(a)

\begin{align} 2x^2 - 8x + 11 & = 2(x^2 - 4x) + 11 \\ & = 2 \left[ x^2 - 4x + \left(4 \over 2\right)^2 - \left(4 \over 2\right)^2 \right] + 11 \\ & = 2 ( x^2 - 4x + 2^2 - 4 ) + 11 \\ & = 2 [ (x - 2)^2 - 4 ] + 11 \\ & = 2 (x - 2)^2 - 8 + 11 \\ & = 2 (x - 2)^2 + 3 \\ \\ y & = 2(x - 2)^2 + 3 \\ \\ \text{Minimum } & \text{point: } (2, 3) \end{align}

(b)

\begin{align} y & = 2x^2 - 8x + 11 \phantom{00} \text{--- (1)} \\ \\ y & = 2x + 3 \phantom{00} \text{--- (2)} \\ \\ \text{Substitute } & \text{(2) into (1),} \\ 2x + 3 & = 2x^2 - 8x + 11 \\ 0 & = 2x^2 - 10x + 8 \\ 0 & = x^2 - 5x + 4 \\ 0 & = (x - 1)(x - 4) \end{align} \begin{align} x - 1 & = 0 && \text{ or } & x - 4 & =0 \\ x & = 1 &&& x & = 4 \\ \\ \text{Substitute } & \text{into (2),} &&& \text{Substitute } & \text{into (2),} \\ y & = 2(1) + 3 &&& y & = 2(4) + 3 \\ y & = 5 &&& y & = 11 \\ \\ \therefore & \phantom{.} (1, 5) &&& \therefore & \phantom{.} (4, 11) \end{align}
\begin{align} \text{Distance between two points} & = \sqrt{ (x_2 - x_1)^2 + (y_2 - y_1)^2 } \\ & = \sqrt{ (4 - 1)^2 + (11 - 5)^2 } \\ & = \sqrt{45} \\ \\ \therefore k & = 45 \end{align}

(a)

\(t\)	\(1\)	\(2\)	\(3\)	\(4\)	\(5\)
\(\ln m\)	\(2.747\)	\(2.493\)	\(2.251\)	\(2.001\)	\(1.740\)

On a phone, tap the diagram to see it full size (and tap again to fit).

(b)

\begin{align} \text{From graph, when } & t = 0, \ln m = 3 \\ \\ \ln m & = 3 \\ \log_e m & = 3 \\ m & = e^3 \\ m & \approx 20.1 \text{ g} \end{align}

(c)

\begin{align} 50\% \text{ of } B, m & = e^3 \times {1 \over 2} \\ & = {1 \over 2} e^3 \\ \\ \ln m & = \ln \left({1 \over 2} e^3 \right) \\ & = 2.3069 \\ & \approx 2.31 \\ \\ \text{From graph, when } & \ln m = 2.31, t = 2.75 \text{ mins} \\ \\ \text{Time taken} & = 2.75 \times 60 \\ & = 165 \text{ seconds} \end{align}

(a)

$$ \require{enclose} \begin{array}{rll} 2 \phantom{00000000000000.}\\ x^3 + 0x^2 + 2x \enclose{longdiv}{ 2x^3 + 5x^2 + 0x + 6\phantom{0}}\kern-.2ex \\ -\underline{( 2x^3 + 0x^2 + 4x){\phantom{000.}}} \\ 5x^2 - 4x + 6 \phantom{0} \end{array} $$
\begin{align} \therefore {2x^3 + 5x^2 + 6 \over x^3 + 2x} & = 2 + {5x^2 - 4x + 6 \over x^3 + 2x} \end{align}

(b)

\begin{align} {2x^3 + 5x^2 + 6 \over x^3 + 2x} & = 2 + {5x^2 - 4x + 6 \over x^3 + 2x} \\ & = 2 + {5x^2 - 4x + 6 \over x(x^2 + 2)} \\ \\ {5x^2 - 4x + 6 \over x(x^2 + 2)} & = {A \over x} + {Bx + C \over x^2 + 2} \\ & = {A(x^2 + 2) \over x(x^2 + 2)} + {x(Bx + C) \over x(x^2 + 2)} \\ & = {A(x^2 + 2) + x(Bx + C) \over x(x^2 + 2)} \\ \\ 5x^2 - 4x + 6 & = A(x^2 + 2) + x(Bx + C) \\ \\ \text{Let } & x =0 , \\ 0 - 0 + 6 & = A(0 + 2) + 0 \\ 6 & = 2A \\ {6 \over 2} & = A \\ 3 & = A \\ \\ 5x^2 - 4x + 6 & = 3(x^2 + 2) + x(Bx + C) \\ 5x^2 - 4x + 6 & = 3x^2 + 6 + Bx^2 + Cx \\ 2x^2 - 4x & = Bx^2 + Cx \\ \\ B & = 2, C = - 4 \\ \\ \\ \therefore {2x^3 + 5x^2 + 6 \over x^3 + 2x} & = 2 + {3 \over x} + {2x - 4 \over x^2 + 2} \end{align}

(a)

\begin{align} y & = 14x - x^2 \\ \\ \text{When } & x = k, \\ y & = 14k - k^2 \\ \\ \therefore & \phantom{.} A(k, 14k - k^2) \\ \\ \text{When } & x = 2k, \\ y & = 14(2k) - (2k)^2 \\ y & = 28k - 4k^2 \\ \\ \therefore & \phantom{.} B(2k, 28k - 4k^2) \\ \\ \\ y \text{-coordinate of } B & > y \text{-coordinate of } A \\ 28k - 4k^2 & > 14k - k^2 \\ -3k^2 + 14k & > 0 \\ 3k^2 - 14k & < 0 \\ k(3k - 14) & < 0 \end{align}

$$ 0 < k < 4{2 \over 3} $$

(b)

\begin{align} y & = 14x - x^2 \\ \\ {dy \over dx} & = 14 - 2x \end{align} \begin{align} \text{When } & x = k, &&& \text{When } & x = 2k, \\ {dy \over dx} & = 14 - 2k &&& {dy \over dx} & = 14 - 2(2k) \\ \\ \text{Gradient at } A & = 14 - 2k &&& \text{Gradient at } B & = 14 - 4k \end{align}
\begin{align} \text{For } 0 < k < 7, \phantom{.} 2k < 4k &\text{ and } 14 - 2k > 14 - 4k \\ \\ \therefore \text{Gradient of } A & > \text{Gradient of } B \end{align}

(a)

\begin{align} \left(2 - {ax \over 2}\right)^5 & = 2^5 + {5 \choose 1} (2)^4 \left(-{ax \over 2}\right) + {5 \choose 2} (2)^3 \left(-{ax \over 2}\right)^2 + {5 \choose 3} (2)^2 \left(-{ax \over 2}\right)^3 + ... \\ & = 32 + (5)(16)\left(-{ax \over 2}\right) + (10)(8)\left[ (ax)^2 \over 2^2 \right] + (10)(4)\left[ - {(ax)^3 \over 2^3} \right] + ... \\ & = 32 - 40ax + 80 \left(a^2 x^2 \over 4\right) + 40 \left(- {a^3 x^3 \over 8}\right) + ... \\ & = 32 - 40ax + 20 a^2 x^2 - 5 a^3 x^3 + ... \end{align}

