\begin{align} y & = 5x + 9 \phantom{00} \text{--- (1)} \\ \\ y & = -2x^2 + 6x + 3 \phantom{00} \text{--- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ 5x + 9 & = -2x^2 + 6x + 3 \\ 0 & = -2x^2 + x - 6 \\ \\ b^2 - 4ac & = (1)^2 - 4(-2)(-6) \\ & = -47 < 0 \\ \\ \implies \text{Equation } & \text{has no real roots} \\ \\ \therefore \text{Line and curve } & \text{oes not intersect each other} \end{align}

(i)

\begin{align} y & = x + 4p \phantom{00} \text{--- (1)} \\ \\ y & = -2x^2 + 5x + 1 \phantom{00} \text{--- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ x + 4p & = -2x^2 + 5x + 1 \\ 0 & = -2x^2 + 4x + 1 - 4p \\ \\ b^2 - 4ac & = (4)^2 - 4(-2)(1-4p) \\ & = 16 + 8(1- 4p) \\ & = 16 + 8 - 32p \\ & = 24 - 32p \\ \\ b^2 - 4ac & < 0 \phantom{000000} [\text{No real roots since line does not meet curve}] \\ 24 -32p & < 0 \\ -32p & < -24 \\ p & > {-24 \over -32} \\ p & > {3 \over 4} \end{align}

(ii)

\begin{align} \text{From (i), } b^2 - 4ac & = 24 - 32p \\ \\ b^2 - 4ac & = 0 \\ 24 -32p & = 0 \\ -32p & = -24 \\ p & = {-24 \over -32} \\ p & = {3 \over 4} \end{align}

\begin{align} -2x^2 & - (4 - p)x + 2p \\ \\ b^2 - 4ac & = [-(4 - p)]^2 - 4(-2)(2p) \\ & = (4 - p)^2 + 8(2p) \\ & = (4 - p)(4 - p) + 16p \\ & = 16 - 4p - 4p + p^2 + 16p \\ & = p^2 + 8p + 16 \\ & = (p + 4)(p + 4) \\ & = (p + 4)^2 \\ \\ \text{For all real } & \text{values of } p, \\ (p + 4)^2 & \ge 0 \\ b^2 - 4ac & \ge 0 \\ \\ \implies \text{Graph of } & y = -2x^2 - (4 - p)x + 2p \text{ will meet the } x \text{-axis at least once} \\ \\ \therefore \text{No real values} &\text{ of } p \text{ for which } -2x^2 - (4 - p)x + 2p \text{ is always negative} \end{align}

(i)

\begin{align} 4ax^2 & + 9x + c \\ \\ 4a & < 0 \phantom{000000} [\text{Graph of } y = 4ax^2 + 9x + c \text{ is max. curve } \cap] \\ a & < 0 \phantom{000000} [\text{Condition 1}] \\ \\ \\ b^2 - 4ac & = (9)^2 - 4(4a)(c) \\ & = 81 - 16ac \\ \\ b^2 - 4ac & < 0 \phantom{000000} [\text{No real roots since curve lies below } x \text{-axis}] \\ 81 - 16ac & < 0 \\ -16ac & < -81 \\ ac & > {-81 \over 16} \\ ac & > {81 \over 16} \\ \\ \\ \text{Conditions: } & a < 0, ac > {81 \over 16} \end{align}

(ii)

\begin{align} \text{Let } & a = -2, \phantom{000000} [\text{to satisfy condition 1}] \\ (-2)c & > {81 \over 16} \\ -2c & > {81 \over 16} \\ c & < {81 \over 16} \div -2 \\ c & < -{81 \over 32} \\ c & < -2{17 \over 32} \\ \\ \\ \text{Possible values: } & a = -2, c = -3 \end{align}

(i)

\begin{align} {k \over 8}x^2 + kx + 2 & > 0 \phantom{000000} [\text{i.e. expression is positive}] \\ \\ {k \over 8} & > 0 \phantom{000000} [\text{Curve of } y = {k \over 8}x^2 + kx + 2 \text{ is minimum curve } \cup] \\ k & > 0 \phantom{000000} [\text{Condition 1}] \\ \\ \\ b^2 - 4ac & = (k)^2 - 4\left(k \over 8\right)(2) \\ & = k^2 - {4k \over 8} (2) \\ & = k^2 - {8k \over 8} \\ & = k^2 - k \\ \\ b^2 - 4ac & < 0 \phantom{000000} [\text{No real roots since curve lies above } x \text{-axis}] \\ k^2 - k & < 0 \\ k(k - 1) & < 0 \end{align}

