\begin{align} \text{Area of rectangle} & = \text{Length} \times \text{Width} \\ 6 + 5 \sqrt{3} & = (3 + 2\sqrt{3}) \times \text{Width} \\ \\ \text{Width} & = {6 + 5 \sqrt{3} \over 3 + 2\sqrt{3}} \\ & = {6 + 5 \sqrt{3} \over 3 + 2 \sqrt{3}} \times {3 - 2\sqrt{3} \over 3 - 2\sqrt{3}} \phantom{00000000000} [\text{Rationalise}] \\ & = {(6 + 5 \sqrt{3})(3 - 2\sqrt{3}) \over (3 + 2\sqrt{3})(3 - 2\sqrt{3})} \\ & = {18 - 12\sqrt{3} + 15 \sqrt{3} - 10(3) \over (3)^2 - (2\sqrt{3})^2} \phantom{000000} [(a + b)(a - b) = a^2 - b^2] \\ & = {18 + 3\sqrt{3} - 30 \over 9 - 12} \\ & = {3 \sqrt{3} - 12 \over -3} \\ & = {3 \sqrt{3} \over -3} - {12 \over -3} \\ & = - \sqrt{3} + 4 \\ & = (4 - \sqrt{3}) \text{ cm} \end{align}

\begin{align} xy + 10 & = 0 \phantom{00} \text{--- (1)} \\ \\ x + 2y + 1 & = 0 \\ x & = - 2y - 1 \phantom{00} \text{--- (2)} \\ \\ \text{Substitute } & \text{(2) into (1),} \\ (-2y - 1)(y) + 10 & = 0 \\ y(-2y - 1) + 10 & = 0 \\ -2y^2 - y + 10 & = 0 \\ 2y^2 + y - 10 & = 0 \\ (2y + 5)(y - 2) & = 0 \end{align} \begin{align} 2y + 5 & = 0 && \text{ or } & y - 2 & = 0 \\ 2y & = -5 &&& y & = 2 \\ y & = -{5 \over 2} \\ \\ \text{Substitute} & \text{ into (2),} &&& \text{Substitute} & \text{ into (2),} \\ x & = -2\left(-{5 \over 2}\right) - 1 &&& x & = -2(2) - 1 \\ & = 4 &&& & = -5 \end{align}

\begin{align} 3 - 12x - 2x^2 & = - 2x^2 - 12x + 3 \\ & = -2(x^2 + 6x) + 3 \\ & = -2 \left[ \left(x + {6 \over 2}\right)^2 - \left(6 \over 2\right)^2 \right] + 3 \phantom{000000} [\text{Complete the square}] \\ & = -2 [(x + 3)^2 - 9] + 3 \\ & = -2 (x + 3)^2 + 18 + 3 \\ & = -2 (x + 3)^2 + 21 \\ \\ \\ y & = 3 - 12x - 2x^2 \\ y & = -2(x + 3)^2 + 21 \\ y & = -2[x - (-3)]^2 + 21 \\ \\ \text{Turning } & \text{point: } (-3, 21) \end{align}

\begin{align} \int {3 \over x^2} + {4 \over 3x - 5} \phantom{.} dx & = \int {3 \over x^2} \phantom{.} dx + \int {4 \over 3x - 5} \phantom{.} dx \\ & = \int 3x^{-2} \phantom{.} dx + 4 \int {1 \over 3x - 5} \phantom{.} dx \\ & = 3 \left(x^{-1} \over -1\right) + 4 \int {1 \over 3x - 5} \phantom{.} dx \phantom{000000000} \text{Use } \int x^n \phantom{.} dx = {x^{n + 1} \over n + 1} \\ & = - 3x^{-1} + 4 \left[ \ln (3x - 5) \over 3\right] \phantom{00000000000} \text{Use } \int {1 \over f(x)} \phantom{.} dx = {\ln [f(x)] \over f'(x)} \\ & = - 3 \left(1 \over x\right) + {4 \over 3} \ln (3x - 5) \\ & = -{3 \over x} + {4 \over 3} \ln (3x - 5) + C \end{align}

\begin{align} {13x - 6 \over x^2 (2x - 3)} & = {A \over x} + {B \over x^2} + {C \over 2x - 3} \\ {13x - 6 \over x^2 (2x - 3)} & = {A(x)(2x - 3) \over x^2 (2x - 3)} + {B(2x - 3) \over x^2 (2x - 3)} + {C(x^2) \over x^2 (2x - 3)} \\ {13x - 6 \over x^2 (2x - 3)} & = {Ax(2x - 3) + B(2x - 3) + Cx^2 \over x^2 (2x - 3)} \\ \\ 13x - 6 & = Ax(2x - 3) + B(2x - 3) + Cx^2 \\ \\ \text{Let } & x = 0, \phantom{0000000000000000} [\text{Elimination method}] \\ 0 - 6 & = 0 + B[2(0) - 3] + 0 \\ -6 & = B(-3) \\ -6 & = -3B \\ {-6 \over -3} & = B \\ 2 & = B \\ \\ 13x - 6 & = Ax(2x - 3) + 2(2x - 3) + Cx^2 \\ \\ \text{Let } & x = 1.5, \\ 13(1.5) - 6 & = 0 + 2(0) + C(1.5)^2 \\ 13.5 & = C(2.25) \\ 13.5 & = 2.25C \\ {13.5 \over 2.25} & = C \\ 6 & = C \\ \\ 13x - 6 & = Ax(2x - 3) + 2(2x - 3) + 6x^2 \\ \\ \text{Let } & x = 1, \\ 13(1) - 6 & = A(1)[2(1) - 3] + 2[2(1) - 3] + 6(1)^2 \\ 7 & = A(1)(-1) + 2(-1) + 6(1) \\ 7 & = -A - 2 + 6 \\ A & = -2 + 6 - 7 \\ A & = -3 \\ \\ \\ \therefore {13x - 6 \over x^2 (2x - 3)} & = {-3 \over x} + {2 \over x^2} + {6 \over 2x - 3} \\ & = -{3 \over x} + {2 \over x^2} + {6 \over 2x - 3} \end{align}