(b)

\begin{align} (2 + 3x)\left(2 - {ax \over 2}\right)^5 & = (2 + 3x)(32 - 40ax + 20 a^2 x^2 - 5 a^3 x^3 + ...) \\ & = ... + (2)(20a^2x^2)+ (3x)(-40ax) + ... \phantom{000000000} [\text{Only look for terms in } x^2] \\ & = ... + 40a^2 x^2 - 120 a x^2 + ... \\ & = ... + (40a^2 - 120a)x^2 + ... \end{align} \begin{align} \text{Since there are} & \text{ no terms in } x^2, \\ 40a^2 - 120a & = 0 \\ 40a(a - 3) & = 0 \end{align}
\begin{align} 40a & = 0 && \text{ or } & a - 3 & =0 \\ a & = 0 \text{ (Rej, since } a > 0) &&& a & = 3 \end{align}

(c)

\begin{align} (2 + 3x)\left(2 - {ax \over 2}\right)^5 & = (2 + 3x)(32 - 40ax + 20 a^2 x^2 - 5 a^3 x^3 + ...) \\ (2 + 3x)\left(2 - {3x \over 2}\right)^5 & = (2 + 3x)[32 - 40(3)x + 20 (3)^2 x^2 - 5 (3)^3 x^3 + ...] \\ & = (2 + 3x)(32 - 120x + 180x^2 - 135x^3 + ...) \\ & = ... + (2)(-135x^3) + (3x)(180x^2) + ... \phantom{000000000} [\text{Only look for terms in } x^3] \\ & = ... - 270x^3 + 540x^3 + ... \\ & = ... + 270x^3 + ... \\ \\ \text{Coefficient} & \text{ of } x^3 = 270 \end{align}

\begin{align} {d^2 y \over dx^2} & = 2 e^{-x} + 6 e^{2x} \\ \\ {dy \over dx} & = \int 2 e^{-x} + 6 e^{2x} \phantom{.} dx \\ & = 2 \left(e^{-x} \over -1\right) + 6 \left(e^{2x} \over 2\right) + c \phantom{000000} \left[ \int e^{f(x)} \phantom{.} dx = {e^{f(x)} \over f'(x)} \right] \\ \\ & = - 2 e^{-x} + 3 e^{2x} + c \\ \\ \text{When } & x = 0 \text{ and } {dy \over dx} = 3, \phantom{000000000000} [\text{Gradient at } P(0, 5) \text{ is } 3] \\ 3 & = -2e^{-0} + 3e^{2(0)} + c \\ 3 & = -2 + 3 + c \\ 3 & = 1 + c \\ 2 & = c \\ \\ {dy \over dx} & = -2 e^{-x} + 3e^{2x} + 2 \\ \\ y & = \int -2 e^{-x} + 3e^{2x} + 2 \phantom{.} dx \\ & = -2 \left(e^{-x} \over -1\right) + 3 \left(e^{2x} \over 2\right) + 2x + d \\ & = 2 e^{-x} + {3 \over 2} e^{2x} + 2x + d \\ \\ \text{Using } & P(0, 5), \\ 5 & = 2 e^{-0} + {3 \over 2}e^{2(0)} + 2(0) + d \\ 5 & = 2 + {3 \over 2} + 0 + d \\ 5 & = {7 \over 2} + d \\ {3 \over 2} & = d \\ \\ \text{Eqn of curve: } & y = 2 e^{-x} + {3 \over 2} e^{2x} + 2x + {3 \over 2} \end{align}

(i)

\begin{align} \text{Since tan} & \text{gents at } P \text{ and at } R \text{ meet at } S, SP = SR \\ \\ \angle SRP & = \angle SPR = x \phantom{0} (\text{Isosceles triangle } SPR \text{ with } SP = SR) \\ \\ \angle PSR & = 180^\circ - x - x \phantom{0} (\text{Angle sum of triangle}) \\ & = 180^\circ - 2x \\ \\ \angle RST & = 180^\circ - (180^\circ - 2x) \phantom{0} (\text{Adjacent angles on a straight line}) \\ & = 180^\circ - 180^\circ + 2x \\ & = 2x \phantom{0} \text{ (Shown)} \end{align}

(ii)

\begin{align} \angle PRQ & = 90^\circ \phantom{0} (\text{Right angle in semicircle}) \\ \\ \angle PRT & = 180^\circ - 90^\circ \phantom{0} (\text{Adjacent angles on a straight line}) \\ & = 90^\circ \\ \\ \angle STR & = 180^\circ - 90^\circ - x \phantom{0} (\text{Angle sum of triangle}) \\ & = 90^\circ - x \\ \\ \angle SRT & = 180^\circ - 2x - (90^\circ - x) \phantom{0} (\text{Angle sum of triangle}) \\ & = 180^\circ - 2x - 90^\circ + x \\ & = 90^\circ - x \\ & = \angle STR \\ \\ \text{Since } & \angle STR = \angle SRT, \text{ triangle } RST \text{ is isosceles (Shown)} \end{align}

(a)

\begin{align} y & = 3\sqrt{x} \phantom{00} \text{--- (1)} \\ \\ 4y & = 3x + 9 \phantom{00} \text{--- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ 4(3 \sqrt{x}) & = 3x + 9 \\ 12 \sqrt{x} & = 3x + 9 \\ (12 \sqrt{x})^2 & = (3x + 9)^2 \\ (12)^2 x & = \underbrace{ (3x)^2 + 2(3x)(9) + 9^2 }_{ (a + b)^2 = a^2 + 2ab + b^2 } \\ 144 x & = 9x^2 + 54x + 81 \\ 0 & = 9x^2 - 90x + 81 \\ 0 & = x^2 - 10x + 9 \\ 0 & = (x - 1)(x - 9) \end{align} \begin{align} x - 1 & = 0 && \text{ or } & x - 9 & = 0 \\ x & = 1 &&& x & = 9 \\ \\ \text{Substitute } & \text{into (1),} &&& \text{Substitute } & \text{into (1),} \\ y & = 3 \sqrt{1} &&& y & = 3 \sqrt{9} \\ y & = 3 &&& y & = 9 \\ \\ \therefore & \phantom{.} P(1, 3) &&& \therefore & \phantom{.} Q(9, 9) \end{align}

