\begin{align} & \underline{\text{For equation } 2x^2 - 5x + 8 = 0:} \\ \\ & \text{Sum of roots, } \alpha + \beta = -{b \over a} \\ & \phantom{000000000000000/} = -\left({-5 \over 2}\right) \\ & \phantom{000000000000000/} = {5 \over 2} \\ \\ & \text{Product of roots, } \alpha \beta = {c \over a} \\ & \phantom{0000000000000000} = {8 \over 2} \\ & \phantom{0000000000000000} = 4 \\ \\ \\ & \underline{\text{For new quadratic equation:}} \\ \\ & \text{Sum of roots} = {\alpha \over \beta} + {\beta \over \alpha} \\ & \phantom{000000000(.} = {\alpha^2 \over \alpha \beta} + {\beta^2 \over \alpha \beta} \\ & \phantom{000000000(.} = {\alpha^2 + \beta^2 \over \alpha \beta} \\ & \phantom{000000000(.} = {(\alpha + \beta)^2 - 2\alpha \beta \over \alpha \beta} \\ & \phantom{000000000(.} = { \left(5 \over 2\right)^2 - 2(4) \over (4) } \\ & \phantom{000000000(.} = -{7 \over 16} \\ \\ & \text{Product of roots} = {\alpha \over \beta} \times {\beta \over \alpha} \\ & \phantom{0000000000000} = {\alpha \beta \over \alpha \beta} \\ & \phantom{0000000000000} = 1 \\ \\ & x^2 - \text{(Sum of roots)}x + \text{(Product of roots)} = 0 \\ & \phantom{00000000000000000(} x^2 - \left(-{7 \over 16}\right)x + 1 = 0 \\ & \phantom{0000000000000000000000}x^2 + {7 \over 16}x + 1 = 0 \end{align}

(a)

\begin{align} \left[ \left(50 \over 3\right)^{-2} \times \sqrt{3^3} \right] \div {5 \over 2} & = \left[ \left( 2 \times 5^2 \over 3\right)^{-2} \times (3^3)^{1 \over 2} \right] \times {2 \over 5} \\ & = \left[ \left( 3 \over 2 \times 5^2 \right)^2 \times 3^{3 \over 2} \right] \times {2 \over 5} \\ & = \left( {3^2 \over 2^2 \times 5^4} \times {3^{3 \over 2} \over 1} \right) \times {2 \over 5} \\ & = {3^{7 \over 2} \over 2^2 \times 5^4} \times {2 \over 5} \\ & = {2 \times 3^{7 \over 2} \over 2^2 \times 5^5} \\ & = 2^{-1} 3^{7 \over 2} 5^{-5} \\ \\ \therefore a & = -1, b = {7 \over 2}, c = -5 \end{align}

(b) Since the value of the painting increases by 7% each year, the value in the current year is 107% (or 1.07 times) of the value in the previous year.

\begin{align} \text{Value at beginning of 2014} & = \$ 1,000,000 \\ \\ \text{Value at beginning of 2015} & = \$ 1,000,000 (1.07) \\ \\ \text{Value at beginning of 2016} & = 1,000,000(1.07) \times 1.07 \\ & = \$ 1,000,000 (1.07)^2 \\ \\ . \\ . \\ . \\ \text{Value at beginning of 2020} & = \$ 1,000,000(1.07)^6 \\ & = \$ 1,500,730.352 \\ & \approx \$ 1, 500, 000 \text{ (2 s.f.)} \end{align}

Note: the fraction is an improper fraction, thus we need to do long division first

$$ \require{enclose} \begin{array}{rll} 2 \phantom{000000000}\\ 2x^2 - x - 3 \enclose{longdiv}{ 4x^2 - 7x + 9 \phantom{0}}\kern-.2ex \\ -\underline{( 4x^2 - 2x - 6 ){\phantom{.}}} \\ -5x + 15 \phantom{} \end{array} $$
\begin{align} {4x^2 - 7x + 9 \over 2x^2 - x - 3} & = 2 + {-5x + 15 \over 2x^2 - x - 3} \\ & = 2 + {-5x + 15 \over (2x - 3)(x + 1)} \end{align}
\begin{align} \text{Consider } {-5x + 15 \over (2x - 3)(x + 1)} & = {A \over 2x - 3} + {B \over x + 1} \\ {-5x + 15 \over (2x - 3)(x + 1)} & = {A(x + 1) \over (2x - 3)(x + 1)} + {B(2x - 3) \over (2x - 3)(x + 1)} \\ {-5x + 15 \over (2x - 3)(x + 1)} & = {A(x + 1) + B(2x - 3) \over (2x - 3)(x + 1)} \\ \\ -5x + 15 & = A(x + 1) + B(2x - 3) \\ \\ \text{Let } & x = -1, \\ -5(-1) + 15 & = A(0) + B[2(-1) - 3] \\ 20 & = 0 + B(-5) \\ 20 & = -5B \\ {20 \over -5} & = B \\ -4 & = B \\ \\ -5x + 15 & = A(x + 1) - 4(2x - 3) \\ \\ \text{Let } & x = 1.5, \\ -5(1.5) + 15 & = A(1.5 + 1) - 4(0) \\ 7.5 & = A(2.5) - 0 \\ 7.5 & = 2.5A \\ {7.5 \over 2.5} & = A \\ 3 & = A \\ \\ \therefore {-5x + 15 \over (2x - 3)(x + 1)} & = {3 \over 2x - 3} + {-4 \over x + 1} \end{align}
\begin{align} \therefore {4x^2 - 7x + 9 \over 2x^2 - x - 3} & = 2 + {-5x + 15 \over (2x - 3)(x + 1)} \\ & = 2 + {3 \over 2x - 3} - {4 \over x + 1} \end{align}

