\begin{align*} y & = 2x + c \phantom{0} \text{--- (1)} \\ y & = x^2 + 3x + 1 \text{ --- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ 2x + c & = x^2 + 3x + 1 \\ 0 & = x^2 + x + 1 - c \\ \\ b^2 - 4ac & = (1)^2 - 4(1)(1 - c) \\ & = 1 - 4(1 - c) \\ & = 1 - 4 + 4c \\ & = -3 + 4c \\ \\ \text{Since line } & \text{is tangent to curve,} \\ b^2 - 4ac & =0 \\ -3 + 4c & = 0 \\ 4c & = 3 \\ c & = {3 \over 4} \end{align*}

\begin{align*} {18 + 11x - 2x^2 \over (x - 1)(x + 2)^2} & = {A \over x - 1} + {B \over x + 2} + {C \over (x + 2)^2} \\ & = {A(x + 2)^2 \over (x - 1)(x + 2)^2} + {B(x - 1)(x + 2) \over (x - 1)(x + 2)^2} + {C(x - 1) \over (x - 1)(x + 2)^2} \\ & = {A(x + 2)^2 + B(x - 1)(x + 2) + C(x - 1) \over (x - 1)(x + 2)^2} \\ \\ 18 + 11x - 2x^2 & = A(x + 2)^2 + B(x - 1)(x + 2) + C(x - 1) \\ \\ \text{Let } & x = -2, \\ 18 + 11(-2) - 2(-2)^2 & = A(0) + B(-3)(0) + C(-3) \\ -12 & = 0 + 0 - 3C \\ -12 & = -3C \\ {-12 \over -3} & = C \\ 4 & = C \\ \\ 18 + 11x - 2x^2 & = A(x + 2)^2 + B(x - 1)(x + 2) + 4(x - 1) \\ \\ \text{Let } & x = 1, \\ 18 + 11(1) - 2(1)^2 & = A(3)^2 + B(0)(3) + 4(0) \\ 27 & = A(9) + 0 + 0 \\ 27 & = 9A \\ {27 \over 9} & = A \\ 3 & = A \\ \\ 18 + 11x - 2x^2 & = 3(x + 2)^2 + B(x - 1)(x + 2) + 4(x - 1) \\ \\ \text{Let } & x = 0, \\ 18 + 0 - 0 & = 3(2)^2 + B(-1)(2) + 4(-1) \\ 18 & = 12 - 2B - 4 \\ 2B & = 12 - 4 - 18 \\ 2B & = -10 \\ B & = {-10 \over 2} \\ B & = -5 \\ \\ \therefore {18 + 11x - 2x^2 \over (x - 1)(x + 2)^2} & = {3 \over x - 1} + {-5 \over x + 2} + {4 \over (x + 2)^2} \\ & = {3 \over x - 1} - {5 \over x + 2} + {4 \over (x + 2)^2} \end{align*}

\begin{align*} \require{cancel} \text{L.H.S} & = { \cos^2 \theta \over (\text{cosec } \theta - 1)(\text{cosec } \theta + 1)} + { \sin^2 \theta \over (\sec \theta - 1)(\sec \theta + 1)} \\ & = { \cos^2 \theta \over \text{cosec}^2 \theta -1 } + {\sin^2 \theta \over \sec^2 \theta - 1} \phantom{000000000.} [(a - b)(a + b) = a^2 - b^2] \\ & = { \cos^2 \theta \over 1 + \cot^2 \theta - 1} + {\sin^2 \theta \over \tan^2 \theta + 1 - 1} \phantom{0000} [1 + \cot^2 A = \text{cosec}^2 A, \tan^2 A + 1 = \sec^2 A] \\ & = { \cos^2 \theta \over \cot^2 \theta } + {\sin^2 \theta \over \tan^2 \theta} \\ & = \cos^2 \theta \div \cot^2 \theta + \sin^2 \theta \div \tan^2 \theta \\ & = \cos^2 \theta \div {\cos^2 \theta \over \sin^2 \theta} + \sin^2 \theta \div {\sin^2 \theta \over \cos^2 \theta} \\ & = \cancel{\cos^2 \theta} \times {\sin^2 \theta \over \cancel{\cos^2 \theta}} + \cancel{\sin^2 \theta} \times {\cos^2 \theta \over \cancel{\sin^2 \theta}} \\ & = \sin^2 \theta + \cos^2 \theta \\ & = 1 \phantom{00000000000000} [\sin^2 A + \cos^2 A = 1] \\ & = \text{R.H.S} \end{align*}

(a)

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

(b)

\begin{align*} \text{From (a), } {d \over dx} (x e^{-2x} ) & = - 2x e^{-2x} + e^{-2x} \\ \\ \implies \int - 2x e^{-2x} + e^{-2x} \phantom{.} dx & = x e^{-2x} \\ \int -2 x e^{-2x} \phantom{.} dx + \int e^{-2x} \phantom{.} dx & = x e^{-2x} \\ \int -2 x e^{-2x} \phantom{.} dx & = x e^{-2x} - \int e^{-2x} \phantom{.} dx \\ \int -2 x e^{-2x} \phantom{.} dx & = x e^{-2x} - {e^{-2x} \over -2} \phantom{000000} \left[ \int e^{f(x)} \phantom{.} dx = {e^{f(x)} \over f'(x)} \right] \\ - 2 \int x e^{-2x} \phantom{.} dx & = x e^{-2x} + {1 \over 2} e^{-2x} \\ \int x e^{-2x} \phantom{.} dx & = -{1 \over 2} \left( x e^{-{2x}} + {1 \over 2} e^{-2x} \right) \\ & = -{1 \over 2} x e^{-{2x}} - {1 \over 4} e^{-2x} + c \end{align*}

