\begin{align*} \text{Area of circle, } A & = \pi r^2 \\ \\ {dA \over dr} & = 2 \pi r \\ \\ {dr \over dt} & = {dr \over dA} \times {dA \over dt} \\ & = {1 \over 2 \pi r } \times {250 \over 1} \\ & = {250 \over 2 \pi r } \\ & = {125 \over \pi r} \\ \\ \text{When } & r = 40, \\ {dr \over dt} & = {125 \over \pi (40)} \\ & = {25 \over 8 \pi }\text{ m per hour} \\ & \approx 0.995 \text{ m per hour} \end{align*}

(a)

\begin{align*} {d \over dx} \sqrt{x^2 + x + 1} & = {d \over dx} (x^2 + x + 1)^{1 \over 2} \\ & = {1 \over 2} (x^2 + x + 1)^{-{1 \over 2}} . (2x + 1) \phantom{000000} [\text{Chain rule}] \\ & = {1 \over 2} \left( { 1 \over \sqrt{x^2 + x + 1} }\right) \left( {2x + 1 \over 1 } \right) \\ & = { 2x + 1 \over 2 \sqrt{x^2 + x + 1} } \end{align*}

(b)

\begin{align*} \text{From (a), } {d \over dx} \sqrt{x^2 + x + 1} & = { 2x + 1 \over 2 \sqrt{x^2 + x + 1} } \\ \\ \implies \int { 2x + 1 \over 2 \sqrt{x^2 + x + 1} } \phantom{.} dx & = \sqrt{x^2 + x + 1} \\ 2 \int { 2x + 1 \over 2 \sqrt{x^2 + x + 1} } \phantom{.} dx & = 2 \sqrt{x^2 + x + 1} \\ \int { 2x + 1 \over \sqrt{x^2 + x + 1} } \phantom{.} dx & = 2 \sqrt{x^2 + x + 1} \\ 8 \int { 2x + 1 \over \sqrt{x^2 + x + 1} } \phantom{.} dx & = 8 (2\sqrt{x^2 + x + 1}) \\ \int { 8(2x + 1) \over \sqrt{x^2 + x + 1} } \phantom{.} dx & = 16 \sqrt{x^2 + x + 1} \\ \int {16x + 8 \over \sqrt{x^2 + x + 1} } \phantom{.} dx & = 16 \sqrt{x^2 + x + 1} + c \end{align*}

\begin{align*} \tan \beta & = {3 \over x} \phantom{000000} \left[ {Opp \over Adj} \right] \\ \\ \tan \alpha & = {2 \over x} \phantom{000000} \left[ {Opp \over Adj} \right] \\ \\ \tan (\alpha + \beta) & = { \tan \alpha + \tan \beta \over 1 - \tan \alpha \tan \beta } \phantom{000000} [\text{Addition formula } \tan (A + B)] \\ & = { {2 \over x} + {3 \over x} \over 1 - \left(2 \over x\right)\left(3 \over x\right) } \\ & = { {5 \over x} \over 1 - {6 \over x^2} } \times {x^2 \over x^2} \\ & = { 5x \over x^2 - 6 } \\ \\ \therefore {5x \over x^2 - 6} & = {2 \over 1} \\ 5x & = 2(x^2 - 6) \\ 5x & = 2x^2 - 12 \\ 0 & = 2x^2 - 5x - 12 \\ 0 & = (x - 4)(2x + 3) \end{align*} \begin{align*} x - 4 & = 0 && \text{ or } & 2x + 3 & = 0 \\ x & = 4 &&& 2x & = -3 \\ & &&& x & = -{3 \over 2} \text{ (Reject, since } x > 0) \end{align*}

\begin{align*} 2y + x & = 1 \\ x & = 1 - 2y \phantom{0} \text{--- (1)} \\ \\ x^2 + y^2 - 2x & = 4 \phantom{0} \text{--- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ (1 - 2y)^2 + y^2 - 2(1 - 2y) & = 4 \\ \underbrace{ 1^2 - 2(1)(2y) + (2y)^2}_{(a - b)^2 = a^2 - 2ab + b^2} + y^2 - 2 + 4y & = 4 \\ 1 - 4y + 4y^2 + y^2 - 2 + 4y & = 4 \\ 5y^2 -1 & = 4 \\ 5y^2 & = 5 \\ y^2 & = 1 \\ y & = \pm \sqrt{1} \\ y & = 1 \text{ or } - 1 \end{align*} \begin{align*} \text{Substitute } & y = 1 \text{ into (1),} &&& \text{Substitute } & y = -1 \text{ into (1),} \\ x & = 1 - 2(1) &&& x & = 1 - 2(-1) \\ x & = -1 &&& x & = 3 \\ \\ \therefore & \phantom{.} A(-1, 1) &&& \therefore & \phantom{.} B(3, -1) \end{align*}
\begin{align*} \text{Midpoint of } AB & = \left( {x_1 + x_2 \over 2}, {y_1 + y_2 \over 2} \right) \\ & = \left( {-1 + 3 \over 2}, {1 + (-1) \over 2} \right) \\ & = (1, 0) \end{align*}

(a)

\begin{align*} \text{General term, } T_{r + 1} & = {n \choose r} (a)^{n - r} (b)^r \\ & = {6 \choose r} (2x)^{6 - r} \left(-{b \over x}\right)^r \\ & = {6 \choose r} (2)^{6 - r} (x)^{6 - r} (-b)^r (x^{-1})^r \phantom{000000} \left[ {1 \over x} = x^{-1}\right] \\ & = {6 \choose r} (2)^{6 - r} (-b)^r (x^{6 - r}) (x^{-r}) \\ & = {6 \choose r} (2)^{6 - r} (-b)^r (x^{6 - r +(-r)}) \phantom{00000000} [ a^m \times a^n = a^{m + n} ] \\ & = {6 \choose r} (2)^{6 - r} (-b)^r (x^{6 - 2r}) \\ \\ \text{Power of } x & = 6 - 2r \\ \\ \text{Let } 6 - 2r & = 0 \\ -2r & = -6 \\ r & = 3 \\ \\ T_{3 + 1} & = {6 \choose 3} (2)^{6 - 3} (-b)^3 (x^{6 - 2(3)}) \\ & = (20) (8) (-b^3) \underbrace{(x^0)}_{1} \phantom{000000000} [(-b)^3 = -b \times -b \times -b = -b^3] \\ & = - 160 b^3 \\ \\ -160b^3 & = -540 \\ b^3 & = {-540 \over -160} \\ b^3 & = {27 \over 8} \\ b & = \sqrt[3]{27 \over 8} \\ b & = {3 \over 2} \end{align*}