\begin{align} y & = x^2 + {4 \over x^2} \\ & = x^2 + 4x^{-2} \\ \\ {dy \over dx} & = 2x + 4(-2)x^{-3} \\ & = 2x - 8x^{-3} \\ & = 2x - {8 \over x^3} \\ \\ \text{For stationary} & \text{ points, } {dy \over dx} = 0 \\ 2x - {8 \over x^3} & = 0 \\ 2x & = {8 \over x^3} \\ 2x^4 & = 8 \\ x^4 & = 4 \\ x & = \pm \sqrt[4]{4} \\ & = \pm (2^2)^{1 \over 4} \\ & = \pm 2^{1 \over 2} \\ & = \pm \sqrt{2} \\ \\ \text{When } & x = \sqrt{2}, \\ y & = (\sqrt{2})^2 + {4 \over (\sqrt{2})^2} \\ & = 4 \\ \\ \text{When } & x = -\sqrt{2}, \\ y & = (-\sqrt{2})^2 + {4 \over (-\sqrt{2})^2} \\ & = 4 \\ \\ \text{Stationary points} & \text{ are } (\sqrt{2}, 4) \text{ and } (-\sqrt{2}, 4) \\ \\ \\ {d^2y \over dx^2} & = 2 - 8(-3)x^{-4} \\ & = 2 + 24x^{-4} \\ & = 2 + {24 \over x^4} \\ \\ \text{For the stationary} & \text{ point } (\sqrt{2}, 4), \\ {d^2 y \over dx^2} & = 2 + {24 \over (\sqrt{2})^4} \\ & = 8 \\ \therefore (\sqrt{2}, 4) & \text{ is a minimum point} \\ \\ \\ \text{For the stationary} & \text{ point } (-\sqrt{2}, 4), \\ {d^2 y \over dx^2} & = 2 + {24 \over (-\sqrt{2})^4} \\ & = 8 \\ \therefore (-\sqrt{2}, 4) & \text{ is a minimum point} \end{align}

(a)

\begin{align} \text{Let } f(x) & = x^3 + ax \\ \\ f(2) & = (2)^3 + a(2) \\ & = 8 + 2a \\ \\ f(-1) & = (-1)^3 + a(-1) \\ & = -1 - a \\ \\ \text{Since remain} & \text{ders are the same,} \\ 8 + 2a & = - 1 - a \\ 2a + a & = -1 - 8 \\ 3a & = -9 \\ a & = {-9 \over 3} \\ a & = -3 \end{align}

(b)

\begin{align} \text{Let } g(x) & = x^3 - 2x^2 - 4x + 3 \\ \\ g(3) & = (3)^3 - 2(3)^2 - 4(3) + 3 \\ & = 0 \\ \\ \text{By Factor } & \text{theorem, }(x - 3) \text{ is a factor of } g(x) \end{align}
$$ \require{enclose} \begin{array}{rll} x^2 + x - 1 \phantom{0000000}\\ x - 3 \enclose{longdiv}{ x^3 - 2x^2 - 4x + 3 \phantom{0}}\kern-.2ex \\ -\underline{( x^3 - 3x^2 ){\phantom{0000000(}}} \\ x^2 - 4x + 3 \phantom{0} \\ -\underline{( x^2 - 3x ){\phantom{000.}}} \\ -x + 3 \phantom{0.} \\ -\underline{( -x + 3 ) \phantom{.}} \\ 0\phantom{0.} \end{array} $$
\begin{align} \text{Polynomial} & = \text{Quotient} \times \text{Divisor} + \text{Remainder} \\ x^3 - 2x^2 - 4x + 3 & = (x^2 + x - 1)(x - 3) + 0 \\ & = (x^2 + x - 1)(x - 3) \\ \\ x^3 - 2x^2 - 4x + 3 & = 0 \\ (x^2 + x - 1)(x - 3) & = 0 \\ \\ x - 3 = 0 \phantom{00} \text{ or } \phantom{00} x^2 + x - 1 & = 0 \\ x = 3 \phantom{00000000000000} x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \phantom{0000000} [\text{Quadratic formula}] \\ & = { -(1) \pm \sqrt{(1)^2 - 4(1)(-1)} \over 2(1)} \\ & = {-1 \pm \sqrt{5} \over 2} \\ & = -{1 \over 2} \pm {\sqrt{5} \over 2} \\ \\ \\ \therefore x & = 3, -{1 \over 2} + {\sqrt{5} \over 2}, -{1 \over 2} - {\sqrt{5} \over 2} \end{align}

(i)

\begin{align} x^2 + y^2 + 4x & \phantom{.} - 6y - 12 = 0 \\ \\ 2g & = 4 \\ g & = {4 \over 2} \\ & = 2 \\ \\ 2h & = -6 \\ h & = {-6 \over 2} \\ & = -3 \\ \\ \therefore \text{Coordinates of } & \text{centre is } (-2, 3) \\ \\ \\ f & = - 12 \\ \\ \text{Radius} & = \sqrt{g^2 + h^2 - f} \\ & = \sqrt{(2)^2 + (-3)^2 - (-12)} \\ & = 5 \text{ units} \end{align}

(ii)

\begin{align} \text{Let } O \text{ denote } & \text{the centre of the circle, } (-2, 3) \\ \text{Let } A \text{ denote } & \text{the point } (1, 7) \\ \\ \text{Gradient of } OA & = {y_2 - y_1 \over x_2 - x_1} \\ & = {7 - 3 \over 1 - (-2)} \\ & = {4 \over 3} \\ \\ \text{Gradient of tangent} \times {4 \over 3} & = -1 \\ \text{Gradient of tangent} & = -1 \div {4 \over 3} \\ & = -{3 \over 4} \\ \\ y - y_1 & = m (x - x_1) \\ y - 7 & = -{3 \over 4}(x - 1) \\ y - 7 & = -{3 \over 4}x + {3 \over 4} \\ y & = -{3 \over 4}x + {31 \over 4} \\ \\ \therefore \text{Equation of tangent} & \text{ at (1, 7) is } y = -{3 \over 4}x + {31 \over 4} \end{align}

(i)

\begin{align} y & = \cos 2x \\ \\ \text{When } & y = -{\sqrt{3} \over 2}, \\ -{\sqrt{3} \over 2} & = \cos 2x \phantom{000000} [\text{2nd or 3rd quadrant}] \end{align}

\begin{align} \text{Basic angle, } \alpha & = \cos^{-1} \left(\sqrt{3} \over 2\right) \\ & = 30^\circ \\ & = {\pi \over 6} \text{ radians} \\ \\ 2x & = \pi - {\pi \over 6}, \pi + {\pi \over 6} \\ & = {5\pi \over 6}, {7\pi \over 6} \\ x & = {5\pi \over 12}, {7\pi \over 12} \\ \\ \therefore x\text{-coordinate of } A & = {5\pi \over 12} \\ x \text{-coordinate of } B & = {7\pi \over 12} \end{align}

(ii)