(a)

\begin{align} \text{Let } P(x) & = 2x^3 - x^2 - 13x + k \\ \\ P(2) & = 2(2)^3 - (2)^2 - 13(2) + k \\ & = 16 - 4 - 26 + k \\ & = -14 + k \\ \\ \text{Since } & P(2) = 6, \phantom{000000} [\text{Remainder theorem}] \\ -14 + k & = 6 \\ k & = 20 \end{align}

(b)

\begin{align} P(x) & = 2x^3 - x^2 - 13x + k \\ & = 2x^3 - x^2 - 13x + (-6) \\ & = 2x^3 - x^2 - 13x - 6 \\ \\ \text{Polynomial} & = \text{Divisor} \times \text{Quotient} + \text{Remainder} \\ \\ 2x^3 - x^2 - 13x - 6 & = (2x^2 + ax - 3)(x - b) + 0, \text{ where } b \text{ is a constant} \\ 2x^3 - x^2 - 13x - 6 & = (2x^2 + ax - 3)(x - b) \\ 2x^3 - x^2 - 13x - 6 & = 2x^3 - 2bx^2 + ax^2 - abx - 3x + 3b \\ 2x^3 - x^2 - 13x - 6 & = 2x^3 + (a - 2b)x^2 + (-ab - 3)x + 3b \\ \\ \text{Comparing } & \text{constants,} \\ -6 & = 3b \\ {-6 \over 3} & = b \\ -2 & = b \\ \\ \text{Comparing } & \text{coefficients of } x^2, \\ -1 & = a - 2b \\ -1 & = a - 2(-2) \\ -1 & = a + 4 \\ -5 & = a \end{align}

(a)

\begin{align} \text{Max. value} & = 2 \\ \\ \text{Min. value} & = -8 \\ \\ \text{Center line: } y & = {2 + (-8) \over 2} \\ & = -3 \\ \\ \therefore c & = -3 \end{align}

(b)

\begin{align} \text{Period} & = {\pi \over 2} \\ {2\pi \over b} & = {\pi \over 2} \\ 2(2\pi) & = \pi b \phantom{000000} [\text{Cross-multiply}] \\ 4\pi & = \pi b \\ {4\pi \over \pi} & = b \\ 4 & = b \end{align}

(c)

\begin{align} \text{Amplitude} & = {2 - (-8) \over 2} \\ & = 5 \\ \\ a & = 5, b = 4, c = -3 \\ \\ y & = a \sin bx + c \\ y & = 5 \sin 4x - 3 \end{align}

(a)

\begin{align} \text{Area of rectangle } ABCD & = \text{Length} \times \text{Breadth} \\ & = 80 \times 50 \\ & = 4000 \\ \\ \text{Area of triangle } ABP & = {1 \over 2} \times AP \times AB \\ & = {1 \over 2} \times x \times 80 \\ & = 40x \\ \\ \text{Area of triangle } BCQ & = {1 \over 2} \times BC \times CQ \\ & = {1 \over 2} \times 50 \times (80 - 2x) \\ & = 25(80 - 2x) \\ & = 2000 - 50x \\ \\ \text{Area of triangle } PDQ & = {1 \over 2} \times PD \times DQ \\ & = {1 \over 2} \times (50 - x) \times 2x \\ & = x(50 - x) \\ & = 50x - x^2 \\ \\ A & = 4000 - 40x - (2000 - 50x) - (50x - x^2) \\ & = 4000 - 40x - 2000 + 50x - 50x + x^2 \\ & = 2000 - 40x + x^2 \\ & = x^2 - 40x + 2000 \phantom{00} \text{ (Shown)} \end{align}

(b)

\begin{align} A & = x^2 - 40x + 2000 \\ \\ {dA \over dx} & = 2x - 40 \\ \\ \text{Let } & {dA \over dx} = 0, \phantom{000000} [\text{Stationary value}] \\ 0 & = 2x - 40 \\ 40 & = 2x \\ {40 \over 2} & = x \\ 20 & = x \\ \\ \text{Substitute } & x = 20 \text{ into } A = x^2 - 40x + 2000, \\ A & = (20)^2 - 40(20) + 2000 \\ & = 1600 \\ \\ {d^2 A \over dx^2} & = 2 > 0 \implies \text{ Minimum value} \\ \\ \\ \therefore \text{Minimum} & \text{ value of } A \text{ is } 1600 \text{ m}^2 \end{align}

(a) In kite ABCD, sides AB = AD and BC = DC

\begin{align} \text{Distance between two points} & = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\ \\ BC & = \sqrt{(6 - 0)^2 + (6 - k)^2} \\ BC & = \sqrt{ 36 + (6)^2 - 2(6)(k) + (k)^2 } \phantom{000000} [(a - b)^2 = a^2 - 2ab + b^2] \\ BC & = \sqrt{ 36 + 36 - 12k + k^2} \\ BC & = \sqrt{ 72 - 12k + k^2} \\ \\ DC & = \sqrt{(6 - h)^2 + (6 - 0)^2} \\ DC & = \sqrt{(6)^2 - 2(6)(h) + (h)^2 + 36} \\ DC & = \sqrt{36 - 12h + h^2 + 36} \\ DC & = \sqrt{72 - 12h + h^2} \\ \\ \text{Since } & BC = DC, \\ \sqrt{ 72 - 12k + k^2} & = \sqrt{72 - 12h + h^2} \\ 72 - 12k + k^2 & = 72 - 12h + h^2 \\ k^2 - 12k & = h^2 - 12h \\ 0 & = h^2 - k^2 - 12h + 12k \\ 0 & = (h + k)(h - k) - 12(h - k) \\ 0 & = (h - k)(h + k - 12) \end{align}
\begin{align} h - k & = 0 \text{ (Reject)} && \text{ or } & h + k - 12 & = 0 \\ & &&& h + k & = 12 \phantom{00} \text{ (Shown)} \end{align}