(b)

\begin{align*} [ \text{Need the constant term } & \text{and the term in } x^2 \text{ to multiply with } (1 + x^2)] \\ \\ \text{From (a), } T_4 & = -540 \text{ and } T_{r + 1} = {6 \choose r} (2)^{6 - r} (-b)^r (x^{6 - 2r}) \\ \\ \text{Let power of } x, 6 - 2r & = 2 \\ -2r & = -4 \\ r & = 2 \\ \\ T_{2 + 1} & = {6 \choose 2} (2)^{6 - 2} (-b)^2 (x^{6 - 2(2)}) \\ & = (15)(16) (b^2) (x^2) \\ & = 240b^2 x^2 \\ & = 240 \left(3 \over 2\right)^2 x^2 \\ & = 540x^2 \\ \\ \therefore (1 + x^2)\left(2x - {b \over x}\right)^6 & = (1 + x^2) ( ... + 540x^2 - 540 + ...) \\ & = ... + (1)(540x^2) + (x^2)(-540) + ... \\ & = ... + 540x^2 - 540x^2 + ... \\ & = ... + 0x^2 + ... \\ \\ \therefore \text{No terms } & \text{ in } x^2 \end{align*}

\begin{align*} \angle CBA & = 90^\circ \phantom{0} \text{(Right angle in a semi-circle)} \\ \\ \angle ABX & = 180^\circ - 90^\circ = 90^\circ \phantom{0} (\text{Adjacent angles on a straight line}) \\ \\ \text{Let } \angle TAB & = x \\ \\ \angle AXB & = 180^\circ - 90^\circ - x \phantom{0} (\text{Angle sum of triangle } ABX) \\ & = 90^\circ - x \\ \\ \text{Since tangents } & TA \text{ and } TB \text{ meet at } T, TA = TB \\ \\ \angle ATB & = 180^\circ - x - x \phantom{0} (\text{Angle sum of isosceles triangle}) \\ & = 180^\circ - 2x \\ & = 2 (90^\circ - x) \\ & = 2 \times \angle AXB \phantom{0} \text{ (Shown)} \end{align*}

\begin{align*} \text{Base area} & = (2 + \sqrt{2})^2 \\ & = \underbrace{ 2^2 + 2(2)(\sqrt{2}) + (\sqrt{2})^2 }_{ (a + b)^2 = a^2 + 2ab + b^2 } \\ & = 4 + 4 \sqrt{2} + 2 \\ & = ( 6 + 4 \sqrt{2} ) \text{ cm}^2 \\ \\ \text{Volume of cuboid} & = \text{Base area} \times \text{Height} \\ \\ \implies \text{Height} & = {\text{Volume} \over \text{Base area}} \\ & = {14 + 8 \sqrt{2} \over 6 + 4 \sqrt{2} } \times { 6 - 4 \sqrt{2} \over 6 - 4 \sqrt{2} } \phantom{000000.0000} [\text{Rationalise}] \\ & = { (14 + 8 \sqrt{2})(6 - 4 \sqrt{2}) \over (6 + 4 \sqrt{2})(6 - 4 \sqrt{2})} \\ & = { 84 - 56 \sqrt{2} + 48 \sqrt{2} - 32(2) \over (6)^2 - (4 \sqrt{2})^2 } \phantom{000000} [ \text{Denominator: } (a + b)(a - b) = a^2 - b^2 ] \\ & = { 20 - 8 \sqrt{2} \over 4 } \\ & = { 20 \over 4} - {8 \sqrt{2} \over 4} \\ & = (5 - 2 \sqrt{2}) \text{ cm} \\ \\ \text{Side area} & = (5 - 2 \sqrt{2})(2 + \sqrt{2}) \\ & = 10 + 5 \sqrt{2} - 4 \sqrt{2} - 2(2) \\ & = (6 + \sqrt{2}) \text{ cm}^2 \\ \\ \text{Total surface area} & = 2 \times \text{Base area} + 4 \times \text{Side area} \\ & = 2(6 + 4 \sqrt{2}) + 4(6 + \sqrt{2}) \\ & = 12 + 8 \sqrt{2} + 24 + 4 \sqrt{2} \\ & = (36 + 12 \sqrt{2}) \text{ cm}^2 \end{align*}

(a)

\begin{align*} V & = l \times b \times h \\ & = x \times x \times y \\ & = x^2 y \phantom{0000000000} [\text{Need to form an equation linking } x \text{ and } y] \\ \\ \text{Total surface area} & = 2 \times \text{Base area} + 4 \times \text{Side area} \\ 600 & = 2 (x)(x) + 4(x)(y) \\ 600 & = 2x^2 + 4xy \\ -4xy & = 2x^2 - 600 \\ 4xy & = 600 - 2x^2 \\ y & = {600 - 2x^2 \over 4x} \\ y & = {600 \over 4x} - {2x^2 \over 4x} \\ y & = {150 \over x} - {x \over 2} \\ \\ \therefore V & = x^2 \left( {150 \over x} - {x \over 2} \right) \\ & = \left(x^2 \over 1\right) \left( {150 \over x} - {x \over 2} \right) \\ & = {150x^2 \over x} - {x^3 \over 2} \\ & = 150x - {x^3 \over 2} \phantom{0} \text{ (Shown)} \end{align*}