(a)

\begin{align*} u & = ax^2 &&& v & = x - a \\ {du \over dx} & = 2ax &&& {dv \over dx} & = 1 \end{align*}
\begin{align*} f'(x) & = {(x - a)(2ax) - (ax^2)(1) \over (x - a)^2 } \phantom{000000} [\text{Quotient rule}] \\ & = {2ax^2 - 2a^2 x - ax^2 \over (x - a)^2 } \\ & = { ax^2 - 2a^2 x \over (x - a)^2 } \end{align*}

(b)

\begin{align*} \require{cancel} g'(x) & = (x - a)^2 f'(x) \\ & = \cancel{ (x - a)^2 } \left[ ax^2 - 2a^2 x \over \cancel{ (x - a)^2} \right] \\ & = ax^2 - 2a^2 x \\ \\ \text{For decreasing} & \text{ function, } g'(x) < 0 \\ ax^2 - 2a^2 x & < 0 \\ ax (x - 2a) & < 0 \\ x (x - 2a) & < {0 \over a} \\ x (x - 2a) & < 0 \end{align*}

$$ 0 < x < 2a $$
\begin{align*} \text{Since } a & > 0 \text{ and } x > a, \\ a & < x < 2a \\ \\ \text{Comparing with } & a < x < 8, \\ 2a & = 8 \\ a & = 4 \end{align*}

\begin{align*} y - 2x & = 12 \\ y & = 2x + 12 \text{ --- (1)} \\ \\ x^2 - xy + y^2 & = 63 \text{ --- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ x^2 - x(2x + 12) + (2x + 12)^2 & = 63 \\ x^2 - 2x^2 - 12x + \underbrace{ (2x)^2 + 2(2x)(12) + (12)^2 }_{(a + b)^2 = a^2 + 2ab + b^2} & = 63 \\ x^2 - 2x^2 - 12x + 4x^2 + 48x + 144 & = 63 \\ 3x^2 + 36x + 144 & = 63 \\ 3x^2 + 36x + 81 & = 0 \\ x^2 + 12x + 27 & = 0 \\ (x + 3)(x + 9) & = 0 \end{align*} \begin{align*} x + 3 & = 0 && \text{ or } & x + 9 & = 0 \\ x & = -3 &&& x & = - 9 \\ \\ \text{Substitute } & \text{into (1),} &&& \text{Substitute } & \text{into (1),} \\ y & = 2(-3) +12 &&& y & = 2(-9) + 12 \\ y & = 6 &&& y & = -6 \\ \\ \therefore & \phantom{.} (-3, 6) &&& \therefore & \phantom{.} (-9, -6) \end{align*}

(a)

\begin{align*} y & = 1 + x - {x^2 \over 2} - 2x^3 \\ \\ {dy \over dx} & = 0 + 1 - {2x \over 2} - 2(3)x^2 \\ & = 1 - x - 6x^2 \\ \\ \text{Let } & {dy \over dx} = 0, \\ 0 & = 1 - x - 6x^2 \\ 0 & = 6x^2 + x - 1 \\ 0 & = (3x - 1)(2x + 1) \end{align*} \begin{align*} 3x - 1 & = 0 && \text{ or } & 2x + 1 & = 0 \\ 3x & = 1 &&& 2x & = - 1 \\ x & = {1 \over 3} &&& x & = -{1 \over 2} \end{align*}

(b)

\begin{align*} \text{From gradient, } {dy \over dx}, & \text{ to be maximum, } {d^2 y \over dx^2} = 0 \text{ and } {d^3 y \over dx^3} < 0 \\ \\ {d^2 y \over dx^2} & = -1 - 6(2)x \\ & = -1 - 12x \\ \\ \text{Let } & {d^2 y \over dx^2} = 0, \\ 0 & = -1 - 12x \\ 12x & = -1 \\ x & = -{1 \over 12} \\ \\ {d^3 y \over dx^3} & = - 12 < 0 \\ \\ \implies \text{At } x = -{1 \over 12}, & \text{ gradient is a maximum} \\ \\ \\ \text{Midpoint} & = \left({x_1 + x_2 \over 2}, {y_1 + y_2 \over 2} \right) \\ \\ x \text{-coordinate of midpoint of } AB & = { {1 \over 3} +\left(-{1 \over 2}\right) \over 2}\\ & = -{1 \over 12} \\ \\ \therefore P \text{ and midpoint of } & AB \text{ have the same } x \text{-coordinates} \end{align*}

(a) By formula

\begin{align*} x^2 + y^2 + 10x - 24y & = 0 \phantom{000000} [x^2 + y^2 + 2gx + 2fy + c = 0] \\ \\ 2g & = 10, 2f = -24, c = 0 \\ \\ g & = 5, f = -12 \\ \\ \text{Centre: } & (-5, 12) \\ \\ \text{Radius} & = \sqrt{g^2 + f^2 - c} \\ & = \sqrt{(5)^2 + (-12)^2 - 0 } \\ & = 13 \text{ units} \end{align*}

(a) By completing the square

\begin{align*} x^2 + y^2 + 10x - 24y & = 0 \\ x^2 + 10x + y^2 - 24y & = 0 \\ \left(x + {10 \over 2}\right)^2 - \left(10 \over 2\right)^2 + \left(y - {24 \over 2}\right)^2 - \left(24 \over 2\right)^2 & = 0 \\ (x + 5)^2 - 25 + (y - 12)^2 - 144 & = 0 \\ (x + 5)^2 + (y - 12)^2 & = 25 + 144 \\ (x + 5)^2 + (y - 12)^2 & = 169 \\ (x + 5)^2 + (y - 12)^2 & = 13^2 \\ \\ \text{Centre: } & (-5, 12) \\ \\ \text{Radius} & = 13 \text{ units} \end{align*}

