(i)

\begin{align} y & = 2x^2 - kx - 4 \\ y & = 2x^2 - 6x - 4 \phantom{00} \text{--- (1)} \\ \\ y + 2x & = 12 \\ y & = -2x + 12 \phantom{00} \text{--- (2)} \\ \\ \text{Substitute } & \text{(2) into (1),} \\ -2x + 12 & = 2x^2 - 6x - 4 \\ 0 & = 2x^2 - 4x - 16 \\ 0 & = x^2 - 2x - 8 \\ 0 & = (x + 2)(x - 4) \end{align}
\begin{align} x + 2 & = 0 && \text{ or } & x - 4 & = 0 \\ x & = -2 &&& x & = 4 \\ \\ \text{Substitute } & \text{into (2),} &&& \text{Substitute } & \text{into (2),} \\ y & = -2(-2) + 12 &&& y & = -2(4) + 12 \\ y & = 16 &&& y & = 4 \\ \\ \therefore & \phantom{.} (-2, 16) &&& \therefore & \phantom{.} (4, 4) \end{align}

(ii)

\begin{align} y & = 2x^2 - kx - 4 \phantom{00} \text{--- (1)} \\ \\ y & = -2x + 12 \phantom{00} \text{--- (2)} \\ \\ \text{Substitute } & \text{(2) into (1),} \\ -2x + 12 & = 2x^2 - kx - 4 \\ 0 & = 2x^2 + 2x - kx - 16 \\ 0 & = 2x^2 + (2 - k)x - 16 \\ \\ b^2 - 4ac & = (2 - k)^2 - 4(2)(-16) \\ & = \underbrace{ 2^2 - 2(2)(k) + k^2 }_{(a - b)^2 = a^2 - 2ab + b^2} + 128 \\ & = 4 - 4k + k^2 + 128 \\ & = k^2 - 4k + 132 \\ & = k^2 - 4k + \left(4 \over 2\right)^2 - \left(4 \over 2\right)^2 + 132 \phantom{000000} [\text{Complete the square}] \\ & = k^2 - 4k + 2^2 - 4 + 132 \\ & = (k - 2)^2 + 128 \\ \\ \text{For all real} & \text{ values of } k, (k - 2)^2 \ge 0 \text{ and } (k - 2)^2 + 128 \ge 128 \\ \\ \therefore b^2 - 4ac > 0 & \text{ and the line intersects the curve at two distinct points for all values of } k \end{align}

\begin{align} \text{Volume} & = \text{Base area} \times \text{Height} \\ 16 + 4 \sqrt{3} & = (\sqrt{6} + \sqrt{2})^2 \times h \\ \\ h & = {16 + 4 \sqrt{3} \over (\sqrt{6} + \sqrt{2})^2 } \\ & = {16 + 4 \sqrt{3} \over (\sqrt{6})^2 + 2(\sqrt{6})(\sqrt{2}) + (\sqrt{2})^2 } \phantom{000000} [(a + b)^2 = a^2 + 2ab + b^2] \\ & = {16 + 4 \sqrt{3} \over 6 + 2 \sqrt{12} + 2 } \\ & = {16 + 4 \sqrt{3} \over 8 + 2 \sqrt{4} \sqrt{3}} \\ & = {16 + 4 \sqrt{3} \over 8 + 4 \sqrt{3}} \times {8 - 4\sqrt{3} \over 8 - 4\sqrt{3}} \phantom{00000000000} [\text{Rationalise denominator}] \\ & = {(16 + 4\sqrt{3})(8 - 4\sqrt{3}) \over (8)^2 - (4\sqrt{3})^2} \phantom{00000000000.} [(a + b)(a - b) = a^2 - b^2] \\ & = {128 - 64 \sqrt{3} + 32\sqrt{3} - (16)(3) \over 64 - 48} \\ & = {128 - 32 \sqrt{3} - 48 \over 16} \\ & = {80 - 32 \sqrt{3} \over 16} \\ & = {80 \over 16} - {32 \sqrt{3} \over 16} \\ & = 5 - 2 \sqrt{3} \phantom{00} \text{(Shown)} \end{align}

(a)(i)

$$ -90^\circ \le \sin^{-1} x \le 90^\circ $$

(a)(ii)

$$ 0^\circ \le \cos^{-1} x \le 180^\circ $$

(b)

\begin{align} y & = a \cos {x \over b} + c \\ y & = a \cos \left({1 \over b}x\right) + c \\ \\ \\ \text{Center line: } & y = 2 \\ \\ c & = 2 \\ \\ \\ \text{Amplitude} & = 3 - 2 = 1 \\ \\ a & = -1 \phantom{000000} [\text{Due to inverted shape}] \\ \\ \\ \text{Period} & = 12\pi \\ \\ {2\pi \over {1 \over b}} & = 12 \pi \\ 2\pi b & = 12 \pi \\ b & = {12\pi \over 2\pi} \\ b & = 6 \\ \\ \therefore a & = -1, b = 6, c = 2 \end{align}

(i)

(ii)

\begin{align} y^2 & = 256x \phantom{00} \text{--- (1)} \\ y & = 2x^2 \phantom{00} \text{--- (2)} \\ \\ \text{Substitute } & \text{(2) into (1),} \\ (2x^2)^2 & = 256x \\ 4x^4 & = 256x \\ x^4 & = 64x \\ x^4 - 64x & = 0 \\ x(x^3 - 64) & = 0 \end{align}
\begin{align} x & = 0 && \text{ or } & x^3 - 64 & = 0 \\ & &&& x^3 & = 64 \\ & &&& x & = \sqrt[3]{64} \\ & &&& x & = 4 \\ \\ \text{Substitute } & \text{into (2),} &&& \text{Substitute } & \text{into (2),} \\ y & = 2(0)^2 &&& y & = 2(4)^2 \\ y & = 0 &&& y & = 32 \\ \\ \therefore & \phantom{.} (0, 0) &&& \therefore & \phantom{.} (4, 32) \end{align}
\begin{align} \text{Gradient} & = {y_2 - y_1 \over x_2 - x_1} \\ & = {32 - 0 \over 4 - 0} \\ & = 8 \\ \\ y & = mx + c \\ y & = 8x + 0 \phantom{000000} [y \text{-intercept is 0 since line passes } O(0, 0)] \\ y & = 8x \end{align}