(b)

\begin{align*} V & = 150x - {x^3 \over 2} \\ V & = 150x - {1 \over 2} x^3 \\ \\ {dV \over dx} & = 150 - {1 \over 2} (3) x^2 \\ & = 150 - {3 \over 2} x^2 \\ \\ \text{Let} & {dV \over dx} = 0, \phantom{000000} [\text{Stationary value}] \\ 0 & = 150 - {3 \over 2}x^2 \\ {3 \over 2}x^2 & = 150 \\ x^2 & = 150 \div {3 \over 2} \\ x^2 & = 100 \\ x & = \pm \sqrt{100} \\ x & = 10 \text{ or } -10 \text{ (Reject, since } x > 0) \\ \\ V & = 150(10) - {1 \over 2}(10)^3 \\ & = 1000 \\ \\ {d^2 V \over dx^2} & = 0 - {3 \over 2}(2)x \\ & = - 3x \\ \\ \text{When } & x = 10, \\ {d^2 V \over dx^2} & = -3(10) \\ & = -30 < 0 \\ \\ \therefore V = 1000 & \text{ is a maximum value} \end{align*}

(a)(i)

\begin{align*} 2x^2 - 6x + 7 & = 2(x^2 - 3x) + 7 \\ & = 2 \left[ x^2 - 3x + \left(3 \over 2\right)^2 - \left(3 \over 2\right)^2 \right] + 7 \\ & = 2 ( \underbrace{x^2 - 3x + 1.5^2}_{a^2 - 2ab + b^2 = (a - b)^2} - 2.25 ) + 7 \\ & = 2 [ (x - 1.5)^2 - 2.25] + 7 \\ & = 2 (x - 1.5)^2 - 4.5 + 7 \\ & = 2 (x - 1.5)^2 + 2.5 \end{align*}

(a)(ii)

\begin{align*} 2x^2 - 6x + 7 & = k \\ 2 (x - 1.5)^2 + 2.5 & = k \\ 2(x - 1.5)^2 & = k - 2.5 \\ (x - 1.5)^2 & = {k - 2.5 \over 2} \\ x - 1.5 & = \pm \sqrt{ k - 2.5 \over 2} \\ \\ \text{If solution has } & \text{1 real root or no real roots,} \\ k - 2.5 & \le 0 \\ k & \le 2.5 \end{align*}

(b)

\begin{align*} \text{Line of symmetry, } x & = {2 + 4 \over 2} \\ x & = 3 \\ \\ \implies \text{Turning point: } & (3, 2) \\ \\ y & = p(x - q)^2 + r \\ y & = p(x - 3)^2 + 2 \\ \\ \text{Using } & (2, 0), \\ 0 & = p(2 - 3)^2 + 2 \\ 0 & = p(1) + 2 \\ 0 & = p + 2 \\ -p & = 2 \\ p & = -2 \\ \\ y & = -2(x - 3)^2 + 2 \\ \\ \therefore p & = -2, q = 3, r = 2 \end{align*}

(a)

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

(b)

\begin{align*} y & = 4 \cos 2x \\ \\ \text{Amplitude} & = 4 \\ \\ \text{Max. value} & = 4, \text{ Min. value} = -4 \\ \\ \text{Period} & = \pi \\ \\ \text{No. of cycles} & = {2\pi \over \pi} = 2 \end{align*}

(c)

\begin{align*} \pi \cos 2x + x & = 0 \\ \pi \cos 2x & = -x \\ \cos 2x & = -{x \over \pi} \\ \cos 2x & = -{1 \over \pi} x \\ \underbrace{4 \cos 2x}_\text{Curve} & = -{4 \over \pi} x \\ \\ \text{Draw the line } & y = -{4 \over \pi}x \phantom{00000} [y = mx + c, \text{ with } c = 0] \\ \\ \text{Let } & x = 0, \\ y & = -{4 \over \pi} (0) \\ y & = 0 \\ \\ \text{Let } & x = \pi, \\ y & = -{4 \over \pi} (\pi) \\ y & = -4 \\ \\ \\ \text{From graph, } & \text{there are 2 solutions to equation} \end{align*}

(a)

\begin{align*} \sin \theta & = {BF \over 3} \phantom{000000} \left[ {Opp \over Hyp} \right] \\ 3 \sin \theta & = BF \\ \\ \cos \theta & = {CN \over 2} \phantom{000000} \left[ {Adj \over Hyp} \right] \\ 2 \cos \theta & = CN \\ \\ CM & = 2 \cos \theta + 3 \sin \theta \phantom{0} \text{ (Shown)} \end{align*}

(b)

\begin{align*} a \cos \theta + b \sin \theta & = R \cos (\theta - \alpha) \\ \\ R & = \sqrt{a^2 + b^2} \\ & = \sqrt{2^2 + 3^2} \\ & = \sqrt{13} \\ \\ \alpha & = \tan^{-1} \left(b \over a\right) \\ & = \tan^{-1} \left(3 \over 2\right) \\ & = 56.31^\circ \\ \\ CM & = 2 \cos \theta + 3 \sin \theta \\ & = \sqrt{13} \cos (\theta - 56.31^\circ) \\ \\ \text{Since } -\sqrt{13} & \le \sqrt{13} \cos (\theta - 56.31^\circ) \le \sqrt{13}, \\ \text{Max. } & \text{length of } CM = \sqrt{13} \text{ m} \end{align*}

(c)

\begin{align*} CM & = \sqrt{13} \cos (\theta - 56.31^\circ) \\ 3.5 & = \sqrt{13} \cos (\theta - 56.31^\circ) \\ {3.5 \over \sqrt{13} } & = \cos (\theta - 56.31^\circ) \phantom{000000} [\text{1st or 4th quadrant since } \cos (\theta - 56.31^\circ) > 0 ] \\ \\ \text{Basic angle, } \alpha & = \cos^{-1} \left(3.5 \over \sqrt{13}\right) \\ & = 13.90^\circ \end{align*}

\begin{align*} \theta - 56.31^\circ & = 13.90^\circ, 360^\circ - 13.90^\circ \text{ (N.A)}, 360^\circ - 13.90^\circ - 360^\circ \\ & = 13.90^\circ, -13.90^\circ \\ \\ \theta & = 70.21^\circ, 42.41^\circ \\ & \approx 70.2^\circ, 42.4^\circ \end{align*}

