(a)

\begin{align} x - 2y + 9 & = 0 \\ -2y & = - x - 9 \\ 2y & = x + 9 \\ y & = {1 \over 2}(x + 9) \\ y & = {1 \over 2}x + {9 \over 2} \phantom{00} \text{--- (1)} \\ \\ x^2 + 11x & = 3xy + 2 \phantom{00} \text{--- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ x^2 + 11x & = 3x \left({1 \over 2}x + {9 \over 2}\right) + 2 \\ x^2 + 11x & = {3 \over 2}x^2 + {27 \over 2}x + 2 \\ 0 & = {1 \over 2}x^2 + {5 \over 2}x + 2 \\ 2(0) & = 2 \left( {1 \over 2}x^2 + {5 \over 2}x + 2 \right) \\ 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 (1),} &&& \text{Substitute } & \text{into (1),} \\ y & = {1 \over 2}(-1) + {9 \over 2} &&& y & = {1 \over 2}(-4) + {9 \over 2} \\ y & = 4 &&& y & = {5 \over 2} \\ \\ x & = -1, y = 4 &&& x & = -4, y = {5 \over 2} \end{align}

(b)

\begin{align} {3 \over x} - {1 \over y} & = 15 \phantom{00} \text{--- (1)} \\ \\ {1 \over xy} & = -12 \\ {1 \over y} & = -12x \phantom{00} \text{--- (2)} \\ \\ \text{Substitute } & \text{(2) into (1),} \\ {3 \over x} - (-12x) & = 15 \\ {3 \over x} + 12x & = 15 \\ x \left({3 \over x} + 12x\right) & = 15x \\ 3 + 12x^2 & = 15x \\ 12x^2 - 15x + 3 & = 0 \\ 4x^2 - 5x + 1 & = 0 \\ (4x - 1)(x - 1) & = 0 \end{align} \begin{align} 4x - 1 & = 0 && \text{ or } & x - 1 & = 0 \\ 4x & = 1 &&& x & = 1 \\ x & = {1 \over 4} \\ \\ \text{Substitute } & \text{into (2),} &&& \text{Substitute } & \text{into (2),} \\ {1 \over y} & = -12\left(1 \over 4\right) &&& {1 \over y} & = -12(1) \\ {1 \over y} & = -3 &&& {1 \over y} & = -12 \\ 1 & = -3y &&& 1 & = -12y \\ -{1 \over 3} & = y &&& -{1 \over 12} & = y \\ \\ x & = {1 \over 4}, y = -{1 \over 3} &&& x & = 1, y = -{1 \over 12} \end{align}

\begin{align} x^2 & = xy + 12 \phantom{00} \text{--- (1)} \\ \\ 2y + x - 6 & = 0 \\ 2y & = -x + 6 \\ y & = {1 \over 2}(-x + 6) \\ y & = -{1 \over 2}x + 3 \phantom{00} \text{--- (2)} \\ \\ \text{Substitute } & \text{(2) into (1),} \\ x^2 & = x \left(-{1 \over 2}x + 3\right) + 12 \\ x^2 & = -{1 \over 2}x^2 + 3x + 12 \\ 0 & = -{3 \over 2}x^2 + 3x + 12 \\ 2(0) & = 2 \left( -{3 \over 2}x^2 + 3x + 12 \right) \\ 0 & = - 3x^2 + 6x + 24 \\ 0 & = 3x^2 - 6x - 24 \\ 0 & = x^2 - 2x - 8 \\ 0 & = (x - 4)(x + 2) \end{align} \begin{align} x - 4 & = 0 && \text{ or } & x + 2 & = 0 \\ x & = 4 &&& x & = -2 \\ \\ \text{Substitute } & \text{into (2),} &&& \text{Substitute } & \text{into (2),} \\ y & = -{1 \over 2}(4) + 3 &&& y & = -{1 \over 2}(-2) + 3 \\ y & = 1 &&& y & = 4 \\ \\ \therefore & \phantom{.} (4, 1) &&& \therefore & \phantom{.} (-2, 4) \end{align}