The shaded area is the difference between the area under the curve and the area of the rectangle

\begin{align} \text{Area bounded by curve and } x \text{-axis} & = - \int_{5\pi \over 12}^{7\pi \over 12} \cos 2x \phantom{.} dx \\ & = - \left[ {\sin 2x \over 2} \right]_{5\pi \over 12}^{7 \pi \over 12} \\ & = - \left\{ \left[ \sin 2\left(7\pi \over 12\right) \over 2\right] - \left[ \sin2\left(5\pi \over12\right) \over 2\right] \right\} \\ & = - \left[ \left( -{1 \over 2} \over 2 \right) - \left( {1 \over 2} \over 2\right) \right] \\ & = {1 \over 2} \text{ square units} \\ \\ \text{Width of rectangle} & = {7\pi \over 12} - {5\pi \over 12} \\ & = {\pi \over 6} \\ \\ \text{Area of rectangle} & = \text{Length} \times \text{Width} \\ & = {\sqrt{3} \over 2} \times {\pi \over 6} \\ & = {\sqrt{3} \pi \over 12} \text{ square units} \\ \\ \therefore \text{Area of shaded region} & = \left( {1 \over 2} - {\sqrt{3} \pi \over 12} \right) \text{ square units} \end{align}

(i)

\begin{align} \text{When } & t = 0, \\ v & = 15 \left({0 \over 20} + e^{k(0)} \right) \\ v & = 15 (0 + 1) \\ v & = 15 \\ \\ \therefore \text{Speed at } A & = 15 \text{ m/s} \\ \\ \text{Speed at } B & = 15 \times 2 \\ & = 30 \text{ m/s} \end{align}

(ii)

\begin{align} v & = 15\left({t \over 20} + e^{kt}\right) \\ v & = {3 \over 4}t + 15e^{kt} \\ \\ \text{When } & t = 10, v = 30, \\ 30 & = {3 \over 4}(10) + 15e^{k(10)} \\ 30 & = 7.5 + 15e^{10k} \\ 22.5 & = 15e^{10k} \\ {22.5 \over 15} & = e^{10k} \\ 1.5 & = e^{10k} \\ \ln 1.5 & = \ln e^{10k} \\ \ln 1.5 & = 10k \ln e \\ \ln 1.5 & = 10k (1) \\ \ln 1.5 & = 10k \\ \\ k & = {1 \over 10} \ln 1.5 \\ \\ \\ s & = \int v \phantom{.} dt \\ & = \int {3 \over 4}t + 15e^{kt} \phantom{.} dt \\ & = { {3 \over 4} t^2 \over 2} + {15e^{kt} \over k} + C \\ & = {3 \over 8} t^2 + {15 \over k} e^{kt} + C \\ \\ \text{When } & t = 0, s = 0, \\ 0 & = {3 \over 8} (0)^2 + {15 \over k} e^{k(0)} + C \\ 0 & = 0 + {15 \over k}(1) + C \\ 0 & = {15 \over k} + C \\ -{15 \over k} & = C \\ \\ s & = {3 \over 8} t^2 + {15 \over k} e^{kt} - {15 \over k} \\ \\ \text{When } & t = 10 \text{ and } k = {1 \over 10} \ln 1.5, \\ s & = {3 \over 8} (10)^2 + {15 \over k} e^{k(10)} - {15 \over k} \\ & = 37{1 \over 2} + {15 \over {1 \over 10} \ln 1.5} e^{\ln 1.5} - {15 \over {1 \over 10} \ln 1.5} \\ & = 222.47276 \\ & \approx 222 \text{ m} \end{align}

(iii)

\begin{align} v & = {3 \over 4}t + 15e^{kt} \\ \\ a & = {dv \over dt} \\ & = {d \over dt} \left( {3 \over 4}t + 15e^{kt} \right) \\ & = {3 \over 4} + 15k e^{kt} \\ \\ \text{When } & t = 2 \text{ and } k = {1 \over 10} \ln 1.5, \\ a & = {3 \over 4} + 15 \left({1 \over 10} \ln 1.5\right) e^{\left({1 \over 10} \ln 1.5\right)(2)} \\ & = 1.40957 \\ & \approx 1.41 \text{ m/s}^2 \end{align}

(i)

\begin{align} \text{L.H.S} & = {\sin (\alpha + \beta) + \sin (\alpha - \beta) \over \cos (\alpha + \beta) + \cos (\alpha - \beta)} \\ & = { \sin \alpha \cos \beta + \cos \alpha \sin \beta + \sin \alpha \cos \beta - \cos \alpha \sin \beta \over \cos \alpha \cos \beta - \sin \alpha \sin \beta + \cos \alpha \cos \beta + \sin \alpha \sin \beta} \\ & = { 2 \sin \alpha \cos \beta \over 2 \cos \alpha \cos \beta} \\ & = { \sin \alpha \over \cos \alpha } \\ & = \tan \alpha \end{align}

(ii)

\begin{align} & \text{Since } \sin (A - B) < 0, \phantom{.} A - B \text{ lies in 4th quadrant such that } -90^\circ < A - B < 0^\circ \\ \\ & \therefore B > A \text{ (such that } A - B < 0 ) \end{align}

(iii)

\begin{align} \sin (A - B) & = -{3 \over 5} \\ {Opp \over Hyp} & = {-3 \over 5} \end{align}

\begin{align} \sqrt{ (5)^2 - (3)^2 } & = 4 \\ \\ \cos (A - B) & = {4 \over 5} \end{align}

(iv)

\begin{align} \tan (A + B) & = {\sin (A + B) \over \cos (A + B)} \\ {24 \over 7} & = {\sin (A + B) \over {7 \over 25}} \\ {7 \over 25} \left(24 \over 7\right) & = \sin ( A+ B) \\ {24 \over 25} & = \sin (A + B) \end{align}

(v)

\begin{align} \text{From (i), } \tan \alpha & = {\sin(\alpha + \beta) + \sin (\alpha - \beta) \over \cos (\alpha + \beta) + \cos (\alpha - \beta) } \\ \\ \therefore \tan A & = { \sin (A + B) + \sin (A - B) \over \cos (A + B) + \cos (A - B) } \\ & = { {24 \over 25} + \left(-{3 \over 5}\right) \over {7 \over 25} + {4 \over 5} } \\ & = {1 \over 3} \phantom{00} \text{ (Shown)} \end{align}