$$ 0 < k < 1 $$

(ii) Coincident roots means equal real roots

\begin{align} b^2 - 4ac & = 0 \\ k^2 - k & = 0 \\ k (k - 1) & = 0 \end{align}
\begin{align} k & = 0 \text{ (Reject, since equation is undefined}) && \text{ or } & k - 1 & = 0 \\ & &&& k & = 1 \\ \end{align}

\begin{align} 2y & = x + 6 \\ y & = {1 \over 2}(x + 6) \\ y & = {1 \over 2}x + 3 \phantom{00} \text{--- (1)} \\ \\ x^2 + xy + y^2 & = 36 \phantom{00} \text{--- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ x^2 + x \left({1 \over 2}x + 3\right) + \left({1 \over 2}x + 3\right)^2 & = 36 \\ x^2 + {1 \over 2}x^2 + 3x + \left({1 \over 2}x\right)^2 + 2\left({1 \over 2}x\right)(3) + (3)^2 & = 36 \phantom{000000} [(a + b)^2 = a^2 + 2ab + b^2] \\ {3 \over 2}x^2 + 3x + {1 \over 4}x^2 + 3x + 9 & = 36 \\ {7 \over 4}x^2 + 6x + 9 & = 36 \\ {7 \over 4}x^2 + 6x - 27 & = 0 \\ 7x^2 + 24x - 108 & = 0 \\ (x + 6)(7x - 18) & = 0 \end{align}
\begin{align} x + 6 & = 0 && \text{ or } & 7x - 18 & = 0 \\ x & = -6 &&& 7x & = 18 \\ & &&& x & = {18 \over 7} \\ \\ \text{Substitute } & \text{into (1),} &&& \text{Substitute } & \text{into (1),} \\ y & = {1 \over 2}(-6) + 3 &&& y & = {1 \over 2}\left(18 \over 7\right) + 3 \\ y & = 0 &&& y & = {30 \over 7} \\ \\ \text{Substitute } & \text{into } 5y = 2x + 12, &&& \text{Substitute } & \text{into } 5y = 2x + 12, \\ 5(0) & = 2x + 12 &&& 5\left(30 \over 7\right) & = 2x + 12 \\ 0 & = 2x + 12 &&& {150 \over 7} & = 2x + 12 \\ -2x & = 12 &&& -2x & = 12 - {150 \over 7} \\ x & = {12 \over -2} &&& -2x & = -{66 \over 7} \\ x & = -6 &&& x & = -{66 \over 7} \div -2 \\ & &&& x & = {33 \over 7} \\ \\ \therefore 5y = 2x + 12 & \text{ passes } (-6, 0) &&& \therefore 5y = 2x + 12 & \text{ does not pass } \left({18 \over 7}, {30 \over 7}\right) \end{align}

\begin{align} y & = 2x^2 + 3kx \phantom{00} \text{--- (1)} \\ \\ y & = 2 - 5x \phantom{00} \text{--- (2)} \\ \\ \text{Substitute } & \text{(2) into (1),} \\ 2 - 5x & = 2x^2 + 3kx \\ 0 & = 2x^2 + 3kx +5x - 2 \\ 0 & = 2x^2 + (3k + 5)x - 2 \\ \\ b^2 - 4ac & = (3k + 5)^2 - 4(2)(-2) \\ & = (3k + 5)^2 + 16 \\ \\ \text{For all real} & \text{ values of } k, \\ (3k + 5)^2 & \ge 0 \\ (3k + 5)^2 + 16 & \ge 16 \\ b^2 - 4ac & \ge 16 \\ \\ \implies 0 = 2x^2 + (3k + 5)x & - 2 \text{ has two real roots for all real values of } k \\ \\ \therefore \text{For all real values of } & k, \text{ graphs will have two points of intersection} \end{align}

\begin{align} \sqrt{x} - 5 & = {36 \over \sqrt{x}} \\ { \sqrt{x} (\sqrt{x} - 5) \over \sqrt{x}} & = {36 \over \sqrt{x}} \\ \sqrt{x}(\sqrt{x} - 5) & = 36 \\ \\ \text{Let } & u = \sqrt{x}, \\ u(u - 5) & = 36 \\ u^2 - 5u & = 36 \\ u^2 - 5u - 36 & = 0 \\ (u - 9)(u + 4) & = 0 \end{align}
\begin{align} u - 9 & = 0 && \text{ or } & u + 4 & = 0 \\ u & = 9 &&& u & = -4 \\ \\ \sqrt{x} & = 9 &&& \sqrt{x} & = -4 \text{ (Reject, since } \sqrt{x} \ge 0) \\ x & = 9^2 \\ x & = 81 \end{align}