(b)

\begin{align*} (x + 5)^2 + (y - 12)^2 & = 13^2 \\ \\ \text{Let } & x = 0, \\ (5)^2 + (y - 12)^2 & = 169 \\ 25 + (y - 12)^2 & = 169 \\ (y - 12)^2 & = 169 - 25 \\ (y - 12)^2 & = 144 \\ y - 12 & = \pm \sqrt{144} \\ y - 12 & = 12 \text{ or } - 12 \\ y & = 24 \text{ or } 0 \\ \\ \therefore & \phantom{.} (0, 24), (0, 0) \end{align*}

(c)

\begin{align*} \overrightarrow{OC} & = {-5 \choose 12} \\ \\ \overrightarrow{OX_1} & = 4 \overrightarrow{OC} \\ & = 4 {-5 \choose 12} \\ & = {-20 \choose 48} \\ \\ \therefore & \phantom{.} X_1 (-20, 48) \\ \\ \overrightarrow{OX_2} & = - \overrightarrow{X_2O} \\ & = - 2 \overrightarrow{OC} \\ & = - 2 {-5 \choose 12} \\ & = {10 \choose -24} \\ \\ \therefore & \phantom{.} X_2 (10, -24) \\ \\ \therefore \text{Possible } & \text{points: } (-20, 48) \text{ or } (10, -24) \end{align*}

(a)

\begin{align*} {dr \over dt} & = {k \over 2t + 1} \\ \\ \text{When } & t = 0 \text{ and } {dr \over dt} = 0.5, \\ 0.5 & = {k \over 2(0) + 1} \\ 0.5 & = {k \over 1} \\ 0.5 & = k \end{align*}

(b)

\begin{align*} {dr \over dt} & = {k \over 2t + 1} \\ & = {0.5 \over 2t + 1} \\ & = { {1 \over 2} \over 2t + 1} \\ & = {1 \over 2} \left(1 \over 2t + 1\right) \\ \\ r & = \int {1 \over 2} \left(1 \over 2t + 1\right) \phantom{.} dt \\ & = {1 \over 2} \int {1 \over 2t + 1} \phantom{.} dt \\ & = {1 \over 2} \left[ \ln (2t + 1) \over 2 \right] + c \phantom{000000} \left[ \int {1 \over f(x)} \phantom{.} dx = {\ln f(x) \over f'(x)} \right] \\ & = {1 \over 4} \ln (2t + 1) + c \\ \\ \text{When } & t = 0 \text{ and } r = 1, \\ 1 & = {1 \over 4} \ln [2(0) + 1] + c \\ 1 & = {1 \over 4} (0) + c \\ 1 & = c \\ \\ \therefore r & = {1 \over 4} \ln (2t + 1) + 1 \end{align*}

(c)

\begin{align*} \text{Let } A & \text{ denote area of the circular patch} \\ \\ {dA \over dt} & = {dA \over dr} \times {dr \over dt} \\ & = {dA \over dr} \times {1 \over 2} \left(1 \over 2t + 1\right) \phantom{000000} [\text{Need to find } {dA \over dr}] \\ \\ A & = \pi r^2 \\ {dA \over dr} & = 2 \pi r \\ \\ {dA \over dt} & = 2 \pi r \times {1 \over 2} \left(1 \over 2t + 1\right) \\ & = \pi r \left(1 \over 2t + 1\right) \\ & = {\pi r \over 2t + 1} \phantom{0000000} [\text{Need to find value of } r \text{ at } t = 3] \\ \\ \text{Substitute } & t = 3 \text{ into eqn of } r, \phantom{000000} [\text{Answer from (b)}] \\ r & = {1 \over 4} \ln [2(3) + 1] + 1 \\ & = {1 \over 4} \ln 7 + 1 \\ \\ \therefore {dA \over dt} & = { \pi \left({1 \over 4} \ln 7 + 1 \right) \over 2(3) + 1 } \\ & = 0.66713 \\ & \approx 0.667 \text{ cm}^2 \text{/s} \end{align*}

(a)

\begin{align*} h & = 1.75 + 5t - 5t^2 \\ \\ \text{Let } & t = 0, \\ h & = 1.75 + 5(0) - 5(0)^2 \\ h & = 1.75 \text{ m} \end{align*}

(b)

\begin{align*} 1.75 + 5t - 5t^2 & = -5t^2 + 5t + 1.75 \\ & = -5 (t^2 - t) + 1.75 \\ & = -5 \left[ \left(t - {1 \over 2}\right)^2 - \left(1 \over 2\right)^2 \right] + 1.75 \phantom{000000} [\text{Complete the square}] \\ & = -5 \left[ (t - 0.5)^2 - 0.25 \right] + 1.75 \\ & = -5 (t - 0.5)^2 + 1.25 + 1.75 \\ & = -5 (t - 0.5)^2 + 3 \\ & = 3 - 5 (t - 0.5)^2 \\ \\ \therefore a & = 3, b = -5, c = -0.5 \end{align*}

(c)

\begin{align*} h & = -5 (t - 0.5)^2 + 3 \\ \\ \text{Max height} & = 3 \text{ m} \text{ when time is } 0.5 \text{ seconds} \end{align*}

