\begin{align} {d^2 y \over dx^2} & = 8 - 6x \\ \\ {dy \over dx} & = \int 8 - 6x \phantom{.} dx \\ & = 8x - 6 \left(x^2 \over 2\right) + c \\ & = 8x - 3x^2 + c \\ \\ \text{When } & x = 2 \text{ and } {dy \over dx} = 3, \phantom{000000} [\text{Gradient at } P (2, 8) \text{ is } 3] \\ 3 & = 8(2) - 3(2)^2 + c \\ 3 & = 16 - 12 + c \\ 3 & = 4 + c \\ -1 & = c \\ \\ {dy \over dx} & = 8x - 3x^2 - 1 \\ \\ y & = \int 8x - 3x^2 - 1 \phantom{.} dx \\ & = 8 \left(x^2 \over 2\right) - 3 \left(x^3 \over 3\right) - x + d \\ & = 4x^2 - x^3 - x + d \\ \\ \text{Using } & P(2, 8), \\ 8 & = 4(2)^2 - (2)^3 - 2 + d \\ 8 & = 16 - 8 - 2 + d \\ 8 & = 6 + d \\ 2 & = d \\ \\ \text{Eqn of curve: } & y = 4x^2 - x^3 - x + 2 \end{align}

(i)

\begin{align} y & = 4x^{1 \over 2} \\ \\ \text{Let } x = 0, & \phantom{.} y = 4(0)^{1 \over 2} = 0 \\ \\ \text{Let } x = 16, & \phantom{.} y = 4 (16)^{1 \over 2} = 16 \end{align}

(ii)

\begin{align} y & = 4x^{1 \over 2} \phantom{00} \text{--- (1)} \\ \\ 4y & = 7x + 4 \phantom{00} \text{--- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ 4 (4x^{1 \over 2}) & = 7x + 4 \\ 16 x^{1 \over 2} & = 7x + 4 \\ 16 x^{1 \over 2} & = 7 (x^{1 \over 2})^2 + 4 \\ \\ \text{Let } & u = x^{1 \over 2}, \\ 16u & = 7u^2 + 4 \\ 0 & = 7u^2 - 16u + 4 \\ 0 & = (7u - 2)(u - 2) \end{align}
\begin{align} 7u - 2 & =0 && \text{ or } & u - 2 & =0 \\ 7u & = 2 &&& u & = 2 \\ u & = {2 \over 7} \\ \\ x^{1 \over 2} & = {2 \over 7} &&& x^{1 \over 2} & = 2 \\ \sqrt{x} & = {2 \over 7} &&& \sqrt{x} & = 2 \\ x & = \left(2 \over 7\right)^2 &&& x & = 2^2 \\ x & = {4 \over 49} &&& x & = 4 \\ \\ \text{Substitute } & \text{into (1),} &&& \text{Substitute } & \text{into (1),} \\ y & = 4 \left(4 \over 49\right)^{1 \over 2} &&& y & = 4(4)^{1 \over 2} \\ y & = {8 \over 7} &&& y & = 8 \\ \\ \therefore & \phantom{.} \left({4 \over 49}, {8 \over 7}\right) &&& \therefore & \phantom{.} (4, 8) \end{align}

\begin{align} \text{Sum of roots, } {1 \over \alpha} + {1 \over \beta} & = -{b \over a} \\ {\beta \over \alpha \beta} + {\alpha \over \alpha \beta} & = -{-3 \over 7} \\ {\alpha + \beta \over \alpha \beta} & = {3 \over 7} \\ \\ \text{Product of roots, } {1 \over \alpha} \times {1 \over \beta} & = {c \over a} \\ {1 \over \alpha \beta} & = {1 \over 7} \\ 7 & = \alpha \beta \\ \\ \therefore {\alpha + \beta \over 7} & = {3 \over 7} \\ \alpha + \beta & = 3 \\ \\ \\ \text{New SOR, } \alpha^2 + \beta^2 & = (\alpha + \beta)^2 - 2\alpha \beta \\ & = (3)^2 - 2(7) \\ & = -5 \\ \\ \text{New POR, } \alpha^2 \beta^2 & = (\alpha \beta)^2 \\ & = (7)^2 \\ & = 49 \\ \\ \\ x^2 - \text{(SOR)}x + \text{POR} & = 0 \\ x^2 - (-5)x + 49 & = 0 \\ x^2 + 5x + 49 & = 0 \end{align}

(i)

\begin{align} \text{L.H.S} & = { \sec x + \text{cosec } x \over \sec x - \text{cosec } x } \\ & = { {1 \over \cos x} + {1 \over \sin x} \over {1 \over \cos x} - {1 \over \sin x}} \times {\sin x \over \sin x} \\ & = { {\sin x \over \cos x} + {\sin x \over \sin x} \over {\sin x \over \cos x} - {\sin x \over \sin x}} \\ & = { \tan x + 1 \over \tan x - 1} \\ & = \text{R.H.S} \end{align}

(ii)

\begin{align} { \sec x + \text{cosec } x \over \sec x - \text{cosec } x } & = {5 \over 2} \\ { \tan x + 1 \over \tan x - 1} & = {5 \over 2} \phantom{000000} [\text{Use identity from (i)}] \\ 2 (\tan x + 1) & = 5(\tan x - 1) \\ 2 \tan x + 2 & = 5 \tan x - 5 \\ -3 \tan x & = -7 \\ \tan x & = {-7 \over -3} \\ \tan x & = {7 \over 3} \phantom{000000} [\text{1st or 3rd quadrant}] \\ \\ \text{Basic angle, } \alpha & = \tan^{-1} \left(7 \over 3\right) \phantom{000000} [\text{Radian mode!}] \\ & = 1.1659 \end{align}

\begin{align} x & = 1.1659, \pi + 1.1659 \\ & = 1.1659, 4.3075 \\ & \approx 1.17, 4.31 \end{align}

(i)

\begin{align} \text{Total area, } A & = 3(x \times y) \\ & = 3xy \phantom{000000} [\text{Need to find relationship between } y \text{ and } x ] \\ \\ \text{Total length of wire netting} & = 4y + 6x \\ 4y + 6x & = 288 \\ 4y & = 288 - 6x \\ y & = {1 \over 4}(288 - 6x) \\ y & = 72 - {3 \over 2}x \\ \\ \therefore A & = 3x \left(72 - {3 \over 2}x\right) \\ & = 216x - {9 \over 2}x^2 \phantom{00} \text{(Shown)} \end{align}