(b)

\begin{align} y & = 3\sqrt{x} \phantom{00} \text{--- (1)} \\ \\ 4y & = 3x + k \phantom{00} \text{--- (3)} \\ \\ \text{Substitute } & \text{(1) into (3),} \\ 4(3 \sqrt{x}) & = 3x + k \\ 12 \sqrt{x} & = 3x + k \\ (12 \sqrt{x})^2 & = (3x + k)^2 \\ 144x & = (3x)^2 + 2(3x)(k) + k^2 \\ 144x & = 9x^2 + 6kx + k^2 \\ 0 & = 9x^2 + (6k - 144)x + k^2 \\ \\ b^2 -4ac & = (6k - 144)^2 - 4(9)(k^2) \\ & = \underbrace{ (6k)^2 - 2(6k)(144) + 144^2}_{(a - b)^2 = a^2 - 2ab + b^2} - 36k^2 \\ & = 36k^2 - 1728k + 20 \phantom{.} 736 - 36k^2 \\ & = - 1728k + 20 \phantom{.} 736 \\ \\ b^2 - 4ac & = 0 \phantom{000000} [\text{One real root since line is tangent to curve}] \\ - 1728k + 20 \phantom{.} 736 & = 0 \\ - 1728k & = -20 \phantom{.} 736 \\ k & = {-20 \phantom{.} 736 \over -1728} \\ k & = 12 \end{align}

(a)

\begin{align} y & = x^3 - ax^2 + bx + 4 \\ \\ {dy \over dx} & = 3x^2 - 2ax + b \\ \\ \text{If } y \text{ is always } & \text{increasing, } {dy \over dx} > 0 \text{ for all values of } x \\ \\ {dy \over dx} & = 3x^2 - 2ax + b \\ & = 3 \left( x^2 - {2a \over 3}x \right) + b \\ & = 3 \left[ x^2 - {2a \over 3}x + \left(a \over 3\right)^2 - \left(a \over 3\right)^2 \right] + b \phantom{000000} [\text{Complete the square}] \\ & = 3 \left[ \left(x - {a \over 3}\right)^2 - {a^2 \over 9} \right] + b \\ & = 3 \left(x - {a \over 3}\right)^2 - {a^2 \over 3} + b \\ \\ \text{Minimum value of } {dy \over dx} & = -{a^2 \over 3} + b \\ \\ \text{For } {dy \over dx} > 0, -{a^2 \over 3} + b & > 0 \\ -{a^2 \over 3} & > -b \\ {a^2 \over 3} & < b \\ a^2 & < 3b \phantom{0} (\text{Shown}) \end{align}

(b)

\begin{align} y & = x^3 - 8x^2 + 10x + 4 \\ \\ \text{Let } & y = 0, \\ 0 & = x^3 - 8x^2 + 10x + 4 \phantom{000000} [\text{Cubic equation}] \\ \\ \text{Let } f(x) & = x^3 - 8x^2 + 10x + 4 \\ \\ f(2) & = (2)^3 - 8(2)^2 + 10(2) + 4 \\ & = 0 \\ \\ \therefore x - 2 & \text{ is a factor of } f(x) \end{align}
$$ \require{enclose} \begin{array}{rll} x^2 - 6x - 2 \phantom{0000000.}\\ x - 2 \enclose{longdiv}{ x^3 - 8x^2 + 10x + 4 \phantom{0}}\kern-.2ex \\ -\underline{( x^3 - 2x^2){\phantom{000000000}}} \\ -6x^2 + 10x + 4 \phantom{0} \\ -\underline{( -6x^2 + 12x){\phantom{000.}}} \\ -2x + 4 \phantom{0} \\ -\underline{( -2x + 4 ){\phantom{.}}} \\ 0 \phantom{.} \end{array} $$
\begin{align} 0 & = x^3 - 8x^2 + 10x + 4 \\ 0 & = (x - 2)(x^2 - 6x - 2) \end{align} \begin{align} x - 2 & = 0 && \text{ or } & x^2 & - 6x - 2 = 0 \\ x & = 2 &&& x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & &&& & = {-(-6) \pm \sqrt{(-6)^2 - 4(1)(-2)} \over 2(1)} \\ & &&& & = {6 \pm \sqrt{44} \over 2} \\ & &&& & = {6 \pm \sqrt{4} \sqrt{11} \over 2} \\ & &&& & = {6 \pm 2 \sqrt{11} \over 2} \\ & &&& & = {6 \over 2} \pm {2\sqrt{11} \over 2} \\ & &&& & = 3 \pm \sqrt{11} \end{align}

(a)(i)

\begin{align} \text{Amplitude} & = 2 \\ \\ \text{Period} & = {2\pi \over 1} = 2\pi \end{align}

(a)(ii)

\begin{align} \text{Amplitude} & = 4 \\ \\ \text{Period} & = {2\pi \over 2} = \pi \end{align}

(b)

\begin{align} y & = 2 \sin x && \phantom{000000} & y = 3 - &4 \cos 2x \\ & &&& y = - 4 & \cos 2x + 3 \phantom{000} [\text{Inverted shape}] \\ \\ \text{Amplitude} & = 2 &&& \text{Amplitude} & = 4 \\ \\ & &&& \text{Center line: } & y = 3 \\ \\ \text{Max. value} & = 2 &&& \text{Max. value} & = 3 + 4 = 7 \\ \\ \text{Min. value} & = -2 &&& \text{Min. value} & = 3 - 4 = -1 \\ \\ \text{Period} & = 2\pi &&& \text{Period} & = \pi \end{align}

(a)

\begin{align} AB & = 10 - 4 = 6 \text{ units} \\ \\ {1 \over 2} \times AB \times h & = 15 \\ {1 \over 2} \times 6 \times h & = 15 \\ 3h & = 15 \\ h & = {15 \over 3} \\ h & = 5 \text{ units} \\ \\ \text{Since } q > 0, & \phantom{.} C \text{ lies above } AB \\ \\ y \text{-coordinate of } C, q & = 2 + 5 \phantom{0} (\text{not } 2 - 5 = -3) \\ & = 7 \end{align}

(b)

\begin{align} \therefore & \phantom{.} C(p, 7) \\ \\ AC & = \sqrt{ (x_2 - x_1)^2 + (y_2 - y_1)^2 } \\ \sqrt{89} & = \sqrt{ (p - 4)^2 + (7 - 2)^2 } \\ 89 & = (p - 4)^2 + 5^2 \\ 89 & = \underbrace{ p^2 - 2(p)(4) + 4^2 }_{(a - b)^2 = a^2 - 2ab + b^2} + 25 \\ 89 & = p^2 - 8p + 16 + 25 \\ 0 & = p^2 - 8p - 48 \\ 0 & = (p + 4)(p - 12) \end{align} \begin{align} p + 4 & = 0 && \text{ or } & p - 12 & = 0 \\ p & = -4 &&& p & = 12 \end{align}

