\begin{align} {x \over a} + {y \over b} & = 1 \\ {y \over b} & = - {x \over a} + 1 \\ {y \over b} & = -{1 \over a} x + 1 \\ y & = b \left( -{1 \over a} x + 1 \right) \\ y & = -{b \over a}x + b \phantom{000000} [y = mx + c] \\ \\ \text{Gradient} & = -{b \over a} \\ -{1 \over 3} & = -{b \over a} \\ {1 \over 3} & = {b \over a} \\ a & = 3b \phantom{00} \text{--- (1)} \\ \\ \\ y & = -{b \over 3b}x + b \\ y & = -{1 \over 3}x + b \\ \\ \text{Let } & x = 0, \\ y & = -{1 \over 3}(0) + b \\ y & = b \\ \\ \therefore & \phantom{.} T(0, b) \\ \\ \\ \text{Let } & y = 0, \\ 0 & = -{1 \over 3}x + b \\ {1 \over 3}x & = b \\ x & = 3b \\ \\ \therefore & \phantom{.} S(3b, 0) \\ \\ \\ ST & = \sqrt{ (x_2 - x_1)^2 + (y_2 - y_1)^2 } \\ \sqrt{40} & = \sqrt{ (3b - 0)^2 + (0 - b)^2 } \\ \sqrt{40} & = \sqrt{ (3b)^2 + (-b)^2 } \\ \sqrt{40} & = \sqrt{ 9b^2 + b^2 } \\ \sqrt{40} & = \sqrt{10b^2} \\ 40 & = 10b^2 \\ {40 \over 10} & = b^2 \\ 4 & = b^2 \\ \\ b & = \pm \sqrt{4} \\ b & = 2 \text{ or } - 2 \text{ (Reject, since } b > 0) \\ \\ \text{Substitute } & b = 2 \text{ into (1),} \\ a & = 3(2) \\ a & = 6 \\ \\ \therefore a & = 6, b = 2 \end{align}

(a)

\begin{align} y & = 3 - 4 \sin 2x \\ y & = - 4 \sin 2x + 3 \phantom{000000} [\text{Inverted shape}] \\ \\ \text{Amplitude} & = 4 \\ \\ \text{Center line: } & y = 3 \\ \\ \text{Max. value} & = 3 + 4 = 7 \\ \\ \text{Min. value} & = 3 - 4 = -1 \end{align}

(b)

\begin{align} \text{From (a), center line: } & y = 3 \\ \text{Max. value} & = 7 \\ \text{Min. value} & = -1 \\ \\ \text{Period} & = {360^\circ \over 2} \\ & = 180^\circ \\ \\ \text{No. of cycles} & = {360^\circ \over 180^\circ} \\ & = 2 \end{align}

(a) Formula for binomial expansion is provided

\begin{align} \left(2 - {x \over 4}\right)^6 & = (2)^6 + {6 \choose 1} (2)^5 \left(-{x \over 4}\right) + {6 \choose 2} (2)^4 \left(-{x \over 4} \right)^2 + ... \\ & = 64 + (6)(32)\left(-{x \over 4}\right) + (15)(16) \left(x^2 \over 16\right) + ... \\ & = 64 - 48 x + 15x^2 + ... \end{align}

(b) Term independent of x refers to the constant term

\begin{align} \text{From (a), } \left(2 - {x \over 4}\right)^6 & = 64 - 48 x + 15x^2 + ... \\ \\ \\ \left({3 \over x} - x \right)^2 & = \left({3 \over x}\right)^2 - 2\left({3 \over x}\right)(x) + (x)^2 \phantom{000000} [(a - b)^2 = a^2 - 2ab + b^2] \\ & = {9 \over x^2} - 6 + x^2 \\ \\ \\ \left(2 - {x \over 4}\right)^6 \left({3 \over x} - x \right)^2 & = (64 - 48 x + 15x^2 + ...) \left( {9 \over x^2} - 6 + x^2 \right) \\ & = ... + (64)(-6) + ... + (15x^2)\left(9 \over x^2\right) + ... \\ & = ... - 384 + ... + 135 + ... \\ & = ... -249 + ... \\ \\ \\ \text{Term indepen} & \text{dent of } x \text{ is } - 249 \end{align}

(a)

\begin{align} u & = x^2 - 4 \phantom{0000000} {du \over dx} = 2x \\ v & = x^2 + 6 \phantom{0000000} {dv \over dx} = 2x \\ \\ \\ f'(x) & = {(x^2 + 6)(2x) - (x^2 - 4)(2x) \over (x^2 + 6)^2} \phantom{00000000} [\text{Quotient rule}] \\ & = {2x(x^2 + 6) - 2x(x^2 - 4) \over (x^2 + 6)^2} \\ & = {2x^3 + 12x - 2x^3 + 8x \over (x^2 + 6)^2} \\ & = {20x \over (x^2 + 6)^2} \\ \\ \text{For } & x > 0, 20x > 0 \text{ and } (x^2 + 6)^2 > 0 \\ \\ \implies & f'(x) > 0 \\ \\ \therefore f \text{ is } & \text{an increasing function for } x > 0 \end{align}

(b)

\begin{align} \text{From (a), } {dy \over dx} & = f'(x) \\ {dy \over dx} & = {20x \over (x^2 + 6)^2} \\ \\ \\ {dx \over dt} & = {dx \over dy} \times {dy \over dt} \phantom{000000} [\text{Chain rule}] \\ & = {(x^2 + 6)^2 \over 20x} \times 0.05 \\ & = {(x^2 + 6)^2 \over 20x} \times {1 \over 20} \\ & = {(x^2 + 6)^2 \over 400x} \\ \\ \text{When } & x = 2, \\ {dx \over dt} & = {[(2)^2 + 6]^2 \over 400(2)} \\ & = 0.125 \text{ units per second} \end{align}