(b) In kite ABCD, diagonals AC and BD are perpendicular

\begin{align} h + k & = 12 \\ 4 + k & = 12 \\ k & = 12 - 4 \\ & = 8 \\ \\ \therefore B(0, 8) & \text{ and } D(4, 0) \phantom{000000} [\text{Update in diagram!}] \\ \\ \text{Gradient} & = {y_2 - y_1 \over x_2 - x_1} \\ \text{Gradient of } BD & = {0 - 8 \over 4 - 0} \\ & = -2 \\ \\ \text{Gradient of } AC & = {-1 \over \text{Gradient of } BD} \\ & = {-1 \over -2} \\ & = {1 \over 2} \\ \\ y & = mx + c \\ y & = {1 \over 2}x + c \\ \\ \text{Using } & C(6, 6) \\ 6 & = {1 \over 2}(6) + c \\ 6 & = 3 + c \\ 3 & = c \\ \\ \text{Eqn of line } & AC : y = {1 \over 2}x + 3 \\ \\ \text{Let } & y = 0, \\ 0 & = {1 \over 2}x + 3 \\ -{1 \over 2}x & = 3 \\ x & = {3 \over -{1 \over 2}} \\ x & = -6 \\ \\ \therefore & \phantom{.} A(-6, 0) \end{align}

(c) For 'shoelace' method, select the coordinates in an anti-clockwise manner and repeat the first coordinates chosen

\begin{align} \text{Area of kite } ABCD & = {1 \over 2} \left| \begin{matrix} -6 & 4 & 6 & 0 & -6 \\ 0 & 0 & 6 & 8 & 0 \end{matrix} \right| \\ & = {1 \over 2} \left[ (-6)(0) + (4)(6) + (6)(8) + (0)(0) \right] - {1 \over 2} \left[ (0)(4) + (0)(6) + (6)(0) + (8)(-6) \right] \\ & = 60 \text{ units}^2 \end{align}

(a) The identity used (sin² A + cos² A = 1) can be found in the provided formula sheet

\begin{align} \text{L.H.S} & = {\sin \theta \over 1 - \cos \theta} - {1 \over \sin \theta} \\ & = {\sin \theta (\sin \theta) \over \sin \theta (1 - \cos \theta) } - {1 - \cos \theta \over \sin \theta (1 - \cos \theta)} \\ & = {\sin^2 \theta - (1 - \cos \theta) \over \sin \theta (1 - \cos \theta)} \\ & = {\sin^2 \theta - 1 + \cos \theta \over \sin \theta (1 - \cos \theta)} \\ & = {1 - \cos^2 \theta - 1 + \cos \theta \over \sin \theta (1 - \cos \theta)} \phantom{000000} [\sin^2 A + \cos^2 A = 1 \rightarrow \sin^2 A = 1 - \sin^2 A] \\ & = { \cos \theta - \cos^2 \theta \over \sin \theta (1 - \cos \theta)} \\ & = { \cos \theta (1 - \cos \theta) \over \sin \theta (1 - \cos \theta) } \\ & = { \cos \theta \over \sin \theta } \\ & = \cot \theta \\ & = \text{R.H.S} \end{align}

(b) The word 'Hence' suggests that we need to use the identity proved in (i)

\begin{align} {\sin 2\theta \over 1 - \cos 2\theta} - {1 \over \sin 2 \theta} & = -2 \\ \\ \text{From (i), } {\sin \theta \over 1 - \cos \theta} - {1 \over \sin \theta} & = \cot \theta \\ \\ \therefore \cot 2 \theta & = -2 \\ {1 \over \tan 2 \theta} & = -2 \\ -{1 \over 2} & = \tan 2 \theta \phantom{000000} [\text{2nd or 4th quadrant since } \tan 2\theta < 0] \\ \\ \text{Basic angle, } \alpha & = \tan^{-1} \left(1 \over 2\right) \\ & = 26.56^\circ \end{align}

\begin{align} 2 \theta & = 180^\circ - 26.56^\circ, 360^\circ - 26.56^\circ \phantom{000000} [0^\circ \le \theta \le 180^\circ] \\ & = 153.44^\circ, 333.44^\circ \phantom{000000000000000} [0^\circ \le 2 \theta \le 360^\circ] \\ \\ \theta & = 76.72^\circ, 166.72^\circ \\ & \approx 76.7^\circ, 166.7^\circ \end{align}

(a)

\begin{align} \text{Let } \angle TCA & = \theta \\ \\ \angle BAC & = \angle TCA \phantom{00000} [\text{Alternate angles, } BA \phantom{.} // \phantom{.} CT] \\ & = \theta \\ \\ \angle ABC & = \angle TCA \phantom{00000} [\text{Alternate segment theorem}] \\ & = \theta \\ \\ \text{Since } \angle ABC & = \angle BAC, \text{ triangle } ABC \text{ is isoceles} \end{align}

(b)

\begin{align} \text{Since tangents } & \text{meet at external point } T, TA = TC \\ \\ \angle TAC & = \angle TCA \phantom{0000000000} [\text{Isosceles triangle}] \\ & = \theta \\ \\ \angle CTA & = 180^\circ - \theta - \theta \phantom{00000} [\text{Angle sum of triangle}] \\ & = 180^\circ - 2 \theta \\ \\ \angle BCA & = 180^\circ - \theta - \theta \phantom{00000} [\text{Angle sum of triangle}] \\ & = 180^\circ - 2 \theta \\ & = \angle CTA \phantom{00} \text{ (Shown)} \end{align}

(a)

\begin{align} 6^x & = 5 \times 3^{x + 1} \\ \\ \text{Take } & \lg \text{ of both sides}, \\ \lg 6^x & = \lg (5 \times 3^{x + 1}) \\ \lg 6^x & = \lg 5 + \lg 3^{x + 1} \phantom{00000000.} [\text{Product law (logarithms)}] \\ x \lg 6 & = \lg 5 + (x + 1)\lg 3 \phantom{00000} [\text{Power law (logarithms)}] \\ x \lg 6 & = \lg 5 + x \lg 3 + \lg 3 \\ x \lg 6 - x \lg 3 & = \lg 5 + \lg 3 \\ x(\lg 6 - \lg 3) & = \lg 5 + \lg 3 \\ x \lg {6 \over 3} & = \lg 5 + \lg 3 \phantom{00000000000.} [\text{Quotient law (logarithms)}] \\ x \lg 2 & = \lg (5 \times 3) \phantom{000000000000.} [\text{Product law (logarithms)}] \\ x \lg 2 & = \lg 15 \\ x & = {\lg 15 \over \lg 2} \phantom{00} (\text{Shown)} \end{align}

