

\begin{align*} f(-1) & = (-1)^3 + p(-1)^2 + q(-1) - 6 \\ & = -1 + p(1) - q - 6 \\ & = p - q - 7 \\ \\ \text{Since } x + 1 \text{ is } & \text{a factor of } f(x), \\ p - q - 7 & = 0 \\ p & = q + 7 \phantom{0} \text{--- (1)} \\ \\ f(-2) & = (-2)^3 + p(-2)^2 + q(-2) - 6 \\ & = -8 + p(4) - 2q - 6 \\ & = 4p - 2q - 14 \\ \\ 4p - 2q - 14 & = 4 \\ 4p - 2q & = 18 \\ 2p - q & = 9 \phantom{0} \text{--- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ 2(q + 7) - q & = 9 \\ 2q + 14 - q & = 9 \\ q & = 9 - 14 \\ q & = -5 \\ \\ \text{Substitute } & q = -5 \text{ into (1),} \\ p & = -5 + 7 \\ p & = 2 \\ \\ \therefore p & = 2, q = -5 \end{align*}

(b)

$$ \require{enclose} \begin{array}{rll} 3x^2 - 5x - 2 \phantom{000000.}\\ 3x - 1 \enclose{longdiv}{ 9x^3 - 18x^2 - x + 2 \phantom{0}}\kern-.2ex \\ -\underline{(9x^3 - 3x^2){\phantom{00000000.}}} \\ - 15x^2 - x + 2 \phantom{0} \\ -\underline{(-15x^2 + 5x){\phantom{00.}}} \\ -6x + 2 \phantom{0} \\ -\underline{(-6x + 2){\phantom{.}}} \\ 0 \phantom{.} \end{array} $$ \begin{align*} \text{Polynomial} & = \text{Divisor} \times \text{Quotient} + \text{Remainder} \\ & = (3x - 1)(3x^2 - 5x - 2) + 0 \\ & = (3x - 1)(3x + 1)(x - 2) \end{align*}

\begin{align*} u & = 1 - e^{-x} &&& v & = 1 + e^{-x} \\ {du \over dx} & = -(-1)e^{-x} &&& {dv \over dx} & = (-1)e^{-x} \phantom{000000} [ {d \over dx} (e^{f(x)}) = f'(x) . e^{f(x)} ] \\ & = e^{-x} &&& & = -e^{-x} \end{align*} \begin{align*} {dy \over dx} & = { v{du \over dx} - u {dv \over dx} \over v^2} \phantom{00000000000000000000000} [\text{Quotient rule}] \\ & = { (1 + e^{-x})(e^{-x}) - (1 - e^{-x})(-e^{-x}) \over (1 + e^{-x})^2} \\ & = { e^{-x} + e^{-x + (-x)} + e^{-x}(1 - e^{-x}) \over (1 + e^{-x})^2 } \\ & = { e^{-x} + e^{-2x} + e^{-x} - e^{-x + (-x)} \over (1 + e^{-x})^2 } \\ & = { e^{-x} + e^{-2x} + e^{-x} - e^{-2x} \over (1 + e^{-x})^2} \\ & = { 2 e^{-x} \over (1 + e^{-x})^2 } \\ \\ \text{For } & \text{all real values of } x, e^{-x} > 0 \text{ and } (1 + e^{-x})^2 > 0 \\ \\ \implies {dy \over dx} & = { 2 e^{-x} \over (1 + e^{-x})^2 } >0 \text{ for all real values of } x \\ \\ \therefore y & \text{ increases for all real values of } x \end{align*}