(d)

\begin{align*} & \text{The ball's path started from 1.75 m above the ground instead of from the ground} \end{align*}

(e)

\begin{align*} h & = 3 - 5(t - 0.5)^2 \\ \\ \text{Let } & h = 2, \\ 2 & = 3 - 5(t - 0.5)^2 \\ 5(t - 0.5)^2 & = 1 \\ (t - 0.5)^2 & = {1 \over 5} \\ t - 0.5 & = \pm \sqrt{1 \over 5} \\ t & = \sqrt{1 \over 5} + 0.5 \text{ or } - \sqrt{1 \over 5} + 0.5 \\ t & = 0.94721 \text{ or } 0.05279 \end{align*}

\begin{align*} \text{Duration} & = 0.94721 - 0.05279 \\ & = 0.89443 \\ & \approx 0.894 \text{ s} \end{align*}

(a)

\begin{align*} \text{Gradient of } AB & = {-7 - h \over -8 - 2} \\ & = {-7 -h \over -10} \\ \\ \text{Gradient of perp. bisector} \times \text{Gradient of } AB & = -1 \\ \text{Gradient of perp. bisector} & = -1 \div \text{Gradient of } AB \\ & = - 1 \div {-7 -h \over -10} \\ & = { -1 \over 1} \times {-10 \over -7 - h} \\ & = {10 \over - 7 - h} \end{align*}

(b)

\begin{align*} \text{Midpoint of } AB & = \left( {x_1 + x_2 \over 2}, {y_1 + y_2 \over 2} \right) \\ & = \left( {2 + (-8) \over 2} , {h + (-7) \over 2} \right) \\ & = \left( -3, {1 \over 2}h - {7 \over 2} \right) \\ \\ \text{Gradient of perp. bisector} & = \text{Gradient between } X \text{ and midpoint of } AB \\ & = {{1 \over 2}h - {7 \over 2} - \left(-{13 \over 2}\right) \over -3 - h} \\ & = {{1 \over 2}h - {7 \over 2} + {13 \over 2} \over - 3 - h} \\ & = {{1 \over 2}h + 3 \over -3 - h} \\ \\ \therefore {{1 \over 2}h + 3 \over - 3 - h} & = {10 \over -7 - h} \phantom{000000} [\text{Answer from (a)}] \\ (-7 - h)\left( {1 \over 2}h + 3 \right) & = 10(-3 - h) \\ -{7 \over 2}h - 21 - {1 \over 2}h^2 - 3h & = -30 -10h \\ -{13 \over 2}h - 21 - {1 \over 2}h^2 & = -30 - 10h \\ 0 & = {1 \over 2}h^2 - {7 \over 2}h - 9 \\ 0 & = h^2 - 7h - 18 \\ 0 & = (h + 2)(h - 9) \end{align*} \begin{align*} h + 2 & = 0 && \text{ or } & h - 9 & = 0 \\ h & = -2 &&& h & = 9 \end{align*}

\begin{align*} \left(2 - {1 \over 2}x\right)^6 & = 2^6 + {6 \choose 1} (2)^5 \left(-{1 \over 2}x\right) + {6 \choose 2} (2)^4 \left(-{1 \over 2}x\right)^2 + {6 \choose 3} (2)^3 \left(-{1 \over 2}x\right)^3 + ... \\ & = 64 + (6)(32)\left(-{1 \over 2}x\right) + (15)(16)\left({1 \over 4}x^2\right) + (20)(8)\left(-{1 \over 8}x^3\right) + ... \\ & = 64 - 96x + 60x^2 - 20x^3 + ... \\ \\ (k + 2x)\left(2 - {1 \over 2}x\right)^6 & = (k + 2x)(64 - 96x + 60x^2 - 20x^3 + ...) \\ & = ... + (k)(-20x^3) + (2x)(60x^2) + ... \phantom{000000} [\text{Only need terms in } x^3] \\ & = ... - 20k x^3 + 120x^3 + ... \\ & = ... + (120 - 20k)x^3 + ... \\ \\ \text{Since } & \text{coefficient of } x^3 = 0, \\ 120 - 20k & = 0 \\ -20k & = -120 \\ k & = {-120 \over -20} \\ k & = 6 \end{align*}

(a)

\begin{align*} \angle BAD & = \angle FAB \phantom{0} (\text{Common angle}) [A] \\ \\ \angle ADB & = \angle ABF \phantom{0} (\text{Alternate segment theorem}) [A] \\ \\ \therefore \text{Triangles } & ABD \text{ and } AFB \text{ are similar } (AA \text{ similarity test}) \end{align*}

(b)

\begin{align*} \angle BXC & = 90^\circ \phantom{0} (\text{Right angle in semicircle}) \\ \\ \angle DBF & = 90^\circ \phantom{0} (\text{Right angle in semicircle}) \\ \\ \text{Since } \angle BXC & + \angle XBF = 180^\circ, \text{ by the converse of interior angles,} \\ \text{lines } BF \text{ and } & EC \text{ are parallel} \\ \\ \text{Since } BF \text{ is } & \text{parallel to } EC, EBFC \text{ is a trapezium} \end{align*}

(a)

\begin{align*} T_c & = 86 e^{-0.06t} \\ \\ \text{Let } & t = 0, \\ T_c & = 86 e^{-0.06(0)} \\ & = 86^\circ \text{C} \end{align*}

