(i)

\begin{align} \cos \theta & = c \\ {Adj \over Hyp} & = {c \over 1} \end{align}

\begin{align} \text{By Pythagoras theorem, } Opp & = \sqrt{1^2 - c^2} \\ & = \sqrt{1 - c^2} \\ \\ \tan \theta & = {Opp \over Adj} \\ & = { \sqrt{1 - c^2} \over c} \end{align}

(ii)

\begin{align} \text{cosec } \theta & = {1 \over \sin \theta} \\ & = {1 \over {Opp \over Hyp}} \\ & = {Hyp \over Opp} \\ & = {1 \over \sqrt{1 - c^2}} \end{align}

\begin{align} y & = Ae^{2x} + Be^{-x} \\ \\ {dy \over dx} & = 2Ae^{2x} + (-1)Be^{-x} \\ & = 2Ae^{2x} - Be^{-x} \\ \\ \text{LHS} & = {dy \over dx} + 4y \\ & = 2A^{2x} - Be^{-x} + 4(Ae^{2x} + Be^{-x}) \\ & = 2A^{2x} - Be^{-x} + 4Ae^{2x} + 4Be^{-x} \\ & = 6A^{2x} + 3Be^{-x} \\ \\ \\ \therefore 6A^{2x} + 3Be^{-x} & = e^{2x} - e^{-x} \\ \\ \text{Comparing} & \text{ coefficient of } e^{2x}, \\ 6A & = 1 \\ A & = {1 \over 6} \\ \\ \text{Comparing} & \text{ coefficient of } e^{-x}, \\ 3B & = -1 \\ B & = -{1 \over 3} \\ \\ \\ \therefore A & = {1 \over 6}, B = -{1 \over 3} \end{align}

\begin{align} {dA \over dt} & = -48 \text{ cm}^2 \text{/s} \phantom{000} [\text{Decreasing}] \\ \\ A & = 6(x)(x) \\ A & = 6x^2 \\ \\ {dA \over dx} & = 2(6x) \\ & = 12x \\ \\ \\ {dx \over dt} & = {dx \over dA} \times {dA \over dt} \\ & = {1 \over 12x} \times -48 \\ & = -{48 \over 12x} \\ & = -{4 \over x} \\ \\ \text{When } & x = 10, \\ {dx \over dt} & = -{4 \over 10} \\ & = -0.4 \\ \\ \therefore \text{Rate of change of } x & = -0.4 \text{ cm/s} \end{align}

(i) Every minute, 21% of existing germs are destroyed, leaving 79% of germs behind

\begin{align} & \text{Initially } (n = 0), \text{ number of germs} = N \\ \\ & \text{After 1 minute } (n = 1), \text{ number of germs} = 0.79N \\ \\ & \text{After 2 minutes } (n = 2), \text{ number of germs} = 0.79(0.79N) = (0.79)^2 N \\ \\ & \text{After 3 minutes } (n = 3), \text{ number of germs} = 0.79 [ (0.79)^2 N ] = (0.79)^3 N \\ & . \\ & . \\ & . \\ \\ & \text{After } n \text{ minutes}, \text{ number of germs} = (0.79)^n N \end{align}

(ii)

\begin{align} \text{From (i), no. of germs alive after } n \text{ minutes} & = (0.79)^n N \\ \\ \text{No. of germs alive after } 20 \text{ minutes} & = (0.79)^{20} N \\ \\ \\ x\% \text{ of germs are destroyed after 20 mins, } & \text{leaving } (100 - x)\% \text{ behind} \\ \\ \text{No. of germs alive after 20 minutes} & = N \times (100 - x) \% \\ & = N \times {100 - x \over 100} \\ & = N\left(100 -x \over 100\right) \\ \\ \\ N\left(100 -x \over 100\right) & = (0.79)^{20} N \\ {100 -x \over 100} & = (0.79)^{20} \\ {100 - x \over 100} & = 0.896 \phantom{.} 482 \phantom{.} 5 \times 10^{-3} \\ 100 - x & = 0.896 \phantom{.} 482 \phantom{.} 5 \\ -x & = -99.103 \phantom{.} 517 \phantom{.} 5 \\ x & = 99.103 \phantom{.} 517 \phantom{.} 5 \\ & \approx 99 \text{ (2 s.f.)} \end{align}

(iii)

\begin{align} \text{From part (i), } \text{no. of germs alive} & = (0.79)^n N \\ \\ \therefore Ne^{kn} & = (0.79)^n N \\ e^{kn} & = 0.79^n \\ (e^k)^n & = (0.79)^n \\ \\ e^k & = 0.79 \\ \ln e^k & = \ln 0.79 \\ [\text{Power law (logarithms)}] \phantom{000} k\ln e & = \ln 0.79 \\ k(1) & = \ln 0.79 \\ k & = \ln 0.79 \\ & \approx -0.236 \end{align}