(ii)

\begin{align} A & = 216x - {9 \over 2}x^2 \\ \\ {dA \over dx} & = 216 - {9 \over 2}(2)(x) \\ & = 216 - 9x \\ \\ \text{Let } & {dA \over dx} = 0, \phantom{000000} [\text{Maximum value of } A] \\ 0 & = 216 - 9x \\ 9x & = 216 \\ x & = {216 \over 9} \\ x & = 24 \\ \\ \text{From (i), } y & = 72 - {3 \over 2}(24) \\ y & = 72 - 36 \\ y & = 36 \\ \\ \text{Dimensions: } & 36 \text{ m} \text{ by 24 m} \end{align}

(i)

\begin{align} \text{Area of triangle} & = {1 \over 2} ab \sin C \\ {1 \over 4}(9 + \sqrt{3}) & = {1 \over 2} (\sqrt{3} + 1) (AC) \sin 60^\circ \\ 9 + \sqrt{3} & = 2 (\sqrt{3} + 1) (AC) \left(\sqrt{3} \over 2\right) \phantom{000000} [\text{Special angle } 60^\circ] \\ 9 + \sqrt{3} & = 2 \left(\sqrt{3} \over 2\right) (\sqrt{3} + 1) (AC) \\ 9 + \sqrt{3} & = \sqrt{3} (\sqrt{3} + 1) (AC) \\ 9 + \sqrt{3} & = (3 + \sqrt{3}) (AC) \\ \\ AC & = {9 + \sqrt{3} \over 3 + \sqrt{3}} \times {3 - \sqrt{3} \over 3 - \sqrt{3}} \phantom{000000} [\text{Rationalise denominator}] \\ & = {(9 + \sqrt{3})(3 - \sqrt{3}) \over (3)^2 - (\sqrt{3})^2} \phantom{000000} [(a + b)(a - b) = a^2 - b^2] \\ & = {27 - 9 \sqrt{3} + 3 \sqrt{3} - 3 \over 6} \\ & = {24 - 6 \sqrt{3} \over 6} \\ & = {24 \over 6} - {6 \sqrt{3} \over 6} \\ & = (4 - \sqrt{3}) \text{ cm} \end{align}

(ii)

\begin{align} BC^2 & = AB^2 + AC^2 - 2(AB)(AC) \cos \angle BAC \phantom{00000000000000000} [\text{Cosine rule}] \\ & = (\sqrt{3} + 1)^2 + (4 - \sqrt{3})^2 - 2(\sqrt{3} + 1)(4 - \sqrt{3}) \cos 60^\circ \\ & = \underbrace{ (\sqrt{3})^2 + 2(\sqrt{3})(1) + 1^2 }_{ (a + b)^2 = a^2 + 2ab + b^2 } + \underbrace{ 4^2 - 2(4)(\sqrt{3}) + (\sqrt{3})^2 }_{ (a - b)^2 = a^2 - 2ab + b^2 } - 2 (4 \sqrt{3} - 3 + 4 - \sqrt{3})\left(1 \over 2\right) \\ & = 3 + 2 \sqrt{3} + 1 + 16 - 8 \sqrt{3} + 3 - 2 \left(1 \over 2\right) (3 \sqrt{3} + 1) \\ & = 23 - 6 \sqrt{3} - (3 \sqrt{3} + 1) \\ & = 23 - 6 \sqrt{3} - 3 \sqrt{3} - 1 \\ & = (22 - 9 \sqrt{3}) \text{ cm}^2 \end{align}

(i)

$$ \require{enclose} \begin{array}{rll} x^2 + 9 \phantom{000000}\\ 3x - 1 \enclose{longdiv}{ 3x^3 - x^2 + 27x - 9 \phantom{0}}\kern-.2ex \\ -\underline{( 3x^3 - x^2){\phantom{000000000}}} \\ 27x - 9 \phantom{0} \\ -\underline{( 27x - 9){\phantom{.}}} \\ 0 \phantom{0} \end{array} $$

(ii)

\begin{align} \text{From (i), } 3x^3 - x^2 + 27x - 9 & = (3x - 1)(x^2 + 9) \\ \\ {6 + 11x - 5x^2 \over (3x - 1)(x^2 + 9)} & = {A \over 3x - 1} + {Bx + C \over x^2 + 9} \\ & = {A(x^2 + 9) \over (3x - 1)(x^2 + 9)} + {(Bx + C)(3x - 1) \over (3x - 1)(x^2 + 9)} \\ & = {A(x^2 + 9) + (Bx + C)(3x - 1) \over (3x - 1)(x^2 + 9)} \\ \\ 6 + 11x - 5x^2 & = A(x^2 + 9) + (Bx + C)(3x - 1) \\ \\ \text{Let } & x = {1 \over 3}, \\ 6 + 11 \left(1 \over 3\right) - 5 \left(1 \over 3\right)^2 & = A \left( {82 \over 9} \right) + \left({1 \over 3}B + C\right)(0) \\ {82 \over 9} & = {82 \over 9}A + 0 \\ 1 & = A \\ \\ 6 + 11x - 5x^2 & = x^2 + 9 + (Bx + C)(3x - 1) \\ \\ \text{Let } & x = 0, \\ 6 & = 9 + (0 + C)(-1) \\ 6 & = 9 + (C)(-1) \\ 6 & = 9 - C \\ C & = 9 - 6 \\ C & = 3 \\ \\ 6 + 11x - 5x^2 & = x^2 + 9 + (Bx + 3)(3x - 1) \\ \\ \text{Let } & x = 1, \\ 6 + 11(1) - 5(1)^2 & = (1)^2 + 9 + (B + 3)(2) \\ 12 & = 10 + 2(B + 3) \\ 12 & = 10 + 2B + 6 \\ 12 - 10 - 6 & = 2B \\ -4 & = 2B \\ {-4 \over 2} & = B \\ -2 & = B \\ \\ \\ {6 + 11x - 5x^2 \over (3x - 1)(x^2 + 9)} & = {1 \over 3x - 1} + {-2x + 3 \over x^2 + 9} \\ & = {1 \over 3x - 1} + {3 - 2x \over x^2 + 9} \end{align}