(a)

\begin{align} N & = 160e^{kt} \\ \\ \text{When } & t = 5 \text{ and } N = 245, \\ 245 & = 160 e^{5k} \\ {245 \over 160} & = e^{5k} \\ {49 \over 32} & = e^{5k} \\ \\ \text{Take } & \ln \text{ of both sides,} \\ \ln {49 \over 32} & = \ln e^{5k} \\ \ln {49 \over 32} & = 5k \ln e \phantom{000000} [\text{Power law (logarithms)}] \\ \ln {49 \over 32} & = 5k (1) \\ \ln {49 \over 32} & = 5k \\ {1 \over 5} \ln {49 \over 32} & = k \\ 0.085 \phantom{.} 216 & = k \\ \\ N & = 160e^{0.085 \phantom{.} 216 t} \\ \\ \text{When } & t = 7, \\ N & = 160e^{0.085 \phantom{.} 216 (7)} \\ & = 290.523 \\ & \approx 290 \end{align}

(b)

\begin{align} N & = 160e^{0.085 \phantom{.} 216 t} \\ \\ \text{When } & N = 400, \\ 400 & = 160e^{0.085 \phantom{.} 216 t} \\ {400 \over 160} & = e^{0.085 \phantom{.} 216 t} \\ 2.5 & = e^{0.085 \phantom{.} 216 t} \\ \\ \text{Take } & \ln \text{ of both sides,} \\ \ln 2.5 & = \ln e^{0.085 \phantom{.} 216 t} \\ \ln 2.5 & = 0.085 \phantom{.} 216 t \ln e \phantom{000000} [\text{Power law (logarithms)}] \\ \ln 2.5 & = 0.085 \phantom{.} 216 t (1) \\ \ln 2.5 & = 0.085 \phantom{.} 216 t \\ \\ t & = {\ln 2.5 \over 0.085 \phantom{.} 216 } \\ & = 10.752 \\ \\ \therefore \text{Influenza } & \text{is declared an epidemic after 11 days} \end{align}

\begin{align} {d^2 y \over dx^2} & = 3 \cos x - 4 \sin 2x \\ \\ {dy \over dx} & = \int 3 \cos x - 4 \sin 2x \phantom{.} dx \\ & = 3 (\sin x) - 4 \left( - \cos 2x \over 2\right) + c \\ & = 3 \sin x + 2 \cos 2x + c \\ \\ \text{When } & x = {\pi \over 2} \text{ and } {dy \over dx} = 5, \phantom{000000} [\text{Gradient at } P = 5] \\ 5 & = 3 \sin {\pi \over 2} + 2 \cos \left[ 2 \left({\pi \over 2}\right) \right] + c \\ 5 & = 3 (1) + 2 \cos \pi + c \\ 5 & = 3 + 2(-1) + c \\ 5 & = 3 - 2 + c \\ 5 & = 1 + c \\ 4 & = c \\ \\ {dy \over dx} & = 3 \sin x + 2 \cos 2x + 4 \\ \\ y & = \int 3 \sin x + 2 \cos 2x + 4 \phantom{.} dx \\ & = 3(-\cos x) + 2 \left(\sin 2x \over 2\right) + 4x + c \\ & = - 3 \cos x + \sin 2x + 4x + c \\ \\ \text{Using } & P \left({\pi \over 2}, 9 \right), \\ 9 & = -3 \cos {\pi \over 2} + \sin \left[ 2 \left({\pi \over 2}\right) \right] + 4 \left({\pi \over 2}\right) + c \\ 9 & = - 3 (0) + \sin \pi + 2 \pi + c \\ 9 & = 0 + 0 + 2 \pi + c \\ 9 & = 2 \pi + c \\ 9 - 2\pi & = c \\ \\ \\ \text{Eqn of curve: } & y = - 3 \cos x + \sin 2x + 4x + 9 - 2 \pi \end{align}

(a) Complete the square: $x^2 \pm bx + c = \left(x \pm {b \over 2}\right)^2 - \left(b \over 2\right)^2 + c$

\begin{align} 2x^2 - 4x + 5 & = 2(x^2 - 2x) + 5 \phantom{0000000000000} [\text{Coefficient of } x^2 \text{ must be 1}] \\ & = 2 \left[ \left(x - {2 \over 2}\right)^2 - \left(2 \over 2\right)^2 \right] + 5 \\ & = 2 [ (x - 1)^2 - 1 ] + 5 \\ & = 2(x - 1)^2 - 2 + 5 \\ & = 2(x - 1)^2 + 3 \\ \\ \\ - x^2 - 4x - 2 & = - (x^2 + 4x) - 2 \\ & = - \left[ \left(x + {4 \over 2}\right)^2 - \left(4 \over 2\right)^2 \right] - 2 \\ & = - [ (x + 2)^2 - 4] - 2 \\ & = - (x + 2)^2 + 4 - 2 \\ & = - (x + 2)^2 + 2 \end{align}

(b)

\begin{align} y & = 2x^2 - 4x + 5 \\ y & = 2(x - 1)^2 + 3 \\ \\ \text{Minimum curve } & (\cup) \text{ with turning point } (1, 3) \\ \\ \\ y & = - x^2 - 4x - 2 \\ y & = - (x + 2)^2 + 2 \\ \\ \text{Maximum curve } & (\cap) \text{ with turning point } (-2, 2) \end{align}

$$ \therefore \text{Curves will not intersect} $$

(a)

\begin{align} \cos 75^\circ & = \cos (30^\circ + 45^\circ) \\ & = \cos 30^\circ \cos 45^\circ - \sin 30^\circ \sin 45^\circ \phantom{000000} [\text{Formula for } \cos (A + B)] \\ & = \left(\sqrt{3} \over 2\right)\left(1 \over \sqrt{2}\right) - \left(1 \over 2\right)\left(1 \over \sqrt{2}\right) \\ & = {\sqrt{3} \over 2 \sqrt{2}} - {1 \over 2 \sqrt{2}} \\ & = {\sqrt{3} - 1 \over 2 \sqrt{2}} \phantom{00} \text{ (Shown)} \end{align}