(a)

\begin{align*} {3x - 5 \over x^2(x - 1)} & = {A \over x} + {B \over x^2} + {C \over x - 1} \\ & = {Ax(x - 1) \over x^2(x - 1)} + {B(x - 1) \over x^2(x - 1)} + {Cx^2 \over x^2(x - 1)} \\ & = {Ax(x - 1) + B(x - 1) + Cx^2 \over x^2(x - 1)} \\ \\ 3x - 5 & = Ax(x - 1) + B(x - 1) + Cx^2 \\ \\ \text{Let } & x = 0 ,\\ 3(0) - 5 & = 0 + B(0 - 1) + 0 \\ -5 & = -B \\ 5 & = B \\ \\ 3x - 5 & = Ax(x - 1) + 5(x - 1) + Cx^2 \\ \\ \text{Let } & x = 1, \\ 3(1) - 5 & = 0 + 0 + C(1)^2 \\ -2 & = C \\ \\ 3x - 5 & = Ax(x - 1) + 5(x - 1) - 2x^2 \\ \\ \text{Let } & x = 2, \\ 3(2) - 5 & = A(2)(2 - 1) + 5(2 - 1) - 2(2)^2 \\ 1 & = 2A + 5 - 8 \\ -2A & = 5 - 8 - 1 \\ -2A & = -4 \\ A & = 2 \\ \\ \therefore {3x - 5 \over x^2(x - 1)} & = {2 \over x} + {5 \over x^2} + {-2 \over x - 1} \\ & = {2 \over x} + {5 \over x^2} - {2 \over x - 1} \end{align*}

(b)

\begin{align*} \int {3x - 5 \over x^2(x - 1)} \phantom{.} dx & = \int {2 \over x} + {5 \over x^2} - {2 \over x - 1} \phantom{.} dx \\ & = \int 2 \left(1 \over x\right) + 5x^{-2} - 2 \left(1 \over x - 1\right) \phantom{.} dx \\ & = 2 \ln x + 5 \left(x^{-1} \over -1\right) - 2 \left[ \ln (x - 1) \over 1\right] \\ & = 2 \ln x - 5x^{-1} - 2 \ln(x - 1) \\ & = 2 \ln x - {5 \over x} - 2 \ln (x - 1) \\ \\ \therefore \int_2^3 {3x - 5 \over x^2(x - 1)} \phantom{.} dx & = \left[ 2 \ln x - {5 \over x} - 2 \ln (x - 1) \right]_2^3 \\ & = 2 \ln 3 - {5 \over 3} - 2 \ln (3 - 1) - \left( 2 \ln 2 - {5 \over 2} - 2 \ln (2 - 1) \right) \\ & = 2 \ln 3 - {5 \over 3} - 2 \ln 2 - 2 \ln 2 + {5 \over 2} + 0 \\ & = 2 \ln 3 - 2 \ln 2 - 2 \ln 2 + {5 \over 6} \\ & = \ln 3^2 - \ln 2^2 - \ln 2^2 + {5 \over 6} \phantom{000000} [\text{Power law (logarithms)}] \\ & = \ln 9 - \ln 4 - \ln 4 + {5 \over 6} \\ & = \ln {9 \over 4} - \ln 4 + {5 \over 6} \phantom{0000000000000} [\text{Quotient law (logarithms)}] \\ & = \ln \left({9 \over 4} \div 4\right) + {5 \over 6} \phantom{000000000000} [\text{Quotient law (logarithms)}] \\ & = \ln {9 \over 16} + {5 \over 6} \\ & = \ln \left(16 \over 9\right)^{-1} + {5 \over 6} \\ & = - \ln {16 \over 9} + {5 \over 6} \phantom{0} \text{ (Shown)} \end{align*}

(a)

\begin{align*} y & = (x - 1)^4 \\ \\ \text{Let } & x = 0, \\ y & = (0 - 1)^4 \\ y & = 1 \\ \\ \therefore & \phantom{.} Q(0, 1) \\ \\ y & = (x - 1)^4 \\ {dy \over dx} & = 4(x - 1)^3 . (1) \phantom{000000} [\text{Chain rule}] \\ & = 4(x - 1)^3 \\ \\ \text{Let } & x = 2, \\ {dy \over dx} & = 4(2 - 1)^3 \\ & = 4 \\ \\ \text{Gradient of tangent at } P(2, 1) & = 4 \\ \\ \text{Gradient of normal at } P(2, 1) & = -1 \div 4 \phantom{00000000000} [m_1 \times m_2 = -1] \\ & = -{1 \over 4} \\ \\ y & = mx + c \\ y & = -{1 \over 4}x + c \\ \\ \text{Using } & P(2, 1), \\ 1 & = -{1 \over 4}(2) + c \\ 1 & = - 0.5 + c \\ -c & = -0.5-1 \\ -c & = -1.5 \\ c & = 1.5 \\ \\ \text{Eqn of normal at } & P (2, 1) : y = -{1 \over 4}x + 1.5 \\ \\ \implies & \phantom{.} S(0, 1.5) \\ \\ QS & = 1.5 - 1 = 0.5 \text{ units} \\ QO & = 1 - 0 = 1 \text{ unit} \\ \\ \text{Since } QS & < QO, \\ \therefore Q(0,1) \text{ is nearer} & \text{ to } S(0, 1.5) \text{ than to } O(0, 0) \end{align*}