(b)

\begin{align*} T_c & = 86 e^{-0.06t} \\ \\ \text{Let } & T_c = 37, \\ 37 & = 86 e^{-0.06t} \\ {37 \over 86} & = e^{-0.06t} \\ \ln {37 \over 86} & = \ln e^{-0.06t} \\ \ln {37 \over 86} & = -0.06t \ln e \phantom{000000} [\text{Power law (logarithms)}] \\ \ln {37 \over 86} & = -0.06t (1) \\ \ln {37 \over 86} & = -0.06t \\ {\ln {37 \over 86} \over -0.06} & = t \\ 14.057 & = t \\ \\ \text{Time taken} & \approx 14.1 \text{ minutes} \end{align*}

(c)(i)

\begin{align*} T_f & = 86 e^{-\lambda t} \\ \\ \text{When } & t = 60 \text{ and } T_f = 82, \phantom{000000} [1 \text{ hour = 60 mins}] \\ 82 & = 86 e^{-\lambda (60)} \\ {82 \over 86} & = e^{-60 \lambda} \\ \ln {82 \over 86} & = \ln e^{-60 \lambda} \\ \ln {82 \over 86} & = -60 \lambda \ln e \phantom{000000} [\text{Power law (logarithms)}] \\ \ln {82 \over 86} & = -60 \lambda (1) \\ \ln {82 \over 86} & = -60 \lambda \\ {\ln {82 \over 86} \over -60} & = \lambda \\ 0.000 \phantom{.} 7938 & = \lambda \\ \\ \lambda & \approx 0.000 \phantom{.} 794 \end{align*}

(c)(ii)

\begin{align*} \text{Temperature of tea in cup} & = {1 \over 2} \times \text{Temperature of tea in flask} \\ 86e^{-0.06t} & = {1 \over 2} \times 86e^{-\lambda t} \\ 86e^{-0.06t} & = 43e^{-\lambda t} \\ {e^{-0.06t} \over e^{-\lambda t}} & = {43 \over 86} \\ e^{-0.06t - (- \lambda t)} & = 0.5 \\ e^{-0.06t + \lambda t} & = 0.5 \\ \ln e^{-0.06t + \lambda t} & = \ln 0.5 \\ (-0.06t + \lambda t) \ln e & = \ln 0.5 \phantom{000000} [\text{Power law (logarithms)}] \\ (-0.06t + \lambda t) (1) & = \ln 0.5 \\ -0.06t + \lambda t & = \ln 0.5 \\ t(-0.06 + \lambda) & = \ln 0.5 \\ t & = {\ln 0.5 \over -0.06 + \lambda} \\ & = {\ln 0.5 \over -0.06 + 0.000 \phantom{.} 7938} \\ & = 11.707 \\ & \approx 11.7 \text{ minutes} \end{align*}

(a)

\begin{align*} \tan 2A & = {2 \tan A \over 1 - \tan^2 A} \\ \\ \text{Let } & A = 15^\circ, \\ \tan 30^\circ & = {2 \tan 15^\circ \over 1 - \tan^2 15^\circ} \\ {1 \over \sqrt{3}} & = {2 \tan 15^\circ \over 1 - \tan^2 15^\circ} \\ 1 - \tan^2 15^\circ & = 2 \sqrt{3} \tan 15^\circ \\ 0 & = \tan^2 15^\circ + 2 \sqrt{3} \tan 15^\circ - 1 \phantom{000000} [\text{Quadratic equation in } \tan 15^\circ] \\ \\ \tan 15^\circ & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {-2 \sqrt{3} \pm \sqrt{(2 \sqrt{3})^2 - 4(1)(-1)} \over 2(1)} \\ & = {-2 \sqrt{3} \pm \sqrt{16} \over 2} \\ & = {-2 \sqrt{3} \pm 4 \over 2} \\ & = {-2 \sqrt{3} + 4 \over 2} \text{ or } {-2 \sqrt{3} - 4 \over 2} \\ & = - \sqrt{3} + 2 \text{ or } - \sqrt{3} - 2 \\ \\ \text{Since } 15^\circ & \text{ lies in the first quadrant, } \tan 15^\circ > 0 \\ \\ \therefore \tan 15^\circ & = 2 - \sqrt{3} \text{ (Shown)} \end{align*}

(b)

\begin{align*} \tan 15^\circ & = 2 - \sqrt{3} \\ {Opp \over Adj} & = {2 - \sqrt{3} \over 1} \end{align*}

\begin{align*} \tan 105^\circ & = {1 \over - (2 - \sqrt{3})} \\ & = {1 \over \sqrt{3} - 2} \times {\sqrt{3} + 2 \over \sqrt{3} + 2} \phantom{000000} [\text{Rationalise denominator}] \\ & = {\sqrt{3} + 2 \over (\sqrt{3})^2 - (2)^2} \phantom{00000000000} [(a - b)(a + b) = a^2 - b^2] \\ & = {\sqrt{3} + 2 \over 3 - 4} \\ & = {\sqrt{3} + 2 \over -1} \\ & = - \sqrt{3} - 2 \\ \\ \text{L.H.S} & = \tan 15^\circ - \tan 105^\circ \\ & = 2 - \sqrt{3} - (-\sqrt{3} - 2) \\ & = 2 - \sqrt{3} + \sqrt{3} + 2 \\ & = 4 \\ & = \text{R.H.S} \end{align*}

Alternative method to obtain tan 105°:

\begin{align*} \tan 105^\circ & = \tan (180^\circ - 75^\circ) \\ & = {\tan 180^\circ - \tan 75^\circ \over 1 + \tan 180^\circ \tan 75^\circ} \phantom{000000} [\tan (A - B) \text{ formula}] \\ & = {0 - \tan 75^\circ \over 1 + (0) (\tan 75^\circ) } \\ & = {- \tan 75^\circ \over 1 } \\ & = - \tan 75^\circ \\ & = - \tan (90^\circ - 15^\circ) \\ & = - {1 \over \tan 15^\circ} \phantom{000000} [\text{Complementary angle: } \tan (90^\circ - \theta) = {1 \over \tan \theta}] \\ & = - {1 \over 2 - \sqrt{3}} \\ & = {1 \over \sqrt{3} - 2} \end{align*}