(vi)

\begin{align} \tan (A + B) & = {\tan A + \tan B \over 1 - \tan A \tan B} \\ {24 \over 7} & = {{1 \over 3} + \tan B \over 1 - {1 \over 3} \tan B} \\ 24 \left(1 - {1 \over 3} \tan B\right) & = 7 \left({1 \over 3} + \tan B\right) \\ 24 - 8 \tan B & = {7 \over 3} + 7 \tan B \\ -8 \tan B - 7 \tan B & = {7 \over 3} - 24 \\ -15 \tan B & = -{65 \over 3} \\ \tan B & = { -{65 \over 3} \over -15} \\ \tan B & = {13 \over 9} \end{align}

(i)

\begin{align} a \cos \theta + b \sin \theta & = R \cos (\theta - \alpha) \\ \\ a = 7, b & = 4 \\ \\ R & = \sqrt{a^2 + b^2} \\ & = \sqrt{ 7^2 + 4^2 } \\ & = \sqrt{ 65 } \\ \\ \alpha & = \tan^{-1} \left(b \over a\right) \\ & = \tan^{-1} \left(4 \over 7\right) \\ & = 0.51915 \\ \\ 7 \cos \theta + 4 \sin \theta & = \sqrt{ 65} \cos (\theta - 0.51915) \\ \\ \\ \sqrt{65} \cos (\theta - 0.51915) & = 6 \\ \cos (\theta - 0.51915) & = {6 \over \sqrt{65}} \phantom{00000} [\text{1st or 4th quadrant}] \\ \\ \text{Basic angle, } \alpha & = \cos^{-1} \left(6 \over \sqrt{65}\right) \\ & = 0.73145 \end{align}

\begin{align} \text{Since } 0 \le \theta \le \pi, & \phantom{0} - 0.51915 \le \theta - 0.51915 \le 2.6224 \\ \\ \theta - 0.51915 & = 0.73145, 2\pi - 0.73145 \\ & = 0.73145, 5.5517 \text{ (N.A.)} \\ & = 0.73145 \\ \\ \theta & = 0.73145 + 0.51915 \\ & \approx 1.25 \end{align}

(ii)

\begin{align} 80 - (7 \cos \theta + 4 \sin \theta)^2 & = 80 - \left[ \sqrt{65} \cos (\theta - 0.51915) \right]^2 \\ \\ \\ \text{Largest value} & = 80 - (0)^2 \\ & = 80 \\ \\ \theta - 0.51915 & = {\pi \over 2}, {3\pi \over 2} \\ \theta & = 2.0899, 5.2315 \text{ (N.A. since } 0 \le \theta \le \pi) \\ \theta & \approx 2.09 \\ \\ \\ \text{Smallest value} & = 80 - \left( \sqrt{65}\right)^2 \\ & = 15 \\ \\ \theta - 0.51915 & = 0, 2\pi \\ \theta & = 0.51915, 6.8023 \text{ (N.A. since } 0 \le \theta \le \pi) \\ \theta & = 0.51915 \\ \theta & \approx 0.519 \end{align}

(a)

\begin{align} y & = 2x^2 - 7 \phantom{00} \text{--- (1)} \\ \\ y & = 3x + 20 \phantom{00} \text{--- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ 2x^2 - 7 & = 3x + 20 \\ 2x^2 - 3x - 27 & = 0 \\ (x + 3)(2x - 9) & = 0 \end{align} \begin{align} x + 3 & = 0 && \text{ or } & 2x - 9 & = 0 \\ x & = -3 &&& 2x & = 9 \\ & &&& x & = {9 \over 2} \\ & &&& x & = 4.5 \\ \\ \text{Substitute } & \text{into (2),} &&& \text{Substitute } & \text{into (2),} \\ y & = 3(-3) + 20 &&& y & = 3(4.5) + 20 \\ y & = 11 &&& y & = 33.5 \end{align}
$$ \therefore x = -3, y = 11 \text{ or } x = 4.5, y = 33.5 $$

(b) For a quadratic expression to be always negative, a < 0 and b² - 4ac < 0

\begin{align} ax^2 & \phantom{.} + 5x - 2 \\ \\ \text{Condition 1: } & a < 0 \\ \\ \\ b^2 - 4ac & < 0 \phantom{000000} [\text{No real roots}] \\ (5)^2 - 4(a)(-2) & < 0 \\ 25 - 4a(-2) & < 0 \\ 25 + 8a & < 0 \\ 8a & < - 25 \\ a & < -{25 \over 8} \\ \text{Condition 2: } a & < -3{1 \over 8} \\ \\ \\ \text{To satisfy both, } & a < -3{1 \over 8} \\ \\ \therefore \text{Greatest value} & \text{ of } a = -4 \end{align}

(c)

\begin{align} y & = 4x + c \phantom{00} \text{--- (1)} \\ \\ y & = x^2 + cx + {21 \over 4} \phantom{00} \text{--- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ 4x + c & = x^2 + cx + {21 \over 4} \\ 0 & = x^2 + cx - 4x + {21 \over 4} - c \\ 0 & = x^2 + (c - 4)x + \left({21 \over 4} - c\right) \\ \\ b^2 - 4ac & = (c - 4)^2 - 4(1)\left({21 \over 4} - c \right) \\ & = \underbrace{(c)^2 - 2(c)(4) + (4)^2}_{(a - b)^2 = a^2 - 2ab + b^2} - 4 \left({21 \over 4} - c\right) \\ & = c^2 - 8c + 16 - 21 + 4c \\ & = c^2 - 4c - 5 \\ \\ b^2 - 4ac & = 0 \phantom{000000} [\text{Line tangent to curve}] \\ c^2 - 4c - 5 & = 0 \\ (c - 5)(c + 1) & = 0 \end{align} \begin{align} c - 5 & = 0 && \text{ or } & c + 1 & = 0 \\ c & = 5 &&& c & = -1 \end{align}

(i)

\begin{align} T_{r + 1} & = {n \choose r} a^{n - r} b^r \\ & = {8 \choose r} \left(3 \over x^2\right)^{8 - r} (x)^r \\ & = {8 \choose r} \left(3^{8 - r} \over x^{16 - 2r}\right) (x^r) \\ & = {8 \choose r} (3^{8 - r}) \left(x^r \over x^{16 - 2r}\right) \\ & = {8 \choose r} (3^{8 - r}) (x^{r - (16 - 2r)}) \\ & = {8 \choose r} (3^{8 - r}) (x^{r - 16 + 2r}) \\ & = {8 \choose r} (3^{8 - r}) (x^{3r - 16}) \\ \\ \text{Let } 3r - 16 & = 0 \\ 3r & = 16 \\ r & = {16 \over 3} \\ r & = 5{1 \over 3} \\ \\ \\ \text{Since } r = 5{1 \over 3} & \text{ is not an integer, there are no terms independent of } x \\ \\ \therefore \text{Every term } & \text{is dependent on } x \end{align}