\begin{align} y & = 3x + 4 \phantom{00} \text{--- (1)} \\ \\ x^2 + y^2 & = 2(x + 2y) \phantom{00} \text{--- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ x^2 + (3x + 4)^2 & = 2[x + 2(3x + 4)] \\ x^2 + (3x + 4)(3x + 4) & = 2(x + 6x + 8) \\ x^2 + 9x^2 + 12x + 12x + 16 & = 2(7x + 8) \\ 10x^2 + 24x + 16 & = 14x + 16 \\ 10x^2 + 24x - 14x + 16 - 16 & = 0 \\ 10x^2 + 10x & = 0 \\ x^2 + x & = 0 \\ x(x + 1) & = 0 \end{align} \begin{align} x & = 0 && \text{ or } & x + 1 & = 0 \\ & &&& x & = -1 \\ \\ \text{Substitute } & \text{into (1),} &&& \text{Substitute } & \text{into (1),} \\ y & = 3(0) + 4 &&& y & = 3(-1) + 4 \\ y & = 4 &&& y & = 1 \\ \\ \therefore & \phantom{.} (0, 4) &&& \therefore & \phantom{.} (-1, 1) \\ \\ \text{Substitute } & x = 0 \text{ into } 7y + 5x = 28, &&& \text{Substitute } & x = -1 \text{ into } 7y + 5x = 28, \\ 7y + 5(0) & = 28 &&& 7y + 5(-1) & = 28 \\ 7y + 0 & = 28 &&& 7y - 5 & = 28 \\ 7y & = 28 &&& 7y & = 33 \\ y & = 4 &&& y & = {33 \over 7} \phantom{0} (\ne 1) \\ \\ \therefore \text{Line } & \text{passes } (0, 4) &&& \therefore \text{Line } & \text{does not pass } (-1, 1) \end{align}

\begin{align} y & = 1 - 5x \phantom{00} \text{--- (1)} \\ \\ y & = x^2 - 4x + 2 \phantom{00} \text{--- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ 1 - 5x & = x^2 - 4x + 2 \\ 0 & = x^2 + x + 1 \\ \\ b^2 - 4ac & = (1)^2 - 4(1)(1) \\ & = -3 < 0 \\ \\ \implies \text{Equation } & \text{has no real roots} \\ \\ \therefore \text{Line & curve} & \text{ has 0 point of intersection} \end{align}

\begin{align} y - 7x - 9 & = 0 \\ y & = 7x + 9 \phantom{00} \text{--- (1)} \\ \\ y & = 2x^2 + bx + 17 \phantom{00} \text{--- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ 7x + 9 & = 2x^2 + bx + 17 \\ 0 & = 2x^2 + bx - 7x + 8 \\ 0 & = 2x^2 + (b - 7)x + 8 \\ \\ b^2 - 4ac & = (b - 7)^2 - 4(2)(8) \\ & = (b - 7)(b - 7) - 64 \\ & = b^2 - 7b - 7b + 49 - 64 \\ & = b^2 - 14b - 15 \\ \\ b^2 - 4ac & = 0 \phantom{000000} [\text{1 real root since line is tangent to curve}] \\ b^2 - 14b - 15 & = 0 \\ (b - 15)(b + 1) & = 0 \end{align}
\begin{align} b - 15 & = 0 && \text{ or } & b + 1 & = 0 \\ b & = 15 &&& b & = -1 \end{align}

(a)

\begin{align} y & = 9x^2 - 6x + k \\ \\ \text{Let } & y = 0, \\ 0 & = 9x^2 - 6x + k \\ \\ b^2 - 4ac & = (-6)^2 - 4(9)(k) \\ & = 36 - 36k \\ \\ b^2 - 4ac & > 0 \\ 36 - 36k & > 0 \\ -36k & > - 36 \\ k & < {-36 \over -36} \\ k & < 1 \end{align}

(b)

\begin{align} y & = {1 \over 2}x^2 + 4x - k \\ \\ \text{Let } & y = 0, \\ 0 & = {1 \over 2}x^2 + 4x - k \\ \\ b^2 - 4ac & = (4)^2 - 4\left(1 \over 2\right)(-k) \\ & = 16 + 2k \\ \\ b^2 - 4ac & = 0 \\ 16 + 2k & = 0 \\ 2k & = -16 \\ k & = {-16 \over 2} \\ k & = -8 \end{align}