(a)

\begin{align*} y & = 3 \cos 3x - 5 \sin 3x \\ \\ {dy \over dx} & = 3(3)(- \sin 3x) - 5(3)(\cos 3x) \phantom{000000} [ {d \over dx} [ \cos f(x) ] = f'(x). - \sin f(x)] \\ & = -9 \sin 3x - 15 \cos 3x \phantom{0000000000000} [ {d \over dx} [ \sin f(x) ] = f'(x). \cos f(x)] \\ \\ {d^2 y \over dx^2} & = -9 (3)(\cos 3x) - 15 (3)(- \sin 3x) \\ & = -27 \cos 3x + 45 \sin 3x \end{align*} \begin{align*} p {d^2 y \over dx^2} + q {dy \over dx} + 14y + 34 \sin 3x & = p (-27 \cos 3x + 45 \sin 3x) + q (-9 \sin 3x - 15 \cos 3x) \\ & \phantom{00} + 14(3 \cos 3x - 5 \sin 3x) + 34 \sin 3x \\ & = - 27p \cos 3x + 45p \sin 3x - 9q \sin 3x - 15q \cos 3x \\ & \phantom{00} + 42 \cos 3x - 70 \sin 3x + 34 \sin 3x \\ & = 42 \cos 3x - 27p \cos 3x - 15q \cos 3x \\ & \phantom{00} - 36 \sin 3x + 45p \sin 3x - 9q \sin 3x \\ & = (42 - 27p - 15q) \cos 3x + (-36 + 45p - 9q) \sin 3x \\ \\ \therefore A & = 42 - 27p - 15q \text{ (Shown)} \\ \\ B & = -36 + 45p - 9q \end{align*}

(b)

\begin{align*} A = 42 - 27p - 15q & = 0 \phantom{0} \text{ --- (1)} \\ \\ B = -36 + 45p - 9q & = 0 \\ -9q & = 36 - 45p \\ 9q & = 45p - 36 \\ q & = 5p - 4 \phantom{0} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(2) into (1),} \\ 42 - 27p - 15(5p - 4) & = 0 \\ 42 - 27p - 75p + 60 & = 0 \\ -102p & = -102 \\ p & = 1 \\ \\ \text{Substitute } & p = 1 \text{ into (2),} \\ q & = 5(1) - 4 \\ q & = 1 \\ \\ \therefore p & = 1, q = 1 \end{align*}

(a)

\begin{align*} \text{Let } f(x) & = 6x^3 - 7x^2 - x + 2 \\ \\ f(-0.5) & = 6(-0.5)^3 - 7(-0.5)^2 - (-0.5) + 2 \\ & = 0 \\ \\ \therefore 2x + 1 & \text{ is a factor of } f(x) \end{align*}
$$ \require{enclose} \begin{array}{rll} 3x^2 - 5x + 2 \phantom{0000.}\\ 2x + 1 \enclose{longdiv}{ 6x^3 - 7x^2 - x + 2 \phantom{0}}\kern-.2ex \\ -\underline{( 6x^3 + 3x^2){\phantom{0000000.}}} \\ -10x^2 - x + 2 \phantom{0} \\ -\underline{( - 10x^2 - 5x ){\phantom{00.}}} \\ 4x + 2 \phantom{0} \\ -\underline{( 4x + 2){\phantom{}}} \\ 0 \phantom{.} \end{array} $$
\begin{align*} \text{Polynomial} & = \text{Divisor} \times \text{Quotient} + \text{Remainder} \\ 6x^3 - 7x^2 - x + 2 & = (2x + 1)(3x^2 - 5x + 2) + 0 \\ & = (2x + 1)(3x - 2)(x - 1) \end{align*}

(b)

\begin{align*} 6(4^y) + 2(2^{-y}) & = 7(2^y) + 1 \\ 6[(2^2)^y] + 2(2^{-y}) & = 7(2^y) + 1 \\ 6( 2^{2y} ) + 2(2^{-y}) & = 7(2^y) + 1 \phantom{000000} [ (a^m)^n = a^{mn} ] \\ 6(2^{2y}) + 2 \left(1 \over 2^y\right) & = 7(2^y) + 1 \\ \\ \text{Let } & u = 2^y, \\ 6u^2 + 2 \left(1 \over u\right) & = 7u + 1 \\ 6u^2 + {2 \over u} & = 7u + 1 \\ u \left( 6u^2 + {2 \over u} \right) & = u(7u + 1) \\ 6u^3 + 2 & = 7u^2 + u \\ 6u^3 - 7u^2 - u + 2 & = 0 \\ (2u + 1)(3u - 2)(u - 1) & = 0 \phantom{000000} [\text{Same polynomial from (a)}] \end{align*} \begin{align*} 2u + 1 & = 0 && \text{ or } & 3u - 2 & = 0 && \text{ or } & u - 1 & = 0 \\ 2u & = -1 &&& 3u & = 2 &&& u & = 1 \\ u & = -{1 \over 2} &&& u & = {2 \over 3} &&& \\ 2^y & = -{1 \over 2} \text{ (No solutions)} &&& 2^y & = {2 \over 3} &&& 2^y & = 1 \\ & &&& \ln 2^y & = \ln {2 \over 3} &&& 2^y & = 2^0 \\ & &&& y \ln 2 & = \ln {2 \over 3} &&& y & = 0 \\ & &&& y & = {\ln {2 \over 3} \over \ln 2} \\ & &&& y & = -0.58496 \\ & &&& y & \approx -0.585 \end{align*}