$$ \require{enclose} \begin{array}{rll} 2 \phantom{00000000000}\\ x^2 + x - 6 \enclose{longdiv}{ 2x^2 + 4x - 31 \phantom{0}}\kern-.2ex \\ -\underline{( 2x^2 + 2x - 12){\phantom{.}}} \\ 2x - 19 \phantom{.} \end{array} $$
\begin{align} {2x^2 + 4x - 31 \over x^2 + x - 6} & = 2 + {2x - 19 \over x^2 + x - 6} \\ \\ {2x - 19 \over x^2 + x - 6} = {2x - 19 \over (x + 3)(x - 2)} & = {A \over x + 3} + {B \over x - 2} \\ & = {A(x - 2) \over (x + 3)(x - 2)} + {B(x + 3) \over (x + 3)(x - 2)} \\ & = {A(x - 2) + B(x + 3) \over (x + 3)(x - 2)} \\ \\ 2x - 19 & = A(x - 2) + B(x + 3) \\ \\ \text{Let } & x= 2, \\ 2(2) - 19 & = A(0) + B(5) \\ -15 & = 5B \\ {-15 \over 5} & = B \\ -3 & = B \\ \\ 2x - 19 & = A(x - 2) - 3(x + 3) \\ \\ \text{Let } & x = 0, \\ -19 & = A(-2) - 3(3) \\ -19 & = -2A - 9 \\ 2A & = -9 + 19 \\ 2A & = 10 \\ A & = {10 \over 2} \\ A & = 5 \\ \\ \\ \therefore {2x - 19 \over x^2 + x - 6} & = {5 \over x + 3} + {-3 \over x - 2} \\ & = {5 \over x + 3} - {3 \over x - 2} \\ \\ \therefore {2x^2 + 4x - 31 \over x^2 + x - 6} & = 2 + {5 \over x + 3} - {3 \over x - 2} \end{align}

(i)

$$ \text{Since } a > 0, \text{ then } y = ax^2 + bx + c \text{ is a minimum curve } (\cup) $$

\begin{align} \implies \text{Line of symmetry, } & x = 4 \\ \\ {p + 6 \over 2} & = 4 \\ p + 6 & = 2(4) \\ p + 6 & = 8 \\ p & = 2 \end{align}

(ii)

$$ x \text{-intercepts: } 2, 6 $$
\begin{align} x & = 2 && \text{ or } & x & = 6 \\ x - 2 & = 0 && \text{ or } & x - 6 & = 0 \end{align}
\begin{align} 0 & = (x - 2)(x - 6) \\ \\ \text{Let } y & = d(x - 2)(x - 6), \text{ where } d \text{ is a constant} \\ \\ \text{Using } & M (4, - 6), \\ -6 & = d(4- 2)(4 - 6) \\ -6 & = d(2)(-2) \\ -6 & = -4d \\ {-6 \over -4} & = d \\ {3 \over 2} & = d \\ \\ y & = {3 \over 2}(x - 2)(x - 6) \\ & = {3 \over 2}(x^2 - 6x - 2x + 12) \\ & = {3 \over 2}(x^2 - 8x + 12) \\ & = {3 \over 2}x^2 - 12x + 18 \\ \\ \\ \therefore a & = {3 \over 2}, b = -12, c = 18 \end{align}

(iii) Note: Question is asking for set of values, not range of values, hence set notation is required

$$ \{ q \in \mathbb{R}, 0 < q < 6 \} $$

(i)

\begin{align} \text{Distance} & = \text{Speed} \times \text{Time} \\ \\ \text{Distance travelled by first cyclist} & = 5 \times t \\ & = 5t \\ \\ \therefore OP & = 5t \text{ m} \\ \\ \\ \text{Distance travelled by second cyclist} & = 10 \times t \\ & = 10t \\ \\ \therefore OQ & = (100 - 10t) \text{ m} \\ \\ \\ \text{By Pytha} & \text{goras theorem,} \\ PQ^2 & = PO^2 + OQ^2 \\ s^2 & = (5t)^2 + (100 - 10t)^2 \\ s^2 & = 25t^2 + \underbrace{ 100^2 - 2(100)(10t) + (10t)^2 }_{ (a - b)^2 = a^2 - 2ab + b^2 } \\ s^2 & = 25t^2 + 10 \phantom{.} 000 - 2000t + 100t^2 \\ s^2 & = 125t^2 - 2000t + 10 \phantom{.} 000 \\ s^2 & = 125 (t^2 - 16t + 80) \\ s & = \sqrt{125(t^2 - 16t + 80)} \phantom{00} \text{(Shown)} \end{align}

(ii)

\begin{align} s & = \sqrt{125(t^2 - 16t + 80)} \\ s & = \sqrt{ 125t^2 - 2000t + 10 \phantom{.} 000 } \\ s & = ( 125t^2 - 2000t + 10 \phantom{.} 000)^{1 \over 2} \\ \\ {ds \over dt} & = {1 \over 2} (125t^2 - 2000t + 10 \phantom{.} 000)^{-{1 \over 2}} . (250t - 2000) \phantom{000000} [\text{Chain rule}] \\ & = {1 \over 2}(250t - 2000) \left(1 \over \sqrt{125t^2 - 2000t + 10 \phantom{.} 000}\right) \\ & = (125t - 1000)\left(1 \over \sqrt{125t^2 - 2000t + 10 \phantom{.} 000}\right) \\ & = {125t - 1000 \over \sqrt{125t^2 - 2000t + 10 \phantom{.} 000}} \end{align}

(iii)