(b)

\begin{align} \sec^2 75^\circ & = (\sec 75^\circ)^2 \\ & = \left(1 \over \cos 75^\circ \right)^2 \\ & = \left(1 \over {\sqrt{3} - 1 \over 2 \sqrt{2}} \right)^2 \phantom{000000} [\text{Use result from (a)}] \\ & = \left(2\sqrt{2} \over \sqrt{3} - 1 \right)^2 \\ & = {8 \over (\sqrt{3} - 1)^2 } \\ & = {8 \over (\sqrt{3})^2 - 2(\sqrt{3})(1) + (1)^2 } \phantom{000000} [(a - b)^2 = a^2 - 2ab + b^2] \\ & = {8 \over 3 - 2 \sqrt{3} + 1 } \\ & = {8 \over 4 - 2\sqrt{3}} \\ & = {8 \over 4 - 2\sqrt{3}} \times {4 + 2 \sqrt{3} \over 4 + 2 \sqrt{3}} \phantom{000000000} [\text{Rationalise denominator}] \\ & = {32 + 16\sqrt{3} \over (4 - 2\sqrt{3})(4 + 2\sqrt{3})} \\ & = {32 + 16\sqrt{3} \over (4)^2 - (2 \sqrt{3})^2} \phantom{000000000000000} [(a + b)(a - b) = a^2 - b^2] \\ & = {32 + 16\sqrt{3} \over 16 - 12} \\ & = {32 + 16\sqrt{3} \over 4} \\ & = {32 \over 4} + {16\sqrt{3} \over 4} \\ & = 8 + 4 \sqrt{3} \end{align}

(a)

\begin{align} \text{Triangles } ABC & \text{ and } RQC \text{ are similar} \\ \\ {AB \over RQ} & = {AC \over RC} \\ {12 \over y} & = {16 \over 16 - x} \\ 12(16 - x) & = 16y \\ {12(16 - x) \over 16} & = y \\ {3 \over 4}(16 - x) & = y \\ 12 - {3 \over 4}x & = y \\ \\ \therefore y & = 12 - {3x \over 4} \phantom{00} \text{ (Shown)} \end{align}

(b)

\begin{align} \text{Area of plot, } A & = AR \times AP \\ A & = xy \\ \\ \text{Since } & y = 12 - {3 \over 4}x, \\ A & = x \left(12 - {3 \over 4}x\right) \\ A & = 12x - {3 \over 4}x^2 \\ \\ {dA \over dx} & = 12 - {3 \over 4}(2)x \\ {dA \over dx} & = 12 - {3 \over 2}x \\ \\ \text{Let } & {dA \over dx} = 0 , \\ 0 & = 12 - {3 \over 2}x \\ {3 \over 2}x & = 12 \\ 3x & = 24 \\ x & = 8 \\ \\ \text{Substitute } & x = 8 \text{ into } A = 12x - {3 \over 4}x^2, \\ A & = 12(8) - {3 \over 4}(8)^2 \\ A & = 48 \\ \\ {d^2 A \over dx^2} & = - {3 \over 2} < 0 \\ \\ \\ \therefore \text{Largest } & \text{area of plot is } 48 \text{ m}^2 \end{align}

(a)

\begin{align} \text{Let } f(x) & = 2x^3 - x^2 + ax + b \\ \\ f(2) & = 2(2)^3 - (2)^2 + a(2) + b \\ 0 & = 16 - 4 + 2a + b \phantom{000000000.} [x - 2 \text{ is a factor}] \\ 0 & = 12 + 2a + b \\ -2a - 12 & = b \phantom{00} \text{--- (1)} \\ \\ f(-2) & = 2(-2)^3 - (-2)^2 + a(-2) + b \\ 12 & = -16 - 4 - 2a + b \phantom{000000000} [\text{Remainder of 12 when divided by } x + 2] \\ 12 & = -20 - 2a + b \\ 32 & = -2a + b \phantom{00} \text{--- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ 32 & = -2a + (-2a - 12) \\ 32 & = -2a - 2a - 12 \\ 44 & = -4a \\ {44 \over -4} & = a \\ -11 & = a \\ \\ \text{Substitute } & a = -11 \text{ into (1),} \\ b & = -2(-11) - 12 \\ b & = 10 \\ \\ \\ \therefore a & = -11, b = 10 \end{align}

(b) From (a), we know that x - 2 is a factor of the polynomial

\begin{align} f(x) & = 2x^3 - x^2 + ax + b \\ & = 2x^3 - x^2 + (-11)x + 10 \\ & = 2x^3 - x^2 - 11x + 10 \end{align}
$$ \require{enclose} \begin{array}{rll} 2x^2 + 3x - 5 \phantom{0000000.}\\ x - 2 \enclose{longdiv}{2x^3 - x^2 - 11x + 10\phantom{0}}\kern-.2ex \\ -\underline{(2x^3 - 4x^2){\phantom{000000000}}} \\ 3x^2 - 11x + 10 \phantom{0} \\ -\underline{(3x^2 - 6x){\phantom{00000.}}} \\ -5x + 10 \phantom{0} \\ -\underline{(-5x + 10){\phantom{}}} \\ 0 \phantom{.} \end{array} $$
\begin{align} \text{Polynomial} & = \text{Divisor} \times \text{Quotient} + \text{Remainder} \\ 2x^3 - x^2 - 11x + 10 & = (x - 2)(2x^2 + 3x - 5) + 0 \\ 2x^3 - x^2 - 11x + 10 & = (x - 2)(x - 1)(2x + 5) \\ \\ \therefore (x - 2)(x - 1)(2x + 5) & = 0 \end{align}
\begin{align} x - 2 & = 0 &&& x - 1 & = 0 &&& 2x + 5 & = 0 \\ x & = 2 &&& x & = 1 &&& 2x & = -5 \\ & &&& & &&& x & = -{5 \over 2} \end{align}

