(a)(i)

\begin{align*} y & = 2x^2 - 4x + 5 \\ y & = 2(x^2 - 2x) + 5 \\ y & = 2 \left[ \left( x - {2 \over 2} \right)^2 -\left( 2 \over 2\right)^2 \right] + 5 \\ y & = 2 [ (x - 1)^2 - 1] + 5 \\ y & = 2(x - 1)^2 - 2 + 5 \\ y & = 2(x - 1)^2 + 3 \\ \\ \text{Mini} & \text{mum point is } (1, 3) \\ \\ \text{Let } & x = 0, \\ y & = 2(0 - 1)^2 + 3 \\ y & = 5 \\ \\ \text{Coor} & \text{dinates of } y\text{-intercept is } (0, 5) \end{align*}

(a)(ii)

\begin{align*} y & = 2x^2 - 4x + 5 \phantom{000} \text{--- (1)} \\ y & = 2x + c \phantom{000} \text{--- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ 2x^2 - 4x + 5 & = 2x + c \\ 2x^2 - 6x + 5 - c & = 0 \\ \\ b^2 - 4ac & = (-6)^2 - 4(2)(5 - c) \\ & = 36 - 8(5 - c) \\ & = 36 - 40 + 8c \\ & = 8c - 4 \\ \\ b^2 - 4ac & = 0 \phantom{000} [\text{Real and equal roots since line is tangent to curve}] \\ 8c - 4 & = 0 \\ 8c & = 4 \\ c & = {4 \over 8} \\ c & = {1 \over 2} \end{align*}

(b)

\begin{align*} y & = px - 3 \phantom{000} \text{--- (1)} \\ y & = 4x^2 + 5 \phantom{000} \text{--- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ px - 3 & = 4x^2 + 5 \\ 0 & = 4x^2 - px + 8 \\ \\ b^2 - 4ac & = (-p)^2 - 4(4)(8) \\ & = p^2 - 128 \\ \\ b^2 - 4ac & \ge 0 \phantom{000} [\text{Real and distinct/equal roots since line intersects curve}] \\ p^2 - 128 & \ge 0 \\ (p)^2 - (\sqrt{128})^2 & \ge 0 \\ (p + \sqrt{128})(p - \sqrt{128}) & \ge 0 \phantom{000} [a^2 - b^2 = (a + b)(a - b)] \end{align*}

$$ p < - \sqrt{128} \text{ or } p > \sqrt{128} $$ $$ \text{Largest negative integer} = -12 $$

(i)

\begin{align*} 3x + 4y & = 8 \\ 4y & = 8 - 3x \phantom{000} \text{--- (1)} \\ (4y + 3)^2 + (6x - 20)^2 & = 200 \phantom{000} \text{--- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ (8 - 3x + 3)^2 + (6x - 20)^2 & = 200 \\ (11 - 3x)^2 + (6x - 20)^2 & = 200 \\ (11)^2 - 2(11)(3x) + (3x)^2 + (6x)^2 - 2(6x)(20) + (20)^2 & = 200 \\ 121 - 66x + 9x^2 + 36x^2 - 240x + 400 & = 200 \\ 45x^2 - 306x + 521 & = 200 \\ 45x^2 - 306x + 321 & = 0 \\ \\ x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {-(-306) \pm \sqrt{(-306)^2 - 4(45)(321)} \over 2(45)} \\ & = {306 \pm \sqrt{35856} \over 90} \\ & = 5.5039 \text{ or } 1.296 \\ \\ \text{Substitute } & x = 5.5039 \text{ into } (1), \\ 4y & = 8 - 3(5.5039) \\ 4y & = -8.5117 \\ y & = -2.127925 \\ \\ \text{Substitute } & x = 1.296 \text{ into } (1), \\ 4y & = 8 - 3(1.296) \\ 4y & = 4.112 \\ y & = 1.028 \\ \\ \therefore x \approx 5.50, y \approx -2.13 & \text{ or } x \approx 1.30, y \approx 1.03 \end{align*}