(b)

\begin{align*} y & = (x - 1)^4 \\ \\ \text{Let } & y = 0, \\ 0 & = (x - 1)^4 \\ \pm \sqrt[4]{0} & = x - 1 \\ 0 & = x - 1 \\ 1 & = x \\ \\ \therefore & \phantom{.} T(1, 0) \\ \\ \text{Area bounded by curve} & = \int_1^2 (x - 1)^4 \phantom{.} dx \\ & = \left[ (x - 1)^5 \over (5)(1) \right]_1^2 \phantom{000000} \left[ \int [f(x)]^n \phantom{.} dx = {[f(x)]^{n + 1} \over (n + 1) f'(x) } \right] \\ & = \left[ (x - 1)^5 \over 5\right]_1^2 \\ & = { (2 - 1)^5 \over 5 } - { (1 - 1)^5 \over 5 } \\ & = 0.2 \text{ units}^2 \\ \\ \text{From (a), gradient of tangent at } P(2, 1) & = 4 \\ \\ y & = mx + c \\ y & = 4x + c \\ \\ \text{Using } & P(2, 1), \\ 1 & = 4(2) + c \\ 1 & = 8 + c \\ -7 & = c \\ \\ \text{Eqn of tangent at } & P(2, 1): y = 4x - 7 \\ \\ \text{Let } & y = 0, \\ 0 & = 4x - 7 \\ -4x & = -7 \\ x & = {-7 \over -4} \\ x & = 1.75 \\ \\ \therefore & \phantom{.} R(1.75, 0) \\ \\ \text{Area of triangle} & = {1 \over 2} \times (2 - 1.75) \times 1 \\ & = 0.125 \text{ units}^2 \\ \\ \text{Area of shaded region} & = 0.2 - 0.125 \\ & = 0.075 \text{ units}^2 \end{align*}

\begin{align*} V & = 1800 e^{-kt} \\ \\ \text{Let } t = 12 & \text{ and } V = 1000, \\ 1000 & = 1800 e^{-k(12)} \\ {1000 \over 1800} & = e^{-12k} \\ \ln {1000 \over 1800} & = \ln e^{-12k} \\ \ln {5 \over 9} & = -12k \ln e \phantom{000000} [\text{Power law (logarithms)}] \\ \ln {5 \over 9} & = -12k (1) \\ \ln {5 \over 9} & = -12k \\ \\ k & = - { \ln {5 \over 9} \over 12} \\ \\ \$ 1800 \times {25 \over 100} & = \$ 450 \\ \\ \text{Let } & V = 450, \\ 450 & = 1800 e^{-kt} \\ {450 \over 1800} & = e^{-kt} \\ \ln {450 \over 1800} & = \ln e^{-kt} \\ \ln {1 \over 4} & = -kt \ln e \phantom{000000} [\text{Power law (logarithms)}] \\ \ln {1 \over 4} & = -kt (1) \\ \ln {1 \over 4} & = -kt \\ \\ t & = - {\ln {1 \over 4} \over k} \\ & = - {\ln {1 \over 4} \over - { \ln {5 \over 9} \over 12} } \\ & = 28.302 \\ & \approx 28.3 \end{align*}

\begin{align*} \log_2 x - \log_4 (x + 3) & = 3 \log_8 2 \\ \log_2 x - \log_4 (x + 3) & = \log_8 2^3 \phantom{000000} [\text{Power law}] \\ \log_2 x - {\log_2 (x + 3) \over \log_2 4} & = \log_8 8 \phantom{0000000} [\text{Change-of-base}] \\ \log_2 x - {\log_2 (x + 3) \over \log_2 2^2} & = 1 \\ \log_2 x - {\log_2 (x + 3) \over 2 \log_2 2} & = 1 \phantom{00000000000} [\text{Power law}] \\ \log_2 x - {\log_2 (x + 3) \over 2(1)} & = 1 \\ \log_2 x - {\log_2 (x + 3) \over 2} & = 1 \\ 2 \left[ \log_2 x - {\log_2 (x + 3) \over 2} \right] & = 2(1) \\ 2 \log_2 x - \log_2 (x + 3) & = 2 \\ \log_2 x^2 - \log_2 (x + 3) & = 2 \phantom{00000000000} [\text{Power law}] \\ \log_2 {x^2 \over x+ 3} & = 2 \phantom{00000000000} [\text{Quotient law}] \\ {x^2 \over x + 3} & = 2^2 \phantom{0000000000} [\text{Change to exponential form}] \\ {x^2 \over x + 3} & = {4 \over 1} \\ x^2 & = 4(x + 3) \\ x^2 & = 4x + 12 \\ x^2 - 4x - 12 & = 0 \\ (x - 6)(x + 2)& = 0 \end{align*} \begin{align*} x - 6 & = 0 && \text{ or } & x + 2 & = 0 \\ x & = 6 &&& x & = -2 \text{ (Reject, since } \log_2 (-2) \text{ is undefined}) \end{align*}

(i)

\begin{align*} u & = 4x &&& v & = 5 - 3x \\ {du \over dx} & = 4 &&& {dv \over dx}& = -3 \end{align*} \begin{align*} {dy \over dx} & = { v {du \over dx} - u {dv \over dx} \over v^2 } \phantom{000000} [\text{Quotient rule}] \\ & = { (5 - 3x)(4) - (4x)(-3) \over (5 - 3x)^2 } \\ & = { 20 - 12x + 12x \over (5 - 3x)^2 } \\ & = {20 \over (5 - 3x)^2} \\ \\ \text{When } & x = 1, \\ {dy \over dx} & = {20 \over [5 - 3(1)]^2 } \\ & = 5 \end{align*}

(ii)

\begin{align*} {dy \over dx} & = {20 \over (5 - 3x)^2} \\ \\ \text{For all real } & \text{values of } x \text{ (except } x = {5 \over 3}), \\ (5 - 3x)^2 & > 0 \\ {20 \over (5 - 3x)^2} & > 0 \\ {dy \over dx} & > 0 \\ \\ \therefore \text{Not possible } & \text{for } {dy \over dx} = 0 \text{ and curve has no stationary point} \end{align*}

\begin{align*} y & = f(x) \\ {dy \over dx}& = f'(x) \\ & = {4 \over \cos^2 x} \\ & = \left(4 \over 1\right) \left(1 \over \cos^2 x\right) \\ & = 4 \sec^2 x \phantom{000000000} \left[ {1 \over \cos A} = \sec A\right] \\ \\ y & = \int 4 \sec^2 x \phantom{.} dx \\ y & = 4 \left( \tan x \over 1\right) \phantom{000000} \left[ \int \sec^2 f(x) \phantom{.} dx = { \tan f(x) \over f'(x) } \right] \\ y & = 4 \tan x + c \\ \\ \text{Using } & \left({\pi \over 4}, 1\right), \\ 1 & = 4 \tan {\pi \over 4} + c \\ 1 & = 4 (1) + c \phantom{000000000} [\text{Use radian mode!}] \\ 1 & = 4 + c \\ -3 & = c \\ \\ \therefore y = f(x) & = 4 \tan x - 3 \end{align*}