(a) In other words, we need to show that the CA is the angle bisector of angle BCE

Note: The alternate segment theorem is also known as the tangent-chord theorem

\begin{align} \angle ECA & = \angle TAE \phantom{00} \text{ (Alternate segment theorem)} \\ & = \theta \\ \\ \angle BAC & = \theta \phantom{00} \text{ (Alternate angles, } BA \phantom{.} // \phantom{.} CEE) \\ \\ \angle BCA & = \theta \phantom{00} \text{ (Isosceles triangle } BAC, AB = BC) \\ \\ \text{Since } \angle BCA & = \angle ECA, CA \text{ bisects } BCE \end{align}

(b)

\begin{align} \text{Let the left} & \text{most point of tangent } AT \text{ be } L \\ \\ \angle BAL & = \angle BCA \phantom{00} \text{ (Alternate segment theorem)} \\ & = \theta \\ \\ \angle CAE & = 180^\circ - \theta - \theta - \theta \phantom{00} \text{ (Adjacent angles on a straight line)} \\ & = 180^\circ - 3 \theta \\ \\ \angle CDE & = 180^\circ - (180^\circ - 3 \theta) \phantom{00} \text{ (Angles in opposite segment)} \\ & = 180^\circ - 180^\circ + 3 \theta \\ & = 3 \theta \phantom{00} \text{ (Shown)} \end{align}

(a) The identity used (in the third last step) is provided in the formula sheet

\begin{align} \text{L.H.S} & = (\text{cosec } x - \cot x)(\sec x + 1) \\ & = \left( {1 \over \sin x} - {\cos x \over \sin x} \right) \left({1 \over \cos x} + {\cos x \over \cos x} \right) \\ & = \left( 1 - \cos x \over \sin x \right) \left(1 + \cos x \over \cos x\right) \\ & = { (1 - \cos x)(1 + \cos x) \over \sin x \cos x} \\ & = { (1)^2 - (\cos x)^2 \over \sin x \cos x} \phantom{000000} [(a - b)(a + b) = a^2 - b^2] \\ & = {1 - \cos^2 x \over \sin x \cos x} \\ & = {\sin^2 x \over \sin x \cos x} \phantom{0000000000} [\sin^2 A + \cos^2 A = 1 \rightarrow \sin^2 A = 1 - \cos^2 A] \\ & = {\sin x \over \cos x} \\ & = \tan x \\ & = \text{R.H.S} \end{align}

(b) Note 0° < x < 180°

\begin{align} (\text{cosec } x - \cot x)(\sec x + 1) & = 4 \cot x \\ \tan x & = 4 \cot x \phantom{000000} [\text{Use identity from (a)}] \\ \tan x & = {4 \over \tan x} \\ \tan^2 x & = 4 \\ \tan x & = \pm \sqrt{4} \\ \tan x & = \pm 2 \phantom{000000} [\text{All 4 quadrants}] \\ \\ \text{Basic angle, } \alpha & = \tan^{-1} (2) \\ & = 63.435^\circ \end{align}

\begin{align} x & = 63.435^\circ, 180^\circ - 63.435^\circ, 180^\circ + 63.435^\circ, 360^\circ - 63.435^\circ \\ & = 63.435^\circ, 116.565^\circ, 243.435^\circ \text{ (Rej)}, 296.565^\circ \text{ (Rej)} \\ & \approx 63.4^\circ, 116.6^\circ \end{align}

(c)

\begin{align} (\text{cosec } x - \cot x)(\sec x + 1) & = \tan 2x \\ \tan x & = \tan 2x\phantom{000000000} [\text{Use identity from (a)}] \\ \tan x & = {2 \tan x \over 1 - \tan^2 x} \phantom{00000} [\text{Double angle formula}] \\ \tan x (1 - \tan^2 x) & = 2 \tan x \\ \tan x - \tan^3 x & = 2 \tan x \\ 0 & = \tan^3 x + \tan x \\ 0 & = \tan x (\tan^2 x + 1) \\ \\ \tan x = 0 \phantom{0} & \text{ or } \phantom{0} \tan^2 x + 1 = 0 \\ \\ \\ \tan x & = 0 \end{align}

\begin{align} \text{No possible values of } & x \phantom{00} (\text{Since } 0^\circ < x < 180^\circ) \\ \\ \\ \tan^2 x + 1 & = 0 \\ \tan^2 x & = -1 \\ \tan x & = \pm \sqrt{-1} \phantom{00} \text{ (No real roots)} \\ \\ \\ \therefore \text{Equation has no } & \text{solutions for } 0^\circ < x < 180^\circ \end{align}

(a)

\begin{align} v & = 0.1t^2 + pt + q \\ \\ \text{When } & t = 0 \text{ and } v = 5, \phantom{000000} [\text{At } A, \text{ speed is 5 m/s}] \\ 5 & = 0.1(0)^2 + p(0) + q \\ 5 & = 0 + 0 + q \\ 5 & = q \phantom{00} \text{ (Shown)} \\ \\ \\ v & = 0.1t^2 + pt + 5 \\ \\ \text{When } & t = 10 \text{ and } v = 20, \\ 20 & = 0.1(10)^2 + p(10) + 5 \\ 20 & = 10 + 10p + 5 \\ 20 & = 15 + 10p \\ 5 & = 10p \\ {5 \over 10} & = p \\ {1 \over 2} & = p \end{align}