(ii)

\begin{align*} 3x + 4y & = -10 \\ 4y & = -10 - 3x \phantom{000} \text{--- (1)} \\ (4y + 3)^2 + (6x - 20)^2 & = 200 \phantom{000} \text{--- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ (-10 - 3x + 3)^2 + (6x - 20)^2 & = 200 \\ (-7 - 3x)^2 + (6x - 20)^2 & = 200 \\ (-7)^2 - 2(-7)(3x) + (3x)^2 + (6x)^2 - 2(6x)(20) + (20)^2 & = 200 \\ 49 + 42x + 9x^2 + 36x^2 - 240x + 400 & = 200 \\ 45x^2 - 198x + 449 & = 200 \\ 45x^2 - 198x + 249 & = 0 \\ 15x^2 - 66x + 83 & = 0 \\ \\ b^2 - 4ac & = (-66)^2 - 4(15)(83) \\ & = -624 \\ \\ \text{Since } b^2 - 4ac < 0, \text{ there are no} & \text{ solutions for the simultaneous equations} \end{align*}

(iii)

The line can only cut the ellipse at most twice (graph on Desmos).

\begin{align*} \text{Let } x & \text{ denote the length of the lawn in m} \\ \\ \text{Width} & = (x - 2) \text{ m} \\ \text{Width} & > 6.4 \text{ m} \\ x - 2 & > 6.4 \\ x & > 8.4 \phantom{00000} [\text{Condition #1}] \\ \\ \text{Area} & < 63 \text{ m}^2 \\ x(x - 2) & < 63 \\ x^2 - 2x & < 63 \\ x^2 - 2x - 63 & < 0 \\ (x + 7)(x - 9) & < 0 \end{align*}

$$ -7 < x < 9 \phantom{00000} [\text{Condition #2}] $$
We need to satisfy condition #1 and condition #2
$$ \therefore 8.4 < x < 9 $$ $$ 8.4 \text{ m} < \text{Length} < 9 \text{ m} $$

\begin{align*} \text{Let } x \text{ denote the length of} & \text{ 'shorter' leg of the triangle in cm} \\ \\ \text{Length of 'longer' leg} & = (x + 7) \text{ cm} \\ \\ \text{Hypotenuse} & \ge 13 \text{ cm} \\ (\text{Hypotenuse})^2 & \ge 13^2 \\ (\text{Hypotenuse})^2 & \ge 169 \\ \\ [\text{Pythagoras theorem}] \phantom{0000} x^2 + (x + 7)^2 & \ge 169 \\ x^2 + (x)^2 + 2(x)(7) + (7)^2 & \ge 169 \\ x^2 + x^2 + 14x + 49 & \ge 169 \\ 2x^2 + 14x - 120 & \ge 0 \\ x^2 + 7x - 60 & \ge 0 \\ (x + 12)(x - 5) & \ge 0 \end{align*}

$$ x \le -12 \text{ or } x \ge 5 $$ $$ \text{Since length of shorter leg } > 0, \text{length of short leg is at least 5 cm} $$

(a)

\begin{align*} x^2 - 5x + 3 & > 5 - 4x \\ x^2 - x - 2 & > 0 \\ (x + 1)(x - 2) & > 0 \end{align*}

$$ x < -1 \text{ or } x > 2 $$

(b)

\begin{align*} 3x^2 - 6x + c & > 4 \\ 3x^2 - 6x + c - 4 & > 0 \phantom{00000} [\text{Expression is always positive}] \\ \\ b^2 - 4ac & = (-6)^2 - 4(3)(c - 4) \\ & = 36 - 12(c - 4) \\ & = 36 - 12c + 48 \\ & = 84 - 12c \\ \\ b^2 - 4ac & < 0 \phantom{00000} [\text{No real roots}] \\ 84 - 12c & < 0 \\ -12c & < -84 \\ 12c & > 84 \\ c & > {84 \over 12} \\ c & > 7 \end{align*}

(i)

\begin{align*} {1 \over 4} < & \phantom{.} x < 1 \\ \\ (4x - 1)(x - 1) & < 0 \\ 4x^2 - 4x - x + 1 & < 0 \\ 4x^2 - 5x + 1 & < 0 \\ -4x^2 + 5x - 1 & > 0 \\ \\ \text{Comparing coefficients, } & p = 4 \text{ and } q = 5 \end{align*}