\begin{align*} a & = (2t + 1)^{-1} + 0.3t + 0.2 \\ a & = {1 \over 2t + 1} + 0.3t + 0.2 \\ \\ v & = \int {1 \over 2t + 1} + 0.3t + 0.2 \phantom{.} dt \\ v & = \underbrace{{ \ln (2t + 1) \over 2}}_{ \int {1 \over f(x)} \phantom{.} dx = {\ln f(x) \over f'(x)} } + {0.3t^2 \over 2} + 0.2t + c \\ v & = {1 \over 2} \ln (2t + 1) + 0.15t^2 + 0.2t + c \\ \\ \text{When } & t = 0 \text{ and } v = -3, \\ -3 & = {1 \over 2} \ln (1) + 0.15(0)^2 + 0.2(0) + c \\ -3 & = 0 + 0 + 0 + c \\ -3 & = c \\ \\ v & = {1 \over 2} \ln (2t + 1) + 0.15t^2 + 0.2t - 3 \\ \\ \text{When } & t = 5, \\ v & = {1 \over 2} \ln [2(5) + 1] + 0.15(5)^2 + 0.2(5) - 3 \\ v & = 2.9489 \\ v & \approx 2.95 \text{ m/s} \end{align*}

(a)

\begin{align*} 2^p & = 5(3^{2 - p}) \\ 2^p & = 5\left(3^2 \over 3^p\right) \phantom{000000} \left[ a^{m - n} = {a^m \over a^n} \right] \\ {2^p \over 1} & = {45 \over 3^p} \\ 2^p \times 3^p & = 45 \\ 6^p & = 45 \\ \lg 6^p & = \lg 45 \\ p \lg 6 & = \lg 45 \phantom{0000000000} [\text{Power law (logarithms)}] \\ p & = {\lg 45 \over \lg 6} \\ p & \approx 2.12 \end{align*}

(b)

\begin{align*} 9(2^x) + 2(3^{y + 1}) & = 38 \\ 9(2^x) & = 38 - 2(3^{y + 1}) \\ 2^x & = {1 \over 9} [38 - 2(3^{y + 1})] \\ 2^x & = {38 \over 9} - {2 \over 9} (3^{y + 1}) \\ \\ 15(2^{x - 1}) - 21(3^y) & = 23 \\ 15 \left(2^x \over 2\right) - 21(3^y) & = 23 \phantom{000000} \left[ a^{m - n} = {a^m \over a^n} \right] \\ {15 \over 2} (2^x) - 21(3^y) & = 23 \\ \\ \text{Since } 2^x & = {38 \over 9} - {2 \over 9} (3^{y + 1}), \\ {15 \over 2} \left[ {38 \over 9} - {2 \over 9} (3^{y + 1}) \right] - 21(3^y) & = 23 \\ {95 \over 3} - {5 \over 3}(3^{y + 1}) - 21(3^y) & = 23 \\ -{5 \over 3}(3^y \times 3^1) - 21(3^y) & = 23 - {95 \over 3} \\ -5 (3^y) - 21(3^y) & = -{26 \over 3} \\ -26(3^y) & = -{26 \over 3} \\ 26(3^y) & = {26 \over 3} \\ 3^y & = {1 \over 3} \\ 3^y & = 3^{-1} \\ y & = -1 \\ \\ \text{Since } 2^x & = {38 \over 9} - {2 \over 9} (3^{y + 1}), \\ 2^x & = {38 \over 9} - {2 \over 9} (3^{-1 + 1}) \\ 2^x & = 4 \\ 2^x & = 2^2 \\ x & = 2 \\ \\ \therefore x & = 2, y = -1 \end{align*}

(a)