(b)

\begin{align} v & = 0.1t^2 + pt + 5 \\ v & = 0.1t^2 + 0.5t + 5 \\ \\ a & = {dv \over dt} \\ & = {d \over dt} (0.1t^2 + 0.5t + 5) \\ & = 0.1 (2) t + 0.5 \\ & = 0.2t + 0.5 \\ \\ \text{Substitute } & v = 11.6 \text{ into } v = 0.1t^2 + 0.5t + 5, \phantom{000000} [\text{To find } t] \\ 11.6 & = 0.1t^2 + 0.5t + 5 \\ 0 & = 0.1t^2 + 0.5t - 6.6 \\ 0 & = t^2 + 5t - 66 \phantom{00000000} [\text{Multiply every term by 10}] \\ 0 & = (t - 6)(t + 11) \end{align} \begin{align} t - 6 & = 0 && \text{ or } & t + 11 & = 0 \\ t & = 6 &&& t & = -11 \text{ (N.A.)} \end{align}
\begin{align} \text{Substitute } & t = 6 \text{ into } a = 0.2t + 0.5, \\ a & = 0.2(6) + 0.5 \\ a & = 1.7 \text{ m/s}^2 \end{align}

(c)

\begin{align} s & = \int v \phantom{.} dt \\ s & = \int 0.1t^2 + 0.5t + 5 \phantom{.} dt \\ s & = 0.1 \left(t^3 \over 3\right) + 0.5 \left(t^2 \over 2\right) + 5t + c \\ s & = {1 \over 30} t^3 + {1 \over 4} t^2 + 5t + c \\ \\ \text{When } & t = 0 \text{ and } s = 0, \phantom{000000} [\text{Start from } A] \\ 0 & = {1 \over 30}(0)^3 + {1 \over 4}(0)^2 + 5(0) + c \\ 0 & = 0 + 0 + 0 + c \\ 0 & = c \\ \\ s & = {1 \over 30}t^3 + {1 \over 4}t^2 + 5t \\ \\ \text{When } & t = 10, \\ s & = {1 \over 30}(10)^3 + {1 \over 4}(10)^2 + 5(10) \\ & = 108{1 \over 3} \\ \\ \text{Distance } & AB = 108{1 \over 3} \text{ m} \end{align}

$$ \require{enclose} \begin{array}{rll} 2 \phantom{000000000.}\\ x^3 + 4x \enclose{longdiv}{2x^3 + 0x - 8\phantom{0}}\kern-.2ex \\ -\underline{(2x^3 + 8x){\phantom{000.}}} \\ -8x - 8\phantom{0} \end{array} $$
\begin{align} { \text{Polynomial} \over \text{Divisor} } & = \text{Quotient} + {\text{Remainder} \over \text{Divisor}} \\ {2x^3 - 8 \over x^3 + 4x} & = 2 + {-8x - 8 \over x^3 + 4x} \\ \\ \\ {-8x - 8 \over x^3 + 4x} & = {-8x - 8 \over x(x^2 + 4)} \\ & = {A \over x} + {Bx + C \over x^2 + 4} \\ & = {A(x^2 + 4) \over x(x^2 + 4)} + {x(Bx + C) \over x(x^2 + 4)} \\ & = {A(x^2 + 4) + x(Bx + C) \over x (x^2 + 4) } \\ \\ -8x - 8 & = A(x^2 + 4) + x(Bx + C) \\ \\ \text{Let } & x = 0, \\ 0 - 8 & = A(0^2 + 4) + 0 \\ -8 & = A(4) \\ {-8 \over 4} & = A \\ -2 & = A \\ \\ -8x - 8 & = -2(x^2 + 4) + x(Bx + C) \\ -8x - 8 & = - 2x^2 - 8 + Bx^2 + Cx \\ -8x - 8 & = Bx^2 - 2x^2 + Cx - 8 \\ -8x - 8 & = (B - 2)x^2 + Cx - 8 \\ \\ \text{Comparing } & \text{coefficients of } x^2, \\ 0 & = B - 2 \\ 2 & = B \\ \\ \text{Comparing } & \text{coefficients of } x, \\ -8 & = C \\ \\ \\ {-8x - 8 \over x^3 + 4x} & = {-2 \over x} + {2x + (-8) \over x^2 + 4} \\ & = -{2 \over x} + {2x - 8 \over x^2 + 4} \\ \\ \\ \therefore {2x^3 - 8 \over x^3 + 4x} & = 2 - {2 \over x} + {2x - 8 \over x^2 + 4} \end{align}

(a)

\begin{align} y x^n & = k \\ \\ \text{Take } \lg & \text{ of both sides,} \\ \lg yx^n & = \lg k \\ \lg y + \lg x^n & = \lg k \phantom{0000000000} [\text{Product law (logarithms)}] \\ \lg y + n \lg x & = \lg k \phantom{0000000000} [\text{Power law (logarithms)}] \\ \lg y & = - n \lg x + \lg k \\ \\ Y & = \lg y \\ m & = - n \\ X & = \lg x \\ c & = \lg k \\ \\ \\ \text{Plot } & \lg y \text{ against } \lg x \\ \\ \text{Gradient} & = -n \\ \\ \text{Vertical intercept, } c & = \lg k \\ c & = \log_{10} k \\ 10^c & = k \end{align}

(b)(i)

\begin{align} t & \propto \sqrt{l} \\ t & = k \sqrt{l} \\ \\ \text{Plot } & t \text{ against } \sqrt{l} \\ \\ \text{Gradient} & = k \end{align}

√l	0.447	0.632	0.774	1
t	0.90	1.27	1.55	2.02

(b)(ii)

\begin{align} \sqrt{l} & = \sqrt{0.8} \\ & = 0.894 \\ \\ \text{From graph, } & \text{when } \sqrt{l} = 0.984, \\ t & = 1.8 \text{ s} \end{align}

(b)(iii)

\begin{align} t & = 2 \pi \sqrt{l \over g} \\ t & = 2 \pi { \sqrt{l} \over \sqrt{g}} \\ t & = {2\pi \over \sqrt{g}} \sqrt{l} \\ \\ \text{From (b)(ii), } t & = k \sqrt{l} \\ \\ \therefore \text{Gradient} & = k \\ {2\pi \over \sqrt{g}} & = {1.8125 - 0.4 \over 0.9 - 0.2} \\ {2\pi \over \sqrt{g}} & = 2.01785 \\ 2\pi & = 2.01785 \sqrt{g} \\ \\ \sqrt{g} & = {2\pi \over 2.01785} \\ g & = \left(2\pi \over 2.01785\right)^2 \\ g & = 9.6956 \\ g & \approx 9.70 \end{align}