(ii)

\begin{align*} y & = -4x^2 + 5x - 1 \\ \\ y & = 1 - 4x \\ \\ 1 - 4x & > -4x^2 + 5x - 1 \phantom{00000} [\text{Curve lies below line}] \\ 0 & > -4x^2 + 9x - 2 \\ 0 & < 4x^2 - 9x + 2 \\ 0 & < (4x - 1)(x - 2) \end{align*}

$$ x < {1 \over 4} \text{ or } x > 2 $$

\begin{align*} x^2 + xy & = 15 \phantom{000} \text{--- (1)} \\ \\ 3x + 5y & = 5 \\ 5y & = 5 - 3x \\ y & = 1 - {3 \over 5}x \phantom{000} \text{--- (2)} \\ \\ \text{Substitute } & \text{(2) into (1),} \\ x^2 + x\left(1 - {3 \over 5}x\right) & = 15 \\ x^2 + x - {3 \over 5}x^2 & = 15 \\ {2 \over 5}x^2 + x & = 15 \\ 2x^2 + 5x & = 75 \\ 2x^2 + 5x - 75 & = 0 \\ (2x + 15)(x - 5) & = 0 \\ \\ 2x + 15 = 0 \phantom{00-} & \text{ or } \phantom{0} x - 5 = 0 \\ 2x = -15 \phantom{0} & \phantom{0000000} x = 5 \\ x = -7.5 \phantom{.} & \\ \\ \text{Substitute } & x = -7.5 \text{ into (2),} \\ y & = 1 - {3 \over 5}(-7.5) \\ y & = 5.5 \\ \\ \text{Substitute } & x = 5 \text{ into (2),} \\ y & = 1 - {3 \over 5}(5) \\ y & = -2 \\ \\ \therefore \text{Coordinates of intersect} & \text{ion are } (-7.5, 5.5) \text{ and } (5, -2) \end{align*}

(a)

\begin{align} (x + 1)^2 & = h(x + 2) \\ x^2 + 2(x) + 1^2 & = hx + 2h \\ x^2 + 2x + 1 & = hx + 2h \\ x^2 + 2x - hx + 1 - 2h & = 0 \\ x^2 + (2 - h)x + (1 - 2h) & = 0 \\ \\ [a = 1, b & = 2 - h, c = 1 - 2h] \\ \\ b^2 - 4ac & \ge 0 \\ (2 - h)^2 - 4(1)(1 - 2h) & \ge 0 \\ 2^2 - 2(2)(h) + h^2 - 4(1 - 2h) & \ge 0 \\ 4 - 4h + h^2 - 4 + 8h & \ge 0 \\ h^2 + 4h & \ge 0 \\ h(h + 4) & \ge 0 \end{align}

$$ h \le -4 \text{ or } h \ge 0 $$ $$ \therefore h \text{ cannot lie between -4 and 0}. $$

(b)(i)

\begin{align} y & = 3x + k \phantom{000} \text{ --- (1)} \\ \\ y^2 & = 1 - x^2 \phantom{000} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ (3x + k)^2 & = 1 - x^2 \\ (3x)^2 + 2(3x)(k) + k^2 & = 1 - x^2 \\ 9x^2 + 6kx + k^2 & = 1 - x^2 \\ 9x^2 + x^2 + 6kx + k^2 - 1 & = 0 \\ 10x^2 + 6kx + (k^2 - 1) & = 0 \\ \\ [a = 10, b & = 6k, c = k^2 - 1] \\ \\ b^2 - 4ac & > 0 \\ (6k)^2 - 4(10)(k^2 - 1) & > 0 \\ 36k^2 - 40(k^2 - 1) & > 0 \\ 36k^2 - 40k^2 + 40 & > 0 \\ -4k^2 + 40 & > 0 \\ \\ 4k^2 - 40 & < 0 \\ k^2 - 10 & < 0 \\ k^2 - (\sqrt{10})^2 & < 0 \\ (k + \sqrt{10})(k - \sqrt{10}) & < 0 \end{align}

$$ -\sqrt{10} < k < \sqrt{10} $$

(b)(ii) For line to be a tangent to the curve, $b^2 - 4ac = 0$.