(b)

\begin{align} \log_3 x + \log_9 (x + 2) & = 2 \\ \log_3 x + {\log_3 (x + 2) \over \log_3 9} & = 2 \phantom{000000} [\text{Change-of-base}] \\ \log_3 x + {\log_3 (x + 2) \over \log_3 3^2} & = 2 \\ \log_3 x + {\log_3 (x + 2) \over 2 \log_3 3} & = 2 \phantom{000000} [\text{Power law (logarithms)}] \\ \log_3 x + {\log_3 (x + 2) \over 2(1)} & = 2 \\ 2 \left[ \log_3 x + {\log_3 (x + 2) \over 2} \right] & = 2(2) \\ 2 \log_3 x + \log_3 (x + 2) & = 4 \\ \log_3 x^2 + \log_3 (x + 2) & = 4 \phantom{000000} [\text{Power law (logarithms)}] \\ \log_3 [x^2 (x + 2)] & = 4 \phantom{000000} [\text{Product law (logarithms)}] \\ \\ x^2(x + 2) & = 3^4 \\ x^3 + 2x^2 & = 81 \\ x^3 + 2x^2 - 81 & = 0 \end{align}

(a) (i)

\begin{align} \text{Volume after 1 min (60 seconds)} & = 500 \times 60 \\ & = 30 \phantom{.} 000 \text{ cm}^3 \\ \\ \text{Let } V \text{ denote the } & \text{volume of the balloon} \\ \\ V & = {4 \over 3} \pi r^3 \\ \\ \text{When } & V = 30 \phantom{.} 000, \\ 30 \phantom{.} 000 & = {4 \over 3} \pi r^3 \\ {30 \phantom{.} 000 \over {4 \over 3} \pi} & = r^3 \\ 7 \phantom{.} 161.972 & = r^3 \\ \sqrt[3]{ 7 \phantom{.} 161.972 } & = r \\ 19.275 & = r \\ \\ \text{Radius} & \approx 19.3 \text{ cm} \end{align}

(a) (ii)

\begin{align} {dr \over dt} & = {dr \over dV} \times {dV \over dt} \\ {dr \over dt} & = \underbrace{dr \over dV}_{\text{Need to find this}} \times 500 \\ \\ V & = {4 \over 3} \pi r^3 \\ \\ {dV \over dr} & = {4 \over 3} (3) \pi r^2 \\ & = 4 \pi r^2 \\ \\ {dr \over dV} & = {1 \over 4\pi r^2} \\ \\ \text{When } & r = 19.275, \\ {dr \over dV} & = {1 \over 4\pi (19.275)^2} \\ & = 0.000 \phantom{.} 214 \phantom{.} 191 \\ \\ \therefore {dr \over dt} & = 0.000 \phantom{.} 214 \phantom{.} 191 \times 500 \\ & = 0.107 \phantom{.} 095 \\ & \approx 0.107 \text{ cm/s} \end{align}

(b)

\begin{align} PV & = k \\ \\ \text{When } & P = 1.2 \text{ and } V = 2, \\ (1.2)(2) & = k \\ 2.4 & = k \\ \\ PV & = 2.4 \\ V & = {2.4 \over P} \\ V & = 2.4 P^{-1} \\ \\ {dV \over dP} & = 2.4 (-1) P^{-2} \\ & = -2.4 \left(1 \over P^2\right) \\ & = -{2.4 \over P^2} \\ \\ \text{When } & V = 2, P = 1.2 \\ \\ \text{Substitute } & P = 1.2 \text{ into } {dV \over dP}, \\ {dV \over dP} & = -{2.4 \over (1.2)^2} \\ & = -{5 \over 3} \text{ litres per atmosphere} \end{align}

(a)

\begin{align} y & = (4 + 3x)^{1 \over 2} \\ \\ {dy \over dx} & = {1 \over 2}(4 + 3x)^{-{1 \over 2}}. (3) \phantom{000000} [\text{Chain rule}] \\ & = {3 \over 2} \left(1 \over \sqrt{4 + 3x}\right) \\ & = {3 \over 2 \sqrt{4 + 3x}} \\ \\ \text{When } & x = 4, \\ {dy \over dx} & = {3 \over 2 \sqrt{4 + 3(4)} } \\ & = {3 \over 8} \\ \\ \text{Gradient of tangent at } P & = {3 \over 8} \\ \\ \text{Gradient of normal at } P & = {-1 \over {3 \over 8}} \\ & = -{8 \over 3} \\ \\ y & = -{8 \over 3}x + c \\ \\ \text{Using } & P(4, 4), \\ 4 & = -{8 \over 3}(4) + c \\ 4 & = -{32 \over 3} + c \\ {44 \over 3} & = c \\ \\ \text{Eqn of normal at } & P: y = -{8 \over 3}x + {44 \over 3} \\ \\ \text{Let } & y = 0, \\ 0 & = -{8 \over 3}x + {44 \over 3} \\ {8 \over 3}x & = {44 \over 3} \\ x & = { {44 \over 3} \over {8 \over 3}} \\ x & = 5{1 \over 2} \\ \\ \therefore & \phantom{.} Q \left(5{1 \over 2}, 0\right) \end{align}

(b)

\begin{align} \text{Area under curve} & = \int_0^4 (4 + 3x)^{1 \over 2} \phantom{.} dx \\ & = \left[ (4 + 3x)^{3 \over 2} \over \left(3 \over 2\right) (3) \right]_0^4 \phantom{00000000000} \text{Use } \int [f(x)]^n \phantom{.} dx = {[f(x)]^{n + 1} \over (n + 1). f'(x)} \phantom{.} dx \\ & = { [4 + 3(4)]^{3 \over 2} \over {9 \over 2} } - { [4 + 3(0)]^{3 \over 2} \over {9 \over 2} } \\ & = 12{4 \over 9} \text{ units}^2 \\ \\ \text{Area under normal at } P & = \text{Area of triangle} \\ & = {1 \over 2} \times \left(5{1 \over 2} - 4\right) \times 4 \\ & = 3 \text{ units}^2 \\ \\ \text{Area of shaded region} & = 12{4 \over 9} + 3 \\ & = 15{4 \over 9} \text{ units}^2 \end{align}