(ii)

\begin{align} \text{From (i), no term independent} & \text{ of } x \text{ for } \left({3 \over x^2} +x\right)^8 \\ \\ \text{From (i), } T_{r + 1} & = {8 \choose r} (3^{8 - r}) (x^{3r - 16}) \\ \\ \\ \text{Let } 3r - 16 & = -1 \phantom{000000} \left[\text{Looking for term in } {1 \over x} \text{ or } x^{-1} \right] \\ 3r & = -1 + 16 \\ 3r & = 15 \\ r & = {15 \over 3} \\ r & = 5 \\ \\ T_{5 + 1} & = {8 \choose 5} (3^{8 - 5})(x^{3(5) - 16}) \\ & = (56) (27) (x^{-1}) \\ & = 1512 x^{-1} \\ & = 1512 {1 \over x} \\ \\ \\ \left({3 \over x^2} +x\right)^8 (5 - 2x) & = \left(... + 1512 {1 \over x} + ... \right) (5 - 2x) \\ & = ... + \left(1512 {1 \over x}\right)(-2x) + ... \\ & = ... - 3024 + ... \\ \\ \\ \text{Term independent} & \text{ of } x = - 3024 \end{align}

\begin{align} \text{Let } \angle CAT & = \theta \\ \\ \angle ABC & = \angle CAT \phantom{000000000000} [\text{Alternate segment theorem}] \\ & = \theta \\ \\ \angle CED & = 180^\circ - \angle DBC \phantom{000000} [\text{Angles in opposite segment}] \\ & = 180^\circ - \theta \\ \\ \angle AED & = 180^\circ - \angle CED \phantom{000000} [\text{Adjacent angles on a straight line}] \\ & = 180^\circ - (180^\circ - \theta) \\ & = 180^\circ - 180^\circ + \theta \\ & = \theta \\ \\ \text{Since } \angle EAT = \angle AED & = \theta, \text{ by the converse of alternate angles, lines } DE \text{ and } ST \text{ are parallel} \end{align}

(b)(i)

(b)(ii)

\begin{align} y & = 3x^{1 \over 3} \phantom{00} \text{--- (1)} \\ \\ y & = {1 \over 3}x^{-3} \phantom{00} \text{--- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ 3x^{1 \over 3} & = {1 \over 3}x^{-3} \\ {x^{1 \over 3} \over x^{-3}} & = { {1 \over 3} \over 3} \\ x^{{1 \over 3}- (-3)} & = {1 \over 9} \\ x^{10 \over 3} & = {1 \over 9} \\ \sqrt[3]{x^{10}} & = {1 \over 9} \\ x^{10} & = \left(1 \over 9\right)^3 \\ x^{10} & = {1 \over 729} \phantom{00} \text{ (Shown)} \end{align}

(i)

x	0	1	2	3	4
ln P	1.2000	1.3218	1.4446	1.5686	1.6919

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

(ii)

\begin{align} \text{Gradient} & = {1.63 - 1.26 \over 3.5 - 0.5} \\ & = {37 \over 300} \\ \\ \\ Y & = mX + c \\ \\ m & = {37 \over 300} \\ \\ c & = 1.2 \\ \\ Y & = {37 \over 300}X + {6 \over 5} \\ \\ \\ \text{Since } & Y = \ln P \text{ and } X = x, \\ \ln P & = {37 \over 300}x + {6 \over 5} \\ \log_e P & = {37 \over 300}x + {6 \over 5} \\ P & = e^{ {37 \over 300}x + {6 \over 5} } \\ & = e^{{37 \over 300}x} \times e^{6 \over 5} \\ & = (e^{6 \over 5}) e^{{37 \over 300}x} \end{align}

(iii)

\begin{align} P & > 8 \\ (e^{6 \over 5}) e^{{37 \over 300}x} & > 8 \\ e^{{37 \over 300}x} & > {8 \over e^{6 \over 5}} \\ e^{{37 \over 300}x} & > 2.4096 \\ \ln e^{{37 \over 300}x} & > \ln 2.4096 \\ {37 \over 300}x \ln e & > \ln 2.4096 \phantom{000000} [\text{Power law (logarithms)}] \\ {37 \over 300}x (1) & > \ln 2.4096 \\ {37 \over 300}x & > \ln 2.4096 \\ x & > {\ln 2.4096 \over {37 \over 300}} \\ x & > 7.1306 \\ \\ \implies P > 8 & \text{ when } x > 7.1306 \\ \\ \text{It first exceeds 8 million in} & \text{ the interval from } x = 7 \text{ to } x = 8 \\ \\ \text{No. of years to start of interval} & = 7 \times 5 \\ & = 35 \\ \\ 1995 + 35 & = 2030 \\ \\ \\ \therefore \text{First year of the } & \text{interval is 2030} \end{align}

(i)

\begin{align} u & = x &&& v & = (3x - 5)^{5 \over 3} \\ \\ {du \over dx} & = 1 &&& {dv \over dx} & = {5 \over 3}(3x - 5)^{2 \over 3}. (3) \phantom{0000000} [\text{Chain rule}] \\ & &&& & = 5(3x - 5)^{2 \over 3} \end{align} \begin{align} {d \over dx} \left[ x (3x - 5)^{5 \over 3} \right] & = (x)\left[ 5 (3x - 5)^{2 \over 3} \right] + (3x - 5)^{5 \over 3} (1) \phantom{000000} [\text{Product rule}] \\ & = 5x (3x - 5)^{2 \over 3} + (3x - 5)^{5 \over 3} \\ & = 5x (3x - 5)^{2 \over 3} + (3x - 5) (3x - 5)^{2 \over 3} \phantom{000000} \left[ 1 + {2 \over 3} = {5 \over 3} \right] \\ & = (3x - 5)^{2 \over 3} \left( 5x + 3x - 5 \right) \\ & = (3x - 5)^{2 \over 3} (8x - 5) \phantom{00} \text{ (Shown)} \end{align}