\begin{align*} \text{Line: } & y = kx \\ \\ \text{When } & x = \alpha, \\ y &= k\alpha \\ \\ \therefore & \phantom{.} P(\alpha, k\alpha) \\ \\ \\ \text{Curve: } & y = x(10 - x) \\ \\ \text{When } & x = \alpha, \\ y & = \alpha(10 - \alpha) \\ \\ \therefore & \phantom{.} P(\alpha, \alpha(10 - \alpha)) \\ \\ \\ \implies k \alpha & = \alpha(10 - \alpha) \\ k & = 10 - \alpha \phantom{000000} [k \text{ in terms of } \alpha] \\ \\ \text{Area of triangle } OQP & = {1 \over 2} \times OQ \times PQ \\ & = {1 \over 2} \times \alpha \times k \alpha \\ & = {1 \over 2} k \alpha^2 \\ & = {1 \over 2} (10 - \alpha) \alpha^2 \\ & = {1 \over 2} (10 \alpha^2 - \alpha^3) \\ \\ \text{Area bounded by curve and } x \text{-axis} & = \int_0^\alpha x(10 - x) \phantom{.} dx \\ & = \int_0^\alpha 10x - x^2 \phantom{.} dx \\ & = \left[ {10x^2 \over 2} - {x^3 \over 3}\right]_0^\alpha \\ & = {10\alpha^2 \over 2} - {\alpha^3 \over 3} - (0 - 0) \\ & = 5 \alpha^2 - {1 \over 3} \alpha^3 \\ \\ R_1 & = 5 \alpha^2 - {1 \over 3} \alpha^3 - {1 \over 2} (10 \alpha^2 - \alpha^3) \\ & = 5 \alpha^2 - {1 \over 3} \alpha^3 - 5\alpha^2 + {1 \over 2} \alpha^3 \\ & = {1 \over 6} \alpha^3 \\ & = {\alpha^3 \over 6} \phantom{0} \text{ (Shown)} \end{align*}

(b)

\begin{align*} R_1 & = R_2 \\ {\alpha^3 \over 6} & = {1 \over 2} (10 \alpha^2 - \alpha^3) \\ 2\alpha^3 & = 6(10 \alpha^2 - \alpha^3) \\ 2\alpha^3 & = 60 \alpha^2 - 6 \alpha^3 \\ 0 & = 60 \alpha^2 - 8 \alpha^3 \\ 0 & = 4\alpha^2 (15 - 2\alpha) \end{align*} \begin{align*} 4 \alpha^2 & = 0 && \text{ or } & 15 - 2\alpha & = 0 \\ \alpha^2 & = 0 &&& -2\alpha & = -15 \\ \alpha & = 0 \text{ (Reject)} &&& \alpha & = {-15 \over -2} \\ & &&& \alpha & = 7.5 \end{align*}
\begin{align*} k & = 10 - \alpha \\ k & = 10 - 7.5 \\ k & = 2.5 \end{align*}

(a)

\begin{align*} \angle AEB & = \angle BEC \phantom{0} \text{ (Common angle) } [A] \\ \\ \angle BAE & = \angle CBE = x \phantom{0} \text{ (Alternate segment theorem) } [A] \\ \\ \therefore \text{Triangles } & ABE \text{ and } BCE \text{ are similar } (AA \text{ similarity test}) \end{align*}

(b)(i)

\begin{align*} \angle ADC & = 180^\circ - \angle ABC \phantom{0} \text{ (Angles in opposite segments)} \\ & = 180^\circ - (180^\circ - x - y) \phantom{0} \text{ (Adjacent angles on a straight line)} \\ & = 180^\circ - 180^\circ + x + y \\ & = x + y \phantom{0} \text{ (Shown)} \end{align*}

(b)(ii)

\begin{align*} \angle ACD & = x \phantom{0} \text{ (Alternate angles, } AB \phantom{.} // \phantom{.} DC) \\ \\ \angle BDC & = x \phantom{0} \text{ (Angles in the same segment)} \\ \\ \angle ADB & = x + y - x = y \\ \\ \angle ACB & = y \phantom{0} \text{ (Angles in the same segment)} \\ \\ \\ \angle ADC & = \angle BCD = x + y \phantom{0} [A] \\ \\ DC & \text{ is a common side } [S] \\ \\ \angle ACD & = \angle BDC = x \phantom{0} [A] \\ \\ \text{Triangles } & ADC \text{ and } BCD \text{ are congruent } (ASA \text{ congruency test}) \\ \\ \implies AD & = BC \text{ (Shown)} \end{align*}

(a)

\begin{align*} & \text{For } 0^\circ < \theta < 90^\circ, 6 \cos \theta > 0 \text{ and } 8 \sin \theta > 0 . \\ \\ & \text{If } 0^\circ < \theta < 90^\circ, 0\text{ the speed of both objects are positive.} \end{align*}