\begin{align*} y & = e^x + e^{-x} \\ \\ {dy \over dx} & = (1)e^x + (-1)e^{-x} \phantom{000000} \left[ {d \over dx} (e^{f(x)}) = f'(x) . e^{f(x)} \right] \\ & = e^x - e^{-x} \\ \\ \text{Let } & x = 0, \\ {dy \over dx} & = e^0 - e^{-(0)} \\ & = 0 \\ \\ \therefore \text{Curve has } & \text{a stationary point at } x = 0 \end{align*}

\begin{align*} e^x + e^{-x} & = 2.9 \\ e^x + {1 \over e^x} & = 2.9 \\ \\ \text{Let } & u = e^x, \\ u + {1 \over u} & = 2.9 \\ 10u \left(u + {1 \over u}\right) & = 10u(2.9) \\ 10u^2 + 10 & = 29u \\ 10u^2 - 29u + 10 & = 0 \\ (5u - 2)(2u - 5) & = 0 \end{align*} \begin{align*} 5u - 2 & = 0 && \text{ or } & 2u - 5 & = 0 \\ 5u & = 2 &&& 2u & = 5 \\ u & = {2 \over 5} &&& u & = {5 \over 2} \\ e^x & = {2 \over 5} &&& e^x & = {5 \over 2} \phantom{0000000} [\text{From earlier, } u = e^x] \\ \ln e^x & = \ln {2 \over 5} &&& \ln e^x & = \ln {5 \over 2} \\ x \ln e & = \ln {2 \over 5} &&& x \ln e & = \ln {5 \over 2} \phantom{00000} [\text{Power law (logarithms)}] \\ x(1) & = \ln {2 \over 5} &&& x (1) & = \ln {5 \over 2} \\ x & = \ln {2 \over 5} &&& x & = \ln {5 \over 2} \end{align*}

(a)

\begin{align*} \text{Let } f(x) & = x^3 + x^2 + ax + 2b \\ \\ f(-3) & = (-3)^3 + (-3)^2 + a(-3) + 2b \\ 0 & = -27 + 9 - 3a + 2b \\ 18 & = -3a + 2b \phantom{0} \text{--- (1)} \\ \\ \text{Let } g(x) & = x^3 + 4x^2 - ax + b \\ \\ g(-3) & = (-3)^3 + 4(-3)^2 - a(-3) + b \\ 0 & = -27 + 36 + 3a + b \\ -b & = 9 + 3a \\ b & = -9 - 3a \phantom{0} \text{--- (2)} \\ \\ \text{Substitute } & \text{(2) into (1),} \\ 18 & = -3a + 2(-9 - 3a) \\ 18 & = -3a - 18 - 6a \\ 3a + 6a & = -18 - 18 \\ 9a & = -36 \\ a & = -4 \\ \\ \text{Substitute } & a = -4 \text{ into (2),} \\ b & = -9 - 3(-4) \\ b & = 3 \\ \\ \therefore a & = -4, b = 3 \end{align*}

(b)

\begin{align*} \text{Let } h(x) & = x^3 + 2x^2 - 11x + 6 \\ \\ h(2) & = (2)^3 + 2(2)^2 - 11(2) + 6 \\ & = 0 \\ \\ \text{By Factor theorem, } & x - 2 \text{ is a factor of } h(x) \end{align*} $$ \require{enclose} \begin{array}{rll} x^2 + 4x - 3 \phantom{0000000.}\\ x - 2 \enclose{longdiv}{ x^3 + 2x^2 - 11x + 6 \phantom{0}}\kern-.2ex \\ -\underline{(x^3 - 2x^2){\phantom{000000000.}}} \\ 4x^2 - 11x + 6 \phantom{0} \\ -\underline{(4x^2 - 8x){\phantom{0000.}}} \\ -3x + 6 \phantom{0} \\ -\underline{(-3x + 6){\phantom{.}}} \\ 0 \phantom{.} \end{array} $$ \begin{align*} \text{Polynomial} & = \text{Divisor} \times \text{Quotient} + \text{Remainder} \\ x^3 + 2x^2 - 11x + 6 & = (x - 2)\underbrace{ (x^2 + 4x - 3) }_\text{Cannot factorise} + 0 \\ \\ \therefore (x - 2)&(x^2 + 4x - 3) = 0 \end{align*} \begin{align*} x - 2 & = 0 && \text{ or } & x^2 + 4x - 3 & = 0 \\ x & = 2 &&& x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & &&& & = {-4 \pm \sqrt{(4)^2 - 4(1)(-3)} \over 2(1)} \\ & &&& & = {-4 \pm \sqrt{28} \over 2} \\ & &&& & = {-4 \pm \sqrt{4} \times \sqrt{7} \over 2} \\ & &&& & = {-4 \pm 2 \sqrt{7} \over 2} \\ & &&& & = {-4 \over 2} \pm {2 \sqrt{7} \over 2} \\ & &&& & = -2 \pm \sqrt{7} \end{align*}

(a)

\begin{align*} \text{Centre of circle} & = \text{Midpoint of } PQ \\ & = \left( {3 + (-5) \over 2}, {1 + 7 \over 2} \right) \phantom{000000} \left[ \text{Formula: } \left( {x_1 + x_2 \over 2}, {y_1 + y_2 \over 2} \right) \right] \\ & = \left( -1, 4 \right) \\ \\ \text{Diameter} & = \text{Length of } PQ \\ & = \sqrt{ [3 - (-5)]^2 + (1 - 7)^2 } \phantom{000.} [ \text{Formula: } \sqrt{ (x_2 - x_1)^2 + (y_2 - y_1)^2 } ] \\ & = 10 \text{ units} \\ \\ \text{Radius} & = {10 \over 2} = 5 \text{ units} \end{align*}