(a)

\begin{align} x & = a \sin nt \\ \\ \text{Amplitude} & = a \implies -a \le x \le a \\ \\ OA = OB & = {120 \over 2} = 60 \text{ cm} \\ \\ \text{Since max. displacement} = 60 &\text{ cm and min. displacement = -60 cm, amplitude} = 60 \\ \\ \therefore a & = 60 \\ \\ \\ x & = a \sin nt \\ \\ \text{Period} & = {2\pi \over n} \\ 6 & = {2\pi \over n} \\ 6n & = 2\pi \\ n & = {2\pi \over 6} \\ n & = {\pi \over 3} = {1 \over 3} \pi \phantom{0} (\text{Shown}) \end{align}

(b)

\begin{align} \text{Velocity, } v & = {dx \over dt} \\ v & = {d \over dt} \left( 60 \sin {1 \over 3}\pi t \right) \\ v & = 60 \left({1 \over 3} \pi \right) \cos {1 \over 3}\pi t \phantom{000000} \left[ {d \over dx} [ \sin f(x)] = f'(x) \cos f(x) \right] \\ v & = 20 \pi \cos {1 \over 3} \pi t \\ \\ \text{Amplitude} & = 20 \pi \\ \\ \therefore \text{Max. speed} & = 20 \pi \text{ cm/s} \end{align}

(c)

\begin{align} \text{Acceleration, } a & = {dv \over dt} \\ & = {d \over dt} \left( 20 \pi \cos {1 \over 3} \pi t \right) \\ & = 20 \pi \left({1 \over 3} \pi\right) \left(- \sin {1 \over 3} \pi t \right) \phantom{000000} \left[ {d \over dx} [ \cos f(x)] = f'(x) . - \sin f(x) \right] \\ & = -{20 \over 3} \pi^2 \sin {1 \over 3} \pi t \\ \\ \text{Period/Time for 1 oscillation} & = 6 \text{ s} \\ \\ \text{Time taken for dot to reach } A & = 6 \times {3 \over 4} \phantom{000000} [\text{Dot goes from } O > B > O > A > O] \\ & = 4.5 \text{ s} \\ \\ \text{When } & t = 4.5, \\ a & = - {20 \over 3} \pi^2 \sin {1 \over 3} \pi (4.5) \\ & = -{20 \over 3} \pi^2 (-1) \phantom{0000000} [\text{Radian mode!}] \\ & = {20 \over 3} \pi^2 \text{ cm/s}^2 \end{align}

(a)

\begin{align} u & = 4x + 2 &&& v & = x + 1 \\ {du \over dx} & = 4 &&& {dv \over dx} & = 1 \end{align} \begin{align} {dy \over dx} & = {v {du \over dx} - u {dv \over dx} \over v^2} \phantom{000000} [\text{Quotient rule}] \\ & = { (x + 1)(4) - (4x + 2)(1) \over (x + 1)^2 } \\ & = { 4x + 4 - 4x - 2 \over (x + 1)^2 } \\ & = {2 \over (x + 1)^2 } \\ \\ \text{For } x > -1, \phantom{.} & (x + 1)^2 > 0 \\ & {2 \over (x + 1)^2} > 0 \\ \\ \therefore \text{For } x > -1, & \phantom{.} {dy \over dx} \ne 0 \text{ and curve has no stationary points} \end{align}

(b)

\begin{align} \text{Gradient of tangent at } P & = {1 \over 2} \\ \\ \text{Gradient of normal at } P & = {-1 \over {1 \over 2}} \phantom{000000} [m_1 \times m_2 = -1] \\ & = -2 \\ \\ [\text{Need to find } & \text{point } P \text{ to form equation of normal}] \\ \\ \text{Let } & {dy \over dx} = {1 \over 2}, \\ {2 \over (x + 1)^2} & = {1 \over 2} \\ 2(2) & = (x + 1)^2 \\ 4 & = (x + 1)^2 \\ \pm \sqrt{4} & = x + 1 \\ \pm 2 & = x + 1 \end{align} \begin{align} x + 1 & = 2 && \text{ or } & x + 1 & = -2 \\ x & = 1 &&& x & = -3 \text{ (Not applicable for } P) \end{align}
\begin{align} \text{Substitute } & x = 1 \text{ into eqn of curve,} \\ y & = {4(1) + 2 \over 1 + 1} \\ y & = 3 \\ \\ \therefore & \phantom{.} P(1, 3) \\ \\ y & = mx + c \\ y & = -2x + c \\ \\ \text{Using } & P(1, 3), \\ 3 & = -2(1) + c \\ 3 & = -2 + c \\ 5 & = c \\ \\ \text{Eqn of normal at } P: & \phantom{0} y = -2x + 5 \\ \\ \implies & \phantom{.} B(0, 5) \\ \\ \text{Let } & y = 0, \\ 0 & = -2x + 5 \\ 2x & = 5 \\ x & = 2.5 \\ \\ \therefore & \phantom{.} A(2.5, 0) \\ \\ \text{Area of triangle } AOB & = {1 \over 2} \times AO \times OB \\ & = {1 \over 2} \times 2.5 \times 5 \\ & = 6.25 \text{ units}^2 \end{align}

(c)

$$ \require{enclose} \begin{array}{rll} 4 \phantom{0000.}\\ x + 1 \enclose{longdiv}{ 4x + 2\phantom{0}}\kern-.2ex \\ -\underline{( 4x + 4){\phantom{.}}} \\ -2 \phantom{0} \end{array} $$
\begin{align} {4x + 2 \over x + 1} & = 4 + {-2 \over x + 1} \\ \\ y & = 4 - {2 \over x + 1} \\ \\ \\ \text{For } x > -1, \phantom{.} x + 1 & > 0 \\ {2 \over x + 1} & > 0 \\ -{2 \over x + 1} & < 0 \\ -{2 \over x + 1} + 4 & < 0 + 4 \\ 4 - {2 \over x + 1} & < 4 \\ \\ \implies \text{For } x > -1, y \text{-coordinates of } & \text{all points on curve are smaller than } 4 \\ \\ \therefore \text{For } c \ge 4, \text{ the line } y = c \text{ does } & \text{not intersect the curve} \end{align}

(a)

\begin{align} N & = 2400 e^{-kt} \\ \\ \text{When } & t = 3 \text{ and } N = 2050, \\ 2050 & = 2400 e^{-k (3)} \\ 2050 & = 2400 e^{-3k} \\ {2050 \over 2400} & = e^{-3k} \\ {41 \over 48} & = e^{-3k} \\ \ln {41 \over 48} & = \ln e^{-3k} \\ \ln {41 \over 48} & = -3k \ln e \phantom{000000} [\text{Power law (logarithms)}] \\ \ln {41 \over 48} & = -3k (1) \\ \ln {41 \over 48} & = -3k \\ \\ k & = -{1 \over 3} \ln {41 \over 48} \\ k & = 0.052 \phantom{.} 543 \\ \\ \text{When } & t = 7, \\ N & = 2400 e^{ - (0.052 \phantom{.} 543)(7)} \\ N & = 1 \phantom{.} 661.41 \\ N & \approx 1661 \text{ (to nearest integer)} \end{align}