(a)

\begin{align} y & = 2x - {16 \over x^2} \\ \\ \text{When } & y = 0, \\ 0 & = 2x - {16 \over x^2} \\ {16 \over x^2} & = 2x \\ 16 & = x^2 (2x) \\ 16 & = 2x^3 \\ {16 \over 2} & = x^3 \\ 8 & = x^3 \\ \sqrt[3]{8} & = x \\ 2 & = x \\ \\ \therefore & \phantom{.} P(2, 0) \\ \\ \\ y & = 2x - 16x^{-2} \\ \\ {dy \over dx} & = 2 - 16(-2)x^{-3} \\ & = 2 + 32 x^{-3} \\ & = 2 + {32 \over x^3} \\ \\ \text{When } & x = 2, \\ {dy \over dx} & = 2 + {32 \over (2)^3} \\ & = 6 \\ \\ \text{Gradient of} & \text{ tangent at } P = 6 \\ \\ y & = mx + c \\ y & = 6x + c \\ \\ \text{Using } & P(2, 0), \\ 0 & = 6(2) + c \\ 0 & = 12 + c \\ -12 & = c \\ \\ \text{Eqn of line } & PQ : y = 6x - 12 \end{align}

(b)

\begin{align} \text{Eqn of line } & PQ : y = 6x - 12 \\ \\ \text{Substitute } & x = 4, \\ y & = 6(4) - 12 \\ y & = 12 \\ \\ \therefore & \phantom{.} Q(4, 12) \\ \\ \\ \text{Area bounded by tangent} & = \text{Area of triangle} \\ & = {1 \over 2} \times (4 - 2) \times 12 \\ & = 12 \text{ units}^2 \\ \\ \text{Area under curve} & = \int_2^4 2x - 16x^{-2} \phantom{.} dx \\ & = \left[ 2 \left(x^2 \over 2\right) - 16 \left(x^{-1} \over -1\right) \right]_2^4 \\ & = \left[ x^2 + 16x^{-1} \right]_2^4 \\ & = \left[ x^2 + {16 \over x} \right]_2^4 \\ & = \left[ (4)^2 + {16 \over 4} \right] - \left[ (2)^2 + {16 \over 2} \right] \\ & = 8 \text{ units}^2 \\ \\ \text{Area of shaded region} & = 12 - 8 \\ & = 4 \text{ units}^2 \end{align}

(a)

\begin{align} y & = {2 \over x} + k \phantom{00} \text{--- (1)} \\ \\ 2x + 3y & = k \phantom{00} \text{--- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ 2x + 3 \left({2 \over x} + k\right) & = k \\ 2x + {6 \over x} + 3k & = k \\ 2x + {6 \over x} + 2k & = 0 \\ x \left(2x + {6 \over x} + 2k\right) & = x(0) \\ 2x^2 + 6 + 2kx & = 0 \\ 2x^2 + 2kx + 6 & = 0 \\ x^2 + kx + 3 & = 0 \\ \\ b^2 - 4ac & = (k)^2 - 4(1)(3) \\ & = k^2 - 12 \\ \\ b^2 - 4ac & < 0 \phantom{000000} [\text{No real roots since there's no intersection}] \\ k^2 - 12 & < 0 \\ k^2 - (\sqrt{12})^2 & < 0 \\ (k + \sqrt{12})(k - \sqrt{12}) & < 0 \phantom{000000} [a^2 - b^2 = (a + b)(a - b)] \end{align}

\begin{align} -\sqrt{12} & \phantom{.} < k < \sqrt{12} \phantom{000000} [\text{This is range of values}] \\ \\ \\ \therefore \{ k : k \in \mathbb{R} & \text{ and } -\sqrt{12} < k < \sqrt{12} \} \end{align}

(b)(i) Two conditions: (1) a < 0 (maximum curve ∩) and (2) b² - 4ac < 0

\begin{align} \text{Discriminant} & = (b)^2 - 4(a)(a) \\ & = b^2 - 4a^2 \\ \\ \\ \text{Condition #1: } & a < 0 \\ \text{Condition #2: } & b^2 - 4a^2 < 0 \end{align}

(b)(ii)

\begin{align} a & = -1, b = -1 \end{align}

(a)

$$ \require{enclose} \begin{array}{rll} 1 \phantom{0000}\\ 2x + 3 \enclose{longdiv}{2x\phantom{0000}}\kern-.2ex \\ -\underline{(2x + 3){\phantom{ }}} \\ -3 \phantom{.} \end{array} $$
\begin{align} { \text{Polynomial} \over \text{Divisor} } & = \text{Quotient} + {\text{Remainder} \over \text{Divisor}} \\ {2x \over 2x + 3} & = 1 + {-3 \over 2x + 3} \\ & = 1 - {3 \over 2x + 3} \\ \\ \\ \int {2x \over 2x + 3} \phantom{.} dx & = \int 1 - {3 \over 2x + 3} \phantom{.} dx \\ & = \int 1 \phantom{.} dx - \int {3 \over 2x + 3} \phantom{.} dx \\ & = x - 3 \int {1 \over 2x + 3} \phantom{.} dx \\ & = x - 3 \left[ \ln (2x + 3) \over 2\right] \phantom{000000000} \text{Use } \int {1 \over f(x)} \phantom{.} dx = {\ln [f(x)] \over f'(x)} \\ & = x - {3 \over 2} \ln (2x + 3) + c \end{align}