$$ \therefore k = \pm \sqrt{10} $$

(a)

\begin{align*} (1 + x)(6 - x) & \le -8 \\ 6 - x + 6x - x^2 & \le - 8 \\ -x^2 + 5x + 6 & \le -8 \\ -x^2 + 5x + 14 & \le 0 \\ x^2 - 5x - 14 & \ge 0 \\ (x - 7)(x + 2) & \ge 0 \end{align*}

$$ x \le -2 \text{ or } x \ge 7 $$

(b)

\begin{align*} 2x (x + 2) & < (x + 1)(x + 3) \\ 2x^2 + 4x & < x^2 + 3x + x + 3 \\ 2x^2 + 4x & < x^2 + 4x + 3 \\ x^2 - 3 & < 0 \\ x^2 - (\sqrt{3})^2 & < 0 \\ (x + \sqrt{3}) (x - \sqrt{3}) & < 0 \phantom{0000000} [a^2 - b^2 = (a + b)(a - b)] \end{align*}

$$ -\sqrt{3} < x < \sqrt{3} $$

(a)

\begin{align} y & = -x^2 + 2(k - 3)x - 25 \\ & = -x^2 + (2k - 6)x - 25 \\ \\ [a = -1, b & = 2k - 6, c = - 25] \\ \\ b^2 - 4ac & < 0 \\ (2k - 6)^2 - 4(-1)(-25) & < 0 \\ (2k)^2 - 2(2k)(6) + 6^2 - 100 & < 0 \\ 4k^2 - 24k + 36 - 100 & < 0 \\ 4k^2 - 24k - 64 & < 0 \\ k^2 - 6k - 16 & < 0 \\ (k + 2)(k - 8) & < 0 \end{align}

$$ -2 < k < 8 $$

(i)

\begin{align} \text{When } & y = 0, \\ 0 & = (x - 3)(x + 1) \\ \\ x - 3 = 0 \phantom{00}&\text{or}\phantom{00} x + 1 = 0 \\ x = 3 \phantom{00} &\phantom{or00+1} x = - 1 \\ \\ \implies \text{Coordinates of } & x\text{-intercepts are } (3, 0) \text{ and } (-1, 0) \end{align}

(ii)

\begin{align*} (x - 3)(x + 1) & = p \\ x^2 + x - 3x - 3 & = p \\ x^2 - 2x - 3 & = p \\ x^2 - 2x - 3 - p & = 0 \\ \\ b^2 - 4ac & = (-2)^2 - 4(1)(-3-p) \\ & = 4 - 4(-3-p) \\ & = 4 +12 + 4p \\ & = 16 + 4p \\ \\ b^2 - 4ac & = 0 \phantom{00000} [\text{Real and equal roots}] \\ 16 + 4p & = 0 \\ 4p & = -16 \\ p & = {-16 \over 4} \\ p & = -4 \\ \\ \text{When } & p = -4, \\ x^2 - 2x - 3 - (-4) & = 0 \\ x^2 - 2x + 1 & = 0 \\ (x)^2 - 2(x)(1) + (1)^2 & = 0 \\ (x - 1)^2 & = 0 \phantom{00000} [a^2 - 2ab + b^2 = (a - b)^2] \\ x - 1 & = \sqrt{0} \\ x - 1 & = 0 \\ x & = 1 \end{align*}

(i)

\begin{align*} y & = 5x^2 \phantom{000} \text{--- (1)} \\ y & = px - 5 \phantom{000} \text{--- (2)} \\ \\ \text{Substitute } & \text{(1) into (2)}, \\ 5x^2 & = px - 5 \\ 5x^2 - px + 5 & = 0 \\ \\ b^2 - 4ac & = (-p)^2 - 4(5)(5) \\ & = p^2 - 100 \\ \\ b^2 - 4ac & < 0 \phantom{00000} [\text{No real roots, since curve lies above line}] \\ p^2 - 100 & < 0 \\ p^2 - 10^2 & < 0 \\ (p + 10)(p - 10) & < 0 \phantom{00000} [a^2 - b^2 = (a + b)(a - b)] \end{align*}

$$ -10 < p < 10 $$

(ii) From part (i), the expression 5x² - px + 5 is always positive (since a = 5 > 0) and b² - 4ac < 0 (since curve lies above the line)

\begin{align*} -10 < & \phantom{.} p < 10 \\ -20 < & \phantom{.} 2p < 20 \\ 5 < 2p & + 25 < 45 \\ \\ \implies 2p + 25 & > 0 \\ \\ \text{For } -10 < \phantom{.} p < 10, & \phantom{.} 2p + 25 > 0 \text{ and } 5x^2 - px + 5 > 0 \\ \\ \therefore {2p + 25 \over 5x^2 - px + 5} & > 0 \end{align*}