(i)

\begin{align} \text{Acceleration, } a & = {dv \over dt} = 10 \\ \\ v & = \int 10 \phantom{.} dt \\ & = 10t + c \\ \\ \text{When } & t = 0 \text{ and } v = 0, \phantom{000000} [\text{'falls vertically from rest'}] \\ 0 & = 10(0) + c \\ 0 & = c \\ \\ v & = 10t \\ \\ \text{When } & t = 4, \\ v & = 10(4) \\ & = 40 \text{ m/s} \end{align}

(ii)

\begin{align} \text{Displacement, } s & = \int v \phantom{.} dt \\ & = \int 10t \phantom{.} dt \\ & = 10 \left(t^2 \over 2\right) + d \\ & = 5t^2 + d \\ \\ \text{When } & t = 0 \text{ and } s = 0, \phantom{000000} [\text{starts from point } O] \\ 0 & = 5(0)^2 + d \\ 0 & = d \\ \\ s & = 5t^2 \\ \\ \text{When } & t = 4, \\ s & = 5(4)^2 \\ & = 80 \\ \\ \therefore OX & = 80 \text{ m} \end{align}

(iii)

\begin{align} \text{New acceleration, } A & = {dV \over dT} = 10 - kT \\ \\ V & = \int 10 - kT \phantom{.} dT \\ & = 10T - k\left(T^2 \over 2\right) + c \\ & = 10T - {1 \over 2}k T^2 + c \\ \\ \text{When } & T = 0 \text{ and } V = 40, \phantom{000000} [\text{from part i, velocity at } X \text{ is 40 m/s}] \\ 40 & = 10(0) - {1 \over 2}k(0)^2 + c \\ 40 & = c \\ \\ V & = 10T - {1 \over 2} k T^2 + 40 \\ \\ \text{When } & T = 3 \text{ and } V = 0, \\ 0 & = 10(3) - {1 \over 2}k (3)^2 + 40 \\ 0 & = 30 - {9 \over 2}k + 40 \\ {9 \over 2}k & = 30 + 40 \\ {9 \over 2}k & = 70 \\ 9k & = 140 \\ k & = {140 \over 9} \phantom{00} \text{(Shown)} \end{align}

(i)

\begin{align} \angle PBA & = \angle ACB \phantom{0} (\text{Alternate segment theorem}) \\ & = \angle DAC \phantom{0} (\text{Alternate angles, } AD \phantom{.} // \phantom{.} BC) \\ \\ \therefore \angle PBA & = \angle DAC \phantom{0} \text{(Shown)} \end{align}

(ii)

\begin{align} \text{Let } \angle PBA & = \angle ACB = \angle DAC = \theta \\ \\ \angle ACD & = 90^\circ \phantom{0} (\text{Right angle in semicircle)} \\ \\ \angle ADC & = 180^\circ - 90^\circ - \theta \phantom{0} (\text{Angle sum of triangle}) \\ & = 90^\circ - \theta \\ \\ \angle ABC & = 180^\circ - (90^\circ - \theta) \phantom{0} (\text{Angles in opposite segments}) \\ & = 180^\circ - 90^\circ + \theta \\ & = 90^\circ + \theta \\ \\ \angle CBT & = 180^\circ - (90^\circ + \theta) - \theta \phantom{0} (\text{Adjacent angles on a straight line}) \\ & = 180^\circ - 90^\circ - \theta - \theta \\ & = 90^\circ - 2 \theta \\ & = 90^\circ - 2 \times \angle PBA \phantom{0} \text{ (Shown)} \end{align}

(i)

\begin{align} u & = 2x + 1 &&& v & = x - 1 \\ {du \over dx} & = 2 &&& {dv \over dx} & = 1 \end{align}
\begin{align} {dy \over dx} & = {v {du \over dx} - u {dv \over dx} \over v^2} \phantom{00000000000000} [\text{Quotient rule}] \\ & = {(x - 1)(2) - (2x + 1)(1) \over (x - 1)^2} \\ & = {2(x - 1) - (2x + 1) \over (x - 1)^2} \\ & = {2x - 2 - 2x - 1 \over (x - 1)^2} \\ & = {-3 \over (x - 1)^2} \\ \\ \text{For all real }& \text{values of } x \text{ and } x \ne 1, \phantom{0} (x - 1)^2 > 0 \text{ and } {-3 \over (x - 1)^2} < 0 \\ \\ \implies {dy \over dx} & \ne 0 \text{ and the curve has no turning points} \end{align}

(ii)

$$ \require{enclose} \begin{array}{rll} 2 \phantom{0000}\\ x - 1 \enclose{longdiv}{ 2x + 1 \phantom{0}}\kern-.2ex \\ -\underline{( 2x - 2){\phantom{.}}} \\ 3 \phantom{0} \end{array} $$

\begin{align} {2x + 1 \over x - 1} & = 2 + {3 \over x - 1} \\ \\ \\ \text{Area bounded by line and } x \text{-axis} & = \text{Area of trapezium} \\ & = {1 \over 2} \times \text{Sum of parallel sides} \times \text{Height} \\ & = {1 \over 2} \times (5 + 3) \times 2 \\ & = 8 \text{ units}^2 \\ \\ \text{Area bounded by curve and } x \text{-axis} & = \int_2^4 {2x + 1 \over x - 1} \phantom{.} dx \\ & = \int_2^4 2 + {3 \over x - 1} \phantom{.} dx \\ & = \int_2^4 2 + 3 \left(1 \over x - 1\right) \phantom{.} dx \\ & = \left[ 2x + 3 \left( \ln (x - 1) \over 1\right) \right]_2^4 \\ & = \left[ 2x + 3 \ln (x - 1) \right]_2^4 \\ & = [2(4) + 3 \ln (4 - 1)] - [2(2) + 3 \ln (2 - 1)] \\ & = 8 + 3 \ln 3 - (4 + 3 \ln 1) \\ & = 8 + 3 \ln 3 - 4 - 3 \ln 1 \\ & = 8 + 3 \ln 3 - 4 - 0 \\ & = 4 + 3 \ln 3 \\ \\ \text{Area of shaded region} & = 8 - (4 + 3 \ln 3) \\ & = 8 - 4 - 3 \ln 3 \\ & = 4 - 3 \ln 3 \\ & \approx 0.704 \text{ units}^2 \end{align}