(c)

\begin{align} y & = kx^2 + 8x - 2 \\ \\ \text{Condition 1: } & k < 0 \phantom{000000} [\text{must be maximum curve } \cap \text{ to lie below } x \text{-axis}] \\ \\ \\ \text{Let } & y = 0, \\ 0 & = kx^2 + 8x - 2 \\ \\ b^2 - 4ac & = (8)^2 - 4(k)(-2) \\ & = 64 + 8k \\ \\ b^2 - 4ac & < 0 \phantom{000000} [\text{No real roots since curve lies below } x \text{-axis}] \\ 64 + 8k & < 0 \\ 8k & < -64 \\ k & < {-64 \over 8} \\ k & < -8 \end{align}

(i)

\begin{align} y & = 2x^2 + (h + 4)x + 2h \\ y & = 2x^2 + (3 + 4)x + 2(3) \\ y & = 2x^2 + 7x + 6 \phantom{00} \text{--- (1)} \\ \\ y & = 19x - 12 \phantom{00} \text{--- (2)} \\ \\ \text{Substitute } & \text{(2) into (1),} \\ 19x - 12 & = 2x^2 + 7x + 6 \\ 0 & = 2x^2 - 12x + 18 \\ 0 & = x^2 - 6x + 9 \\ \\ b^2 - 4ac & = (-6)^2 - 4(1)(9) \\ & = 0 \\ \\ \implies \text{Line } & \text{is tangent to curve} \\ \\ \\ 0 & = x^2 - 6x + 9 \\ 0 & = (x - 3)(x - 3) \\ 0 & = (x - 3)^2 \\ 0 & = x - 3 \\ 3 & = x \\ \\ \text{Substitute } & x = 3 \text{ into (2),} \\ y & = 19(3) - 12 \\ y & = 45 \\ \\ \text{Point: } & (3, 45) \end{align}

(ii) If y cannot be a negative value, then y must be positive or zero. This means the curve is entirely above the x-axis or meets the x-axis only once!

\begin{align} y & = 2x^2 + (h + 4)x + 2h \\ \\ \text{Let } & y = 0, \\ 0 & = 2x^2 + (h + 4)x + 2h \\ \\ b^2 - 4ac & = (h + 4)^2 - 4(2)(2h) \\ & = (h + 4)(h + 4) - 16h \\ & = h^2 + 4h + 4h + 16 - 16h \\ & = h^2 - 8h + 16 \\ & = (h - 4)(h - 4) \\ & = (h - 4)^2 \\ \\ \text{If } h = 4, b^2 - 4ac & = 0 \implies \cup \text{ curve meets } x \text{-axis once and will not be negative} \\ \\ \text{If } h \ne 4, b^2 - 4ac & > 0 \implies \cup \text{ curve meets } x \text{-axis twice and it will be negative} \end{align}

(i)

\begin{align} y & = 16x^2 + (1 - 2k)x + 1 + c \phantom{00} \text{--- (1)} \\ \\ y & = -kx + c \phantom{00} \text{--- (2)} \\ \\ \text{Substitute } & \text{(2) into (1),} \\ -kx + c & = 16x^2 + (1 - 2k)x + 1 + c \\ -kx + c & = 16x^2 + x - 2kx + 1 + c \\ 0 & = 16x^2 + x - kx + 1 \\ 0 & = 16x^2 + (1 - k)x + 1 \\ \\ b^2 - 4ac & = (1 - k)^2 - 4(16)(1) \\ & = (1 - k)(1 - k) - 64 \\ & = 1 - k - k + k^2 - 64 \\ & = k^2 - 2k - 63 \\ \\ b^2 - 4ac & = 0 \phantom{000000} [\text{Line tangent to curve}] \\ k^2 - 2k - 63 & = 0 \\ (k - 9)(k + 7) & = 0 \end{align} \begin{align} k - 9 & = 0 && \text{ or } & k + 7 & = 0 \\ k & = 9 &&& k & = - 7 \end{align}
$$ \text{Positive value of } k = 9 $$