(i)

\begin{align} \angle QTP & = 90^\circ \phantom{000} (\text{Right angle in semi-circle}) \\ \\ \angle PTR & = 180^\circ - 90^\circ \phantom{000} (\text{Adjacent angles on a straight line}) \\ & = 90^\circ \\ \\ \text{Let } \angle PQR & = x^\circ \\ \\ \angle PRQ & = \angle PQR \phantom{000} (\text{Base angles of isosceles triangle } PQR) \\ & = x^\circ \\ \\ \angle PTS & = \angle PQR \phantom{000} (\text{Alternate segment theorem}) \\ & = x^\circ \\ \\ \angle RTS & = (90 - x)^\circ \\ \\ \angle TSR & = 180 - x - (90 - x) \phantom{000} (\text{Angle sum of triangle}) \\ & = 180 - x - 90 + x \\ & = 90^\circ \phantom{0} \text{ (Shown)} \end{align}

(ii)

\begin{align} & \text{Since } \angle TSR = 90^\circ, RT \text{ is the diameter of the (new) circle} \\ & \text{Thus, the midpoint of } TR \text{ is the centre of the (new) circle} \end{align}

(i)

\begin{align} \left(2 - {x \over 8}\right)^6 & = 2^6 + {6 \choose 1} (2)^5 \left(-{x \over 8}\right) + {6 \choose 2} (2)^4 \left(-{x \over 8}\right)^2 + ... \\ & = 64 + 6(32)\left(-{x \over 8}\right) + 15(16)\left(x^2 \over 64\right) + ... \\ & = 64 - 24x + 3.75x^2 + ... \end{align}

(ii)

\begin{align} (4 + kx + x^2) \left(2 - {x \over 8}\right)^6 & = (4 + kx + x^2) (64 - 24x + 3.75x^2 + ...) \\ & = 4(64) - 4(24x) + 4(3.75x^2) + kx(64) - kx(24x) + ... + x^2 (64) + ... \\ & = 256 - 96x + 15x^2 + 64kx - 24kx^2 + 64x^2 + ... \\ & = 256 + 64kx - 96x + 79x^2 - 24kx^2 + ... \\ & = 256 + (64k - 96)x + (79 - 24k)x^2 + ... \\ \\ \text{Coefficient of } x & = 64k - 96 \\ \\ \text{Coefficient of } x^2 & = 79 -24k \\ \\ (64k - 96) + (79 - 24k) & = 0 \\ 64k - 96 + 79 - 24k & = 0 \\ 40k - 17 & = 0 \\ 40k & = 17 \\ k & = {17 \over 40} \end{align}

(i)

\begin{align} u & = 2x + 5 &&& v & = x - 2 \\ {du \over dx} & = 2 &&& {dv \over dx} & = 1 \end{align}
\begin{align} {dy \over dx} & = 1 + \underbrace{ (x - 2)(2) - (2x + 5)(1) \over (x - 2)^2 }_\text{Quotient rule} \\ & = 1 + {2(x - 2) - (2x + 5) \over (x - 2)^2} \\ & = 1 + {2x - 4 - 2x - 5 \over (x - 2)^2} \\ & = 1 + { - 9 \over (x - 2)^2} \\ & = 1 - {9 \over (x - 2)^2} \\ & = 1 - 9(x - 2)^{-2} \\ \\ {d^2 y \over dx^2} & = 0 - (-2)(9)(x - 2)^{-3} \phantom{00000} [\text{Chain rule}] \\ & = 18(x - 2)^{-3} \\ & = {18 \over (x - 2)^3} \end{align}

(ii)

\begin{align} \text{At stationary points, } & {dy \over dx} = 0 \\ \\ 0 & = 1 - {9 \over (x - 2)^2} \\ {9 \over (x - 2)^2} & = 1 \\ 9 & = (x - 2)^2 \\ \pm \sqrt{9} & = x - 2 \\ \pm 3 & = x - 2 \end{align} \begin{align} x - 2 & = 3 && \text{ or } & x - 2 & = -3 \\ x & = 5 &&& x & = -1 \end{align}

(iii)

\begin{align} \text{When } x = 5, \phantom{.} {d^2y \over dx^2} & = {18 \over (5 - 2)^3} \\ & = {2 \over 3} > 0 \\ \\ \therefore \text{The point where } x & = 5 \text{ is a minimum point} \\ \\ \\ \text{When } x = -1, \phantom{.} {d^2y \over dx^2} & = {18 \over (-1 - 2)^3} \\ & = -{2 \over 3} < 0 \\ \\ \therefore \text{The point where } x & = -1 \text{ is a maximum point} \\ \end{align}