\begin{align} 3e^x + 5 & = 2e^{-x} \\ 3e^x + 5 & = 2 \left(1 \over e^x\right) \\ 3e^x + 5 & = {2 \over e^x} \\ \\ \text{Let } & u = e^x, \\ 3u + 5 & = {2 \over u} \\ u(3u + 5) & = u\left(2 \over u\right) \\ 3u^2 + 5u & = 2 \\ 3u^2 + 5u - 2 & = 0 \\ (3u - 1)(u + 2) & = 0 \end{align} \begin{align} 3u - 1 & = 0 && \text{ or } & u + 2 & = 0 \\ 3u & = 1 &&& u & = -2 \\ u & = {1 \over 3} \\ \\ \text{Since } u & = e^x, &&& \text{Since } u & = e^x, \\ e^x & = {1 \over 3} &&& e^x & = -2 \text{ (Reject, since } e^x > 0) \\ \\ \text{Take } & \ln \text{ of both sides,} \\ \ln e^x & = \ln {1 \over 3} \\ x \ln e & = \ln {1 \over 3} \\ x (1) & = \ln {1 \over 3} \\ x & = -1.09861 \\ x & \approx -1.1 \text{ (2 s.f.)} \end{align}
$$ \therefore \text{Only solution of equation is } x = -1.1 $$

\begin{align} \text{Let } f(x) & = 3x^3 + 4x^2 - x - 2 \\ \\ f(-1) & = 3(-1)^3 + 4(-1)^2 - (-1) - 2 \\ & = 0 \\ \\ \text{By Factor} & \text{ theorem, } x + 1 \text{ is a factor of } f(x) \\ \therefore x = - 1 & \text{ is a solution of the equation} \\ \\ \text{Polynomial} & = \text{Divisor} \times \text{Quotient} + \text{Remainder} \\ 3x^3 + 4x^2 - x - 2 & = (x + 1)(ax^2 + bx + c) + 0 \\ 3x^3 + 4x^2 - x - 2 & = (x + 1)(ax^2 + bx + c) \\ 3x^3 + 4x^2 - x - 2 & = ax^3 + bx^2 + cx + ax^2 + bx + c \\ 3x^3 + 4x^2 - x - 2 & = ax^3 + (a + b)x^2 + (b + c)x + c \\ \\ \text{Comparing } & \text{constants,} \\ -2 & = c \\ \\ \text{Comparing } & \text{coefficients of } x^3, \\ 3 & = a \\ \\ \text{Comparing } & \text{coefficients of } x^2, \\ 4 & = a + b \\ 4 & = 3 + b \\ 1 & = b \\ \\ 3x^3 + 4x^2 - x - 2 & = (x + 1)[(3)x^2 + (1)x + (-2)] \\ & = (x + 1)(3x^2 + x - 2) \\ & = (x + 1)(x + 1)(3x - 2) \\ & = (x + 1)^2 (3x - 2) \\ \\ (x + 1)^2 (3x - 2) & = 0 \\ \\ x + 1 & = 0 \phantom{0} \text{ or } \phantom{0} 3x - 2 = 0 \\ x & = -1 \phantom{000000.} 3x = 2 \\ & \phantom{000000000000.} x = {2 \over 3} \end{align}

(a)

\begin{align} {dy \over dx} & = {v {du \over dx} - u {dv \over dx} \over v^2} \phantom{0000000000000000000} [\text{Quotient rule}] \\ & = {(2x + 1)^{1 \over 2} (1) - (x)(2x + 1)^{-{1 \over 2}} \over [(2x + 1)^{1 \over 2}]^2 } \\ & = {(2x + 1)^{1 \over 2} - x (2x + 1)^{-{1 \over 2}} \over (2x + 1)} \\ & = { (2x + 1)^{-{1 \over 2}} (2x + 1 - x) \over (2x + 1)} \\ & = { x + 1 \over (2x + 1)^{1 \over 2} (2x + 1) } \\ & = { x + 1 \over (2x + 1)^{3 \over 2} } \phantom{00} \text{ (Shown)} \end{align}

(b)

\begin{align} \text{From (i), } {d \over dx} \left[x \over (2x + 1)^{1 \over 2}\right] & = { x + 1 \over (2x + 1)^{3 \over 2} } \\ \\ \implies \int { x + 1 \over (2x + 1)^{3 \over 2} } \phantom{.} dx & = {x \over (2x + 1)^{1 \over 2}} \\ \\ \int {x \over (2x + 1)^{3 \over 2} } \phantom{.} dx + \int {1 \over (2x + 1)^{3 \over 2} } \phantom{.} dx & = {x \over (2x + 1)^{1 \over 2}} \\ \int {x \over (2x + 1)^{3 \over 2} } \phantom{.} dx & = {x \over (2x + 1)^{1 \over 2}} - \int {1 \over (2x + 1)^{3 \over 2} } \phantom{.} dx \\ & = {x \over (2x + 1)^{1 \over 2}} - \int (2x + 1)^{-{3 \over 2}} \phantom{.} dx \\ & = {x \over (2x + 1)^{1 \over 2}} - { (2x + 1)^{-{1 \over 2}} \over \left(-{1 \over 2}\right)(2) } \phantom{000000} \text{Use } \int [f(x)]^n \phantom{.} dx = {[f(x)]^{n + 1} \over (n+1). f'(x)} \\ & = {x \over (2x + 1)^{1 \over 2}} - { (2x + 1)^{-{1 \over 2}} \over -1} \\ & = {x \over (2x + 1)^{1 \over 2}} + {1 \over \sqrt{2x + 1}} \\ \\ \\ \therefore \int_0^4 {x \over (2x + 1)^{3 \over 2} } \phantom{.} dx & = \left[ {x \over (2x + 1)^{1 \over 2}} + {1 \over \sqrt{2x + 1}} \right]_0^4 \\ & = \left[ {4 \over [2(4) + 1]^{1 \over 2}} + {1 \over \sqrt{2(4) + 1}} \right] - \left[ {0 \over [2(0) + 1]^{1 \over 2}} + {1 \over \sqrt{2(0) + 1}} \right] \\ & = {2 \over 3} \end{align}