(ii)

\begin{align} y & = 16x^2 + (1 - 2k)x + 1 + c \\ y & = 16x^2 + [1 - 2(9)]x + 1 + c \\ y & = 16x^2 - 17x + 1 + c \\ \\ \text{Using } & (-2, 104), \\ 104 & = 16(-2)^2 - 17(-2) + 1 + c \\ 104 & = 64 + 34 + 1 + c \\ 104 & = 99 + c \\ 5 & = c \\ \\ y & = 16x^2 - 17x + 1 + 5 \\ y & = 16x^2 - 17x + 6 \\ \\ \text{From (i), } 0 & = 16x^2 + (1 - k)x + 1 \\ 0 & = 16x^2 + (1 - 9)x + 1 \\ 0 & = 16x^2 - 8x + 1 \\ 0 & = (4x - 1)(4x - 1) \\ 0 & = (4x - 1)^2 \\ 0 & = 4x - 1 \\ -4x & = -1 \\ x & = {-1 \over -4} \\ x & = {1 \over 4} \\ \\ \text{Substitute } & x = {1 \over 4} \text{ into } y = 16x^2 - 17x + 6, \\ y & = 16 \left(1 \over 4\right)^2 - 17 \left(1 \over 4\right) + 6 \\ y & = 2{3 \over 4} \\ \\ \therefore & \phantom{.} \left({1 \over 4}, 2{3 \over 4}\right) \end{align}

(iii)

\begin{align} & L \text{ is a vertical line} \end{align}

(i)

\begin{align} y & = 2x^2 - kx - 1 \\ y & = 2x^2 -(-6)x - 1 \\ y & = 2x^2 + 6x - 1 \phantom{00} \text{--- (1)} \\ \\ y + 3x & = 10 \phantom{00} \text{--- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ 2x^2 + 6x - 1 + 3x & = 10 \\ 2x^2 + 9x - 1 & = 10 \\ 2x^2 + 9x - 11 & = 0 \\ (2x + 11)(x - 1) & = 0 \end{align} \begin{align} 2x + 11 & = 0 && \text{ or } & x - 1 & = 0 \\ 2x & = -11 &&& x & = 1 \\ x & = -{11 \over 2} \\ x & = -5{1 \over 2} \\ \\ \text{Substitute } & \text{into (2),} &&& \text{Substitute } & \text{into (2),} \\ y + 3 \left(-5{1 \over 2}\right) & = 10 &&& y + 3(1) & = 10 \\ y - 16{1 \over 2} & = 10 &&& y + 3 & = 10 \\ y & = 26{1 \over 2} &&& y & = 7 \end{align}
$$ \text{Points: } \left(-5{1 \over 2}, 26{1 \over 2}\right), (1, 7) $$

(ii)

\begin{align} y & = 2x^2 - kx - 1 \phantom{00} \text{--- (1)} \\ \\ y + 3x & = 10 \phantom{00} \text{--- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ 2x^2 - kx - 1 + 3x & = 10 \\ 2x^2 + 3x - kx - 11 & = 0 \\ 2x^2 + (3 - k)x - 11 & = 0 \\ \\ b^2 - 4ac & = (3 - k)^2 - 4(2)(-11) \\ & = (3 - k)(3 - k) + 88 \\ & = 9 - 3k - 3k + k^2 + 88 \\ & = k^2 - 6k + 97 \\ & = k^2 - 6k + \left(6 \over 2\right)^2 - \left(6 \over 2\right)^2 + 97 \\ & = k^2 - 6k + 3^2 - 3^2 + 97 \\ & = (k - 3)^2 + 88 \\ \\ \text{For all real} & \text{ values of } k, \\ (k - 3)^2 & \ge 0 \\ (k - 3)^2 + 88 & \ge 88 \\ \implies b^2 - 4ac & \ge 88 \\ \\ \therefore \text{Line intersects curve } & \text{at two distinct points for all values of } k \end{align}

(i)

\begin{align} ax^2 + 6y - 4b & = 23 \\ \\ \text{Let } x = -1 \text{ and } & y = 4, \\ a(-1)^2 + 6(4) - 4b & = 23 \\ a(1) + 24 - 4b & = 23 \\ a & = 4b - 1 \phantom{00} \text{--- (1)} \\ \\ \\ 6by - ax - 4 & = 23 \\ \\ \text{Let } x = -1 \text{ and } & y = 4, \\ 6b(4) - a(-1) - 4 & = 23 \\ 24b + a - 4 & = 23 \\ 24b + a & = 27 \phantom{00} \text{--- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ 24 + 4b - 1 & = 27 \\ 23 + 4b & = 27 \\ 4b & = 4 \\ b & = 1 \\ \\ \text{Substitute } & \text{into (1),} \\ a & = 4(1) - 1 \\ a & = 3 \\ \\ \\ \therefore a & = 3, b = 1 \end{align}