(i)

\begin{align} & \text{Since length of } AB \text{ is 4 units } \Rightarrow B(0, -2) \\ \\ & \text{Since } P \text{ lies on } x \text{-axis, } y \text{-coordinate of } P \text{ is 0} \\ \\ & \text{Since } P \text{ is midpoint of } AC, \text{ let } y\text{-coordinate of } C \text{ be } c \\ \\ &\phantom{000000000000000000000.} {-2 + c \over 2} = 0 \\ &\phantom{000000000000000000000.}-2 + c = 0 \\ &\phantom{0000000000000000000000000(-} c = -2 \\ \\ & \text{Since } B \text{ and } C \text{ have the same } y \text{-coordinate, } BC \text{ is a horizontal line and parallel to the } x\text{-axis} \end{align}

(ii)

\begin{align} \text{Since } AB = BC = 4 \text{ units}, & \phantom{0} C(4, -2) \end{align}

(iii)

\begin{align} \text{Let } (x , y) & \text{ denote coordinates of } D \\ \\ \text{Area of kite} & = {1 \over 2} \left| \begin{matrix} x & 0 & 0 & 4 & x \\ y & 2 & - 2 & -2 & y \end{matrix} \right| \\ 48 & = {1 \over 2} \left[ (2x + 0 + 0 + 4y) - (0 + 0 - 8 - 2x) \right] \\ 48 & = {1 \over 2} (2x + 4y + 8 + 2x) \\ 48 & = {1 \over 2} (4x + 4y + 8) \\ 96 & = 4x + 4y + 8 \\ 88 & = 4x + 4y \\ 12 & = x + y \\ \\ y & = -x + 12 \phantom{000000} [\text{Coordinates of } D \text{ satisfies this equation}] \\ \\ \\ \text{Coordinates of } P & = \left( {0 + 4 \over 2}, 0 \right) \\ & = (2, 0) \\ \\ \text{Gradient of } BP & = {-2 - 0 \over 0 - 2} \\ & = 1 \\ \\ y & = m x + c \\ y & = x + c \\ \\ \text{Using } & P(2, 0), \\ \\ 0 & = 2 + c \\ c & = -2 \\ \\ \text{Equation of line } BD: & \phantom{.} y = x - 2 \\ \\ \text{Substitute } y = x - 2 & \text{ into } y = -x + 12, \\ x - 2 & = -x + 12 \\ 2x & = 14 \\ x & = {14 \over 2} \\ x & = 7 \\ \\ \text{Substitute } x = 7 & \text{ into } y = -x + 12, \\ y & = -(7) + 12 \\ y & = 5 \\ \\ \\ \therefore & \phantom{.} D(7, 5) \end{align}

\begin{align} 3^{x + y} & = \sqrt[3]{27} \\ 3^{x + y} & = 3 \\ 3^{x + y} & = 3^1 \\ \\ \therefore x + y & = 1 \\ y & = 1 - x \phantom{000} \text{ --- (1)} \\ \\ \\ {4^y \over 2^x} & = \left(1 \over 2\right)^{-3} \\ {2^{2y} \over 2^x} & = 2^3 \\ 2^{2y - x} & = 2^3 \\ \\ \therefore 2y - x & = 3 \phantom{000} \text{ --- (2)} \\ \\ \\ \text{Substitute } & \text{(1) into (2),} \\ 2(1 - x) - x & = 3 \\ 2 - 2x - x & = 3 \\ -3x & = 1 \\ x & = -{1 \over 3} \\ \\ \text{Substitute } & x = -{1 \over 3} \text{ into (1)}, \\ y & = 1 - \left(-{1 \over 3}\right) \\ & = {4 \over 3} \end{align}

\begin{align} \text{Volume of cylinder} & = \pi r^2 h \\ \\ (3\sqrt{7} - 6)\pi & = \pi r^2 (2 + \sqrt{7}) \\ 3 \sqrt{7} - 6 & = r^2 (2 + \sqrt{7}) \\ \\ r^2 & = {3\sqrt{7} - 6 \over 2 + \sqrt{7}} \\ & = {3\sqrt{7} - 6 \over 2 + \sqrt{7}} \times {2 - \sqrt{7} \over 2 - \sqrt{7}} \phantom{000000} [\text{Rationalise denominator}] \\ & = {6\sqrt{7} - 3(7) - 6(2) + 6\sqrt{7} \over 4 - 2\sqrt{7} + 2\sqrt{7} - 7} \\ & = {12\sqrt{7} -33 \over -3} \\ & = {12\sqrt{7} \over -3} + {-33 \over -3} \\ & = -4\sqrt{7} + 11 \\ & = 11 - 4\sqrt{7} \end{align}

(i) When the dot is at rest, the velocity of the dot is equals to 0 (i.e. v = 0)

\begin{align} s & = t^3 - 6t^2 + 9t \\ \\ v & = {ds \over dt} \\ & = 3t^2 - 2(6t) +9 \\ & = 3t^2 - 12t + 9 \\ \\ \text{When } & v = 0, \\ 0 & = 3t^2 - 12t + 9 \\ 0 & = t^2 - 4t + 3 \\ 0 & = (t - 1)(t - 3) \end{align} \begin{align} t - 1 & = 0 && \text{ or } & t - 3 & = 0 \\ t & = 1 &&& t & = 3 \end{align}