(b)

\begin{align*} (x - a)^2 + (y - b)^2 & = r^2 \\ [x - (-1)]^2 + (y - 4)^2 & = 5^2 \\ (x + 1)^2 + (y - 4)^2 & = 25 \\ \underbrace{ x^2 + 2(x)(1) + 1^2 }_{ (a + b)^2 = a^2 + 2ab + b^2 } + \underbrace{ y^2 - 2(y)(4) + 4^2 }_{ (a - b)^2 = a^2 - 2ab + b^2 } & = 25 \\ x^2 + 2x + 1 + y^2 - 8y + 16 & = 25 \\ x^2 + y^2 + 2x - 8y + 1 + 16 - 25 & = 0 \\ x^2 + y^2 + 2x - 8y - 8 & = 0 \end{align*}

(c)

\begin{align*} 3y & = 4x - 34 \\ y & = {4 \over 3}x - {34 \over 3} \phantom{000000} [y = mx + c] \\ \\ \text{Gradient of line} & = {4 \over 3} \\ \\ \text{Gradient of } QP & = {7 - 1 \over -5 - 3} \\ & = -{3 \over 4} \\ \\ {4 \over 3} \times -{3 \over 4} & = -1 \\ \\ \implies \text{Line } QP & \text{ is perpendicular to the line } 3y = 4x - 34 \end{align*}

Since \(PQ\) is a diameter, line \(QP\) passes through the centre of the circle. The point on the line that is closest to the circle lies on the perpendicular from the centre to the line.

\(\therefore\) Point \(R\) on the line \(3y = 4x - 34\) lies on \(QP\) extended.

(d)

\begin{align*} 3y & = 4x - 34 \phantom{0} \text{--- (1)} \\ \\ \text{Gradient of } QP & = -{3 \over 4} \\ \\ y & = mx + c \\ y & = -{3 \over 4}x + c \\ \\ \text{Using } & Q(-5, 7), \\ 7 & = -{3 \over 4}(-5) + c \\ 7 & = 3.75 + c \\ 3.25 & = c \\ \\ \text{Eqn of line } QP &: \phantom{.} y = -{3 \over 4}x + 3.25 \text{ --- (2)} \\ \\ \text{Substitute } & \text{(2) into (1),} \\ 3\left(-{3 \over 4}x + 3.25\right) & = 4x - 34 \\ -2.25x + 9.75 & = 4x - 34 \\ -2.25x - 4x & = -34 - 9.75 \\ -6.25x & = -43.75 \\ x & = {-43.75 \over -6.25} \\ x & = 7 \\ \\ \text{Substitute } & x = 7 \text{ into (2),} \\ y & = -{3 \over 4}(7) + 3.25 \\ y & = -2 \\ \\ \therefore & \phantom{.} R(7, -2) \end{align*}

(a)

\begin{align*} v & = A t^{2.5} + 11t + C \\ \\ \text{When } t = 0 & \text{ and } v = 25, \phantom{000000} [\text{Speed at } P ] \\ 25 & = A(0)^{2.5} + 11(0) + C \\ 25 & = A(0) + 0 + C \\ 25 & = C \\ \\ \\ v & = A t^{2.5} + 11t + 25 \\ \\ \text{When } t = 4 & \text{ and } v = 85, \phantom{000000} [\text{Speed at } Q] \\ 85 & = A(4)^{2.5} + 11(4) + 25 \\ 85 & = A(32) + 44 + 25 \\ -32A & = 44 + 25 - 85 \\ -32A & = -16 \\ A & = {-16 \over -32} \\ A & = {1 \over 2} \end{align*}

(b)

\begin{align*} v & = A t^{2.5} + 11t + C \\ & = {1 \over 2} t^{2.5} + 11t + 25 \\ \\ a & = {dv \over dt} \\ & = {1 \over 2}(2.5) t^{1.5} + 11 \\ & = 1.25 t^{1.5} + 11 \\ \\ \text{When } & t = 4, \\ a & = 1.25 (4)^{1.5} + 11 \\ & = 21 \text{ m/s}^2 \end{align*}

(c)

\begin{align*} s & = \int v \phantom{.} dt \\ & = \int {1 \over 2} t^{2.5} + 11t + 25 \phantom{.} dt \\ & = {1 \over 2} \left(t^{3.5} \over 3.5 \right) + 11 \left(t^2 \over 2\right) + 25t + c \\ & = {1 \over 7} t^{3.5} + 5.5 t^2 + 25t + c \\ \\ \text{When } t = 0 & \text{ and } s = 0, \\ 0 & = {1 \over 7} (0)^{3.5} + 5.5 (0)^2 + 25(0) + c \\ 0 & = c \\ \\ s & = {1 \over 7} t^{3.5} + 5.5t^2 + 25t \\ \\ \text{When } & t = 4, \\ s & = {1 \over 7} (4)^{3.5} + 5.5(4)^2 + 25(4) \\ & = 206.286 \\ & \approx 206 \text{ m} \end{align*}

(a)

\begin{align*} \text{L.H.S} & = \tan x + \cot x \\ & = {\sin x \over \cos x} + {\cos x \over \sin x} \\ & = {\sin^2 x \over \sin x \cos x} + {\cos^2 x \over \sin x \cos x} \\ & = { \sin^2 x + \cos^2 x \over \sin x \cos x} \\ & = {1 \over \sin x \cos x} \times {2 \over 2} \phantom{000000} [ \sin^2 A + \cos^2 A = 1] \\ & = {2 \over 2 \sin x \cos x} \\ & = {2 \over \sin 2x} \phantom{0000000000000.} [\sin 2A = 2 \sin A \cos A] \\ & = \left(2 \over 1\right) \left(1 \over \sin 2x\right) \\ & = 2 \text{cosec } 2x \\ & = \text{R.H.S} \end{align*}