(i) Plan: 1) Form equation of line AB 2) Solve simultaneous equations using lines AB and BC to find the coordinates of B (since both lines meet at B)

\begin{align} \text{Eqn of } BC: \phantom{0} 2y + 3x & = 45 \\ 2y & = -3x + 45 \\ y & = -{3 \over 2}x + {45 \over 2} \phantom{000000} [y = mx + c] \\ \\ \text{Gradient of } BC & = -{3 \over 2} \\ \\ \text{Gradient of } AB & = {-1 \over -{3 \over 2}} \phantom{000000000000} [m_1 \times m_2 = -1] \\ & = {2 \over 3} \\ \\ y & = mx + c \\ y & = {2 \over 3}x + c \\ \\ \text{Using } & A(-2, 6), \\ 6 & = {2 \over 3}(-2) + c \\ 6 & = -{4 \over 3} + c \\ {22 \over 3} & = c \\ \\ \text{Eqn of } AB: & \phantom{0} y = {2 \over 3}x + {22 \over 3} \phantom{00} \text{--- (1)} \\ \\ \text{Eqn of } BC: & \phantom{0} y = -{3 \over 2}x + {45 \over 2} \phantom{00} \text{--- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ {2 \over 3}x + {22 \over 3} & = -{3 \over 2}x + {45 \over 2} \\ {2 \over 3}x + {3 \over 2}x & = {45 \over 2} - {22 \over 3} \\ {13 \over 6}x & = {91 \over 6} \\ 13x & = 91 \\ x & = {91 \over 13} \\ x & = 7 \\ \\ \text{Substitute } & \text{into (1),} \\ y & = {2 \over 3}(7) + {22 \over 3} \\ y & = 12 \\ \\ \therefore & \phantom{.} B(7, 12) \end{align}

(ii)

\begin{align} \text{Eqn of } BC: & \phantom{0} 2y + 3x = 45 \\ \\ \text{Let } & y = 0, \\ 2(0) + 3x & = 45 \\ 3x & = 45 \\ x & = {45 \over 3} \\ x & = 15 \\ \\ \therefore & \phantom{.} C(15, 0) \\ \\ \text{Midpoint of } AC, M & = \left({x_1 + x_2 \over 2}, {y_1 + y_2 \over 2}\right) \\ & = \left( {-2 + 15 \over 2}, {6 + 0 \over 2} \right) \\ & = (6.5, 3) \\ \\ \\ \overrightarrow{AB} & = \overrightarrow{OB} - \overrightarrow{OA} \\ & = {7 \choose 12} - {-2 \choose 6} \\ & = {9 \choose 6} \\ \\ \overrightarrow{CD} & = \overrightarrow{BA} \\ & = - \overrightarrow{AB} \\ & = {-9 \choose -6} \\ \\ \overrightarrow{OD} & = \overrightarrow{OC} + \overrightarrow{CD} \\ & = {15 \choose 0} + {-9 \choose -6} \\ & = {6 \choose -6} \\ \\ \therefore & \phantom{.} D(6, - 6) \end{align}

(i)

\begin{align} y & = 2 - x^2 - {16 \over x^2} \\ y & = 2 - x^2 - 16 x^{-2} \\ \\ {dy \over dx} & = 0 - 2x - 16(-2)x^{-3} \\ & = -2x + 32x^{-3} \\ & = -2x + {32 \over x^3} \\ \\ \text{Let } & {dy \over dx} = 0, \\ 0 & = -2x + {32 \over x^3} \\ 2x & = {32 \over x^3} \\ x^3(2x) & = 32 \\ 2x^4 & = 32 \\ x^4 & = {32 \over 2} \\ x^4 & = 16 \\ x & = \pm \sqrt[4]{16} \\ x & = \pm 2 \end{align}
\begin{align} \text{Substitute } & x = 2 \text{ into eqn of curve,} &&& \text{Substitute } & x = -2 \text{ into eqn of curve,} \\ y & = 2 - 2^2 - {16 \over 2^2} &&& y & = 2 - (-2)^2 - {16 \over (-2)^2} \\ y & = -6 &&& y & = -6 \\ \\ \therefore & \phantom{.} (2, -6) &&& \therefore & \phantom{.} (-2, -6) \end{align}

(ii)

\begin{align} {dy \over dx} & = -2x + 32 x^{-3} \\ \\ {d^2 y \over dx^2} & = -2 + 32(-3)x^{-4} \\ & = -2 - 96 x^{-4} \\ & = -2 - {96 \over x^4} \end{align}
\begin{align} \text{Substitute } & x = 2 \text{ into } {d^2 y \over dx^2}, &&& \text{Substitute } & x = -2 \text{ into } {d^2 y \over dx^2}, \\ {d^2 y \over dx^2} & = -2 - {96 \over (2)^4} &&& {d^2 y \over dx^2} & = -2 - {96 \over (-2)^4} \\ & = -8 < 0 &&& & = -8 < 0 \\ \\ \therefore \phantom{.} (2, -6) & \text{ is a maximum point} &&& \therefore \phantom{.} (-2, -6) & \text{ is a maximum point} \end{align}

(i)

\begin{align} u & = \ln x &&& v & = x^2 \\ {du \over dx} & = {1 \over x} &&& {dv \over dx} & = 2x \end{align}
\begin{align} {d \over dx} \left(u \over v\right) & = {v {du \over dx} - u {dv \over dx} \over v^2} \phantom{000000} [\text{Quotient rule}] \\ \\ {d \over dx} \left(\ln x \over x^2\right) & = { (x^2)\left(1 \over x\right) - (\ln x)(2x) \over (x^2)^2} \\ & = {x - 2x \ln x \over x^4} \\ & = {x \over x^4} - {2x \ln x \over x^4} \\ & = {1 \over x^3} - {2 \ln x \over x^3} \phantom{00} \text{(Shown)} \end{align}

(ii)

\begin{align} \text{From (i), } {d \over dx} \left(\ln x \over x^2\right) & = {1 \over x^3} - {2 \ln x \over x^3} \\ \\ \therefore \int {1 \over x^3} - {2 \ln x \over x^3} \phantom{.} dx & = {\ln x \over x^2} \\ \int {1 \over x^3} \phantom{.} dx - \int {2 \ln x \over x^3} \phantom{.} dx & = {\ln x \over x^2} \\ - \int {2 \ln x \over x^3} \phantom{.} dx & = {\ln x \over x^2} - \int {1 \over x^3} \phantom{.} dx \\ -2 \int {\ln x \over x^3} \phantom{.} dx & = {\ln x \over x^2} - \int x^{-3} \phantom{.} dx \\ -2 \int {\ln x \over x^3} \phantom{.} dx & = {\ln x \over x^2} - {x^{-2} \over -2} \\ -2 \int {\ln x \over x^3} \phantom{.} dx & = {\ln x \over x^2} + {1 \over 2x^2} \\ \int {\ln x \over x^3} \phantom{.} dx & = -{1 \over 2} \left( {\ln x \over x^2} + {1 \over 2x^2} \right) \\ \int {\ln x \over x^3} \phantom{.} dx & = -{\ln x \over 2 x^2} - {1 \over 4x^2} + c \end{align}