\begin{align} 3x - 5y + 14 & = 0 \\ 3x & = 5y - 14 \phantom{00} \text{--- (1)} \\ \\ {3 \over x} \times {3 \over 3} - {5 \over y} & = {1 \over 4} \\ {9 \over 3x} - {5 \over y} & = {1 \over 4} \phantom{00} \text{--- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ {9 \over 5y - 14} - {5 \over y} & = {1 \over 4} \\ {9y \over y(5y - 14)} - {5(5y - 14) \over y(5y - 14)} & = {1 \over 4} \\ {9y - 5(5y -14) \over y(5y - 14)} & = {1 \over 4} \\ {9y - 25y + 70 \over y(5y - 14)} & = {1 \over 4} \\ {-16y + 70 \over y(5y - 14)} & = {1 \over 4} \\ 4(-16y + 70) & = y(5y - 14) \\ -64y + 280 & = 5y^2 - 14y \\ 0 & = 5y^2 + 50y - 280 \\ 0 & = y^2 + 10y - 56 \\ 0 & = (y + 14)(y - 4) \end{align}
\begin{align} y + 14 & = 0 && \text{ or } & y - 4 & = 0 \\ y & = -14 &&& y & = 4 \\ \\ \text{Substitute } & \text{into (1),} &&& \text{Substitute } & \text{into (1),} \\ 3x & = 5(-14) - 14 &&& 3x & = 5(4) - 14 \\ 3x & = -84 &&& 3x & = 6 \\ x & = {-84 \over 3} &&& x & = {6 \over 3} \\ x & = -28 &&& x & = 2 \\ \\ \\ \therefore x & = -28, y = -14 &&& \therefore x & = 2, y = 4 \end{align}

(i)

\begin{align} y & = -0.25x^2 + 0.2x + 0.1 \\ \\ \text{Let } & x = 0, \\ y & = -0.25(0)^2 + 0.2(0)+ 0.1 \\ y & = 0.1 \\ \\ \text{Initially,} & \text{ the frog is 0.1 m above the ground} \end{align}

(ii)

\begin{align} y & = -0.25x^2 + 0.2x + 0.1 \phantom{00} \text{--- (1)} \\ \\ y & = 0.45x - 0.24 \phantom{00} \text{--- (2)} \\ \\ \text{Substitute } & \text{(2) into (1),} \\ 0.45x - 0.24 & = -0.25x^2 + 0.2x + 0.1 \\ 0 & = -0.25x^2 - 0.25x + 0.34 \\ \\ b^2 - 4ac & = (-0.25)^2 - 4(-0.25)(0.34) \\ & = 0.4025 > 0 \\ \\ \implies 0 = -0.25x^2 &- 0.25x + 0.34 \text{ has two distinct roots} \\ \\ \therefore \text{Frog } & \text{catches the dragonfly} \end{align}

(iii)

\begin{align} y & = -0.25x^2 + 0.2x + 0.1 \phantom{000000} [\text{Maximum curve } \cap] \\ y & = -0.25(x^2 - 0.8x) + 0.1 \\ y & = -0.25 \left[ x^2 - 0.8x + \left(0.8 \over 2\right)^2 - \left(0.8 \over 2\right)^2 \right] + 0.1 \\ y & = -0.25 ( x^2 - 0.8x + 0.4^2 - 0.4^2 ) + 0.1 \\ y & = -0.25 [ (x - 0.4)^2 - 0.16 ] + 0.1 \\ y & = -0.25(x - 0.4)^2 + 0.04 + 0.1 \\ y & = -0.25(x - 0.4)^2 + 0.14 \\ \\ \text{Turning } & \text{point: } (0.4, 0.14) \\ \\ \text{From (i), } & y \text{-intercept} = 0.1 \\ \\ \\ y & = 0.45x - 0.24 \\ \\ \text{Let } & x = 0, \\ y & = 0.45(0) - 0.24 \\ y & = -0.24 \\ \\ \text{Let } & x = 0.4, \\ y & = 0.45(0.4) - 0.24 \\ y & = -0.06 \end{align}

(i)

\begin{align} (6x + 17)(x - 2)(x^2 + 2) & = (6x^2 - 12x + 17x - 34)(x^2 + 2) \\ & = (6x^2 + 5x - 34)(x^2 + 2) \\ & = 6x^4 + 12x^2 + 5x^3 + 10x - 34x^2 - 68 \\ & = 6x^4 + 5x^3 - 22x^2 + 10x - 68 \phantom{00} (\text{Shown}) \end{align}

(ii)

\begin{align} 6x^4 + 5x^3 - 22x^2 + 10x & < 68 \\ 6x^4 + 5x^3 - 22x^2 + 10x - 68 & < 0 \\ (6x + 17)(x - 2)(x^2 + 2) & < 0 \\ \\ \text{Since } x^2 + 2 > 0 \text{ for all } & \text{value of } x, \\ (6x + 17)(x - 2) & < 0 \end{align}

$$ -{17 \over 6} < x < 2 $$