(ii)

\begin{align} ax^2 + 6y - 4b & = 23 \\ \\ \text{Let } a = 3 \text{ and } & b = 1, \\ 3x^2 + 6y - 4(1) & = 23 \\ 3x^2 + 6y - 4 & = 23 \\ 3x^2 + 6y & = 27 \phantom{00} \text{--- (3)} \\ \\ \\ 6by - ax - 4 & = 23 \\ \\ \text{Let } a = 3 \text{ and } & b = 1, \\ 6(1)y - 3x - 4 & = 23 \\ 6y - 3x & = 27 \\ 6y & = 3x + 27 \\ y & = {1 \over 6}(3x + 27) \\ y & = {1 \over 2}x + {9 \over 2} \phantom{00} \text{--- (4)} \\ \\ \text{Substitute } & \text{(4) into (3),} \\ 3x^2 + 6 \left( {1 \over 2}x + {9 \over 2} \right) & = 27 \\ 3x^2 + 3x + 27 & = 27 \\ 3x^2 + 3x & = 0 \\ x^2 + x & = 0 \\ x(x + 1) & = 0 \end{align} \begin{align} x & = 0 && \text{ or } & x + 1 & = 0 \\ & &&& x & = -1 \text{ (Provided value}) \\ \\ \text{Substitute} & \text{ into (4),} \\ y & = {1 \over 2}(0) + {9 \over 2} \\ y & = {9 \over 2} \\ \\ \therefore x & = 0, y = {9 \over 2} \end{align}

\begin{align} v_P & = v_Q \\ 0.4t + 3 & = -2t^2 + 7t - 1 \\ 2t^2 - 6.6t + 4 & = 0 \\ t^2 - 3.3t + 2 & = 0 \\ 10t^2 - 33t + 20 & = 0 \\ \\ b^2 - 4ac & = (-33)^2 - 4(10)(20) \\ & = 289 > 0 \\ \\ \implies \text{Equation } & \text{has two distinct real roots} \\ \\ \therefore \text{Particles } & \text{travel at the same speed at two timings} \\ \\ \\ 10t^2 - 33t + 20 & = 0 \\ (5t - 4)(2t - 5) & = 0 \end{align} \begin{align} 5t - 4 & = 0 && \text{ or } & 2t - 5 & = 0 \\ 5t & = 4 &&& 2t & = 5 \\ t & = {4 \over 5} &&& t & = {5 \over 2} \end{align}

\begin{align} 16x^2 + y^2 & = 8xy + 9 \phantom{00} \text{--- (1)} \\ \\ (x - y)^2 & = 9 \\ (x - y)(x - y) & = 9 \\ x^2 - xy - xy + y^2 & = 9 \\ x^2 - 2xy + y^2 & = 9 \phantom{00} \text{--- (2)} \\ \\ \text{(1)} & - (2), \\ 15x^2 + 2xy & = 8xy \\ 15x^2 - 6xy & = 0 \\ 3x(5x - 2y) & = 0 \end{align} \begin{align} 3x & = 0 && \text{ or } & 5x - 2y & = 0 \\ x & = 0 &&& 5x & = 2y \\ & &&& x & = {2 \over 5}y \\ \\ \text{Substitute } & \text{into (2),} &&& \text{Substitute } & \text{into (2),} \\ (0 - y)^2 & = 9 &&& \left({2 \over 5}y - y\right)^2 & = 9 \\ (-y)^2 & = 9 &&& \left(-{3 \over 5}y\right)^2 & = 9 \\ y^2 & = 9 &&& {9 \over 25}y^2 & = 9 \\ y & = \pm \sqrt{9} &&& 9y^2 & = 25(9) \\ y & \pm 3 &&& y^2 & = 25 \\ & &&& y & = \pm \sqrt{25} \\ & &&& y & = \pm 5 \\ & &&& x & = {2 \over 5}(\pm 5) \\ & &&& x & = \pm 2 \end{align}
$$ x = 0, y = 3 \text{ or } x = 0, y = -3 \text{ or } x = 2, y = 5 \text{ or } x = -2, y = -5 $$