(i)

\begin{align} x^2 + y^2 + 8x - 24y + 96 & = 0 \\ x^2 + 8x + y^2 - 24y & = - 96 \\ \left(x + {8 \over 2}\right)^2 - \left(8 \over 2\right)^2 + \left(y - {24 \over 2}\right)^2 - \left(24 \over 2\right)^2 & = -96 \phantom{000000} [\text{Complete the square}] \\ (x + 4)^2 - 16 + (y - 12)^2 - 144 & = -96 \\ (x + 4)^2 + (y - 12)^2 & = -96 + 16 + 144 \\ (x + 4)^2 + (y - 12)^2 & = 64 \\ [x - (- 4)]^2 + (y - 12)^2 & = 8^2 \\ \\ \text{Centre: } & (-4, 12) \\ \\ \text{Radius} & = 8 \text{ units} \end{align}

\begin{align} \text{Eqn of normal: } & 3y + 4x = k \\ \\ \text{Using centre } & (-4, 12), \phantom{0000000} [\text{Normal passes through centre of circle}] \\ 3(12) + 4(-4) & = k \\ 36 - 16 & = k \\ 20 & = k \end{align}

(ii)

\begin{align} \text{Eqn of normal: } & 3y + 4x = 20 \\ \\ \text{Let } & y = 0, \\ 3(0) + 4x & = 20 \\ 4x & = 20 \\ x & = {20 \over 4} \\ x & = 5 \\ \\ \therefore & \phantom{.} S(5, 0) \\ \\ \\ 3y + 4x & = 20 \\ 3y & = 20 - 4x \\ y & = {20 \over 3} - {4 \over 3}x \\ \\ \text{Substitute } & \text{into eqn of circle,} \\ (x + 4)^2 + \left({20 \over 3} - {4 \over 3}x - 12\right)^2 & = 64 \\ (x + 4)^2 + \left(-{4 \over 3}x - {16 \over 3}\right)^2 & = 64 \\ x^2 + 2(x)(4) + (4)^2 + \left(-{4 \over 3}x\right)^2 - 2 \left(-{4 \over 3}x\right)\left(16 \over 3\right) + \left(16 \over 3\right)^2 & = 64 \\ x^2 + 8x + 16 + {16 \over 9}x^2 + {128 \over 9}x + {256 \over 9} & = 64 \\ {25 \over 9}x^2 + {200 \over 9}x + {400 \over 9} & = 64 \\ {25 \over 9}x^2 + {200 \over 9}x - {176 \over 9} & = 0 \\ 25x^2 + 200x - 176 & = 0 \\ (5x - 4)(5x + 44) & = 0 \end{align}
\begin{align} 5x - 4 & = 0 && \text{ or } & 5x + 44 & = 0 \\ 5x & = 4 &&& 5x & = -44 \\ x & = {4 \over 5} &&& x & = -{44 \over 5} \text{ (N.A.)} \\ \\ \text{Substitute } & \text{into } y = {20 \over 3} - {4 \over 3}x, \\ y & = {20 \over 3} - {4 \over 3}\left(4 \over 5\right) \\ y & = 5{3 \over 5} \\ \\ \therefore & \phantom{.} R \left({4 \over 5}, 5{3 \over 5}\right) \end{align}
\begin{align} RS & = \sqrt{ (x_2 - x_1)^2 + (y_2 - y_1)^2 } \\ & = \sqrt{ \left(5 - {4 \over 5}\right)^2 + \left(0 - 5{3 \over 5}\right)^2 } \\ & = 7 \text{ units} \end{align}

\begin{align} u & = e^{-x} &&& v & = x^2 \\ \left[ {d \over dx}(e^{f(x)}) = f'(x) e^{f(x)} \right] \phantom{000000} {du \over dx} & = (-1)e^{-x} &&& {dv \over dx} & = 2x \\ & = - e^{-x} \end{align}
\begin{align} {dy \over dx} & = u {dv \over dx} + v {du \over dx} \phantom{000000} [\text{Product rule}] \\ & = (e^{-x})(2x) + (x^2)(-e^{-x}) \\ & = 2x e^{-x} - x^2 e^{-x} \\ & = e^{-x} (2x - x^2) \end{align}
\begin{align} u & = e^{-x} &&& v & = 2x - x^2 \\ {du \over dx} & = (-1)e^{-x} &&& {dv \over dx} & = 2 - 2x \\ & = - e^{-x} \end{align}
\begin{align} {d^2 y \over dx^2} & = (e^{-x})(2 - 2x) + (2x - x^2)(-e^{-x}) \phantom{000000} [\text{Product rule}] \\ & = 2e^{-x} - 2x e^{-x} - e^{-x} (2x - x^2) \\ & = 2e^{-x} - 2x e^{-x} - 2x e^{-x} + x^2 e^{-x} \\ & = 2e^{-x} - 4x e^{-x} + x^2 e^{-x} \\ \\ \\ e^{x} \left( {d^2 y \over dx^2} + 2 {dy \over dx} + y \right) & = e^{x} \left[ 2e^{-x} - 4x e^{-x} + x^2 e^{-x} + 2e^{-x} (2x - x^2) + e^{-x} x^2 \right] \\ & = e^{x} ( 2e^{-x} - 4x e^{-x} + x^2 e^{-x} + 4x e^{-x} - 2x^2 e^{-x} + x^2 e^{-x} ) \\ & = e^{x} ( 2e^{-x} ) \\ & = 2e^{-x + x} \\ & = 2e^{0} \\ & = 2(1) \\ & = 2 \\ \\ \therefore k & = 2 \end{align}

(i)

\begin{align} {d \over dx} (\tan x - x) & = \sec^2 x - 1 \\ & = \tan^2 x \phantom{000000} [\tan^2 A + 1 = \sec^2 A \implies \tan^2 A = \sec^2 A - 1] \end{align}