(ii)

\begin{align} a & = {dv \over dt} \\ & = 2(3t) - 12 \\ & = 6t - 12 \\ \\ \text{When } & t = 1, \\ a & = 6(1) - 12 \\ & = -6 \text{ cm/s}^2 \end{align}

(iii)

\begin{align} & \text{Finding the value of } s \text{ when } t = 4 \text{ gives the displacement of the dot relative to fixed point } O \text{ when } t = 4. \\ \\ & \text{Since the dot is instantaneously at rest when } t = 1 \text{ or } t = 3, \text{ the dot may have changed direction and the total distance} \\ & \text{travelled is larger than the displacement when } t = 4. \end{align}

(iv)

\begin{align} \text{When } t = 0, & \phantom{.} s = 0^3 - 6(0)^2 + 9(0) = 0 \text{ cm} \\ \\ \text{When } t = 1, & \phantom{.} s = (1)^3 - 6(1)^2 + 9(1) = 4 \text{ cm} \\ \\ \text{When } t = 3, & \phantom{.} s = (3)^3 - 6(3)^2 + 9(3) = 0 \text{ cm} \\ \\ \text{When } t = 4, & \phantom{.} s = (4)^3 - 6(4)^2 + 9(4) = 4 \text{ cm} \end{align}

\begin{align} \text{Total distance} & = 4 + 4 + 4 \\ & = 12 \text{ cm} \end{align}

(i)

\begin{align} \text{Amplitude of } f(x) & = 2 \\ \\ \text{Least value of } f(x) & = -2 \\ \\ \text{Greatest value of } f(x) & = 2 \end{align}

(ii)

\begin{align} \text{Amplitude of } g(x) & = 3 \\ \\ \text{Center line: } & y = -1 \\ \\ \text{Least value of } g(x) & = -1 - 3 \\ & = -4 \\ \\ \text{Greatest value of } g(x) & = -1 + 3 \\ & = 2 \end{align}

(iii)

\begin{align} \text{Period of } f(x) & = {2\pi \over 2} \\ & = \pi = 180^\circ \end{align}

(iv)

\begin{align} g(x) & = 3 \cos \left(x \over 2\right) - 1 \\ & = 3 \cos \left({1 \over 2}x\right) - 1 \\ \\ \text{Period of } g(x) & = {2\pi \over {1 \over 2}} \\ & = 4\pi = 720^\circ \end{align}

(v)

\begin{align} \text{For } y = 2 \sin 2x, \text{ max. value} & = 2 \\ \text{Min. value} & = -2 \\ \text{Period} & = 180^\circ \\ \text{No. of cycles} & = {360^\circ \over 180^\circ} = 2 \\ \\ \\ \text{For } y = 3 \cos \left(x \over 2\right) - 1, \text{ center line: } & y = -1 \\ \text{Max. value} & = 2 \\ \text{Min. value} & = -4 \\ \text{Period} & = 720^\circ \\ \text{No. of cycles} & = {360^\circ \over 720^\circ} = {1 \over 2} \end{align}

(vi)

\begin{align} 2\sin 2x + 1 & = 3\cos \left(x \over 2\right) \\ \underbrace{2\sin 2x}_{y = f(x)} & = \underbrace{3\cos \left(x \over 2\right) - 1}_{y = g(x)} \\ \\ \text{From the sketch, } & \text{graphs meet at 3 points} \\ \\ \therefore \text{No. } & \text{of solutions} = 3 \end{align}

(i)

\begin{align} u & = x^2 &&& v & = \ln x \\ {du \over dx} & = 2x &&& {dv \over dx} & = {1 \over x} \end{align}
\begin{align} {d \over dx} (x^2 \ln x) & = (x^2)\left(1 \over x\right) + (\ln x)(2x) \phantom{00000} [\text{Product rule}] \\ & = x + 2x \ln x \end{align}

(ii)

\begin{align} \text{Since } {d \over dx} (x^2 \ln x) = x + 2x \ln x, & \phantom{0} \int (x + 2x \ln x) \phantom{.} dx = x^2 \ln x \\ \\ \int x \phantom{.} dx + \int 2x \ln x \phantom{.} dx & = x^2 \ln x \\ \int 2x \ln x \phantom{.} dx & = x^2 \ln x - \int x \phantom{.} dx \\ \int 2x \ln x \phantom{.} dx & = x^2 \ln x - {1 \over 2}x^2 \\ 2 \int x \ln x \phantom{.} dx & = x^2 \ln x - {1 \over 2}x^2 \\ \int x \ln x \phantom{.} dx & = {1 \over 2} \left(x^2 \ln x - {1 \over 2}x^2\right) \\ & = {1 \over 2}x^2 \ln x - {1 \over 4}x^2 + C \end{align}