(b)

\begin{align} u & = x \phantom{0000000000000} {du \over dx} = 1 \\ v & = \ln (2x + 3) \phantom{000000} {dv \over dx} = {2 \over 2x + 3} \\ \\ \\ {dy \over dx} & = x \left(2 \over 2x + 3\right) + [\ln (2x + 3)](1) \phantom{000000} [\text{Product rule}] \\ & = {2x \over 2x + 3} + \ln (2x + 3) \end{align}

(c)

\begin{align} \text{From (i), } \int {2x \over 2x + 3} \phantom{.} dx & = x - {3 \over 2} \ln (2x + 3) \\ \\ \\ \text{From (ii), } {d \over dx} [x \ln (2x + 3)] & = {2x \over 2x + 3} + \ln (2x + 3) \\ \\ \implies \int {2x \over 2x + 3} + \ln (2x + 3) \phantom{.} dx & = x \ln (2x + 3) \\ \int {2x \over 2x + 3} \phantom{.} dx + \int \ln (2x + 3) \phantom{.} dx & = x \ln (2x + 3) \\ \int \ln (2x + 3) \phantom{.} dx & = x \ln (2x + 3) - \underbrace{\int {2x \over 2x + 3} \phantom{.} dx}_\text{Use part (i) result} \\ \int \ln (2x + 3) \phantom{.} dx & = x \ln (2x + 3) - \left[x - {3 \over 2}\ln (2x + 3) \right] \\ \int \ln (2x + 3) \phantom{.} dx & = x \ln (2x + 3) - x + {3 \over 2} \ln (2x + 3) + c \end{align}

(a)

\begin{align} \sin \angle BAE & = {BE \over BA} \\ \sin \theta & = {BE \over 8} \\ 8 \sin \theta & = BE \\ \\ \\ \angle ABE & = 180^\circ - 90^\circ - \theta^\circ \phantom{000000} [\text{Angle sum of triangle}] \\ & = (90 - \theta)^\circ \\ \\ \angle FBC & = 90^\circ - (90 - \theta)^\circ \\ & = 90^\circ - 90^\circ + \theta^\circ \\ & = \theta^\circ \\ \\ \cos \angle CBF & = {BF \over BC} \\ \cos \theta & = {BF \over 5} \\ 5 \cos \theta & = BF \\ \\ \\ x & = EF \\ & = BE - BF \\ & = 8 \sin \theta - 5 \cos \theta \phantom{00} \text{ (Shown)} \end{align}

(b)

\begin{align} a \sin \theta - b \cos \theta & = R \sin (\theta - \alpha) \\ \\ a & = 8, b = 5 \\ \\ R & = \sqrt{a^2 + b^2} \\ & = \sqrt{8^2 + 5^2} \\ & = \sqrt{89} \\ \\ \alpha & = \tan^{-1} \left(b \over a\right) \\ & = \tan^{-1} \left(5 \over 8\right) \\ & = 32.00^\circ \\ \\ \\ 8 \sin \theta - 5 \cos \theta & = \sqrt{89} \sin (\theta - 32^\circ) \end{align}

(c)

\begin{align} x & = 8 \sin \theta - 5 \cos \theta \\ x & = \sqrt{89} \sin (\theta - 32^\circ) \\ \\ \text{When } & x = 2, \\ 2 & = \sqrt{89} \sin (\theta - 32^\circ) \\ {2 \over \sqrt{89}} & = \sin (\theta - 32^\circ) \phantom{00000} [\text{1st or 2nd quadrant}] \\ \\ \text{Basic angle, } \alpha & = \sin^{-1} \left(2 \over \sqrt{89}\right) \\ & = 12.24^\circ \end{align}

\begin{align} \theta - 32^\circ & = 12.24^\circ, 180^\circ - 12.24^\circ \\ & = 12.24^\circ, 167.76^\circ \\ \\ \theta & = 44.24^\circ, 199.76^\circ \text{ (Reject, since } 0^\circ < \theta < 90^\circ) \\ & \approx 44.2^\circ \end{align}

(a)

\begin{align} y & = e^{x^2 - 4x} \\ \\ {dy \over dx} & = (2x - 4)e^{x^2 - 4x} \phantom{000000000000000} \left[{d \over dx}(e^{f(x)}) = f'(x) . e^{f(x)}\right] \\ \\ \\ u & = 2x - 4 \phantom{00000000} {du \over dx} = 2 \\ v & = e^{x^2 - 4x} \phantom{0000000.-} {dv \over dx} = (2x - 4)e^{x^2 - 4x} \\ \\ {d^2 y \over dx^2} & = (2x - 4)[(2x - 4)e^{x^2 - 4x}] + (e^{x^2 - 4x}) (2) \phantom{000000} [\text{Product rule}] \\ & = (2x - 4)^2 e^{x^2 - 4x} + 2 e^{x^2 - 4x} \\ & = e^{x^2 - 4x} [ (2x - 4)^2 + 2 ] \\ & = e^{x^2 - 4x} [ (2x)^2 - 2(2x)(4) + (4)^2 + 2 ] \\ & = e^{x^2 - 4x} (4x^2 - 16x + 16 + 2) \\ & = e^{x^2 - 4x} (4x^2 - 16x + 18) \end{align}

(b)

\begin{align} \text{Let } & {dy \over dx} = 0, \phantom{000000} [\text{Stationary point}] \\ 0 & = (2x - 4)e^{x^2 - 4x} \end{align} \begin{align} 2x - 4 & = 0 && \text{ or } & e^{x^2 - 4x} = 0 \text{ (Reject, since } e^{x^2 - 4x} > 0) \\ 2x & = 4 \\ x & = 2 \\ \\ \text{Substitute } & x = 2 \text{ into eqn of curve,} \\ y & = e^{(2)^2 - 4(2)} \\ y & = e^{-4} \\ y & = {1 \over e^4} \\ \\ \therefore \text{Stationary } & \text{point: } \left(2, {1 \over e^4}\right) \end{align}