(ii)

\begin{align} \text{From (i), } {d \over dx} \left[ x (3x - 5)^{5 \over 3} \right] & = (3x - 5)^{2 \over 3} (8x - 5) \\ \\ \implies \int (8x - 5)(3x - 5)^{2 \over 3} \phantom{.} dx & = x (3x - 5)^{5 \over 3} \\ \int \underbrace{8x(3x - 5)^{2 \over 3} - 5(3x - 5)^{2 \over 3}}_\text{Expansion} \phantom{.} dx & = x(3x - 5)^{5 \over 3} \\ \int 8x(3x - 5)^{2 \over 3} \phantom{.} dx - \int 5(3x - 5)^{2 \over 3} \phantom{.} dx & = x(3x - 5)^{5 \over 3} \\ \\ \int 8x(3x - 5)^{2 \over 3} \phantom{.} dx & = x(3x - 5)^{5 \over 3} + \int 5(3x - 5)^{2 \over 3} \phantom{.} dx \\ \int 8x(3x - 5)^{2 \over 3} \phantom{.} dx & = x(3x - 5)^{5 \over 3} + 5 \int (3x - 5)^{2 \over 3} \phantom{.} dx \\ \int 8x(3x - 5)^{2 \over 3} \phantom{.} dx & = x(3x - 5)^{5 \over 3} + 5 \left[ (3x - 5)^{5 \over 3} \over \left(5 \over 3\right) (3) \right] \phantom{000000} \text{Use } \int [f(x)]^n \phantom{.} dx = {[f(x)]^{n + 1} \over (n + 1)[f'(x)]} \\ \int 8x(3x - 5)^{2 \over 3} \phantom{.} dx & = x(3x - 5)^{5 \over 3} + 5 \left[(3x - 5)^{5 \over 3} \over 5\right] \\ \int 8x(3x - 5)^{2 \over 3} \phantom{.} dx & = x(3x - 5)^{5 \over 3} + (3x - 5)^{5 \over 3} \\ \int 8x(3x - 5)^{2 \over 3} \phantom{.} dx & = (3x - 5)^{5 \over 3} (x + 1) \phantom{000000000000} [\text{Factorise } (3x - 5)^{5 \over 3}] \\ 8 \int x(3x - 5)^{2 \over 3} \phantom{.} dx & = (3x - 5)^{5 \over 3} (x + 1) \\ \int x(3x - 5)^{2 \over 3} \phantom{.} dx & = {1 \over 8}(3x - 5)^{5 \over 3} (x + 1) + c \end{align}

(iii)

(a)

\begin{align} e^x (1 + e^x) & = {3 \over 4} \\ \\ \text{Let } & u = e^x, \\ u (1 + u) & = {3 \over 4} \\ 4u (1 + u) & = 4 \left(3 \over 4\right) \\ 4u + 4u^2 & = 3 \\ 4u^2 + 4u - 3 & = 0 \\ (2u - 1)(2u + 3) & = 0 \end{align} \begin{align} 2u - 1 & = 0 && \text{ or } & 2u + 3 & = 0 \\ 2u & = 1 &&& 2u & = -3 \\ u & = {1 \over 2} &&& u & = -{3 \over 2} \\ \\ e^x & = {1 \over 2} &&& e^x & = -{3 \over 2} \phantom{00} \text{ (Reject, since } e^x > 0) \\ \ln e^x & = \ln {1 \over 2} \\ [\text{Power law (logarithms)}] \phantom{000000} x \ln e & = \ln {1 \over 2} \\ x (1) & = \ln {1 \over 2} \\ x & = \ln {1 \over 2} \\ x & \approx -0.693 \end{align}

(b)

\begin{align} 1 + \log_2 y + {1 \over \log_8 2} & = \log_2 (y + 3) \\ 1 + {1 \over \log_8 2} & = \log_2 (y + 3) - \log_2 y \\ 1 + {1 \over \log_8 2} & = \log_2 {y + 3 \over y} \phantom{0000000000} [\text{Quotient law (logarithms)}] \\ 1 + {1 \over {\log_2 2 \over \log_2 8}} & = \log_2 {y + 3 \over y} \phantom{0000000000} [\text{Change-of-base}] \\ 1 + {\log_2 8 \over \log_2 2} & = \log_2 {y + 3 \over y} \\ 1 + {\log_2 8 \over 1} & = \log_2 {y + 3 \over y} \\ 1 + \log_2 8 & = \log_2 {y + 3 \over y} \\ 1 + \log_2 2^3 & = \log_2 {y + 3 \over y} \\ 1 + 3 \log_2 2 & = \log_2 {y + 3 \over y} \phantom{0000000000} [\text{Power law (logarithms)}] \\ 1 + 3 (1) & = \log_2 {y + 3 \over y} \\ 4 & = \log_2 {y + 3 \over y} \\ \\ {y + 3 \over y} & = 2^4 \\ {y + 3 \over y} & = 16 \\ y + 3 & = 16y \\ 3 & = 16y - y \\ 3 & = 15y \\ {3 \over 15} & = y \\ {1 \over 5} & = y \end{align}

(c)

\begin{align} x & = \ln \left(2x + 7 \over 3\right)^2 \\ x & = 2 \ln \left(2x + 7 \over 3 \right) \phantom{000000} [\text{Power law (logarithms)}] \\ {x \over 2} & = \ln \left(2x + 7 \over 3\right) \\ {x \over 2} & = \log_e \left(2x + 7 \over 3\right) \\ e^{x \over 2} & = {2x + 7 \over 3} \phantom{000000000000} [e^{x \over 2} \text{ the subject of equation}] \\ \\ \\ e^{x \over 2} & = {2x + 7 \over 3} \\ 3 (e^{x \over 2}) & = 3 \left(2x + 7 \over 3\right) \\ 3e^{x \over 2} & = 2x + 7 \\ 3e^{x \over 2} + 4 & = 2x + 7 + 4 \\ \underbrace{3e^{x \over 2} + 4}_\text{Curve} & = \underbrace{2x + 11}_\text{Line} \\ \\ \therefore \text{Draw } & y = 2x + 11 \end{align}