(c)

\begin{align*} y & = {\ln {2 \over 3} \over \ln 2} \\ & = {\ln 2 - \ln 3 \over \ln 2} \phantom{000000} [\text{Quotient law (logarithms)}] \\ & = {\ln 2 \over \ln 2} - {\ln 3 \over \ln 2} \\ & = 1 - {\ln 3 \over \ln 2} \\ & = 1 - {\log_e 3 \over \log_e 2} \\ & = 1 - \log_2 3 \phantom{00000000} [\text{Change-of-base: } \log_a b = {\log_c b \over \log_c a}] \\ \\ \therefore a & = 2, b =3 \end{align*}

(a)

\begin{align*} \text{Amplitude} & = 10, \text{Period}= \pi \\ \\ {7 \pi \over 2} - \pi & = {5 \pi \over 2} \\ {5 \pi \over 2} - \pi & = {3 \pi \over 2} \\ {3 \pi \over 2} - \pi & = {\pi \over 2} \\ {\pi \over 2} - \pi & = -{\pi \over 2} \\ \\ \implies \text{Graph starts from } & \text{minimum point and } a < 0 \\ \\ \therefore a & = -10 \\ \\ \text{Period} & = {2\pi \over b} \\ \pi & = {2\pi \over b} \\ \pi b & = 2\pi \\ b & = {2\pi \over \pi} \\ b & = 2 \\ \\ \therefore a & = -10, b = 2 \end{align*}

(b)(i)

\begin{align*} v = 5 \sin& 0.2t + c \\ \\ \text{For particle to not } & \text{change direction, } v \ge 0 \\ \\ \therefore \text{Minimum value of } v & \ge 0 \\ c - 5 & \ge 0 \\ c & \ge 5 \\ \\ \text{Smallest value of } c & = 5 \end{align*}

(b)(ii) Note: Distance travelled in third second (from t = 2 to t = 3) is not the distance travelled in three seconds

\begin{align*} \text{Since } c = 8 > 5, & \text{ particle does not change direction and moves in a straight line} \\ \\ v & = 5 \sin 0.2t + 8 \\ \\ s & = \int 5 \sin 0.2t + 8 \phantom{.} dt \\ & = 5 \left(-\cos 0.2t \over 0.2\right) + 8t + d \phantom{000000} \left[ \int \sin f(x) dx = { - \cos f(x) \over f'(x) } \right] \\ & = - 25 \cos 0.2t + 8t + d \\ \\ \text{When } & t = 2, \\ s & = -25 \cos [0.2(2)] + 8(2) + d \\ & = - 25 \cos 0.4 + 16 + d \\ \\ \text{When } & t = 3, \\ s & = -25 \cos [0.2(3)] + 8(3) + d \\ & = - 25 \cos 0.6 + 24 + d \\ \\ \text{Distance travelled in third second} & = - 25 \cos 0.6 + 24 + d - (-25 \cos 0.4 + 16 + d) \\ & = -25 \cos 0.6 + 24 + d + 25 \cos 0.4 - 16 - d \\ & = -25 \cos 0.6 + 8 + 25 \cos 0.4 \\ & = 10.393 \\ & \approx 10.4 \text{ m} \end{align*}

(a) Note: The wooden board is symmetrical, which means angle BAD = angle CDA, sides AE = ED and so on...

\begin{align*} \cos \angle BAF & = {AF \over AB} \phantom{000000} \left[ {Adj \over Hyp} \right] \\ \cos \theta & = {AF \over r} \\ r \cos \theta & = AF \\ \\ \sin \angle BAF & = {BF \over AB} \\ \sin \theta & = {BF \over r} \\ r \sin \theta & = BF \\ \\ \text{Area of trapezium} & = {1 \over 2} \times \text{Sum of parallel sides} \times \text{Height} \\ & = {1 \over 2} \times (2r \cos \theta + 4 r \cos \theta) \times (r \sin \theta) \\ & = {1 \over 2} \times (6r \cos \theta) \times (r \sin \theta) \\ & = {1 \over 2} (6r^2 \sin \theta \cos \theta) \\ & = {3 \over 2}r^2 (2 \sin \theta \cos \theta) \\ & = {3 \over 2}r^2 \sin 2 \theta \phantom{000000} [\sin 2A = 2 \sin A \cos A] \end{align*}

(b)

\begin{align*} \text{Area of trapezium} & = {3 \over 2}r^2 \sin 2 \theta \\ \\ \text{Since } -{3 \over 2} r^2 \le {3 \over 2} r^2 \sin 2 \theta & \le {3 \over 2}r^2, \text{ maximum area} = {3 \over 2} r^2 \\ \\ \\ {3 \over 2}r^2 \sin 2 \theta & = {3 \over 2} r^2 \\ \sin 2 \theta & = { {3 \over 2} r^2 \over {3 \over 2} r^2} \\ \sin 2 \theta & = 1 \\ \\ \text{Basic angle, } \alpha & = \sin^{-1} (1) \\ & = 90^\circ \\ \\ 2 \theta & = 90^\circ, 180^\circ - 90^\circ \\ & = 90^\circ \\ \\ \theta & = 45^\circ \end{align*}