(a)

\begin{align} \sin \angle BAE & = {BE \over BA} \\ \sin \theta & = {BE \over 6} \\ 6 \sin \theta & = BE \\ \\ \angle AGF & = 180^\circ - 90^\circ - \theta \phantom{00000} [\text{Angle sum of triangle}] \\ & = 90^\circ - \theta \\ \\ \angle CGB & = 90^\circ - \theta \phantom{00000} [\text{Vertically opposite angles}] \\ \\ \angle GCB & = 180^\circ - 90^\circ - (90^\circ - \theta) \phantom{00000} [\text{Angle sum of triangle}] \\ & = 90^\circ - 90^\circ + \theta \\ & = \theta \\ \\ \cos \angle BCD & = {CD \over CB} \\ \cos \theta & = {CD \over 4} \\ 4 \cos \theta & = CD \\ \\ h & = CD + DF \\ & = 4 \cos \theta + BE \\ & = 4 \cos \theta + 6 \sin \theta \phantom{00} \text{ (Shown)} \end{align}

(b)

\begin{align} a \sin \theta + b \cos \theta & = R \sin (\theta + \alpha) \\ \\ a & = 6, b = 4 \\ \\ R & = \sqrt{a^2 + b^2} \\ & = \sqrt{6^2 + 4^2} \\ & = \sqrt{52} \\ \\ \alpha & = \tan^{-1} \left(b \over a\right) \\ & = \tan^{-1} \left(4 \over 6\right) \\ & = 33.69^\circ \\ \\ h & = \sqrt{52} \sin (\theta + 33.69^\circ) \\ h & \approx \sqrt{52} \sin (\theta + 33.7^\circ) \end{align}

(c) The maximum value of the sine function R sin (x + α) is R and it occurs when x + α = 90°.

\begin{align} h = \sqrt{52} & \sin (\theta + 33.69^\circ) \\ \\ \text{Max. value of } h & = \sqrt{52} \\ \\ \theta + 33.69^\circ & = 90^\circ \\ \theta & = 90^\circ - 33.69^\circ \\ \theta & = 56.31^\circ \\ & \approx 56.3^\circ \end{align}

(a)

\begin{align} y & > 11 \phantom{000000} [\text{Lies above}] \\ 2x^2 - 6x + 3 & > 11 \\ 2x^2 - 6x - 8 & > 0 \\ x^2 - 3x - 4 & > 0 \\ (x + 1)(x - 4) & > 0 \end{align}

$$ x < -1 \phantom{0} \text{ or } \phantom{0} x > 4 $$

$$ \{ x : x \in \mathbb{R} \phantom{.} | \phantom{0} x < -1 \text{ or } \phantom{0} x > 4 \} $$

(b)

\begin{align} y & = 2x^2 - 6x + 3 \phantom{00} \text{--- (1)} \\ \\ y & = 2x + k \phantom{00} \text{--- (2)} \\ \\ \text{Substitute } & \text{(2) into (1),} \\ 2x + k & = 2x^2 - 6x + 3 \\ 0 & = 2x^2 - 8x + (3 - k) \\ \\ b^2 - 4ac & = (-8)^2 - 4(2)(3 - k) \\ & = 64 - 8(3 - k) \\ & = 64 - 24 + 8k \\ & = 40 + 8k \\ \\ b^2 - 4ac & = 0 \phantom{00} [\text{Line meets curve only once}] \\ 40 + 8k & = 0 \\ 8k & = -40 \\ k & = {-40 \over 8} \\ k & = -5 \end{align}

(c)

\begin{align} \text{From (b), } 0 & = 2x^2 - 8x + (3 - k) \\ 0 & = 2x^2 - 8x + [3 - (-5)] \\ 0 & = 2x^2 - 8x + 8 \\ 0 & = x^2 - 4x + 4 \\ 0 & = (x - 2)^2 \\ 0 & = x - 2 \\ 2 & = x \\ \\ \text{Substitute } & x = 2 \text{ into } (1), \\ y & = 2(2)^2 - 6(2) + 3 \\ y & = -1 \\ \\ \therefore & \phantom{.} P(2, -1) \end{align}

(a)

\begin{align} \left( 2 - {3 \over x}\right)^6 & = (2)^6 + {6 \choose 1} (2)^5 \left(-{3 \over x}\right) + {6 \choose 2} (2)^4 \left(-{3 \over x}\right)^2 + {6 \choose 3} (2)^3 \left(-{3 \over x}\right)^3 + ... \\ & = ... + (15)(16)\left(9 \over x^2\right) + (20)(8)\left(-{27 \over x^3}\right) + ... \\ & = ... + {2160 \over x^2} - {4320 \over x^3} + ... \\ & = ... + 2160 \left(1 \over x^2\right) - 4320 \left(1 \over x^3\right) + ... \end{align}

(b)

\begin{align} (x^2 + ax)\left( 2 - {3 \over x}\right)^6 & = (x^2 + ax) \left[ ... + 2160 \left(1 \over x^2\right) - 4320 \left(1 \over x^3\right) + ... \right] \\ & = ... + (x^2) (-4320) \left(1 \over x^3\right) + (ax)(2160)\left(1 \over x^2\right) + ... \\ & = ... - 4320 \left(1 \over x\right) + 2160 a \left(1 \over x\right) + ... \\ \\ \text{Coefficient of } {1 \over x} & = 2160a - 4320 \\ \\ \text{Since there } & \text{is no terms in } {1 \over x}, \\ 2160a - 4320 & = 0 \\ 2160a & = 4320 \\ a & = {4320 \over 2160} \\ a & = 2 \end{align}

(c)

\begin{align} \left( 2 - {3 \over x}\right)^6 & = (2)^6 + {6 \choose 1} (2)^5 \left(-{3 \over x}\right) + ... \\ & = 64 + (6)(32) \left(-{3 \over x}\right) + ... \\ & = 64 - {576 \over x} + ... \\ & = 64 - 576 \left(1 \over x\right) + ... \\ \\ (x^2 + 2x)\left( 2 - {3 \over x}\right)^6 & = (x^2 + 2x)\left[64 - 576 \left(1 \over x\right) + ...\right] \\ & = ... + (x^2)(-576)\left(1 \over x\right) + (2x)(64) + ... \\ & = ... - 576x + 128x + ... \\ & = ... - 448x + ... \\ \\ \text{Coefficient} & \text{ of } x = -448 \end{align}