(i)

\begin{align} \text{Length of } AB & = \sqrt{ (x_2 - x_1)^2 + (y_2 - y_1)^2 } \\ & = \sqrt{ [7 - (-5)]^2 + (5 - 0)^2 } \\ & = 13 \text{ units} \\ \\ \text{Perimeter} & = 2AB + 2AD \\ 46 & = 2(13) + 2AD \\ 46 & = 26 + 2AD \\ 20 & = 2AD \\ {20 \over 2} & = AD \\ 10 & = AD \\ \\ AD & = 10 \text{ units} \\ \\ \\ \text{Gradient of } BD & = \text{Gradient of } BE \\ & = {y_2 - y_1 \over x_2 - x_1} \\ & = {6 - 5 \over 5 - 7} \\ & = -{1 \over 2} \\ \\ y & = mx + c \\ y & = -{1 \over 2}x + c \\ \\ \text{Using } & B(7, 5), \\ 5 & = -{1 \over 2}(7) + c \\ 5 & = -{7 \over 2} + c \\ {17 \over 2} & = c \\ \\ \text{Eqn of } & BD: \phantom{0} y = -{1 \over 2}x + {17 \over 2} \\ \\ \\ \text{Since } D \text{ lies on } BD, & \phantom{0} D \left(x, -{1 \over 2}x + {17 \over 2} \right) \\ \\ AD & = \sqrt{ [x - (-5)]^2 + \left(-{1 \over 2}x + {17 \over 2} - 0\right)^2 } \\ 10 & = \sqrt{ (x + 5)^2 + \left({17 \over 2} - {1 \over 2}x\right)^2 } \\ 10 & = \sqrt{ x^2 + 2(x)(5) + (5)^2 + \left(17 \over 2\right)^2 - 2 \left(17 \over 2\right)\left({1 \over 2}x\right) + \left({1 \over 2}x\right)^2 } \\ 10 & = \sqrt{ x^2 + 10x + 25 + {289 \over 4} - {17 \over 2}x + {1 \over 4}x^2 } \\ 10 & = \sqrt{ {5 \over 4}x^2 + {3 \over 2}x + {389 \over 4} } \\ 10^2 & = {5 \over 4}x^2 + {3 \over 2}x + {389 \over 4} \\ 100 & = {5 \over 4}x^2 + {3 \over 2}x + {389 \over 4} \\ 4(100) & = 4 \left( {5 \over 4}x^2 + {3 \over 2}x + {389 \over 4} \right) \\ 400 & = 5x^2 + 6x + 389 \\ 400 - 389 & = 5x^2 + 6x \\ 11 & = 5x^2 + 6x \phantom{00} \text{ (Shown)} \end{align}

(ii)

\begin{align} 5x^2 + 6x & = 11 \\ 5x^2 + 6x - 11 & = 0 \\ (x - 1)(5x + 11) & = 0 \end{align} \begin{align} x - 1 & = 0 && \text{ or } & 5x + 11 & = 0 \\ x & = 1 &&& 5x & = -11 \\ & &&& x & = -{11 \over 5} \text{ (Reject: from the diagram, } x > 0) \end{align}
\begin{align} \text{Substitute } & x = 1 \text{ into eqn of line } BD: y = -{1 \over 2}x + {17 \over 2}, \\ y & = -{1 \over 2}(1) + {17 \over 2} \\ y & = 8 \\ \\ \therefore & \phantom{.} D(1, 8) \\ \\ \\ \text{Without the diagram, } & \text{we do not know if } x \text{-coordinate of } D \text{ is positive or negative} \end{align}

(iii) In a parallelogram, diagonals have a common midpoint

\begin{align} \text{Let coordinates } & \text{of } C \text{ be } (x, y) \\ \\ \text{Midpoint of BD} & = \text{Midpoint of } AC \\ \left({7 + 1 \over 2}, {5 + 8 \over 2}\right) & = \left({-5 + x \over 2}, {0 + y \over 2}\right) \\ \left(4, {13 \over 2}\right) & = \left({x - 5 \over 2}, {y \over 2} \right) \end{align} \begin{align} {x - 5 \over 2} & = 4 &&& {y \over 2} & = {13 \over 2} \\ x - 5 & = 2(4) &&& y & = 13 \\ x - 5 & = 8 \\ x & = 13 \end{align}
$$ \therefore C(13, 13) $$

(i) Note: If the line \(x = {1 \over 2}\) is normal to the curve at the point where \(y = {1 \over 4}\), then the coordinates of the point are \(\left({1 \over 2}, {1 \over 4}\right)\)

\begin{align} f''(x) & = 48x^2 + 2e^{2x - 1} \\ \\ f'(x) & = \int f''(x) \phantom{.} dx \\ f'(x) & = \int 48x^2 + 2e^{2x - 1} \phantom{.} dx \\ f'(x) & = 48 \left(x^3 \over 3\right) + 2 \left(e^{2x - 1} \over 2\right) \phantom{00000000} \text{Use } \int e^{f(x)} \phantom{.} dx = {e^{f(x)} \over f'(x)} \\ f'(x) & = 16x^3 + e^{2x - 1} + c \\ \\ \\ \text{Vertical line } & x = {1 \over 2} \text{ is normal to curve at } \left({1 \over 2}, {1 \over 4}\right) \\ \\ \implies \text{Horizontal line } & y = {1 \over 4} \text{ is tangent to curve at same point} \\ \\ \text{Gradient of tangent} & = 0 \\ \\ \\ \text{Let } & x = {1 \over 2} \text{ and } f'(x) = 0, \\ 0 & = 16\left(1 \over 2\right)^3 + e^{2\left(1 \over 2\right) - 1} + c \\ 0 & = 2 + e^0 + c \\ 0 & = 2 + 1 + c \\ 0 & = 3 + c \\ -3 & = c \\ \\ f'(x) & = 16x^3 + e^{2x - 1} - 3 \end{align}

(ii)

\begin{align} f(x) & = \int f'(x) \phantom{.} dx \\ f(x) & = \int 16x^3 + e^{2x - 1} - 3 \phantom{.} dx \\ f(x) & = 16 \left(x^4 \over 4\right) + {e^{2x - 1} \over 2} - 3x + c \\ f(x) & = 4x^4 + {1 \over 2} e^{2x - 1} - 3x + c \\ \\ \text{Using } & \left({1 \over 2}, {1 \over 4}\right), \\ {1 \over 4} & = 4 \left(1 \over 2\right)^4 + {1 \over 2} e^{2 \left(1 \over 2\right) - 1} - 3 \left(1 \over 2\right) + c \\ {1 \over 4} & = {1 \over 4} + {1 \over 2} e^0 - {3 \over 2} + c \\ {1 \over 4} & = {1 \over 4} + {1 \over 2} - {3 \over 2} + c \\ {1 \over 4} & = -{3 \over 4} + c \\ {1 \over 4} + {3 \over 4} & = c \\ 1 & = c \\ \\ f(x) & = 4x^4 + {1 \over 2} e^{2x - 1} - 3x + 1 \end{align}