(iii)

\begin{align} y & = f(x) \\ y & = \int f'(x) \phantom{.} dx \\ y & = \int {\ln x \over x^3} \phantom{.} dx \\ y & = -{\ln x \over 2 x^2} - {1 \over 4x^2} + c \phantom{000000} [\text{Use result from (ii)}] \\ \\ \text{Using } & \left(1, {3 \over 4}\right), \\ {3 \over 4} & = - {\ln 1 \over 2(1)^2} - {1 \over 4(1)^2} + c \\ {3 \over 4} & = - 0 - {1 \over 4} + c \\ -c & = -{1 \over 4} - {3 \over 4} \\ -c & = -1 \\ c & = 1 \\ \\ \therefore f(x) & = -{\ln x \over 2x^2} - {1 \over 4x^2} + 1 \end{align}

(a)

\begin{align} \require{cancel} \text{L.H.S} & = (\sec \theta - \tan \theta)^2 \\ & = \left(\frac{1}{\cos \theta} - \frac{\sin \theta}{\cos \theta} \right)^2 \\ & = \left(1 - \sin \theta \over \cos \theta\right)^2 \\ & = { (1 - \sin \theta)^2 \over \cos^2 \theta } \\ & = { (1 - \sin \theta)^2 \over 1 - \sin^2 \theta} \phantom{0000000000000.} [\sin^2 A + \cos^2 A = 1 \implies \cos^2 A = 1 - \sin^2 A] \\ & = { (1 - \sin \theta)^\cancel{2} \over (1 + \sin \theta)\cancel{(1 - \sin \theta)} } \phantom{00000} [a^2 - b^2 = (a + b)(a - b)] \\ & = {1 - \sin \theta \over 1 + \sin \theta} \\ & = \text{R.H.S} \end{align}

(b)(i)

\begin{align} \text{Period} & = 60 \text{ mins} \phantom{000000} [\text{60 mins/1 hour for one revolution}] \\ \\ {2\pi \over k} & = 60 \\ 2\pi & = 60k \\ \\ k & = {2\pi \over 60} \\ k & = {\pi \over 30} \text{ radians per minute} \end{align}

(b)(ii)

\begin{align} d & = 80 \sin {\pi \over 30}t \\ \\ \text{Let } & d = \pm 40, \\ \pm 40 & = 80 \sin {\pi \over 30}t \\ \pm {40 \over 80} & = \sin {\pi \over 30}t \\ \pm {1 \over 2} & = \sin {\pi \over 30}t \phantom{000000} [\text{All 4 quadrants}] \\ \\ \text{Basic angle, } \alpha & = \sin^{-1} \left(1 \over 2\right) \\ & = 30^\circ \\ & = {\pi \over 6} \end{align}

\begin{align} {\pi \over 30}t & = {\pi \over 6}, \pi - {\pi \over 6}, \pi + {\pi \over 6}, 2\pi - {\pi \over 6} \\ & = {\pi \over 6}, {5\pi \over 6}, {7\pi \over 6}, {11 \pi \over 6} \\ \\ \pi t & = 5\pi, 25\pi, 35\pi, 55\pi \\ \\ t & = 5, 25, 35, 55 \\ \\ \text{Total duration} & = (25 - 5) + (55- 35) \\ & = 40 \text{ mins} \end{align}

(i)

\begin{align} \int f(x) \phantom{.} dx & = \sin x + k \cos 2x + c \\ \\ \int_0^{\pi \over 6} f(x) \phantom{.} dx & = \left[ \sin x + k \cos 2x \right]_0^{\pi \over 6} \\ & = \left[ \sin {\pi \over 6} + k \cos {\pi \over 3} \right] - \left[ \sin 0 + k \cos 0 \right] \\ & = \left[ {1 \over 2} + k \left(1 \over 2\right)\right] - [0 + k(1)] \phantom{000000000} \left[ \text{Special angle } {\pi \over 6} = 30^\circ \right] \\ & = {1 \over 2} + {1 \over 2}k - k \\ & = {1 \over 2} - {1 \over 2}k \\ \\ \therefore {1 \over 2} - {1 \over 2}k & = {3 \over 4} \\ -{1 \over 2}k & = {3 \over 4} - {1 \over 2} \\ -{1 \over 2}k & = {1 \over 4} \\ k & = {1 \over 4} \div -{1 \over 2} \\ k & = -{1 \over 2} \phantom{00} \text{(Shown)} \end{align}

(ii)

\begin{align} \int f(x) \phantom{.} dx & = \sin x + k \cos 2x + c \\ & = \sin x - {1 \over 2} \cos 2x + c \\ \\ f(x) & = {d \over dx} \left( \sin x - {1 \over 2} \cos 2x + c \right) \\ & = \cos x - {1 \over 2} (2) (- \sin 2x) \\ & = \cos x - (- \sin 2x) \\ & = \cos x + \sin 2x \end{align}