(ii)

\begin{align} \text{From (i), } {d \over dx} (\tan x - x) & = \tan^2 x \\ \\ \implies \int \tan^2 x \phantom{.} dx & = \tan x - x \\ \\ \\ \int \sec^2 x + 5 \tan^2 x \phantom{.} dx & = \tan x + 5 (\tan x - x) + c \\ & = \tan x + 5 \tan x - 5x + c \\ & = 6 \tan x - 5x + c \\ \\ \\ \int_{\pi \over 6}^{\pi \over 3} \sec^2 x + 5 \tan^2 x \phantom{.} dx & = \left[ 6 \tan x - 5x \right]_{\pi \over 6}^{\pi \over 3} \\ & = \left[ 6 \tan {\pi \over 3} - 5\left(\pi \over 3\right) \right] - \left[ 6 \tan {\pi \over 6} - 5\left(\pi \over 6\right) \right] \\ & = \left( 6 \sqrt{3} - {5\pi \over 3} \right) - \left[ 6 \left(1 \over \sqrt{3}\right) - {5\pi \over 6}\right] \\ & = 6 \sqrt{3} - {5\pi \over 3} - {6 \over \sqrt{3}} + {5\pi \over 6} \\ & = 6 \sqrt{3} - {6 \over \sqrt{3}} \times {\sqrt{3} \over \sqrt{3}} - {5\pi \over 3} + {5\pi \over 6} \\ & = 6 \sqrt{3} - {6 \sqrt{3} \over 3} - {5\pi \over 6} \\ & = 6 \sqrt{3} - 2 \sqrt{3} - {5 \over 6} \pi \\ & = 4 \sqrt{3} - {5 \over 6} \pi \\ \\ \\ \therefore a & = 4, b = -{5 \over 6} \end{align}

(i)

\begin{align} \text{General term, } T_{r + 1} & = {n \choose r} a^{n - r} b^r \\ & = {9 \choose r} (px^3)^{9 - r} \left(1 \over x\right)^r \\ & = {9 \choose r} (p^{9 - r}) (x^3)^{9 - r} (x^{-1})^r \\ & = {9 \choose r} (p^{9 - r}) (x^{3(9 - r)}) (x^{-r}) \\ & = {9 \choose r} (p^{9 - r}) (x^{27 - 3r})(x^{-r}) \\ & = {9 \choose r} (p^{9 - r}) (x^{27 - 3r + (-r)}) \\ & = {9 \choose r} (p^{9 - r}) (x^{27 - 4r}) \\ \\ \text{Power of } x & = 27 - 4r \\ \\ {27 - 4r \over 2} & = {27 \over 2} - {4r \over 2} = 13.5 - 2r \\ \\ \\ \therefore \text{Since power of } x & \text{ is not divisible by 2, there are no even powers of } x \end{align}

(ii)

\begin{align} \text{General term, } T_{r + 1} & = {9 \choose r} (p^{9 - r}) (x^{27 - 4r}) \\ \\ \text{Power of } x & = 27 - 4r \\ \\ \\ \text{Let } 27 - 4r & = 11 \\ -4r & = -16 \\ r & = {-16 \over -4} \\ r & = 4 \\ \\ T_{4 + 1} & = {9 \choose 4} (p^{9 - 4})(x^{27 - 4(4)}) \\ & = (126)(p^5)(x^{11}) \\ & = 126 p^5 x^{11} \\ \\ \\ \text{Let } 27 - 4r & = 7 \\ -4r & = -20 \\ r & = {-20 \over -4} \\ r & = 5 \\ \\ T_{5 + 1} & = {9 \choose 5} (p^{9 - 5}) (x^{27 - 4(5)}) \\ & = (126)(p^4)(x^7) \\ & = 126 p^4 x^7 \\ \\ \\ \text{Coefficient of } x^{11} & = 2 \times \text{Coefficient of } x^{7} \\ 126p^5 & = 2 \times 126p^4 \\ 126p^5 & = 252 p^4 \\ {p^5 \over p^4} & = {252 \over 126} \\ p & = 2 \end{align}

(i)

\begin{align} y & = {6 \over \sqrt{x}} + x \\ & = 6 x^{-{1 \over 2}} + x \\ \\ {dy \over dx} & = 6 \left(-{1 \over 2}\right) x^{-{3 \over 2}} + 1 \\ & = -3 x^{-{3 \over 2}} + 1 \\ \\ {d^2 y \over dx^2} & = -3 \left(-{3 \over 2}\right) x^{-{5 \over 2}} \\ & = {9 \over 2} x^{-{5 \over 2}} \\ \\ \text{Let } & {dy \over dx} = 0, \phantom{000000} [\text{minimum point } M] \\ 0 & = -3x^{-{3 \over 2}} + 1 \\ 3x^{-{3 \over 2}} & = 1 \\ {3 \over \sqrt{x^3}} & = 1 \\ 3 & = \sqrt{x^3} \\ 3^2 & = x^3 \\ 9 & = x^3 \\ \sqrt[3]{9} & = x \\ \\ \\ {d^2 y \over dx^2} & = {9 \over 2} (\sqrt[3]{9})^{-{5 \over 2}} \\ & = 0.72112 > 0 \\ \\ \\ \therefore M \text{ is a minimum point} & \text{ and the } x \text{-coordinate } \sqrt[3]{9} \text{ satisfies the equation } x^3 = 9 \end{align}