(iii)

\begin{align} y & = f(x) \\ y & = 4x^4 + {1 \over 2} e^{2x - 1} - 3x + 1 \\ \\ \text{Let } & x = 0, \\ y & = 4(0)^4 + {1 \over 2} e^{2(0) - 1} - 3(0) + 1 \\ y & = 0 + {1 \over 2}e^{-1} - 0 + 1 \\ y & = {1 \over 2}e^{-1} + 1 \\ y & = {1 \over 2e} + 1 \\ \\ y \text{-intercept: } & \left(0, {1 \over 2e} + 1 \right) \\ \\ \\ {dy \over dx} & = f'(x) \\ & = 16x^3 + e^{2x - 1} - 3 \\ \\ \text{When } & x = 0, \\ {dy \over dx} & = 16(0)^3 + e^{2(0) - 1} - 3 \\ & = 0 + e^{-1} - 3 \\ & = {1 \over e} - 3 \\ \\ \text{Gradient of tangent} & = {1 \over e} - 3 \\ \\ y & = mx + c \\ y & = \left({1 \over e} - 3\right)x + c \\ \\ \text{Using } & \left(0, {1 \over 2e} + 1 \right), \\ {1 \over 2e} + 1 & = \left({1 \over e} - 3\right)(0) + c \\ {1 \over 2e} + 1 & = 0 + c \\ {1 \over 2e} + 1 & = c \\ \\ y & = \left({1 \over e} - 3\right)x + {1 \over 2e} + 1 \\ y & = {1 \over e}x - 3x + {1 \over 2e} + 1 \\ e(y) & = e \left({1 \over e}x - 3x + {1 \over 2e} + 1 \right) \phantom{000000} [\text{Multiply } e \text{ on both sides}] \\ ey & = x - 3ex + {1 \over 2} + e \\ ey + 3ex - e & = x + {1 \over 2} \\ e(y + 3x - 1) & = x + {1 \over 2} \\ 2e(y + 3x - 1) & = 2 \left(x + {1 \over 2}\right) \phantom{00000000000000000} [\text{Multiply } 2 \text{ on both sides}] \\ 2e(y + 3x - 1) & = 2x + 1 \phantom{00} \text{ (Shown)} \end{align}

(i) Note: The 'big cone' refers to the entire cone while the 'small cone' refers to the portion of the container which is empty

\begin{align} \text{Let } r \text{ repre} & \text{sent the radius of 'small' cone} \\ \\ \text{Volume of liquid, } V & = \text{'Big' cone} - \text{'Small' cone} \\ & = {1 \over 3} \pi (15)^2 (25) - {1 \over 3} \pi (r)^2 x \\ & = {5625 \over 3} \pi - {1 \over 3} \pi r^2 x \\ & = 1875 \pi - {1 \over 3} \pi r^2 x \\ \\ \text{By simi} & \text{lar triangles,} \\ {r \over 15} & = {x \over 25} \\ 25r & = 15x \\ r & = {15x \over 25} \\ r & = {3x \over 5} \\ r & = {3 \over 5}x \\ \\ \text{Substitute } & r = {3 \over 5}x \text{ into } V = 1875 \pi - {1 \over 3} \pi r^2 x, \\ V & = 1875 \pi - {1 \over 3} \pi \left({3 \over 5}x\right)^2 x \\ & = 1875 \pi - {1 \over 3} \pi \left({9 \over 25}x^2\right)x \\ & = 1875 \pi - {3 \over 25} \pi x^3 \\ & = 3\pi \left( 625 - {1 \over 25} x^3 \right) \phantom{00} \text{ (Shown)} \end{align}

(ii)

\begin{align} V & = 3\pi \left( 625 - {1 \over 25} x^3 \right) \\ V & = 1875 \pi - {3 \over 25} \pi x^3 \\ \\ {dV \over dx} & = - {3 \over 25} \pi (3) x^2 \\ & = - {9 \over 25} \pi x^2 \\ \\ {dx \over dt} & = {dx \over dV} \times {dV \over dt} \\ & = {1 \over -{9 \over 25}\pi x^2} \times kx^2 \\ & = {1 \over -{9\pi x^2 \over 25}} \times kx^2 \\ & = - {25 \over 9\pi x^2} \times {kx^2 \over 1} \\ & = - {25kx^2 \over 9\pi x^2} \\ & = - {25k \over 9\pi} \\ \\ {dt \over dx} & = {1 \over -{25k \over 9\pi}} \\ & = - {9\pi \over 25k} \end{align}

(iii)

\begin{align} {dt \over dx} & = - {9\pi \over 25k} \\ \\ t & = \int -{9\pi \over 25k} \phantom{.} dx \\ t & = -{9\pi \over 25k}x + c \\ \\ \text{When } & t = 0 \text{ and } x = 25, \phantom{000000} [\text{No water initially}] \\ 0 & = -{9\pi \over 25k}(25) + c \\ 0 & = -{9\pi \over k} + c \\ {9 \pi \over k} & = c \\ \\ t & = -{9\pi \over 25k}x + {9 \pi \over k} \\ \\ \\ \text{At } t = 72\pi, & \text{ height of liquid} = 12 \\ \\ \implies 25 - x & = 12 \\ - x & = 12 - 25 \\ - x & = -13 \\ x & = 13 \\ \\ \text{Substitute } x = 13 & \text{ and } t = 72\pi \text{ into } t = -{9\pi \over 25k}x + {9 \pi \over k}, \\ 72 \pi & = -{9\pi \over 25k} (13) + {9 \pi \over k} \\ 72 \pi & = -{117\pi \over 25k} + {9\pi \over k} \\ 72 \pi & = {-117\pi \over 25k} + {9\pi (25) \over 25k} \\ 72 \pi & = {-117\pi + 9\pi (25) \over 25k} \\ 72 \pi & = {-117\pi + 225 \pi \over 25k } \\ {72 \pi \over 1} & = {108 \pi \over 25k} \\ 25k (72\pi) & = 108\pi \\ 25k & = {108 \pi \over 72\pi} \\ 25k & = 1.5 \\ k & = {1.5 \over 25} \\ k & = 0.06 \end{align}