(iii)

\begin{align} y = f(x) & = \cos x + \sin 2x \\ \\ \text{When } & x = {\pi \over 6}, \\ y & = \cos {\pi \over 6} + \sin {\pi \over 3} \\ & = {\sqrt{3} \over 2} + {\sqrt{3} \over 2} \\ & = {2 \sqrt{3} \over 2} \\ & = \sqrt{3} \\ \\ \text{Coordinates } & \text{of point: } \left({\pi \over 6}, \sqrt{3}\right) \\ \\ \\ {dy \over dx} & = {d \over dx} (\cos x + \sin 2x) \\ & = - \sin x + (2)(\cos 2x) \\ & = - \sin x + 2 \cos 2x \\ \\ \text{Let } & x = {\pi \over 6}, \\ {dy \over dx} & = - \sin {\pi \over 6} + 2 \cos {\pi \over 3} \\ & = - {1 \over 2} + 2 \left(1 \over 2\right) \\ & = {1 \over 2} \\ \\ \text{Gradient of normal} & = {-1 \over {1 \over 2}} \phantom{000000} [m_1 \times m_2 = -1] \\ & = -2 \\ \\ y & = mx + c \\ y & = -2x + c \\ \\ \text{Using } & \left({\pi \over 6}, \sqrt{3}\right), \\ \sqrt{3} & = -2 \left(\pi \over 6\right) + c \\ \sqrt{3} & = - {\pi \over 3} + c \\ \sqrt{3} + {\pi \over 3} & = c \\ \\ \\ \therefore \text{Eqn of normal: } & y = -2x + \sqrt{3} + {\pi \over 3} \end{align}

(i)

\begin{align} P & = P_0 e^{kt} \\ \ln P & = \ln (P_0 e^{kt}) \\ \ln P & = \ln P_0 + \ln e^{kt} \phantom{000000.} [\text{Product law (logarithms)}] \\ \ln P & = \ln P_0 + kt \ln e \phantom{000000} [\text{Power law (logarithms)}] \\ \ln P & = \ln P_0 + kt (1) \\ \ln P & = \ln P_0 + kt \\ \ln P & = kt + \ln P_0 \phantom{000000000} [Y = mX + c] \\ \\ \text{Plot } & \ln P \text{ against } t \end{align}

$$t$$	$$ 0$$	$$ 5$$	$$ 10$$	$$ 15$$
$$\ln P$$	$$ 0.693$$	$$ 0.892$$	$$ 1.099$$	$$ 1.295$$

$$ \text{Since a straight line is obtained, model is valid} $$

(ii)

\begin{align} \text{Gradient, } m & = \frac{1.295 - 0.693}{15 - 0} \\ k & = 0.040 \phantom{.} 133 \\ & \approx 0.0401 \\ \\ \text{Vertical intercept, } c & = 0.693 \\ \ln P_0 & = 0.693 \\ \log_e P_0 & = 0.693 \\ P_0 & = e^{0.693} \\ & = 1.9997 \\ & \approx 2.00 \end{align}

(iii)

\begin{align} P & = P_0 e^{kt} \\ & = 1.9997 e^{0.040 \phantom{.} 133 t} \\ \\ \text{When } & t = 20, \\ P & = 1.9997 e^{0.040 \phantom{.} 133 (20)} \\ & = 4.4623 \\ & \approx \$ 4.46 \end{align}

(i)

\begin{align} (1 - 2x)^2 & = 1^2 - 2(1)(2x) + (2x)^2 \phantom{000000} [(a - b)^2 = a^2 - 2ab + b^2] \\ & = 1 - 4x + 4x^2 \\ \\ (1 - px)^6 & = 1^6 + {6 \choose 1} (1)^5 (-px) + {6 \choose 2} (1)^4 (-px)^2 + ... \\ & = 1 + (6)(1)(-px) + (15)(1)(p^2 x^2) + ... \\ & = 1 - 6px + 15 p^2 x^2 + ... \\ \\ (1 - 2x)^2 (1 - px)^6 & = (1 - 4x + 4x^2) (1 - 6px + 15 p^2 x^2 + ...) \\ & = ... + (1)(15p^2 x^2) + (-4x)(-6px) + (4x^2)(1) + ... \\ & = 15p^2 x^2 + 24px^2 + 4x^2 + ... \\ \\ \text{Coefficient of } x^2 & = 15p^2 + 24p + 4 \\ 16 & = 15p^2 + 24p + 4 \\ 0 & = 15p^2 + 24p - 12 \\ 0 & = 5p^2 + 8p - 4 \\ 0 & = (p + 2)(5p - 2) \end{align} \begin{align} p + 2 & = 0 && \text{ or } & 5p - 2 & =0 \\ p & = -2 &&& 5p & = 2 \\ & &&& p & = {2 \over 5} \end{align}

(ii)

\begin{align} (1 - px)^6 & = ... + {6 \choose 3} (1)^3 (-px)^3 + ... \\ & = (20)(1)(-p^3 x^3) + ... \\ & = - 20p^3 x^3 + ... \end{align}
\begin{align} \text{For } & p = -2, &&& \text{For } & p = {2 \over 5}, \\ \text{Coefficient of } x^3 & = -20 (-2)^3 &&& \text{Coefficient of } x^3 & = -20 \left(2 \over 5\right)^3 \\ & = 160 &&& & = -1.28 \end{align}

(i)

\begin{align} \cos 3x & = \cos (2x + x) \\ & = \cos 2x \cos x - \sin 2x \sin x \phantom{000000000000000000} [\cos (A + B) = \cos A \cos B - \sin A \sin B] \\ & = (1- 2 \sin^2 x) \cos x - (2 \sin x \cos x) \sin x \phantom{000000} [\cos 2A = 1 - 2 \sin^2 A, \sin 2A = 2 \sin A \cos A] \\ & = \cos x - 2 \sin^2 x \cos x - 2 \sin^2 x \cos x \\ & = \cos x - 4 \sin^2 x \cos x \\ & = \cos x (1 - 4 \sin^2 x) \phantom{00} \text{ (Shown)} \end{align}