(a)

\begin{align} d & = av^2 + bv \\ {1 \over v}(d) & = {1 \over v}(av^2 + bv) \\ {d \over v} & = av + b \phantom{000000} [Y = mX + c]\\ \\ \text{1) Plot } & {d \over v} \text{ against } v \\ \text{2) Grad} & \text{ient of line is equals to } a \\ \text{3) Verti} & \text{cal intercept is equals to } b \end{align}

(b)(i)

\begin{align} n & = ab^t \\ \\ \text{Take } & \lg \text{ of both sides}, \\ \lg n & = \lg (ab^t) \\ \lg n & = \lg a + \lg b^t \phantom{000000} [\text{Product law (logarithms)}] \\ \lg n & = \lg a + t \lg b \phantom{000000} [\text{Power law (logarithms)}] \\ \\ \lg n & = (\lg b)(t) + \lg a \phantom{000000} [Y = mX + c]\\ \\ \text{Plot } & \lg n \text{ against } t \end{align}

t	1	2	3	4
lg n	2.908	2.653	2.380	2.130

(b)(ii) For the next decade (2000-2009), t = 5

\begin{align} \text{From graph, } \lg n & = 1.87 \\ \log_{10} n & = 1.87 \\ n & = 10^{1.87} \\ & \approx 74.1 \end{align}

(b)(iii)

\begin{align} \text{Efforts may be undertaken to reduce the decrease, such as harsh penalties on poachers} \end{align}

(a)

\begin{align} v & = 24e^{-{t \over 6}} \\ \\ \text{When } & t = 0, \\ v & = 24e^{-{0 \over 6}} \\ v & = 24(1) \\ v & = 24 \\ \\ \text{Velocity at } A & = 24 \text{ m/s} \\ \\ \text{Velocity at } B & = {1 \over 2} \times 24 \\ & = 12 \text{ m/s} \\ \\ \text{Substitute } & v = 12 \text{ into } v = 24e^{-{t \over 6}} \\ 12 & = 24e^{-{t \over 6}} \\ {12 \over 24} & = e^{-{t \over 6}} \\ {1 \over 2} & = e^{-{t \over 6}} \\ \\ \text{Take } \ln & \text{ of both sides,} \\ \ln {1 \over 2} & = \ln e^{-{t \over 6}} \\ \ln {1 \over 2} & = -{t \over 6} \ln e \phantom{000000} [\text{Power law (logarithms)}] \\ \ln {1 \over 2} & = -{t \over 6} (1) \\ \ln {1 \over 2} & = -{t \over 6} \\ -6 \ln {1 \over 2} & = t \\ \\ \text{Time taken} & = -6 \ln {1 \over 2} \\ & = 4.1588 \\ & \approx 4.16 \text{ seconds} \end{align}

(b)

\begin{align} a & = {dv \over dt} \\ & = {d \over dt} (24 e^{-{1 \over 6}t}) \\ & = 24 \left(-{1 \over 6}\right) e^{-{1 \over 6}t} \phantom{00000000} \text{Use } {d \over dx}[e^{f(x)}] = f'(x) . e^{f(x)} \\ & = -4 e^{-{1 \over 6}t} \\ \\ \text{When } & t = 4.1588, \\ a & = -4e^{-{1 \over 6}(4.1588)} \\ & = -2 \text{ m/s}^2 \end{align}

(c)

\begin{align} s & = \int v \phantom{.} dt \\ s & = \int 24 e^{-{t \over 6}} \phantom{.} dt \\ s & = 24 \left( e^{-{t \over 6}} \over -{1 \over 6} \right) + c \\ s & = - 144 e^{-{t \over 6}} + c \\ \\ \text{When } & t = 0 \text{ and } s = 0, \phantom{000000} [\text{Initially at } A] \\ 0 & = -144e^{-{0 \over 6}} + c \\ 0 & = -144(1) + c \\ 0 & = -144 + c \\ 144 & = c \\ \\ s & = - 144 e^{-{t \over 6}} + 144 \\ \\ \text{When } & t = 4.1588, \\ s & = - 144 e^{-{4.1588 \over 6}} + 144 \\ & = 71.999 \\ & \approx 72.0 \text{ m} \end{align}

(a)

\begin{align} (x + 2a)^2 + (y - a)^2 & = ka^2 \\ [x - (-2a)]^2 + (y - a)^2 & = 4a^2 \\ [x - (-2a)]^2 + (y - a)^2 & = (2a)^2 \\ \\ \text{Center: } & (-2a, a) \\ \\ \text{Radius} & = 2a \text{ units} \end{align}

Note a is a positive value, so circle lies to the left of the y-axis:

\begin{align} y \text{-axis meets circle at } (0, a) & \text{ and is perpendicular to radius} \\ \\ \therefore y \text{-axis is tangent to the circ} & \text{le at } (0, a) \end{align}

(b) Refer to the sketch in part (a) - the two tangents are marked in blue

\begin{align} a + 2a & = 3a \\ \\ a - 2a & = -a \\ \\ \text{Points: } & (-2a, -a) \text{ and } (-2a, 3a) \end{align}

(c)

\begin{align} (x + 2a)^2 + (y - a)^2 & = ka^2 \\ (x + 2a)^2 + (y - a)^2 & = 5a^2 \\ \\ \text{Let } & x = 0, \\ (0 + 2a)^2 + (y - a)^2 & = 5a^2 \\ (2a)^2 + (y - a)^2 & = 5a^2 \\ 4a^2 + (y - a)^2 & = 5a^2 \\ (y - a)^2 & = a^2 \\ y - a & = \pm \sqrt{a^2} \\ y - a & = a \text{ or } - a \\ y & = 2a \text{ or } 0 \\ \\ \therefore \text{Circle passes through } & \text{origin (0, 0)} \end{align}