(ii)

\begin{align} {dy \over dx} & = - 3x^{-{3 \over 2}} + 1 \\ & = -{3 \over \sqrt{x^3}} + 1 \\ \\ \text{Let } & x = 1, \\ {dy \over dx} & = -{3 \over \sqrt{1^3}} + 1 \\ & = -2 \\ \\ \text{Gradient at } A & = -2 \\ \\ y & = -2x + c \\ \\ \text{Using } & A(1, 7), \\ 7 & = -2(1) + c \\ 7 & = -2 + c \\ 9 & = c \\ \\ \text{Tangent at } A: & \phantom{.} y = -2x + 9 \phantom{00} \text{--- (1)} \\ \\ \\ \text{Let } & x = 4, \\ {dy \over dx} & = -{3 \over \sqrt{4^3}} + 1 \\ & = {5 \over 8} \\ \\ \text{Gradient at } B & = {5 \over 8} \\ \\ y & = {5 \over 8}x + c \\ \\ \text{Using } & B(4, 7), \\ 7 & = {5 \over 8}(4) + c \\ 7 & = {5 \over 2} + c \\ {9 \over 2} & = c \\ \\ \text{Tangent at } B: & \phantom{.} y = {5 \over 8}x + {9 \over 2} \phantom{00} \text{--- (2)} \\ \\ \\ \text{Substitute } & \text{(1) into (2),} \\ -2x + 9 & = {5 \over 8}x + {9 \over 2} \\ -2x - {5 \over 8}x & = {9 \over 2} - 9 \\ -{21 \over 8}x & = -{9 \over 2} \\ x & = -{9 \over 2} \div -{21 \over 8} \\ x & = 1{5 \over 7} \phantom{000000} [\text{this is } x \text{-coordinate of } P] \\ \\ \\ \text{Since } \sqrt[3]{9} > 1{5 \over 7}, & \phantom{.} x\text{-coordinate of } P \text{ is less than } x\text{-coordinate of } M \end{align}

(i)

\begin{align} \log_5 (x - 1) - \log_5 (x + 1) & = 1 + \log_5 {1 \over 7} \\ \log_5 {x - 1 \over x + 1} & = \log_5 5 + \log_5 {1 \over 7} \phantom{000000} [\text{Quotient law (logarithms), } \log_a a = 1] \\ \log_5 {x - 1 \over x + 1} & = \log_5 \left(5 \times {1 \over 7}\right) \phantom{0000000.} [\text{Product law (logarithms)}] \\ \log_5 {x - 1 \over x + 1} & = \log_5 {5 \over 7} \\ \\ {x - 1 \over x + 1} & = {5 \over 7} \\ 7(x - 1) & = 5(x + 1) \\ 7x - 7 & = 5x + 5 \\ 7x - 5x & = 5 + 7 \\ 2x & = 12 \\ x & = {12 \over 2} \\ x & = 6 \end{align}

(ii)

\begin{align} \log_y 100 & = \lg y \\ {\log_{10} 100 \over \log_{10} y} & = \lg y \phantom{00000000} [\text{Change-of-base}] \\ \\ {\lg 100 \over \lg y} & = \lg y \\ \lg 100 & = (\lg y)(\lg y) \\ \lg 100 & = (\lg y)^2 \\ \lg 10^2 & = (\lg y)^2 \\ 2 \lg 10 & = (\lg y)^2 \phantom{000000} [\text{Power law (logarithms)}] \\ 2 (1) & = (\lg y)^2 \\ 2 & = (\lg y)^2 \\ \pm \sqrt{2} & = \lg y \end{align}
\begin{align} \lg y & = \sqrt{2} && \text{ or } & \lg y & = -\sqrt{2} \\ \log_{10} y & = \sqrt{2} &&& \log_{10} y & = - \sqrt{2} \\ y & = 10^{\sqrt{2}} &&& y & = 10^{-\sqrt{2}} \\ & \approx 26 \text{ (2 s.f.)} &&& & \approx 0.039 \text{ (2 s.f.)} \end{align}

(i)

\begin{align} y & = 9x^2 + (2m + 1)x + 1 + c \phantom{00} \text{--- (1)} \\ \\ y & = mx + c \phantom{00} \text{--- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ mx + c & = 9x^2 + (2m + 1)x + 1 + c \\ 0 & = 9x^2 + (2m + 1)x - mx + 1 \\ 0 & = 9x^2 + (m + 1)x + 1 \\ \\ b^2 - 4ac & = (m + 1)^2 - 4(9)(1) \\ & = \underbrace{ m^2 + 2(m)(1) + 1^2}_{ (a + b)^2 = a^2 + 2ab + b^2 } - 36 \\ & = m^2 + 2m + 1 - 36 \\ & = m^2 + 2m - 35 \\ \\ b^2 - 4ac & = 0 \phantom{000000} [\text{1 real root since line is tangent to curve}] \\ m^2 + 2m - 35 & = 0 \\ (m - 5)(m + 7) & = 0 \end{align}
\begin{align} m - 5 & = 0 && \text{ or } & m + 7 & = 0 \\ m & = 5 &&& m & = -7 \text{ (Reject, since } m > 0) \end{align}

(ii)

\begin{align} y & = 9x^2 + (2m + 1)x + 1 + c \\ y & = 9x^2 + [2(5) + 1]x + 1 + c \\ y & = 9x^2 + 11x + 1 + c \\ \\ \text{Using } & (-2, 19), \\ 19 & = 9(-2)^2 + 11(-2) + 1 + c \\ 19 & = 15 + c \\ 4 & = c \\ \\ y & = 9x^2 + 11x + 1 + 4 \\ y & = 9x^2 + 11x + 5 \\ \\ \\ \text{From (i), } 0 & = 9x^2 + (m + 1)x + 1 \\ 0 & = 9x^2 + (5 + 1)x + 1 \\ 0 & = 9x^2 + 6x + 1 \\ 0 & = (3x + 1)(3x + 1) \\ 0 & = (3x + 1)^2 \\ 0 & = 3x + 1 \\ -3x & = 1 \\ x & = -{1 \over 3} \\ \\ \text{Substitute } & \text{into } y = 9x^2 + 11x + 5, \\ y & = 9\left(-{1 \over 3}\right)^2 + 11 \left(-{1 \over 3}\right) + 5 \\ y & = 2{1 \over 3} \\ \\ \therefore & \phantom{.} P\left(-{1 \over 3}, 2{1 \over 3}\right) \end{align}