(b)

\begin{align*} \tan x + \cot x & = 3 \\ {2 \over \sin 2x} & = 3 \phantom{000000000000} [\text{From part (a)}] \\ 2 & = 3 \sin 2x \\ {2 \over 3} & = \sin 2x \phantom{00000000.} [\text{1st or 2nd quadrant since } \sin 2x > 0] \\ \\ \text{Basic angle, } \alpha & = \sin^{-1} \left(2 \over 3\right) \phantom{0000} [\text{Radian mode}] \\ & = 0.7297 \end{align*}

\begin{align*} 0 & \le x \le 2\pi \implies 0 \le 2x \le 4 \pi \\ \\ 2x & = 0.7297, \pi - 0.7297, 0.7297 + 2\pi, \pi - 0.7297 + 2\pi \\ & = 0.7297, 2.412, 7.013, 8.695 \\ \\ x & = 0.36485, 1.206, 3.5065, 4.3475 \\ & \approx 0.365, 1.21, 3.51, 4.35 \end{align*}

(c)

\begin{align*} \tan x + \cot x & = k \\ {2 \over \sin 2x} & = k \phantom{000000000000} [\text{From part (a)}] \\ 2 & = k \sin 2x \\ \\ \text{Consider graph of } & y = 2 \text{ and } y = k \sin 2x \end{align*}

\begin{align*} \text{For equation to have no solutions, } & \text{the graphs of } y = 2 \text{ and } y = k \sin 2x \text{ must not meet} \\ \\ \therefore -2 < k & < 2 \end{align*}

(a)(i)

\begin{align*} R & = A v^k \\ \lg R & = \lg (A v^k) \\ \lg R & = \lg A + \lg v^k \phantom{000000.} [\text{Product law (logarithms)}] \\ \lg R & = \lg A + k \lg v \phantom{000000} [\text{Power law (logarithms)}] \\ \lg R & = k \lg v + \lg A \phantom{000000} [Y = mX + c] \\ \\ \text{Plot } & \lg R \text{ against } \lg v \end{align*}

\(\lg v\)	\(0.699\)	\(1\)	\(1.176\)	\(1.301\)	\(1.398\)
\(\lg R\)	\(1.322\)	\(1.740\)	\(2.041\)	\(2.164\)	\(2.301\)

(a)(ii)

\begin{align*} \text{Incorrect value of } R & = 110 \phantom{000000} [\text{Third value}] \\ \\ \text{From graph, } \lg R & = 1.98 \\ \log_{10} R & = 1.98 \\ R & = 10^{1.98} \\ R & \approx 95.5 \end{align*}

(a)(iii)

\begin{align*} \lg R & = k \lg v + \lg A \phantom{000000} [Y = mX + c] \\ \\ k & = \text{Gradient, } m \\ & = {2.301 - 1.322 \over 1.398 - 0.699} \\ & = 1.4006 \\ & \approx 1.40 \end{align*}

(b)

\begin{align*} y & = ax + {b \over x} \\ xy & = x \left( ax + {b \over x} \right) \\ xy & = ax^2 + b \phantom{000000} [Y = mX + c] \\ \\ 1. & \text{ Plot } xy \text{ against } x^2 \\ 2. & \text{ Gradient of straight line} = a \\ 3. & \text{ Vertical intercept of graph} = b \end{align*}

(a)

\begin{align*} \text{For all real values of } & x, e^{-2x} > 0 \end{align*}

\begin{align*} \text{For } 0 < x < \pi, & \phantom{.} \sin x > 0 \\ \\ \therefore \text{For } 0 < x < \pi, & \phantom{.} e^{-2x} \sin x > 0 \text{ and curve is completely above } x \text{-axis} \end{align*}

(b)

\begin{align*} u & = e^{-2x} &&& v & = \sin x \\ {du \over dx} & = -2 e^{-2x} &&& {dv \over dx} & = \cos x \end{align*} \begin{align*} {dy \over dx} & = u {dv \over dx} + v {du \over dx} \phantom{000000000} [\text{Product rule}] \\ & = (e^{-2x})(\cos x) + (\sin x)(-2e^{-2x}) \\ & = e^{-2x} \cos x - 2 e^{-2x} \sin x \\ \\ \text{Let } & {dy \over dx} = 0, \phantom{000000000000000} [\text{Stationary point}] \\ 0 & = e^{-2x} \cos x - 2 e^{-2x} \sin x \\ 0 & = e^{-2x} (\cos x - 2 \sin x) \end{align*} \begin{align*} e^{-2x} & = 0 \text{ (No solution since } e^{-2x} > 0) \\ \\ \\ \cos x - 2 \sin x & = 0 \\ -2 \sin x & = - \cos x \\ {-2 \sin x \over - \cos x} & = 1 \\ 2 \tan x & = 1 \\ \tan x & = {1 \over 2} \\ x & = \tan^{-1} \left(1 \over 2\right) \phantom{0} \text{ (Shown)} \end{align*}

(c)

\begin{align*} {dy \over dx} & = e^{-2x} \cos x - 2 e^{-2x} \sin x \\ & = e^{-2x} (\cos x - 2 \sin x) \end{align*} \begin{align*} u & = e^{-2x} &&& v & = \cos x - 2 \sin x \\ {du \over dx} & = -2 e^{-2x} &&& {dv \over dx} & = - \sin x - 2 \cos x \end{align*} \begin{align*} {d^2 y \over dx^2} & = (e^{-2x}) (- \sin x - 2 \cos x) + (\cos x - 2 \sin x) (-2e^{-2x}) \phantom{000000} [\text{Product rule}] \\ & = - e^{-2x} \sin x - 2 e^{-2x} \cos x - 2 e^{-2x} \cos x + 4 e^{-2x} \sin x \\ & = 3 e^{-2x} \sin x - 4 e^{-2x} \cos x \\ \\ \text{L.H.S} & = {d^2 y \over dx^2} + 4 {dy \over dx} + 5y \\ & = 3 e^{-2x} \sin x - 4 e^{-2x} \cos x + 4 (e^{-2x} \cos x - 2 e^{-2x} \sin x) + 5 (e^{-2x} \sin x) \\ & = 3 e^{-2x} \sin x - 4 e^{-2x} \cos x + 4 e^{-2x} \cos x - 8 e^{-2x} \sin x + 5 e^{-2x} \sin x \\ & = 0 \\ & = \text{R.H.S} \end{align*}