(i)

\begin{align} \require{cancel} \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)} \\ & = {(1 - \cos^2 \theta) - 1 + \cos \theta \over \sin \theta (1 - \cos \theta)} \phantom{00000} [\sin^2 A + \cos^2 A = 1 \implies \sin^2 A = 1 - \cos^2 A] \\ & = {\cos \theta - \cos^2 \theta \over \sin \theta (1 - \cos \theta)} \\ & = {\cos \theta \cancel{( 1 - \cos \theta)} \over \sin \theta \cancel{(1 - \cos \theta)}} \\ & = {\cos \theta \over \sin \theta} \\ & = \cot \theta \\ & = \text{R.H.S} \end{align}

(ii)

\begin{align} \underbrace{{\sin \theta \over 1 - \cos \theta} - {1 \over \sin \theta}}_\text{from (i)} & = 3 \tan \theta \\ \cot \theta & = 3 \tan \theta \\ {1 \over \tan \theta} & = 3 \tan \theta \\ 1 & = 3 \tan^2 \theta \\ {1 \over 3} & = \tan^2 \theta \\ \\ \tan \theta & = \pm \sqrt{1 \over 3} \phantom{00000} [\text{All 4 quadrants}] \\ \text{Basic angle, } \alpha & = \tan^{-1} \sqrt{1 \over 3} \\ & = 30^\circ \end{align}

\begin{align} \theta & = 30^\circ, 180^\circ - 30^\circ, 180^\circ + 30^\circ, 360^\circ - 30^\circ \\ & = 30^\circ, 150^\circ, 210^\circ (\text{Reject since } 0^\circ \le \theta \le 180^\circ), 330^\circ (\text{Reject since } 0^\circ \le \theta \le 180^\circ) \\ & = 30^\circ, 150^\circ \end{align}

(i)

\begin{align} \text{Let } f(x) & = x^3 + x^2 - x - 1 \\ \\ f(1) & = (1)^3 + (1)^2 - (1) - 1 \\ & = 0 \\ \\ \text{By Fac} & \text{tor theorem, } x - 1 \text{ is a factor of } f(x) \end{align}

(ii)

$$ \require{enclose} \begin{array}{rll} x^2 + 2x + 1 \phantom{0000}\\ x - 1 \enclose{longdiv}{x^3 + x^2 - x - 1 \phantom{0}}\kern-.2ex \\ -\underline{(x^3 - x^2){\phantom{0000000}}} \\ 2x^2 -x - 1 \phantom{0} \\ -\underline{(2x^2 - 2x){\phantom{0.-}}} \\ x - 1 \phantom{0} \\ -\underline{(x - 1){\phantom{.}}} \\ 0 \phantom{0} \end{array} $$
\begin{align} x^3 + x^2 - x - 1 & = (x^2 + 2x + 1)(x - 1) \\ & = (x + 1)(x + 1)(x - 1) \\ & = (x + 1)^2(x - 1) \\ \\ {4 \over x^3 + x^2 - x - 1} & = {4 \over (x + 1)^2 (x - 1)} \\ & = {A \over x - 1} + {B \over x + 1} + {C \over (x + 1)^2} \\ & = {A(x + 1)^2 \over (x + 1)^2 (x - 1)} + {B(x - 1)(x + 1) \over (x + 1)^2 (x - 1)} + {C(x - 1) \over (x + 1)^2 (x - 1)} \\ & = {A(x + 1)^2 + B(x - 1)(x + 1) + C(x - 1) \over (x + 1)^2 (x - 1)} \\ \\ 4 & = A(x + 1)^2 + B(x - 1)(x + 1) + C(x - 1) \\ \\ \text{Let } x = -1, \phantom{0} 4 & = A(0)^2 + B(-2)(0) + C(-2) \\ 4 & = -2C \\ -2 & = C \\ \\ \text{Let } x = 1, \phantom{0} 4 & = A(2)^2 + B(0)(2) + (-2)(0) \\ 4 & = 4A \\ 1 & = A \\ \\ \text{Let } x = 0, \phantom{0} 4 & = (1)(1)^2 + B(-1)(1) + (-2)(-1) \\ 4 & = 1 - B + 2 \\ B & = 1 + 2 - 4 \\ B & = -1 \\ \\ \\ \therefore {4 \over x^3 + x^2 - x - 1} & = {1 \over x - 1} - {1 \over x + 1} - {2 \over (x + 1)^2} \end{align}