(iii)

$$ L \text{ is a vertical line} $$

(a)(i)

\begin{align} P & = 100 e^{-kt} \\ \\ \text{Let } & t = 0, \\ P & = 100 e^{-k(0)} \\ & = 100 e^{0} \\ & = 100 \\ \\ \text{Initial amount} & = 100\% \\ \\ \text{When } & t = 5730 \text{ and } P = 50, \\ 50 & = 100 e^{-k(5730)} \\ {50 \over 100} & = e^{-5730k} \\ {1 \over 2} & = e^{-5730k} \\ \ln {1 \over 2} & = \ln e^{-5730k} \\ \ln {1 \over 2} & = -5730k \ln e \phantom{000000} [\text{Power law (logarithms)}] \\ \ln {1 \over 2} & = -5730k (1) \\ \ln {1 \over 2} & = -5730k \\ \\ k & = -{\ln {1 \over 2} \over 5730} \\ & = 0.000 \phantom{.} 120 \phantom{.} 968 \\ & \approx 0.000 \phantom{.} 121 \end{align}

(a)(ii)

\begin{align} P & = 100 e^{-kt} \\ \\ \text{When } & t = 8000, \\ P & = 100 e^{-k(8000)} \\ & = 100 e^{-8000k} \\ & = 100 e^{-8000(0.000 \phantom{.} 120 \phantom{.} 968)} \\ & = 37.994 \\ & \approx 38.0 \% \end{align}

(b)

\begin{align} S & = \lg {I \over c} \\ & = \lg I - \lg c \phantom{000000} [\text{Quotient law (logarithms)}] \\ \\ \text{When } & S = 2.4, \\ 2.4 & = \lg I - \lg c \\ -\lg I & = -\lg c - 2.4 \\ \lg I & = \lg c + 2.4 \\ \log_{10} I & = \lg c + 2.4 \\ I & = 10^{\lg c + 2.4} \phantom{0000000} [\text{Exponential form}] \\ \\ \text{When } I & = 50 \times 10^{\lg c + 2.4}, \phantom{00000000000000} [50 \text{ times }] \\ S & = \lg (50 \times 10^{\lg c + 2.4}) - \lg c \\ S & = \lg 50 + \lg 10^{\lg c + 2.4} - \lg c \phantom{0000000} [\text{Product law (logarithms)}] \\ S & = \lg 50 + (\lg c + 2.4) \lg 10 - \lg c \phantom{00.} [\text{Power law (logarithms)}] \\ S & = \lg 50 + (\lg c + 2.4) (1) - \lg c \\ S & = \lg 50 + \lg c + 2.4 - \lg c \\ S & = \lg 50 + 2.4 \\ & = 4.0990 \\ & \approx 4.1 \text{ (1 d.p.)} \end{align}

(a)

\begin{align} {dx \over dt} & = {dx \over dy} \times {dy \over dt} \\ & = \underbrace{ {dx \over dy} }_\text{Need this} \times 0.06 \\ \\ \\ y & = \ln (3x - 1) \\ \\ {dy \over dx} & = {3 \over 3x - 1} \phantom{000000} \left[ {d \over dx}[ \ln f(x)] = {f'(x) \over f(x)} \right] \\ \\ {dx \over dy} & = {1 \over {3 \over 3x - 1} } \\ & = {3x - 1 \over 3} \\ \\ {dx \over dt} & = {3x - 1 \over 3} \times 0.06 \\ & = {3x - 1 \over 3} \times {3 \over 50} \\ & = {9x - 3 \over 150} \\ \\ \text{When } & x = 7, \\ {dx \over dt} & = {9(7) - 3 \over 150} \\ & = 0.4 \text{ units per second} \end{align}

(b)(i)

(b)(ii)

\begin{align} \text{Substitute } & x = -{1 \over 2} \text{ into } y = 8 - (2x + 1)^3, \\ y & = 8 - \left[ 2 \left(-{1 \over 2}\right) + 1 \right]^3 \\ & = 8 \\ \\ \therefore & \phantom{.} \left(-{1 \over 2}, 8\right) \end{align}

(i)

\begin{align} \text{Gradient of } AB & = {y_2 - y_1 \over x_2 - x_1} \\ & = {p - 1 \over 0 - (-2)} \\ & = {p - 1 \over 2} \\ \\ \text{Gradient of } CB & = {p - 3 \over 0 - 1} \\ & = {p - 3 \over -1} \\ \\ \tan \angle ABO & = {2 \over p - 1} \phantom{000000} \left[ {Opp \over Adj} \right] \\ \\ \tan \angle CBO & = {1 \over p - 3} \\ \\ \text{Since } \angle ABO & = \angle CBO, \\ \tan \angle ABO & = \tan \angle CBO \\ {2 \over p - 1} & = {1 \over p - 3} \\ 2(p - 3) & = p -1 \\ 2p - 6 & = p - 1 \\ 2p - p & = -1 + 6 \\ p & = 5 \phantom{0} (\text{Shown}) \end{align}

(ii)

\begin{align} \text{Gradient of } AB & = {p - 1 \over 2} \\ & = {5 - 1 \over 2} \\ & = 2 \\ \\ \text{Gradient of } AD & = {-1 \over 2} \phantom{000000} [m_1 \times m_2 = -1] \\ & = -{1 \over 2} \\ \\ y & = mx + c \\ y & = -{1 \over 2}x + c \\ \\ \text{Using } & A(-2, 1), \\ 1 & = -{1 \over 2}(-2) + c \\ 1 & = 1 + c \\ 0 & = c \\ \\ \text{Eqn of } AD: & \phantom{.} y = -{1 \over 2}x \phantom{00} \text{--- (1)} \\ \\ \text{Gradient of } DC & = \text{Gradient of } AB \\ & = 2 \\ \\ y & = mx + c \\ y & = 2x + c \\ \\ \text{Using } & C(1, 3), \\ 3 & = 2(1) + c \\ 3 & = 2 + c \\ 1 & = c \\ \\ \text{Eqn of } DC: & \phantom{.} y = 2x + 1 \phantom{00} \text{--- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ -{1 \over 2}x & = 2x + 1 \\ -{5 \over 2}x & = 1 \\ x & = 1 \div -{5 \over 2} \\ x & = -{2 \over 5} \\ \\ \text{Substitute } & \text{into (1),} \\ y & = -{1 \over 2} \left(-{2 \over 5}\right) \\ y & = {1 \over 5} \\ \\ \therefore & \phantom{.} D \left(-{2 \over 5}, {1 \over 5}\right) \end{align}