(ii)

\begin{align} 2 \cos 3x & = 15 \sin x \cos x \\ 2 [ \underbrace{ \cos x ( 1 - 4 \sin^2 x) }_\text{From (i)}] & = 15 \sin x \cos x \\ 2 \cos x (1 - 4 \sin^2 x) & = 15 \sin x \cos x \\ 2 \cos x (1 - 4 \sin^2 x) - 15 \sin x \cos x & = 0 \\ \cos x [ 2(1 - 4 \sin^2 x) - 15 \sin x] & = 0 \\ \cos x (2 - 8 \sin^2 x - 15 \sin x) & = 0 \\ \cos x (-8 \sin^2 x - 15 \sin x + 2) & =0 \\ \cos x (- \sin x - 2)(8 \sin x - 1) & = 0 \\ \\ \cos x = 0 \phantom{0} \text{ or } \phantom{0} - \sin x - 2 = 0 & \phantom{0} \text{ or } \phantom{0} 8 \sin x - 1 = 0 \\ \\ \\ \cos x & = 0 \end{align}

\begin{align} x & = 90^\circ, 270^\circ \\ \\ \\ - \sin x - 2 & =0 \\ - \sin x & = 2 \\ \sin x & = -2 \phantom{0} \text{ (No solutions, since } -1 \le \sin x \le 1) \\ \\ \\ 8 \sin x - 1 & =0 \\ 8 \sin x & = 1 \\ \sin x & = {1 \over 8} \phantom{000000} [\text{1st or 2nd quadrant since } \sin x > 0] \\ \\ \text{Basic angle, } \alpha & = \sin^{-1} \left(1 \over 8\right) \\ & = 7.18^\circ \end{align}

\begin{align} x & = 7.18^\circ, 180^\circ - 7.18^\circ \\ & = 7.18^\circ, 172.82^\circ \\ & \approx 7.2^\circ, 172.8^\circ \\ \\ \\ \therefore x & = 7.2^\circ, 90^\circ, 172.8^\circ, 270^\circ \end{align}

(i)

\begin{align} \text{Sum of roots, } \alpha + \beta & = -{b \over a} \\ & = -{2 \over 1} \\ & = -2 \\ \\ \text{Product of roots, } \alpha \beta & = {c \over a} \\ & = {5 \over 1} \\ & = 5 \\ \\ \\ \alpha^3 + \beta^3 & = (\alpha + \beta)[ (\alpha + \beta)^2 - 3 \alpha \beta ] \\ & = (-2)[ (-2)^2 - 3(5) ] \\ & = 22 \end{align}

(ii)

\begin{align} \text{New sum of roots, } {\alpha \over \beta^2} + {\beta \over \alpha^2} & = {\alpha^3 \over \alpha^2 \beta^2} + {\beta^3 \over \alpha^2 \beta^2} \\ & = {\alpha^3 + \beta^3 \over (\alpha \beta)^2} \\ & = {22 \over (5)^2} \\ & = {22 \over 25} \\ \\ \text{New product of roots, } \frac{\alpha}{\beta^2} \times \frac{\beta}{\alpha^2} & = {\alpha \beta \over \alpha^2 \beta^2} \\ & = {1 \over \alpha \beta} \\ & = {1 \over 5} \\ \\ \\ x^2 - \text{(SOR)}x + \text{(POR)} & = 0 \\ x^2 - {22 \over 25}x + {1 \over 5} & = 0 \end{align}

(i)

\begin{align} \text{Let } \angle ACB & = \theta \\ \\ \angle DBC & = \theta \phantom{0} (\text{Isosceles triangle } DBC) \\ \\ \angle CDB & = 180^\circ - \theta - \theta \phantom{0} (\text{Angle sum of triangle}) \\ & = 180^\circ - 2 \theta \\ \\ \angle ADB & = 180^\circ - (180^\circ - 2\theta) \phantom{0} (\text{Adjacent angles on a straight line}) \\ & = 180^\circ - 180^\circ + 2 \theta \\ & = 2 \theta \\ \\ \\ \angle ABP & = \angle ACB = \theta \phantom{0} (\text{Alternate segment theorem}) \\ \\ \angle BAP & = \angle ACB = \theta \phantom{0} (\text{Alternate segment theorem}) \\ \\ \angle APB & = 180^\circ - \theta - \theta \phantom{0} (\text{Angle sum of triangle}) \\ & = 180^\circ - 2 \theta \\ \\ \\ \angle ADB + \angle APB & = 2 \theta + 180^\circ - 2 \theta \\ & = 180^\circ \phantom{00} \text{(Shown)} \end{align}

(ii)

\begin{align} \angle PDB & = \angle PAB = \theta \phantom{0} (\text{Angles in the same segment}) \\ \\ \text{Since } \angle PDB = \angle & DBC = \theta, \text{ by the converse of alternate angles, } PD \phantom{.} // \phantom{.} BC \end{align}

(i)

\begin{align} u & = x - 2 &&& v & = (2x - 5)^3 \\ {du \over dx} & = 1 &&& {dv \over dx} & = 3(2x - 5)^2 (2) \phantom{000000} [\text{Chain rule}] \\ & &&& & = 6(2x - 5)^2 \end{align} \begin{align} {d \over dx} (uv) & = u {dv \over dx} + v {du \over dx} \phantom{000000} [\text{Product rule}] \\ {dy \over dx} & = (x - 2)[6(2x - 5)^2] + (2x - 5)^3 (1) \\ & = 6(x - 2)(2x - 5)^2 + (2x - 5)^3 \\ & = (2x - 5)^2 [ 6(x - 2) + 2x - 5] \\ & = (2x - 5)^2 (6x - 12 + 2x - 5) \\ & = (2x - 5)^2 (8x - 17) \end{align}

(ii)

\begin{align} \text{For decreasing function, } & {dy \over dx} < 0 \\ \\ (2x - 5)^2 (8x - 17) & < 0 \\ \\ \text{Since } (2x - 5)^2 \ge 0 & \text{ for all real values of } x, \\ 8x - 17 & < 0 \\ 8x & < 17 \\ x & < {17 \over 8} \end{align}

(iii)

\begin{align} {dx \over dt} & = {dx \over dy} \times {dy \over dt} \\ & = {1 \over (2x - 5)^2 (8x - 17)} \times 0.35 \\ & = {1 \over (2x - 5)^2 (8x - 17)} \times {7 \over 20} \\ & = {7 \over 20(2x - 5)^2 (8x - 17)} \\ \\ \text{When } & x = 3, \\ {dx \over dt} & = {7 \over 20[2(3) - 5]^2 [8(3) - 17]} \\ & = 0.05 \text{ units per second} \end{align}

(iv)

\begin{align} z & = y^2 \\ {dz \over dy} & = 2y \\ \\ {dz \over dt} & = {dz \over dy} \times {dy \over dt} \\ & = 2y \times 0.35 \\ & = 0.7y \phantom{000000} [\text{Need to find value of } y \text{ when } x = 3] \\ \\ \text{When } & x = 3, \\ y & = (3 - 2)[2(3) - 5]^3 \\ y & = 1 \\ \\ \text{When } & y = 1, \\ z & =(1)^2 \\ z & = 1 \\ \\ \therefore {dz \over dt} & = 0.7 (1) \\ & = 0.7 \\ & = 2(0.35) \\ & = 2 \times {dy \over dt} \phantom{00} \text{(Shown)} \end{align}