(a)

\begin{align} \log_2 x + \log_{16} x & = -2.5 \\ \log_2 x + {\log_2 x \over \log_2 16} & = -2.5 \phantom{00000} [\text{Change-of-base}] \\ \log_2 x + {\log_2 x \over \log_2 2^4} & = -2.5 \\ \log_2 x + {\log_2 x \over 4\log_2 2} & = -2.5 \phantom{00000} [\text{Power law}] \\ \log_2 x + {\log_2 x \over 4(1)} & = -2.5 \\ {4\log_2 x \over 4} + {\log_2x \over 4} & = -2.5 \\ {5\log_2 x \over 4} & = -2.5 \\ 5\log_2 x & = 4(-2.5) \\ 5\log_2 x & = -10 \\ \log_2 x & = -{10 \over 5} \\ \log_2 x & = -2 \\ x & = 2^{-2} \\ x & = {1 \over 4} \end{align}

(b)(i)

\begin{align} \lg z - \lg y & = \lg (z + y) \\ [\text{Quotient law}] \phantom{00000} \lg \left( z \over y\right) & = \lg (z + y) \\ \\ \therefore {z \over y} & = z + y \\ z & = yz + y^2 \\ z - yz & = y^2 \\ z(1 - y) & = y^2 \\ z & = {y^2 \over 1 - y} \end{align}

(b)(ii)

\begin{align} & \text{For } \lg z \text{ to be defined, } z \text{ is a positive real number, i.e. } z > 0 \\ \\ & \text{For } 0 < y < 1, \phantom{0} (1 - y) \text{ is positive and } z = {y^2 \over 1 - y} \text{ is a positive real number} \end{align}

\begin{align} f''(x) & = 3\cos 3x - 4\sin 2x \\ \\ f'(x) & = \int 3 \cos 3x - 4\sin 2x \phantom{.} dx \\ & = {3\sin 3x \over 3} - {-4\cos 2x \over 2} + C \\ & = \sin 3x + 2\cos 2x + C \\ \\ f(x) & = \int \sin 3x + 2\cos 2x + C \phantom{.} dx \\ & = {-\cos 3x \over 3} + {2\sin 2x \over 2} + Cx + D \\ & = -{1 \over 3} \cos 3x + \sin 2x + Cx + D \\ \\ \text{Since } & f(0) = 0, \\ 0 & = -{1 \over 3} \cos 0 + \sin 0 + 0 + D \\ 0 & = -{1 \over 3} + 0 + 0 + D \\ {1 \over 3} & = D \\ \\ f(x) & = -{1 \over 3} \cos 3x + \sin 2x + Cx + {1 \over 3} \\ \\ \text{Since } & f\left({\pi \over 2}\right) = {5 \over 6}, \\ {5 \over 6} & = -{1 \over 3} \cos \left[3\left(\pi \over 2\right)\right] + \sin \left[2\left(\pi \over 2\right)\right] + C\left(\pi \over 2\right) + {1 \over 3} \\ {5 \over 6} & = -{1 \over 3} (0) + 0 + {\pi \over 2} C + {1 \over 3} \\ {5 \over 6} & = {\pi \over 2} C + {1 \over 3} \\ {1 \over 2} & = {\pi \over 2} C \\ 1 & = \pi C \\ {1 \over \pi} & = C \\ \\ f(x) & = -{1 \over 3} \cos 3x + \sin 2x + {1 \over \pi}x + {1 \over 3} \\ \\ f\left(\pi \over 3\right) & = -{1 \over 3} \cos \left[3\left({\pi \over 3}\right)\right] + \sin \left[ 2 \left(\pi \over 3\right)\right] + {1 \over \pi}\left(\pi \over 3\right) + {1 \over 3} \\ & = -{1 \over 3}(-1) + {\sqrt{3} \over 2} + {1 \over 3} + {1 \over 3} \\ & = {1 \over 3} + {\sqrt{3} \over 2} + {2 \over 3} \\ & = 1 + {\sqrt{3} \over 2} \text{ (Shown)} \end{align}

(i)

The normal (dotted line) to the circle at (5, 5) meets the other normal 3y = x + 2 at the centre of the circle

\begin{align} \text{Gradient of tangent at } (5, 5) & = {5 - 0 \over 5 - 0} \\ & = 1 \\ \\ \text{Gradient of normal at } (5, 5) & = -1 \div 1 \phantom{000000} [m_1 \times m_2 = -1] \\ & = -1 \\ \\ y & = mx + c \\ y & = -x + c \\ \\ \text{Using } & (5, 5), \\ 5 & = -(5) + c \\ 10 & = c \\ \\ \text{Equation of normal at } (5,5): & \phantom{0} y = -x + 10 \\ \\ \text{Substitute } y = -x + 10 & \text{ into } 3y = x + 2, \\ 3(-x + 10) & = x + 2 \\ -3x + 30 & = x + 2 \\ -4x & = -28 \\ x & = {-28 \over -4} \\ x & = 7 \\ \\ \text{Substitute } x = 7 & \text{ into } y = -x + 10, \\ y & = -(7) + 10 \\ y & = 3 \\ \\ \text{Centre of circle: } & (7, 3) \\ \\ \text{Radius of circle} & = \sqrt{(7 - 5)^2 + (3 - 5)^2} \\ & = \sqrt{8} \\ \\ (x - 7)^2 + (y - 3)^2 & = (\sqrt{8})^2 \\ \text{Eqn: } (x - 7)^2 + (y - 3)^2 & = 8 \end{align}