(c)

\begin{align*} \text{Length of wood used} & = 4r + 6r \cos \theta \\ 8r \sin \theta & = 4r + 6r \cos \theta \\ 8r \sin \theta - 6r \cos \theta & = 4r \\ 2r (4 \sin \theta - 3 \cos \theta) & = 4r \\ 4 \sin \theta - 3 \cos \theta & = {4r \over 2r} \\ 4 \sin \theta - 3 \cos \theta & = 2 \phantom{0} \text{ (Shown)} \end{align*}

(d)

\begin{align*} a \sin \theta - b \cos \theta & = R \sin (\theta - \alpha) \\ \\ R & = \sqrt{a^2 + b^2} \\ & = \sqrt{4^2 + 3^2} \\ & = 5 \\ \\ \alpha & = \tan^{-1} \left(b \over a\right) \\ & = \tan^{-1} \left(3 \over 4\right) \\ & = 36.870^\circ \\ \\ 4 \sin \theta - 3 \cos \theta & = 5 \sin (\theta - 36.870^\circ) \\ \\ \therefore 5 \sin (\theta - 36.870^\circ) & = 2 \\ \sin (\theta - 36.870^\circ) & = {2 \over 5} \\ \\ \text{Basic angle, } \alpha & = \sin^{-1} \left(2 \over 5\right) \\ & = 23.578^\circ \\ \\ \theta - 36.870^\circ & = 23.578^\circ, 180^\circ - 23.578^\circ \\ & = 23.578^\circ, 156.422^\circ \\ \\ \theta & = 60.448^\circ, 193.292^\circ \text{ (Reject since } 0^\circ < \theta < 90^\circ) \\ & \approx 60.4^\circ \end{align*}

(a)

\begin{align*} \text{Time} & = { \text{Distance} \over \text{Speed} } \\ \\ \text{Time taken on road} & = {x \over 5} \\ \\ \text{By Py} & \text{thagoras theorem,} \\ CX^2 & = CB^2 + BX^2 \\ & = 8^2 + (15 - x)^2 \\ & = 64 + \underbrace{ (15)^2 - 2(15)(x) + (x)^2}_{(a - b)^2 = a^2 - 2ab + b^2} \\ & = 64 + 225 - 30x + x^2 \\ & = x^2 - 30x + 289 \\ CX & = \sqrt{x^2 - 30x + 289} \\ \\ \text{Time taken on rough ground} & = { \sqrt{x^2 - 30x + 289} \over 3} \\ \\ \text{Total time, } T & = {x \over 5} + { \sqrt{x^2 - 30x + 289} \over 3} \phantom{0} \text{ (Shown)} \end{align*}

(b)

\begin{align*} T & = {x \over 5} + { \sqrt{x^2 - 30x + 289} \over 3} \\ & = {1 \over 5}x + {1 \over 3} \sqrt{x^2 - 30x + 289} \\ & = {1 \over 5}x + {1 \over 3} (x^2 - 30x + 289)^{1 \over 2} \\ \\ {d T \over dx} & = {1 \over 5} + \underbrace{ {1 \over 3}\left(1 \over 2\right)(x^2 - 30x + 289)^{-{1 \over 2}} (2x - 30)}_\text{Chain rule} \\ & = {1 \over 5} + {1 \over 6} \left( {1 \over \sqrt{x^2 - 30x + 289}} \right) (2x - 30) \\ & = {1 \over 5} + {2x - 30 \over 6 \sqrt{x^2 - 30x + 289}} \\ & = {1 \over 5} + {2(x - 15) \over 6 \sqrt{x^2 - 30x + 289}} \\ & = {1 \over 5} + {x - 15 \over 3 \sqrt{x^2 - 30x + 289}} \\ \\ \text{Let } & {d T \over dx} = 0, \\ 0 & = {1 \over 5} + {x - 15 \over 3 \sqrt{x^2 - 30x + 289}} \\ -{1 \over 5} & = {x - 15 \over 3 \sqrt{x^2 - 30x + 289}} \\ -3 \sqrt{x^2 - 30x + 289} & = 5(x - 15) \\ \left( -3 \sqrt{x^2 - 30x + 289} \right)^2 & = [5(x - 15)]^2 \\ (-3)^2 \left( \sqrt{x^2 - 30x + 289} \right)^2 & = (5)^2 (x - 15)^2 \\ 9 (x^2 - 30x + 289) & = 25[ \underbrace{ (x)^2 - 2(x)(15) + (15)^2 }_{(a - b)^2 = a^2 - 2ab + b^2} ] \\ 9x^2 - 270x + 2601 & = 25(x^2 - 30x + 225) \\ 9x^2 - 270x + 2601 & = 25x^2 - 750x + 5625 \\ 0 & = 16x^2 - 480x + 3024 \\ 0 & = x^2 - 30x + 189 \phantom{0} \text{ (Shown)} \end{align*}

(c)

\begin{align*} 0 & = x^2 - 30x + 189 \\ 0 & = (x - 9)(x - 21) \\ \\ x - 9 & = 0 \phantom{0} \text{ or } \phantom{0} x - 21 = 0 \\ x & = 9 \phantom{00000000..} x = 21 \text{ (Reject since } x < 15) \\ \\ T & = {x \over 5} + {\sqrt{x^2 - 30x + 289} \over 3} \\ & = {9 \over 5} + {\sqrt{(9)^2 - 30(9) + 289} \over 3} \\ & = 5{2 \over 15} \text{ hours} \\ & = 5 \text{ hours } \left({2 \over 15} \times 60\right) \text{ minutes} \\ & = 5 \text{ hours } 8 \text{ minutes} \end{align*}