(b)

\begin{align} \text{When } & N = 1200, \\ 1200 & = 2400 e^{-kt} \\ {1200 \over 2400} & = e^{-kt} \\ {1 \over 2} & = e^{-kt} \\ \ln {1 \over 2} & = \ln e^{-kt} \\ \ln {1 \over 2} & = -kt \ln e \phantom{000000} [\text{Power law (logarithms)}] \\ \ln {1 \over 2} & = -kt (1) \\ \ln {1 \over 2} & = -kt \\ \\ t & = {\ln {1 \over 2} \over -k} \\ & = {\ln {1 \over 2} \over -(0.052 \phantom{.} 543)} \\ & = 13.192 \\ \\ \text{When } & t = 13, \\ N & = 2400 e^{-(0.052 \phantom{.} 543) (13)} = 1212.17 \\ \\ \text{When } & t = 14, \\ N & = 2400 e^{- (0.052 \phantom{.} 543) (14)} = 1150.12 \\ \\ \\ \therefore \text{Species } & \textbf{first} \text{ falls under threat in } 2023 \end{align}

(a)

\begin{align} f(-2) & = (-2)^3 + a(-2)^2 - 3(-2) + b \\ & = -8 + a(4) + 6 + b \\ & = -2 + 4a + b \\ \\ \text{Since } & x + 2 \text{ is a factor,} \\ 0 & = -2 + 4a + b \\ -b & = -2 + 4a \\ b & = 2 - 4a \phantom{00} \text{--- (1)} \\ \\ f(3) & = (3)^3 + a(3)^2 - 3(3) + b \\ 30 & = 27 + a(9) - 9 + b \\ 30 & = 18 + 9a + b \\ 12 & = 9a + b \phantom{00} \text{--- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ 12 & = 9a + 2 - 4a \\ 10 & = 5a \\ {10 \over 5} & = a \\ 2 & = a \\ \\ \text{Substitute } & a = 2 \text{ into (1),} \\ b & = 2 - 4(2) \\ b & = -6 \\ \\ \therefore a & = 2, b = -6 \end{align}

(b)

\begin{align} f(x) & = x^3 + 2x^2 - 3x - 6 \end{align}
$$ \require{enclose} \begin{array}{rll} x^2 - 3\phantom{000000}\\ x + 2 \enclose{longdiv}{ x^3 + 2x^2 - 3x - 6\phantom{0}}\kern-.2ex \\ -\underline{( x^3 + 2x^2){\phantom{00000000}}} \\ -3x - 6 \phantom{0} \\ -\underline{( - 3x - 6){\phantom{.}}} \\ 0 \phantom{0} \end{array} $$
\begin{align} f(x) & = (x^2 - 3)(x + 2) \\ & = [ x^2 - (\sqrt{3})^2] (x + 2) \\ & = (x + \sqrt{3})(x - \sqrt{3})(x + 2) \phantom{000000} [a^2 - b^2 = (a + b)(a - b)] \end{align}

(a)

\begin{align} \require{cancel} \text{L.H.S} & = \tan 2 \theta (2 \cos \theta - \sec \theta) \\ & = \left( \sin 2 \theta \over \cos 2 \theta \right) \left( {2 \cos \theta \over 1} - {1 \over \cos \theta} \right) \\ & = \left( \sin 2 \theta \over \cos 2 \theta \right) \left( {2 \cos^2 \theta \over \cos \theta} - {1 \over \cos \theta} \right) \\ & = \underbrace{ \left( 2 \sin \theta \cos \theta \over 2\cos^2 \theta - 1 \right) }_\text{Double angle formulae} \left( 2 \cos^2 \theta - 1 \over \cos \theta \right) \\ & = { (2 \sin \theta \cancel{\cos \theta})\cancel{(2 \cos^2 \theta - 1)} \over \cancel{(2 \cos^2 \theta - 1)} \cancel{(\cos \theta)} } \\ & = {2 \sin \theta \over 1} \\ & = 2 \sin \theta \\ & = \text{R.H.S} \end{align}

(b)

\begin{align} \tan 2 \theta (2 \cos \theta - \sec \theta) & = {1 \over 2} \sec \theta \\ \underbrace{ 2 \sin \theta }_\text{Use identity from (a)} & = {1 \over 2} \sec \theta \\ 2 \sin \theta & = {1 \over 2} \left(1 \over \cos \theta\right) \\ 2 \sin \theta \cos \theta & = {1 \over 2} \\ \underbrace{ \sin 2 \theta }_\text{Double angle formula} & = {1 \over 2} \phantom{00000} [\text{1st or 2nd quadrant since } \sin 2 \theta > 0] \\ \\ \text{Basic angle, } \alpha & = \sin^{-1} \left(1 \over 2\right) \\ & = 30^\circ \end{align}

\begin{align} \text{Since } 0^\circ & \le \theta \le 360^\circ, \phantom{.} 0^\circ \le 2\theta \le 720^\circ \\ \\ 2 \theta & = 30^\circ, 180^\circ - 30^\circ \\ & = 30^\circ, 150^\circ, 30^\circ + 360^\circ, 150^\circ + 360^\circ \\ & = 30^\circ, 150^\circ, 390^\circ, 510^\circ \\ \\ \theta & = 15^\circ, 75^\circ, 195^\circ, 255^\circ \end{align}

(a)

\begin{align} \text{Total surface area, } A & = 2lb + 2 lh + 2 bh \\ & = 2(3x)(x) + 2(3x)(h) + 2(x)(h) \\ & = 6x^2 + 6hx + 2hx \\ & = 6x^2 + 8hx \phantom{000000} [\text{Need to find a relationship between } h \text{ and } x ] \\ \\ \text{Volume} & = l \times b \times h \\ 2000 & = 3x \times x \times h \\ 2000 & = 3h x^2 \\ \\ h & = {2000 \over 3x^2} \\ \\ \therefore A & = 6x^2 + 8 \left(2000 \over 3x^2\right)(x) \\ & = 6x^2 + \left(8 \over 1\right)\left(2000 \over 3x^2\right)\left(x \over 1\right) \\ & = 6x^2 + {16 \phantom{.} 000 x \over 3x^2 } \\ & = 6x^2 + {16 \phantom{.} 000 \over 3x } \phantom{0} \text{ (Shown)} \end{align}

(b)

\begin{align} A & = 6x^2 + {16 \phantom{.} 000 \over 3x} \\ A & = 6x^2 + {16 \phantom{.} 000 \over 3} x^{-1} \\ \\ {dA \over dx} & = 6(2)x + {16 \phantom{.} 000 \over 3} (-1) (x^{-2}) \\ & = 12x - {16 \phantom{.} 000 \over 3} \left(1 \over x^2\right) \\ & = 12x - {16 \phantom{.} 000 \over 3x^2} \\ \\ \text{Let } & {dA \over dx} = 0 , \\ 0 & = 12x - {16 \phantom{.} 000 \over 3x^2} \\ {16 \phantom{.} 000 \over 3x^2} & = 12x \\ 16 \phantom{.} 000 & = (3x^2)(12x) \\ 16 \phantom{.} 000 & = 36x^3 \\ {16 \phantom{.} 000 \over 36} & = x^3 \\ {4000 \over 9} & = x^3 \\ \\ x & = \sqrt[3]{ 4000 \over 9} \\ & = 7.6314 \\ & \approx 7.63 \end{align}