(ii)

Since the centre is (7, 3) and the radius is √8 units, the circle is completely above the x-axis

\begin{align} y\text{-coordinate of lowest point of circle} & = 3 - \sqrt{8} \\ \\ \therefore \text{Required coordinate is } & (7, 3- \sqrt{8}) \end{align}

(i)

\begin{align} \text{(Minimum) quadratic curve: } y & = x^2 - x - 6 \\ \\ \text{Let } x = 0, \phantom{.} y & = (0)^2 - (0) - 6 \\ y & = -6 \phantom{000000} [y \text{-intercept}] \\ \\ \text{Let } y = 0, \phantom{.} 0 & = x^2 - x - 6 \\ 0 & = (x - 3)(x + 2) \\ x & = 3 \text{ or } - 2 \phantom{000000} [x \text{-intercepts}] \\ \\ \text{Line of symmetry, } x & = {3 + (-2) \over 2} \\ x & = 0.5 \\ \\ \text{When } & x = 0.5, \\ y & = (0.5)^2 - (0.5) - 6 \\ y & = -6.25 \\ \\ \Rightarrow \text{Minimum turning} & \text{ point is } (0.5, -6.25) \end{align}

(ii)

\begin{align} \text{When } & x = 2, \\ y & = |(2)^2 - (2) - 6| \\ y & = |-4| \\ y & = 4 \\ \\ \text{Line is tangent } & \text{to curve at the point } (2, 4) \\ \\ \text{Substitute } & (2, 4) \text{ into } y = k - 3x, \\ 4 & = k - 3(2) \\ 4 & = k - 6 \\ 10 & = k \text{ (Shown)} \end{align}

(iii)

\begin{align} y & = 10 -3x \\ \\ \text{Let } y = 0 , \phantom{.} 0 & = 10 - 3x \\ 3x & = 10 \\ x & = {10 \over 3} \\ x & = 3{1 \over 3} \phantom{000000} [x \text{-intercept}] \end{align}

(i)

$$t$$	$$2$$	$$4$$	$$6$$	$$8$$
$$\ln (v + 3)$$	$$2.24$$	$$2.00$$	$$1.74$$	$$1.48$$

(ii)

\begin{align} v & = Ae^{t \over k} - 3 \\ v + 3 & = Ae^{t \over k} \\ \ln (v + 3) & = \ln Ae^{t \over k} \\ \ln (v + 3) & = \ln A + \ln e^{t \over k} \phantom{00000} [\text{Product law}] \\ \ln (v + 3) & = \ln A + {t \over k} (\ln e) \phantom{000.} [\text{Power law}] \\ \ln (v + 3) & = \ln A + {t \over k} (1) \\ \ln (v + 3) & = \ln A + {1 \over k} t \\ \ln (v + 3) & = {1 \over k}t + \ln A \phantom{0000000} [ Y = mX + c] \\ \\ \text{From graph, vertical intercept, } c & = 2.5 \\ \ln A & = 2.5 \\ \log_e A & = 2.5 \\ A & = e^{2.5} \\ A & = 12.1824 \\ A & \approx 12.2 \\ \\ \\ \text{Gradient, } m & = {2.5 - 1.6 \over 0 - 7} \\ {1 \over k} & = -{9 \over 70} \\ 70 & = -9k \\ \\ k & = {70 \over -9} \\ k & = -7{7 \over 9} \end{align}

(iii)

\begin{align} v & = Ae^{t \over k} - 3 \\ \\ \text{When } & t = 0, \\ v & = Ae^0 - 3 \\ v & = A - 3 \\ v & = e^{2.5} - 3 \\ v & \approx 9.18 \text{ m/s} \end{align}

(iv)

\begin{align} \ln (v + 3) & = \ln (5 + 3) \\ & = 2.0794 \\ & \approx 2.08 \\ \\ \text{1) On graph, locate the } & \text{point on the line where } \ln (v + 3) = 2.08 \\ \\ \text{2) Locate the correspon} & \text{ding value of } t \end{align}

(v)

\begin{align} \text{Linear equation} : \ln (v + 3) & = {1 \over k} t + \ln A \\ \\ \text{When } & v = 0, \\ \ln (0 + 3) & = {1 \over k} t + \ln A \\ \ln 3 & = {1 \over k} t + \ln A \\ \ln 3 - \ln A & = {1 \over k} t \\ \\ t & = k (\ln 3 - \ln A) \\ & = -7{7 \over 9} (\ln 3 - \ln e^{2.5}) \\ & = 10.899 \\ & \approx 10.9 \end{align}