(i)

\begin{align} 2^{2x - 1} & = 2^{x + 2} - 6 \\ {2^{2x} \over 2^1} & = (2^x)(2^2) - 6 \phantom{000000} \left[ {a^m \over a^n} = a^{m - n}, a^m \times a^n = a^{m + n} \right] \\ {2^{2x} \over 2} & = 4(2^x) - 6 \\ \\ \text{Let } & u = 2^x, \\ {u^2 \over 2} & = 4u - 6 \\ u^2 & = 2(4u - 6) \\ u^2 & = 8u - 12 \\ u^2 - 8u + 12 & = 0 \end{align}

(ii)

\begin{align} u^2 - 8u + 12 & = 0 \\ (u - 2)(u - 6) & = 0 \end{align} \begin{align} u - 2 & = 0 && \text{ or } & u - 6 & = 0 \\ u & = 2 &&& u & = 6 \\ \\ 2^x & = 2 &&& 2^x & = 6 \\ 2^x & = 2^1 &&& \lg 2^x & = \lg 6 \\ x & = 1 &&& x \lg 2 & = \lg 6 \phantom{000000} [\text{Power law (logarithms)}] \\ & &&& x & = {\lg 6 \over \lg 2} \\ & &&& x & \approx 2.6 \text{ (1 d.p.)} \end{align}

(iii)

\begin{align} 2^{2x - 1} & = 2^{x + 2} - k \\ {2^{2x} \over 2^1} & = (2^x)(2^2) - k \\ {2^{2x} \over 2} & = 4(2^x) - k \\ \\ \text{Let } & u = 2^x, \\ {u^2 \over 2} & = 4u - k \\ u^2 & = 2(4u - k) \\ u^2 & = 8u - 2k \\ u^2 - 8u + 2k & = 0 \\ \\ b^2 - 4ac & = (-8)^2 - 4(1)(2k) \\ & = 64 - 8k \\ \\ \text{If } k > 8, & \text{ then } b^2 - 4ac < 0 \\ \\ \therefore \text{If } k > 8, & \text{ equation has no solutions} \end{align}

(i)

\begin{align} f(3) & = (3)^3 - 3(3)^2 + 4(3) - 12 \\ & = 0 \\ \\ \therefore x - 3 & \text{ is a factor of } f(x) \end{align}
$$ \require{enclose} \begin{array}{rll} x^2 + 4 \phantom{000000.}\\ x - 3 \enclose{longdiv}{ x^3 - 3x^2 + 4x - 12\phantom{0}}\kern-.2ex \\ -\underline{( x^3 - 3x^2){\phantom{000000000}}} \\ 4x - 12 \phantom{0} \\ -\underline{( 4x - 12){\phantom{.}}} \\ 0 \phantom{0} \end{array} $$
\begin{align} x^3 - 3x^2 + 4x - 12 & = (x - 3)(x^2 + 4) \end{align}

(ii)

\begin{align} f(x) & = 0 \\ x^3 - 3x^2 + 4x - 12 & = 0 \\ (x - 3)(x^2 + 4) & = 0 \\ \end{align} \begin{align} x - 3 & = 0 && \text{ or } & x^2 + 4 & = 0 \\ x & = 3 &&& x^2 & = -4 \\ & &&& x & = \pm \sqrt{-4} \phantom{0} (\text{No real roots}) \end{align}
$$ \text{Only real root is } x = 3 $$

(iii)

\begin{align} y & = f(x) + kx \\ & = x^3 - 3x^2 + 4x - 12 + kx \\ & = x^3 - 3x^2 + (4 + k)x - 12 \\ \\ {dy \over dx} & = 3x^2 - 3(2)x + 4 + k \\ & = 3x^2 - 6x + 4 + k \\ \\ {d^2 y \over dx^2} & = 3(2)x - 6 \\ 0 & = 6x - 6 \\ 6 & = 6x \\ {6 \over 6} & = x \\ 1 & = x \phantom{000000} [\text{This is } x \text{-coordinate of the stationary point}] \\ \\ \\ \text{Substitute } & x = 1 \text{ and } {dy \over dx} = 0 \text{ into } {dy \over dx}, \\ 0 & = 3(1)^2 - 6(1) + 4 + k \\ 0 & = 3 - 6 + 4 + k \\ -k & = 3 - 6 + 4 \\ -k & = 1 \\ k & = -1 \end{align}

(i)

\begin{align} {dy \over dx} & = 3x^2 + 2(2)x - 3 \\ & = 3x^2 + 4x - 3 \\ \\ \text{When } & x = {2 \over 3}, \\ {dy \over dx} & = 3\left(2 \over 3\right)^2 + 4\left(2 \over 3\right) - 3 \\ & = 1 \\ \\ \text{Gradient at } A & = 1 \end{align}

(ii)

\begin{align} \text{Gradient at } B & = \text{Gradient at } A \phantom{000000} [m_1 = m_2] \\ & = 1 \\ \\ \text{Substitute } & {dy \over dx} = 1 \text{ into } {dy \over dx}, \\ 1 & = 3x^2 + 4x - 3 \\ 0 & = 3x^2 + 4x - 4 \\ 0 & = (x + 2)(3x - 2) \end{align} \begin{align} x + 2 & = 0 && \text{ or } & 3x - 2 & = 0 \\ x & = -2 &&& 3x & = 2 \\ & &&& x & = {2 \over 3} \phantom{00} [\text{Point } A] \end{align}
$$ \therefore x \text{-coordinate of } B = -2 $$