(d)

\begin{align} (x + 2a)^2 + (y - a)^2 & = 5a^2 \\ [x - (-2a)]^2 + (y - a)^2 & = 5a^2 \\ \\ \text{Centre: } & (-2a, a) \\ \\ \text{Let } (b, c) \text{ denote } & \text{coordinates of } P \\ \\ \text{Midpoint} & = \left( {x_1 + x_2 \over 2}, {y_1 + y_2 \over 2} \right) \\ \\ \text{Midpoint of } OP & = \left( {0 + b \over 2}, {0 + c \over 2} \right) \\ & = \left( {b \over 2}, {c \over 2}\right) \\ \\ \text{Since } OP \text{ is the diameter, } & \text{midpoint is the centre of the circle} \\ \\ (-2a, a) & = \left( {b \over 2}, {c \over 2}\right) \\ \\ {b \over 2} & = -2a \\ b & = -4a \\ \\ {c \over 2} & = a \\ c & = 2a \\ \\ \therefore & \phantom{.} P(-4a, 2a) \\ \\ \\ \text{Gradient of } OP & = {2a - 0 \over -4a - 0} \\ & = {2a \over -4a} \\ & = -{1 \over 2} \\ \\ \text{Gradient of tangent at } P & = {-1 \over -{1 \over 2}} \\ & = 2 \\ \\ y & = mx + c \\ y & = 2x + c \\ \\ \text{Using } & P(-4a, 2a), \\ 2a & = 2(-4a) + c \\ 2a & = -8a + c \\ 10a & = c \\ \\ \text{Eqn of tangent at } & P : y = 2x + 10a \\ \\ \text{Let } & y = 0, \\ 0 & = 2x + 10a \\ -2x & = 10a \\ x & = {10a \over -2} \\ x & = -5a \\ \\ \therefore \text{Coordinates of } & \text{point: } (-5a, 0) \end{align}

Thought process:

1. The shaded area is the difference between the area bounded by the curve and the area bounded by the line OM (triangle)
2. To find the respective areas, we need the coordinates of point M
3. To find the x-coordinates of maximum point M, we need to find ^dy/_dx and equate it to 0, then solve for the y-coordinate

\begin{align} y & = 4 \sin {x \over 2} - x \\ y & = 4 \sin \left({1 \over 2}x \right) - x \\ \\ {dy \over dx} & = 4 \left({1 \over 2}\right) \cos \left({1 \over 2}x \right) - 1 \phantom{000000000} \text{Use } {d \over dx}[\sin f(x)] = f'(x). \cos f(x) \\ & = 2 \cos \left({1 \over 2}x \right) - 1 \\ \\ \text{Let } & {dy \over dx} = 0, \phantom{000000} [\text{Maximum point } M] \\ 0 & = 2 \cos \left({1 \over 2}x \right) - 1 \\ -2 \cos \left({1 \over 2}x \right) & = -1 \\ \cos \left({1 \over 2}x \right) & = {1 \over 2} \phantom{000000000} \left[\text{1st or 4th quadrant since } \cos \left({1 \over 2}x\right) > 0 \right] \\ \\ \text{Basic angle, } \alpha & = \cos^{-1} \left(1 \over 2\right) \\ & = 60^\circ \\ & = {\pi \over 3} \end{align}

\begin{align} {1 \over 2}x & = {\pi \over 3}, 2\pi - {\pi \over 3} \phantom{00000000000} [0 \le x \le \pi] \\ & = {\pi \over 3}, {5\pi \over 3} \text{ (Reject)} \phantom{0000000} \left[ 0 \le {1 \over 2}x \le {1 \over 2}\pi \right] \\ \\ x & = 2 \left(\pi \over 3\right) \\ x & = {2\pi \over 3} \\ \\ \text{Substitute } & x = {2\pi \over 3} \text{ into } y = 4 \sin \left({1 \over 2}x\right) - x, \\ y & = 4 \sin \left[ {1 \over 2} \left(2\pi \over 3\right) \right] - {2\pi \over 3} \\ y & = 4 \sin {\pi \over 3} - {2\pi \over 3} \\ y & = 4 \left(\sqrt{3} \over 2\right) - {2\pi \over 3} \\ y & = 2 \sqrt{3} - {2\pi \over 3} \\ \\ \therefore & \phantom{.} M \left({2\pi \over 3}, 2 \sqrt{3} - {2\pi \over 3} \right) \\ \\ \\ \text{Area under curve} & = \int_0^{2\pi \over 3} 4 \sin \left({1 \over 2}x\right) - x \phantom{.} dx \\ & = \left[ 4 \left(- \cos \left({1 \over 2}x\right) \over {1 \over 2} \right) - {x^2 \over 2} \right]_0^{2\pi \over 3} \phantom{000000} \text{Use } \int \sin f(x) \phantom{.} dx = {- \cos f(x) \over f'(x)} \\ & = \left[ - 8 \cos \left({1 \over 2}x\right) - {1 \over 2} x^2 \right]_0^{2\pi \over 3} \\ & = \left[ - 8 \cos \left[{1 \over 2}\left(2\pi \over 3\right)\right] - {1 \over 2} \left(2\pi \over 3\right)^2 \right] - \left[ - 8 \cos \left[{1 \over 2}(0)\right] - {1 \over 2} (0)^2 \right] \\ & = \left[ - 8 \cos {\pi \over 3} - {1 \over 2} \left(4 \pi^2 \over 9\right) \right] - \left[ - 8 \cos 0 - 0 \right] \\ & = \left[ - 8 \left(1 \over 2\right) - {2\pi^2 \over 9} \right] - [- 8(1)] \\ & = - 4 - {2\pi^2 \over 9} + 8 \\ & = 4 - {2\pi^2 \over 9} \\ \\ \text{Area under line } OM & = \text{Area of triangle} \\ & = {1 \over 2} \times {2\pi \over 3} \times \left(2\sqrt{3} - {2\pi \over 3}\right) \\ & = {\pi \over 3} \left(2\sqrt{3} - {2\pi \over 3}\right) \\ & = {2\pi \sqrt{3} \over 3} - {2\pi^2 \over 9} \\ \\ \text{Area of shaded region} & = 4 - {2\pi^2 \over 9} - \left( {2\pi \sqrt{3} \over 3} - {2\pi^2 \over 9} \right) \\ & = 4 - {2\pi^2 \over 9} - {2\pi \sqrt{3} \over 3} + {2\pi^2 \over 9} \\ & = 4 - {2\pi \sqrt{3} \over 3} \text{ units}^2 \phantom{00} \text{ (Shown)} \end{align}