(i)

\begin{align} x^2 & = 3x - 4 \\ \\ \text{Substitute } & x = \alpha, \\ \alpha^2 & = 3\alpha - 4 \\ \\ \text{Multiply} & \text{ by } \alpha \text{ on both sides, } \\ \alpha^3 & = \alpha(3\alpha - 4) \\ \alpha^3 & = 3\alpha^2 - 4\alpha \\ \\ \text{Since } & \alpha^2 = 3\alpha - 4, \\ \alpha^3 & = 3(3\alpha - 4) - 4\alpha \\ & = 9\alpha - 12 - 4\alpha \\ & = 5\alpha - 12 \text{ (Shown)} \end{align}

(ii)

\begin{align} \beta^3 & = 5\beta - 12 \end{align}

(iii)

\begin{align} \alpha^3 + \beta^3 & = (5\alpha - 12) + (5\beta - 12) \\ & = 5\alpha + 5\beta - 24 \\ & = 5(\alpha + \beta) - 24 \\ \\ x^2 & = 3x - 4 \\ x^2 - 3x + 4 & = 0 \\ \\ [a = 1, b & = -3, c = 4] \\ \\ \text{Sum of roots, } \alpha + \beta & = -{b \over a} \\ & = -{-3 \over 1} \\ & = 3 \\ \\ \text{Product of roots, } \alpha \beta & = {c \over a} \\ & = {4 \over 1} \\ & = 4 \\ \\ \\ \therefore \alpha^3 + \beta^3 & = 5(3) -24 \\ & = -9 \end{align}

(iv)

\begin{align} \text{Sum of roots} & = {\alpha^2 \over \beta} + {\beta^2 \over \alpha} \\ & = {\alpha^3 \over \alpha \beta} + {\beta^3 \over \alpha \beta} \\ & = {\alpha^3 + \beta^3 \over \alpha \beta} \\ & = {-9 \over 4} \\ & = -{9 \over 4} \\ \\ \text{Product of roots} & = {\alpha^2 \over \beta} \times {\beta^2 \over \alpha} \\ & = {\alpha^2 \beta^2 \over \alpha \beta} \\ & = \alpha \beta \\ & = 4 \\ \\ \\ x^2 - (SOR)x + POR & = 0 \\ x^2 - \left(-{9 \over 4}\right)x + 4 & = 0 \\ x^2 + {9 \over 4}x + 4 & = 0 \\ 4x^2 + 9x + 16 & = 0 \end{align}

(i)

\begin{align} \cos \theta & = {Adj \over Hyp} \\ \cos \theta & = {JN \over 12} \\ \\ JN & = 12 \cos \theta \text{ cm} \\ \\ \sin \theta & = {Opp \over Hyp} \\ \sin \theta & = {JM \over 12} \\ \\ JM & = 12 \sin \theta \text{ cm} \\ \\ \\ P & = 12 \sin \theta + 5 + 12 + (5 + 12 \cos \theta) \\ & = 22 + 12\cos \theta + 12\sin \theta \text{ (Shown)} \end{align}

(ii)

\begin{align} a\cos \theta + b\sin \theta & = R \cos (\theta - \alpha) \\ 12\cos \theta + 12\sin \theta & = R\cos (\theta - \alpha) \\ \\ R & = \sqrt{a^2 + b^2} \\ & = \sqrt{12^2 + 12^2} \\ & = \sqrt{2(12)^2} \\ & = 12\sqrt{2} \\ \\ \alpha & = \tan^{-1} \left(b \over a\right) \\ & = \tan^{-1} \left( 12 \over 12 \right) \\ & = {\pi \over 4} \\ \\ \therefore 12 \cos \theta + 12 \sin \theta & = 12\sqrt{2} \cos \left( \theta - {\pi \over 4} \right) \\ \\ P & = 22 + 12 \cos \theta + 12 \sin \theta \\ & = 22 + 12\sqrt{2} \cos \left( \theta - {\pi \over 4} \right) \\ \\ \\ \text{Since } -12\sqrt{2} \le 12\sqrt{2} & \cos \left( \theta - {\pi \over 4} \right) \phantom{0} \le 12 \sqrt{2}, \\ \\ \text{Maximum value of } P & = (22 + 12\sqrt{2}) \text{ cm} \end{align}

(iii)

\begin{align} \text{Area of trapezium} & = {1 \over 2} (\text{Sum of parallel sides}) (\text{Height}) \\ A & = {1 \over 2} (5 + 5 + 12\cos \theta) (12\sin\theta) \\ & = {1 \over 2} (10 + 12\cos \theta) (12\sin \theta) \\ & = (5 + 6\cos \theta)(12\sin \theta) \\ & = 60\sin \theta + 72 \sin \theta \cos \theta \\ & = 60\sin \theta + 36(2\sin \theta \cos \theta) \phantom{00000} [\text{Double angle formula, } \sin 2A = 2 \sin A \cos A] \\ & = 60\sin \theta + 36\sin 2 \theta \text{ (Shown)} \end{align}

(iv)