(c)

\begin{align} {dA \over dx} & = 12x - {16 \phantom{.} 000 \over 3} x^{-2} \\ \\ {d^2 A \over dx^2} & = 12 - {16 \phantom{.} 000 \over 3} (-2) (x^{-3}) \\ & = 12 + {32 \phantom{.} 000 \over 3} \left(1 \over x^3\right) \\ & = 12 + {32 \phantom{.} 000 \over 3 x^3 } \\ \\ \text{When } & x = 7.6314, \\ {d^2 A \over dx^2} & = 12 + {32 \phantom{.} 000 \over 3 (7.6314)^3 } \\ & = 36 > 0 \\ \\ \therefore \text{When } x = 7.6314, & \text{ the total surface area of container is a minimum value} \end{align}

(a)

\begin{align} \tan 75^\circ & = \tan (30^\circ + 45^\circ) \\ & = {\tan 30^\circ + \tan 45^\circ \over 1 - \tan 30^\circ \tan 45^\circ} \phantom{000000} \left[ \tan (A + B) = {\tan A + \tan B \over 1 - \tan A \tan B} \right] \\ & = { {1 \over \sqrt{3}} + 1 \over 1 - \left(1 \over \sqrt{3}\right)(1)} \\ & = { {1 \over \sqrt{3}} + 1 \over 1 - {1 \over \sqrt{3}} } \times {\sqrt{3} \over \sqrt{3}} \\ & = { 1 + \sqrt{3} \over \sqrt{3} - 1} \times { \sqrt{3} +1 \over \sqrt{3} +1} \phantom{000000} [\text{Rationalise denominator}] \\ & = { (1 + \sqrt{3})^2 \over (\sqrt{3})^2 - (1)^2 } \phantom{0000000000} [\text{Denominator: } (a + b)(a - b) = a^2 - b^2] \\ & = { 1^2 + 2(1)(\sqrt{3}) + (\sqrt{3})^2 \over 2 } \phantom{00} [\text{Numerator: } (a + b)^2 = a^2 + 2ab + b^2 ] \\ & = {1 + 2 \sqrt{3} + 3 \over 2} \\ & = {4 + 2 \sqrt{3} \over 2} \\ & = {4 \over 2} + {2 \sqrt{3} \over 2} \\ & = 2 + \sqrt{3} \phantom{0} \text{ (Shown)} \end{align}

(b)

\begin{align} \sec^2 75^\circ & = 1 + \tan^2 75^\circ \phantom{0000000000000000000} [\text{Identity: } \tan^2 A + 1 = \sec^2 A] \\ & = 1 + (2 + \sqrt{3})^2 \phantom{00000000000000000.} [\text{Use result from (a)}] \\ & = 1 + \underbrace{ (2)^2 + 2(2)(\sqrt{3}) + (\sqrt{3})^2}_{ (a + b)^2 = a^2 + 2ab + b^2 } \\ & = 1 + 4 + 4 \sqrt{3} + 3 \\ & = 8 + 4 \sqrt{3} \end{align}

(c)

\begin{align} \text{From (b), } \sec^2 75^\circ & = 8 + 4 \sqrt{3} \\ {1 \over \cos^2 75^\circ} & = 8 + 4 \sqrt{3} \\ 1 & = ( \cos^2 75^\circ ) (8 + 4 \sqrt{3}) \\ \\ \cos^2 75^\circ & = {1 \over 8 + 4 \sqrt{3}} \times {8 - 4 \sqrt{3} \over 8 - 4 \sqrt{3}} \phantom{000000} [\text{Rationalise denominator}] \\ & = {8 - 4 \sqrt{3} \over (8)^2 - (4 \sqrt{3})^2 } \phantom{00000000000} [(a + b)(a - b) = a^2 - b^2] \\ & = {8 - 4 \sqrt{3} \over 16} \\ & = {8 \over 16} - {4 \sqrt{3} \over 16} \\ & = {1 \over 2} - {\sqrt{3} \over 4} \\ & = {1 \over 2} - {1 \over 4} \sqrt{3} \end{align}

(a)

\begin{align} 2^x + 4^{x -1 } & = 15 \\ 2^x + (2^2)^{x - 1} & = 15 \\ 2^x + 2^{2(x - 1)} & = 15 \phantom{000000} [(a^m)^n = a^{mn}]\\ 2^x + 2^{2x - 2} & = 15 \\ 2^x + {2^{2x} \over 2^2 } & = 15 \phantom{000000} \left[ {a^m \over a^n} = a^{m - n} \right] \\ \\ \text{Let } & u = 2^x, \\ u + {u^2 \over 4} & = 15 \\ 4 \left(u + {u^2 \over 4}\right) & = 4(15) \\ 4u + u^2 & = 60 \\ u^2 + 4u - 60 & = 0 \\ (u - 6)(u + 10)& = 0 \end{align} \begin{align} u - 6 & = 0 && \text{ or } & u + 10 & = 0 \\ u & = 6 &&& u & = -10 \\ \\ 2^x & = 6 &&& 2^x & = -10 \text{ (Reject, since } 2^x > 0) \\ \lg 2^x & = \lg 6 \\ \underbrace{ x \lg 2 }_{\text{Power law (logarithms)}} & = \lg 6 \\ x & = {\lg 6 \over \lg 2} \\ x & \approx 2.58 \end{align}

(b)(i)

\begin{align} y & = 5 - e^{2x} \\ \\ \text{Let } & x = 0, \\ y & = 5 - e^{2(0)} \\ y & = 5 - 1 \\ y & = 4 \\ \\ \therefore & \phantom{.} B (0, 4) \\ \\ \text{Let } & y = 0, \\ 0 & = 5 - e^{2x} \\ e^{2x} & = 5 \\ \ln e^{2x} & = \ln 5 \\ 2x \ln e & = \ln 5 \phantom{000000} [\text{Power law (logarithms)}] \\ 2x (1) & = \ln 5 \\ 2x & = \ln 5 \\ x & = {1 \over 2} \ln 5 \\ \\ \therefore & \phantom{.} A \left({1 \over 2} \ln 5, 0\right) \end{align}

(b)(ii)

\begin{align} y & = 5 - e^{2x} \\ \\ {dy \over dx} & = - 2 e^{2x} \phantom{000000} \left[ {d \over dx} [ e^{f(x)}] = f'(x). e^{f(x)} \right] \\ \\ \text{When } & x = {1 \over 2} \ln 5, \\ {dy \over dx} & = - 2 e^{ 2 \left({1 \over 2} \ln 5 \right)} \\ & = -2 e^{ \ln 5 } \\ & = - 2(5) \\ & = -10 \end{align}

