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

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

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

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

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

(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}$$