(iii)

\begin{align} \text{Area above } x \text{-axis} & = \int_{-2}^0 x^3 + 2x^2 - 3x \phantom{.} dx \\ & = \left[ {x^4 \over 4} + 2 \left(x^3 \over 3\right) - 3 \left(x^2 \over 2\right) \right]_{-2}^0 \\ & = \left[ {1 \over 4}x^4 + {2 \over 3}x^3 - {3 \over 2}x^2 \right]_{-2}^0 \\ & = \left[ {1 \over 4}(0)^4 + {2 \over 3}(0)^3 - {3 \over 2}(0)^2 \right] - \left[ {1 \over 4}(-2)^4 + {2 \over 3}(-2)^3 - {3 \over 2}(-2)^2 \right] \\ & = 7{1 \over 3} \text{ units}^2 \\ \\ \text{Area below } x \text{-axis} & = - \int_0^{2 \over 3} x^3 + 2x^2 - 3x \phantom{.} dx \\ & = -\left[ {1 \over 4}x^4 + {2 \over 3}x^3 - {3 \over 2}x^2 \right]_0^{2 \over 3} \\ & = -\left\{\left[ {1 \over 4}\left(2 \over 3\right)^4 + {2 \over 3}\left(2 \over 3\right)^3 - {3 \over 2}\left(2 \over 3\right)^2 \right] - \left[ {1 \over 4}(0)^4 + {2 \over 3}(0)^3 - {3 \over 2}(0)^2 \right] \right\} \\ & = - \left(-{34 \over 81}\right) \\ & = {34 \over 81} \text{ units}^2 \\ \\ \text{Total area of shaded region} & = 7{1 \over 3} + {34 \over 81} \\ & = 7 {61 \over 81} \text{ units}^2 \end{align}

(i)

\begin{align} v & = 30 e^{25t} + 20 \\ \\ \text{When } & t = 0, \\ v & = 30e^{25(0)} + 20 \\ v & = 50 \\ \\ \therefore p & = 50 \end{align}

(ii)

\begin{align} v & = 30 e^{25t} + 20 \\ \\ \text{When } & v = 80, \\ 80 & = 30 e^{25t} + 20 \\ 60 & = 30 e^{25t} \\ {60 \over 30} & = e^{25t} \\ 2 & = e^{25t} \\ \ln 2 & = \ln e^{25t} \\ \ln 2 & = 25t \ln e \phantom{000000} [\text{Power law (logarithms)}] \\ \ln 2 & = 25t (1) \\ \ln 2 & = 25t \\ \\ t & = {\ln 2 \over 25} \\ & = 0.027 \phantom{.} 726 \text{ hours} \\ & = 99.81 \text{ s} \\ & \approx 100 \text{ s (to nearest second)} \end{align}

(iii)

\begin{align} s & = \int v \phantom{.} dt \\ & = \int 30e^{25t} + 20 \phantom{.} dt \\ & = 30 \left(e^{25t} \over 25\right) + 20t + c \phantom{000000} \left[ \int e^{f(x)} \phantom{.} dx = {e^{f(x)} \over f'(x)}\right] \\ & = {6 \over 5} e^{25t} + 20t + c \\ \\ \text{When } & t = 0 \text{ and } s = 0, \phantom{000000} [\text{Initially, car starts from } A] \\ 0 & = {6 \over 5} e^{25(0)} + 20(0) + c \\ 0 & = {6 \over 5} + 0 + c \\ -{6 \over 5} & = c \\ \\ s & = {6 \over 5} e^{25t} + 20t -{6 \over 5} \\ \\ \text{When } & t = 0.027 \phantom{.} 726, \\ s & = {6 \over 5} e^{25(0.027 \phantom{.} 726)} + 20(0.027 \phantom{.} 726) -{6 \over 5} \\ & = 1.7545 \\ & \approx 1.75 \text{ km} \end{align}

(iv)

\begin{align} a & = {dv \over dt} \\ & = {d \over dt} (30 e^{25t} + 20) \\ & = 30 (25) e^{25t} \phantom{000000} \left[ {d \over dx}[ e^{f(x)}] = f'(x) e^{f(x)} \right] \\ & = 750 e^{25t} \end{align}

(i)

\begin{align} x^2 + y^2 - 4x - 2y & = 95 \\ x^2 - 4x + y^2 - 2y & = 95 \\ \left(x - {4 \over 2}\right)^2 - \left(4 \over 2\right)^2 + \left(y - {2 \over 2}\right)^2 - \left(2 \over 2\right)^2 & = 95 \phantom{000000} [\text{Complete the square}] \\ (x - 2)^2 - 4 + (y - 1)^2 - 1 & = 95 \\ (x - 2)^2 + (y - 1)^2 & = 100 \\ (x - 2)^2 + (y - 1)^2 & = 10^2 \\ \\ \text{Centre: } & (2, 1) \\ \\ \text{Radius} & = 10 \text{ units} \end{align}

(ii)

\begin{align} (x - 2)^2 & + (y - 1)^2 = 10^2 \\ \\ \text{When } x = 10 & \text{ and } y = 7,\\ \text{L.H.S} & = (10 - 2)^2 + (7 - 1)^2 \\ & = 100 \\ & = 10^2 \\ & = \text{R.H.S} \\ \\ \therefore P(10, 7) & \text{ lies on } C_1 \end{align}

(iii)

The tangent at P is perpendicular to the radius AP (circle property)

\begin{align} \text{Gradient of } AP & = {7 - 1 \over 10 - 2} \\ & = {3 \over 4} \\ \\ \text{Gradient of tangent} & = {-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 } & P(10, 7), \\ 7 & = -{4 \over 3}(10) + c \\ 7 & = -{40 \over 3} + c \\ {61 \over 3} & = c \\ \\ \\ \text{Eqn of tangent: } & y = -{4 \over 3}x + {61 \over 3} \end{align}

(iv)

\begin{align} \text{Length of } AP & = \text{Radius of } C_1 \\ & = 10 \text{ units} \\ \\ \text{Radius of } C_2 & = {10 \over 2} = 5 \text{ units} \\ \\ \text{Centre of } C_2 & = \text{Midpoint of } AP \\ & = \left( {2 + 10 \over 2}, {1 + 7 \over 2} \right) \\ & = (6, 4) \\ \\ \\ \text{Eqn: } & (x - 6)^2 + (y - 4)^2 = 5^2 \end{align}

(v) The tangent from (iii) is tangent to both circles

$$ y = -{4 \over 3}x + {61 \over 3} $$