(a)

\begin{align} \sin \angle CAF & = {FC \over AC} \phantom{000000} \left[ {Opp \over Hyp} \right] \\ \sin \theta & = {FC \over 30} \\ 30 \sin \theta & = FC \\ \\ \angle FCA & = 180^\circ - 90^\circ - \theta \phantom{0} (\text{Angle sum of triangle}) \\ & = 90^\circ - \theta \\ \\ \angle GCD & = 90^\circ - (90^\circ - \theta) \\ & = 90^\circ - 90^\circ + \theta \\ & = \theta \\ \\ \cos \angle GCD & = {GC \over CD} \phantom{000000} \left[ {Adj \over Hyp} \right] \\ \cos \theta & = {GC \over 20} \\ 20 \cos \theta & = GC \\ \\ \text{Shortest distance from } D \text{ to border} & = ED \\ & = FG \\ & = FC - GC \\ & = (30 \sin \theta - 20 \cos \theta) \text{ km} \end{align}

(b)

\begin{align} a \sin \theta - b \cos \theta & = R \sin (\theta - \alpha) \\ \\ a = 30, b & = 20 \\ \\ R & = \sqrt{a^2 + b^2} \\ & = \sqrt{30^2 + 20^2} \\ & = \sqrt{1300} \\ \\ \alpha & = \tan^{-1} \left(b \over a\right) \\ & = \tan^{-1} \left(20 \over 30\right) \\ & = 33.69^\circ \\ \\ \therefore 30 \sin \theta - 20 \cos \theta & = \sqrt{1300} \sin (\theta - 33.69^\circ) \\ & \approx \sqrt{1300} \sin (\theta - 33.7^\circ) \end{align}

(c)

\begin{align} ED & = \sqrt{1300} \sin (\theta - 33.69^\circ) \\ \\ \text{When } & ED = 10, \\ 10 & = \sqrt{1300} \sin (\theta - 33.69^\circ) \\ {10 \over \sqrt{1300}} & = \sin (\theta - 33.69^\circ) \phantom{000000} [\text{1st or 2nd quadrant}] \\ \\ \alpha & = \sin^{-1} \left(10 \over \sqrt{1300}\right) \\ & = 16.10^\circ \end{align}

\begin{align} \theta - 33.69^\circ & = 16.10^\circ, 180^\circ - 16.10^\circ \\ & = 16.10^\circ, 163.9^\circ \\ \\ \theta & = 49.79^\circ, 197.59^\circ \text{ (Not applicable since } \theta \text{ is acute}) \\ & \approx 49.8^\circ \end{align}

(a)

\begin{align} u & = x - 3 &&& v & = \sqrt{2x + 1} \\ & &&& & = (2x + 1)^{1 \over 2} \\ \\ {du \over dx} & = 1 &&& {dv \over dx} & = \underbrace{ {1 \over 2} (2x + 1)^{-{1 \over 2}} . (2) }_\text{Chain rule} \\ & &&& & = (2x + 1)^{-{1 \over 2}} \\ & &&& & = {1 \over \sqrt{2x + 1}} \end{align} \begin{align} {dy \over dx} & = u {dv \over dx} + v {du \over dx} \phantom{000000} [\text{Product rule}] \\ & = (x - 3) \left(1 \over \sqrt{2x + 1}\right) + (\sqrt{2x + 1})(1) \\ & = \left(x - 3 \over 1\right) \left(1 \over \sqrt{2x + 1}\right) + {\sqrt{2x + 1} \over 1} \\ & = {x - 3 \over \sqrt{2x + 1}} + {\sqrt{2x + 1} \sqrt{2x + 1} \over \sqrt{2x + 1}} \\ & = {x - 3 \over \sqrt{2x + 1}} + {2x + 1 \over \sqrt{2x + 1}} \\ & = {3x - 2 \over \sqrt{2x + 1}} \phantom{0} \text{ (Shown)} \end{align}

(b)

\begin{align} {dy \over dt} & = {dy \over dx} \times {dx \over dt} \\ & = {3x - 2 \over \sqrt{2x + 1}} \times 0.15 \\ & = {3x - 2 \over \sqrt{2x + 1}} \times {3 \over 20} \\ & = {9x - 6 \over 20 \sqrt{2x + 1}} \\ \\ \text{When } & x = 4, \\ {dy \over dt} & = {9(4) - 6 \over 20 \sqrt{2(4) + 1}} \\ & = 0.5 \text{ units per second} \end{align}

(c)

\begin{align} \text{From (a), } {d \over dx} [(x - 3) \sqrt{2x + 1}] & = {3x - 2 \over \sqrt{2x + 1}} \\ \\ \implies \int {3x - 2 \over \sqrt{2x + 1}} \phantom{.} dx & = (x - 3) \sqrt{2x + 1} \phantom{000000} [\text{*}] \\ \\ \\ \int {x \over \sqrt{2x + 1}} \phantom{.} dx & = {1 \over 3} \int {3x \over \sqrt{2x + 1}} \phantom{.} dx \\ & = {1 \over 3} \int {3x - 2 \over \sqrt{2x + 1}} + {2 \over \sqrt{2x + 1}} \phantom{.} dx \\ & = {1 \over 3} \int {3x - 2 \over \sqrt{2x + 1}} \phantom{.} dx + {1 \over 3} \int {2 \over \sqrt{2x + 1}} \phantom{.} dx \\ & = {1 \over 3} \underbrace{ (x - 3)\sqrt{2x + 1} }_\text{Use result in second line *} + {1 \over 3} \int 2 (2x + 1)^{-{1 \over 2}} \phantom{.} dx \\ & = {1 \over 3} (x - 3)\sqrt{2x + 1} + {2 \over 3} \int (2x + 1)^{-{1 \over 2}} \phantom{.} dx \\ & = {1 \over 3} (x - 3)\sqrt{2x + 1} + {2 \over 3} \left[ (2x + 1)^{1 \over 2} \over \left(1 \over 2\right)(2) \right] \phantom{000000} \left[ \int [f(x)]^n \phantom{.} dx = { [f(x)]^{n + 1} \over (n + 1)[f'(x)] } \right] \\ & = {1 \over 3} (x - 3)\sqrt{2x + 1} + {2 \over 3} (2x + 1)^{1 \over 2} \\ & = {1 \over 3} (x - 3)\sqrt{2x + 1} + {2 \over 3} \sqrt{2x + 1} \\ \\ \int_0^4 {x \over \sqrt{2x + 1} } \phantom{.} dx & = \left[ {1 \over 3} (x - 3) \sqrt{2x + 1} + {2 \over 3} \sqrt{2x + 1} \right]_0^4 \\ & = \left[ {1 \over 3} (4 - 3) \sqrt{8 + 1} + {2 \over 3} \sqrt{8 + 1} \right] - \left[ {1 \over 3} (0 - 3) \sqrt{0 + 1} + {2 \over 3} \sqrt{0 + 1} \right] \\ & = {10 \over 3} \end{align}