(b)

\begin{align*} \text{Distance travelled by } A/B & = {30 \over 2} = 15 \text{ cm} \\ \\ \text{Time taken by } A & = {15 \over 6 \cos \theta} \text{ s} \\ \\ \text{Time taken by } B & = {15 \over 8 \sin \theta} \text{ s} \\ \\ [\text{When } A \text{ and } B \text{ collides at the} & \text{ midpoint, time taken by each object is the same}] \\ \\ {15 \over 6 \cos \theta} & = {15 \over 8 \sin \theta} \\ 15(8 \sin \theta) & = 15(6 \cos \theta) \\ 8 \sin \theta & = 6 \cos \theta \\ {\sin \theta \over \cos \theta} & = {6 \over 8} \\ \tan \theta & = {3 \over 4} \phantom{000000} [\text{1st & 3rd quadrant}] \\ \\ \text{Basic angle, } \alpha & = \tan^{-1} \left(3 \over 4\right) \\ & = 36.87^\circ \end{align*}

\begin{align*} \theta & = 36.87^\circ, 180^\circ + 36.87^\circ \\ & = 36.87^\circ, 216.87^\circ \text{ (NA, since } \theta \text{ is acute}) \\ \\ \text{Time taken} & = {15 \over 6 \cos 36.87^\circ} \\ & = 3.125 \\ & \approx 3.13 \text{ seconds} \end{align*}

(c)(i)

\begin{align*} \text{Distance travelled by } A & = 6 \cos \theta \times 4 \\ & = 24 \cos \theta \\ \\ \text{Distance travelled by } B & = 8 \sin \theta \times 4 \\ & = 32 \sin \theta \\ \\ \text{Total distance travelled by } A \text{ and } B & = 24 \cos \theta + 32 \sin \theta \\ 30 & = 24 \cos \theta + 32 \sin \theta \\ 30 & = 8(3 \cos \theta + 4 \sin \theta) \\ {30 \over 8} & = 3 \cos \theta + 4 \sin \theta \\ 3.75 & = 3 \cos \theta + 4 \sin \theta \phantom{0} \text{ (Shown)} \end{align*}

(c)(ii)

\begin{align*} R \text{-formula: } a \cos \theta & + b \sin \theta = R \cos (\theta - \alpha) \\ \\ R & = \sqrt{a^2 + b^2} \\ R & = \sqrt{ 3^2 + 4^2 } \\ R & = 5 \\ \\ \alpha & = \tan^{-1} \left(b \over a\right) \\ & = \tan^{-1} \left(4 \over 3\right) \\ & = 53.13^\circ \\ \\ 3 \cos \theta + 4 \sin \theta & = 5 \cos (\theta - 53.13^\circ) \\ \\ 5 \cos (\theta - 53.13^\circ) & = 3.75 \phantom{000000} [\text{from (i)}] \\ \cos (\theta - 53.13^\circ) & = {3.75 \over 5} \\ & = 0.75 \phantom{000000} [\text{1st & 4th quadrant}] \\ \\ \text{Basic angle, } \alpha & = \cos^{-1} (0.75) \\ & = 41.41^\circ \end{align*}

\begin{align*} [\text{If } 0^\circ < \theta < 90^\circ, & \text{ then } -53.13^\circ < \theta - 53.13^\circ < 36.87^\circ] \\ \\ \theta - 53.13^\circ & = 41.41^\circ, 360^\circ - 41.41^\circ \\ & = 41.41^\circ \text{ (NA), } 318.59^\circ \text{ (NA), } 318.59^\circ -360^\circ \\ & = -41.41^\circ \\ \\ \theta & = -41.41^\circ + 53.13^\circ \\ & = 11.72^\circ \\ \\ \text{Speed of object } A & = 6 \cos 11.72^\circ \\ & \approx 5.87 \text{ cm/s} \\ \\ \text{Speed of object } B & = 8 \sin 11.72^\circ \\ & \approx 1.63 \text{ cm/s} \end{align*}