(iii)

\begin{align} \text{Area of } ABCD & = {1 \over 2} \left| \begin{matrix} 1 & 0 & -2 & -{2 \over 5} & 1 \\ 3 & 5 & 1 & {1 \over 5} & 3 \end{matrix} \right| \phantom{000000} [\text{Anti-clockwise order: } C, B, A, D, C] \\ & = {1 \over 2} \left[ (1)(5) + (0)(1) + (-2) \left(1 \over 5\right) + \left(-{2 \over 5}\right)(3) \right] \\ & \phantom{==} -{1 \over 2} \left[ (3)(0) + (5)(-2) + (1) \left(-{2 \over 5}\right) + \left(1 \over 5\right)(1) \right] \\ & = 6{4 \over 5} \text{ units}^2 \end{align}

(i)(a)

\begin{align} y_1 & = 2 \sin x + 1 \\ \\ \text{Amplitude} & = 2 \\ \\ \text{Period} & = {360^\circ \over 1} \\ & = 360^\circ \end{align}

(i)(b)

\begin{align} y_2 & = - \cos 2x \\ \\ \text{Amplitude} & = 1 \\ \\ \text{Period} & = {360^\circ \over 2} \\ & = 180^\circ \end{align}

(ii)

\begin{align} y_1 & = y_2 \\ 2 \sin x + 1 & = - \cos 2x \\ 2 \sin x + 1 & = - (1 - 2 \sin^2 x) \phantom{000000} [\cos 2A = 1 - 2 \sin^2 A] \\ 2 \sin x + 1 & = -1 + 2 \sin^2 x \\ 0 & = 2 \sin^2 x - 2 \sin x - 2 \\ 0 & = \sin^2 x - \sin x - 1 \\ \\ \sin x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {- (-1) \pm \sqrt{(-1)^2 - 4(1)(-1)} \over 2(1)} \\ & = {1 \pm \sqrt{5} \over 2} \\ & = 1.618 \text{ (Reject, since } -1 \le \sin x \le 1) \text{ or } -0.61803 \\ \\ \\ \sin x & = -0.61803 \phantom{000000} [\text{3rd or 4th quadrant since } \sin x < 0] \\ \\ \text{Basic angle, } \alpha & = \sin^{-1} (0.61803) \\ & = 38.173^\circ \end{align}

\begin{align} x & = 180^\circ + 38.173^\circ, 360^\circ - 38.173^\circ \\ & = 218.173^\circ, 321.827^\circ \\ & \approx 218.2^\circ, 321.8^\circ \end{align}

(iii)

\begin{align} y_1 & = 2 \sin x + 1 &&& y_2 = - \cos 2x & \phantom{000000} [\text{Inverted shape}] \\ \\ \text{Center line: } & y = 1 \\ \\ \text{Amplitude} & = 2 &&& \text{Amplitude} & = 1 \\ \\ \text{Max. value} & = 1 + 2 = 3 &&& \text{Max. value} & = 1 \\ \\ \text{Min. value} & = 1 - 2 = -1 &&& \text{Min. value} & = -1 \\ \\ \text{Period} & = 360^\circ &&& \text{Period} & = 180^\circ \end{align}

(iv)

\begin{align} y_1 - y_2 & > 0 \\ y_1 & > y_2 \\ \\ \text{From graph, } 0 \le x < 218.2^\circ & \text{ or } 321.8^\circ < x \le 360^\circ \end{align}

(i)

\begin{align} \angle POX & = 90^\circ - \theta \\ \\ \angle OPQ & = 180^\circ - 90^\circ - (90^\circ - \theta) \phantom{000000} [\text{Angle sum of triangle}] \\ & = 180^\circ - 90^\circ - 90^\circ + \theta \\ & = \theta \\ \\ \cos \angle OPQ & = {PQ \over PO} \phantom{000000} \left[ {Adj \over Hyp} \right] \\ \cos \theta & = {PQ \over 4} \\ 4 \cos \theta & = PQ \\ \\ \\ \sin \angle POZ & = {PZ \over PO} \phantom{000000} \left[ {Opp \over Hyp} \right] \\ \sin \theta & = {PZ \over 4} \\ 4 \sin \theta & = PZ \end{align}

(ii)

\begin{align} \text{Area of triangle } YOX & = {1 \over 2} \times 5 \times 3 \\ & = 7.5 \text{ km}^2 \\ \\ 7.5 & = \text{Area of triangle } POY + \text{Area of triangle } POX \\ 7.5 & = {1 \over 2} \times 3 \times 4 \sin \theta + {1 \over 2} \times 5 \times 4 \cos \theta \\ 7.5 & = 6 \sin \theta + 10 \cos \theta \\ 2(7.5) & = 2(6 \sin \theta + 10 \cos \theta) \\ 15 & = 12 \sin \theta + 20 \cos \theta \phantom{00} \text{(Shown)} \end{align}

(iii)

\begin{align} a \cos \theta + b \sin \theta & = R \cos (\theta - \alpha) \\ \\ a = 20, b & = 12 \\ \\ R & = \sqrt{a^2 + b^2} \\ & = \sqrt{20^2 + 12^2} \\ & = \sqrt{544} \\ \\ \alpha & = \tan^{-1} \left(b \over a\right) \\ & = \tan^{-1} \left(12 \over 20\right) \\ & = 30.964^\circ \\ \\ \therefore 20 \cos \theta + 12 \sin \theta & = \sqrt{544} \cos (\theta - 30.964^\circ) \end{align}

(iv)

\begin{align} \text{Since } 20 \cos \theta + 12 \sin \theta & = 15, \\ \sqrt{544} \cos (\theta - 30.964^\circ) & = 15 \\ \cos (\theta - 30.964^\circ) & = {15 \over \sqrt{544}} \\ \cos (\theta - 30.964^\circ) & = 0.64312 \phantom{000000} [\text{1st or 4th quadrant}] \\ \\ \text{Basic angle, } \alpha & = \cos^{-1} (0.64312) \\ & = 49.975^\circ \end{align}

\begin{align} \theta - 30.964^\circ & = 49.975^\circ, 360^\circ - 49.975^\circ \\ & = 49.975^\circ, 310.025^\circ \\ \\ \theta & = 80.939^\circ, 340.989^\circ \text{ (N.A.)} \\ & \approx 80.9^\circ \end{align}