(i)

\begin{align*} h & = -{1 \over 5}t^2 + 5t \\ & = -{1 \over 5} (t^2 - 25t) \\ & = -{1 \over 5} \left[ \left(t - {25 \over 2}\right)^2 - \left(25 \over 2\right)^2 \right] \\ & = -0.2 [ (t - 12.5)^2 - 156.25] \\ & = -0.2(t - 12.5)^2 + 31.25 \\ \\ \text{Max. height} & = 31.25 \text{ m} \\ \\ \therefore \text{Projectile } & \text{can reach a height of 30 m} \end{align*}

(ii)

$$ t = 12.5 \text{ s} $$

(iii) The path of the projectile resembles the shape of a maximum curve ($\cap$)

\begin{align*} h & = -0.2(t - 12.5)^2 + 31.25 \\ \\ \text{When } & h = 10, \\ 10 & = -0.2(t - 12.5)^2 + 31.25 \\ 0.2(t - 12.5)^2 & = 21.25 \\ (t- 12.5)^2 & = {21.25 \over 0.2} \\ (t - 12.5)^2 & = 106.25 \\ t - 12.5 & = \pm \sqrt{106.25} \\ t - 12.5 & = \pm 10.307 \\ t & = 10.307 + 12.5 \text{ or } -10.307 + 12.5 \\ t & = 22.807 \text{ or } 2.193 \\ t & \approx 22.8 \text{ or } 2.19 \\ \\ \therefore 2.19 \le & \phantom{.} t \le 22.8 \end{align*}

(a)(i)

\begin{align*} f(x) & = ax^2 + bx + 5 \\ & = a \left(x^2 + {b \over a}x \right) + 5 \\ & = a \left[ \left(x + {b \over 2a}\right)^2 - \left(b \over 2a\right)^2 \right] + 5 \\ & = a \left[ \left(x + {b \over 2a}\right)^2 - {b^2 \over 4a^2} \right] + 5 \\ & = a \left(x + {b \over 2a}\right)^2 - {b^2 \over 4a} + 5 \\ \\ {b \over 2a} & = -3 \\ b & = -6a \\ \\ -{b^2 \over 4a} + 5 & = -4 \\ -{b^2 \over 4a} & = - 9 \\ {b^2 \over 4a} & = 9 \\ {(-6a)^2 \over 4a} & = 9 \\ {36a^2 \over 4a} & = 9 \\ 9a & = 9 \\ a & = {9 \over 9} \\ a & = 1 \\ \\ b & = -6(1) \\ b & = -6 \end{align*}

(a)(ii)

Since $f(x)$ has a minimum value of $-4$, the graph of $y = f(x)$ will meet the $x$-axis at two distinct points

(b)

\begin{align*} y & = 2x^2 - 8x + 9 \\ \\ b^2 - 4ac & = (-8)^2 - 4(2)(9) \\ & = -8 \\ \\ \text{Since } a > 0 \text{ and } & b^2 - 4ac < 0 , \text{ the expression is always positive} \end{align*}

(a)

\begin{align*} x^2 - 2x + 2 - k & = 0 \\ \\ b^2 - 4ac & = (-2)^2 - 4(1)(2 - k) \\ & = 4 - 4(2 - k) \\ & = 4 - 8 + 4k \\ & = 4k - 4 \\ \\ \text{For } k & < 1, \\ 4k & < 4 \\ 4k - 4 & < 4 -4 \\ 4k - 4 & < 0 \\ \implies b^2 - 4ac & < 0 \\ \\ \text{Since } b^2 - 4ac < 0 \text{ for } k < 1, & \text{ the roots of the equation are not real} \end{align*}

(b)

\begin{align*} y + px & = 8 \\ y & = -px + 8 \phantom{000} \text{--- (1)} \\ \\ x^2 + 4y & = 20 \phantom{000} \text{--- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ x^2 + 4(-px + 8) & = 20 \\ x^2 - 4px + 32 & = 20 \\ x^2 - 4px + 12 & = 0 \\ \\ b^2 - 4ac & = (-4p)^2 - 4(1)(12) \\ & = 16p^2 - 48 \\ \\ b^2 - 4ac & \ge 0 \phantom{00000} [\text{Real and distinct/equal roots since line intersects curve}] \\ 16p^2 - 48 & \ge 0 \\ p^2 - 3 & \ge 0 \\ p^2 - (\sqrt{3})^2 & \ge 0 \\ (p + \sqrt{3})(p - \sqrt{3}) & \ge 0 \end{align*}

$$ p \le -\sqrt{3} \text{ or } p \ge \sqrt{3} $$