(c)

\begin{align} \text{Substitute } & x = 2 \text{ into } {d^2 y \over dx^2}, \phantom{000000} [\text{Second derivative test}] \\ {d^2y \over dx^2} & = e^{(2)^2 - 4(2)} [4(2)^2 - 16(2) + 18] \\ & = e^{-4} (2) \\ & = {2 \over e^4} > 0 \\ \\ \therefore \left(2, {1 \over e^4}\right) & \text{ is a minimum point} \end{align}

(a)

\begin{align} \text{Midpoint of } PQ & = \left({-2 + 6 \over 2}, {7 + 1 \over 2} \right) \\ & = (2, 4) \\ \\ \text{Gradient of } PQ & = {7 - 1 \over -2 - 6} \\ & = -{3 \over 4} \\ \\ \text{Gradient of } \perp \text{ bisector} & = {-1 \over -{3 \over 4}} \phantom{000000} [m_1 \times m_2 = -1] \\ & = {4 \over 3} \\ \\ y & = mx + c \\ y & = {4 \over 3}x + c \\ \\ \text{Using } & \text{midpoint } (2, 4), \\ 4 & = {4 \over 3}(2) + c \\ 4 & = {8 \over 3} + c \\ {4 \over 3} & = c \\ \\ \text{Eqn of perp. bisector: } & y = {4 \over 3}x + {4 \over 3} \end{align}

(b) Circle property: Perpendicular bisector of chord PQ (found in part a) will pass through the centre

Both lines will meet at the center of the circle

\begin{align} \text{Eqn of perp. bisector of } PQ: & y = {4 \over 3}x + {4 \over 3} \phantom{00} \text{--- (1)} \\ \\ & y = x + 4 \phantom{00} \text{--- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ {4 \over 3}x + {4 \over 3} & = x + 4 \\ {1 \over 3}x & = {8 \over 3} \\ x & = 8 \\ \\ \text{Substitute } & x = 8 \text{ into (2),} \\ y & = 8 + 4 \\ y & = 12 \\ \\ \therefore \text{Center, } C & (8, 12) \\ \\ \\ \text{Radius} & = \text{Length of } QC \\ & = \sqrt{ (8 - 6)^2 + (12 - 1)^2 } \\ & = \sqrt{125} \\ \\ \\ (x - 8)^2 & + (y - 12)^2 = (\sqrt{125})^2 \\ \text{Eqn of circle: } (x - 8)^2 & + (y - 12)^2 = 125 \end{align}

(c)

Since PR is the diameter, C is the midpoint of PR

\begin{align} \text{Let coord} & \text{inates of } R \text{ be } (a, b) \\ \\ \text{Midpoint of } PR & = \left({-2 + a \over 2}, {7 + b \over 2} \right) \\ \\ \text{Centre, } C & = \text{Midpoint of } PR \\ (8, 12) & = \left({-2 + a \over 2}, {7 + b \over 2} \right) \end{align} \begin{align} 8 & = {-2 + a \over 2} && \text{ or } & 12 & = {7 + b \over 2} \\ 2(8) & = -2 + a &&& 2(12) & = 7 + b \\ 16 & = -2 + a &&& 24 & = 7 + b \\ 18 & = a &&& 17 & = b \end{align}
$$ \therefore R(18, 17) $$

(a)

\begin{align} 2^{2x} + 2^{x + 2} & = 5 \\ 2^{2x} + (2^x)(2^2) & = 5 \phantom{000000} [a^{m + n} = a^m \times a^n] \\ (2^x)^2 + 4(2^x) & = 5 \\ \\ \text{Let } & u = 2^x, \\ u^2 + 4u & = 5 \\ u^2 + 4u - 5 & = 0 \\ (u + 5)(u - 1) & = 0 \end{align} \begin{align} u - 5 & = 0 && \text{ or } & u + 1 & = 0 \\ u & = 5 &&& u & = -1 \\ \\ 2^x & = 5 &&& 2^x & = - 1 \text{ (Rej, since } 2^x > 0) \\ \ln 2^x & = \ln 5 \\ [\text{Power law}] \phantom{00000} x \ln 2 & = \ln 5 \\ x & = {\ln 5 \over \ln 2} \\ x & \approx 2.32 \end{align}

(b)

\begin{align} \log_2 x + \log_8 x & = \log_5 25 \\ \log_2 x + {\log_2 x \over \log_2 8} & = \log_5 25 \phantom{0000000} [\text{Change-of-base}] \\ \log_2 x + {\log_2 x \over \log_2 2^3} & = \log_5 5^2 \\ \log_2 x + {\log_2 x \over 3\log_2 2} & = 2 \log_5 5 \phantom{000000.} [\text{Power law}] \\ \log_2 x + {\log_2 x \over 3(1)} & = 2 (1) \\ \log_2 x + {\log_2 x \over 3} & = 2 \\ 3 \left(\log_2 x + {\log_2 x \over 3}\right) & = 3(2) \\ 3 \log_2 x + \log_2 x & = 6 \\ 4 \log_2 x & = 6 \\ \log_2 x & = {6 \over 4} \\ \log_2 x & = {3 \over 2} \\ x & = 2^{3 \over 2} \\ \\ \therefore k & = {3 \over 2} \end{align}

(c)

\begin{align} \log_3 (4x - 11) - \log_3 (x - 3) & = 1 \\ [\text{Quotient law}] \phantom{000000} \log_3 \left(4x - 11 \over x - 3\right) & = 1 \\ {4x - 11 \over x - 3} & = 3^1 \\ {4x - 11 \over x - 3} & = 3 \\ 4x - 11 & = 3(x - 3) \\ 4x - 11 & = 3x - 9 \\ x & = 2 \\ \\ \text{For } x = 2, \log_3 (4x - 11) \text{ and } & \log_3 (x - 3) \text{ are undefined} \\ \\ \therefore \text{Equation has no real solutions} & \end{align}