(a)

\begin{align} x^2 + y^2 - 16x - 2y + 40 & = 0 \\ x^2 - 16x + y^2 - 2y & = -40 \\ x^2 - 16x + \left(16 \over 2\right)^2 - \left(16 \over 2\right)^2 + y^2 - 2y + \left(2 \over 2\right)^2 - \left(2 \over 2\right)^2 & = -40 \phantom{000000} [\text{Complete the square}] \\ (x - 8)^2 - 64 + (y - 1)^2 - 1 & = -40 \\ (x - 8)^2 + (y - 1)^2 & = 25 \\ (x - 8)^2 + (y - 1)^2 & = 5^2 \\ \\ \text{Centre: } (8, 1) & \\ \\ \text{Radius} = 5 \text{ units} & \end{align}

(b)

The shortest distance from centre B(8, 1) to the line is also the perpendicular distance from B to the line

\begin{align} y & = 2x + 0 \phantom{000000} [y = mx + c] \\ \\ \text{Gradient of } OA & = 2 \\ \\ \text{Gradient of } AB & = {-1 \over 2} \phantom{000000} [m_1 \times m_2 = -1] \\ & = -{1 \over 2} \\ \\ y & = mx + c \\ y & = -{1 \over 2}x + c \\ \\ \text{Using } & B(8, 1), \\ 1 & = -{1 \over 2}(8) + c \\ 1 & = -4 + c \\ 5 & = c \\ \\ \text{Eqn of } AB: & \phantom{0} y = -{1 \over 2}x + 5 \phantom{00} \text{--- (1)} \\ \\ & \phantom{0} y = 2x \phantom{00} \text{--- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ -{1 \over 2}x + 5 & = 2x \\ -{5 \over 2}x & = -5 \\ x & = -5 \div -{5 \over 2} \\ x & = 2 \\ \\ \text{Substitute } &x = 2 \text{ into (1),} \\ y & = -{1 \over 2}(2) + 5 \\ y & = 4 \\ \\ \therefore & \phantom{.} A(2, 4) \\ \\ \text{Shortest distance} & = \text{Length of } AB \\ & = \sqrt{ (x_2 - x_1)^2 + (y_2 - y_1)^2} \\ & = \sqrt{ (8 - 2)^2 + (1 - 4)^2} \\ & = \sqrt{45} \\ & = \sqrt{9} \sqrt{5} \\ & = 3 \sqrt{5} \text{ units} \\ \\ \\ \text{Since } AB \text{ is longer than } & \text{radius of circle, circle does not intersect the line } y = 2x \end{align}

(c)

\begin{align} \text{Let } (a, b) \text{ denote } & \text{the coordinates of } D \\ \\ \text{Centre of circle} & = \text{Midpoint of } CD \\ (8,1 ) & = \left( {11 + a \over 2}, {5 + b \over 2} \right) \end{align} \begin{align} {11 + a \over 2} & = 8 &&& {5 + b \over 2} & = 1 \\ 11 + a & = 2(8) &&& 5 + b & = 2(1) \\ a & = 16 - 11 &&& b & = 2 - 5 \\ a & = 5 &&& b & = -3 \\ \\ & && \therefore D(5, -3) \end{align}

(a)

\begin{align*} y & = {9 \over 3 - 2x} \\ \\ \text{Let } & x = 0, \\ y & = {9 \over 3 - 2(0)} \\ y & = 3 \\ \\ \therefore & \phantom{.} A(0, 3) \\ \\ \\ y & = {9 \over 3 - 2x} \\ y & = 9 (3 - 2x)^{-1} \\ \\ {dy \over dx} & = \underbrace{ 9 (-1) (3 - 2x)^{-2} (-2) }_\text{Chain rule} \\ & = 18 (3 - 2x)^{-2} \\ & = {18 \over (3 - 2x)^2 } \\ \\ \text{When } & x = -3, \\ {dy \over dx} & = {18 \over [3 - 2(-3)]^2 } \\ & = {2 \over 9} \\ \\ \text{Gradient of tangent at } P & = {2 \over 9} \\ \\ y & = mx + c \\ y & = {2 \over 9}x + c \\ \\ \text{Using } & P(-3, 1), \\ 1 & = {2 \over 9}(-3) + c \\ 1 & = -{2 \over 3} + c \\ {5 \over 3} & = c \\ \\ \text{Eqn of tangent at } P: & \phantom{0} y = {2 \over 9}x + {5 \over 3} \\ \\ \text{Let } & x = 0, \\ y & = {2 \over 9}(0) + {5 \over 3} \\ y & = 1{2 \over 3} \\ \\ \implies & \phantom{.} B \left(0, 1{2 \over 3}\right) \\ \\ OB & = 1{2 \over 3} \text{ units} \\ \\ AB & = 3 - 1{2 \over 3} = 1{1 \over 3} \text{ units} \\ \\ \\ \therefore B \text{ is near} & \text{er to } A \text{ than to } O \end{align*}

(b)

\begin{align} \text{Area bounded by curve and } x \text{-axis} & = \int_{-3}^0 {9 \over 3 - 2x} \phantom{.} dx \\ & = \int_{-3}^0 9 \left(1 \over 3 - 2x\right) \phantom{.} dx \\ & = \left[ 9 \left[ \ln (3 - 2x) \over -2 \right] \right]_{-3}^0 \phantom{000000} \left[ \int {1 \over f(x)} \phantom{.} dx = { \ln [f(x)] \over f'(x)} \right] \\ & = \left[ -{9 \over 2} \ln (3 - 2x) \right]_{-3}^0 \\ & = \left[ -{9 \over 2} \ln (3 - 0) \right] - \left[ -{9 \over 2} \ln (3 + 6) \right] \\ & = -{9 \over 2} \ln 3 + {9 \over 2} \ln 9 \\ & = -{9 \over 2} \ln 3 + {9 \over 2} \ln 3^2 \\ & = -{9 \over 2} \ln 3 + {9 \over 2} (2) \ln 3 \phantom{0000000} [\text{Power law (logarithms)}] \\ & = -{9 \over 2} \ln 3 + 9 \ln 3 \\ & = {9 \over 2} \ln 3 \text{ units}^2 \\ \\ \text{Area bounded by line and } x \text{-axis} & = \text{Area of trapezium} \\ & = {1 \over 2} \times \text{Sum of parallel sides} \times \text{Height} \\ & = {1 \over 2} \times \left(1 + 1{2 \over 3}\right) \times 3 \\ & = 4 \text{ units}^2 \\ \\ \text{Area of shaded region} & = \left({9 \over 2} \ln 3 - 4 \right) \text{ units}^2 \\ \\ \\ \therefore a & = {9 \over 2}, b = -4 \end{align}