\begin{align} x + 2y & = 5 \\ x & = 5 - 2y \phantom{000} \text{ --- (1)} \\ \\ 2x + y & = 2xy \phantom{000} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ 2(5 - 2y) + y & = 2(5 - 2y)y \\ 10 - 4y + y & = 2y(5 - 2y) \\ 10 - 3y & = 10y - 4y^2 \\ 0 & = -4y^2 + 13y - 10 \\ 0 & = 4y^2 - 13y + 10 \\ 0 & = (4y - 5)(y - 2) \end{align} \begin{align} 4y - 5 & = 0 \phantom{00}&\text{or}\phantom{0000} y - 2 & = 0 \\ 4y & = 5 &\phantom{or0000-2} y & = 2 \\ y & = {5 \over 4} \\ \\ \text{Substitute } & \text{into (1),} & \text{Substitute } & \text{into (1),} \\ x & = 5 - 2\left(5 \over 4\right) & x & = 5 - 2(2) \\ & = {5 \over 2} & & = 1 \end{align}
$$ x = {5 \over 2}, y = {5 \over 4} \text{ or } x = 1, y = 2 $$

\begin{align} \text{Let } x \text{ denote the } & \text{length of the rectangle.} \\ \text{Let } y \text{ denote the } & \text{breadth of the rectangle.} \\ \\ 2x + 2y & = 36 \\ x + y & = 18 \\ y & = 18 - x \phantom{000} \text{ --- (1)} \\ \\ \text{By Py} & \text{thgoras theorem,} \\ x^2 + y^2 & = 164 \phantom{000} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ x^2 + (18 - x)^2 & = 164 \\ x^2 + 18^2 - 2(18)(x) + x^2 & = 164 \\ x^2 + 324 - 36x + x^2 & = 164 \\ 2x^2 - 36x + 324 & = 164 \\ 2x^2 - 36x + 160 & = 0 \\ x^2 - 18x + 80 & = 0 \\ (x - 8)(x - 10) & = 0 \end{align} \begin{align} x - 8 & = 0 \phantom{00} & \text{or }\phantom{00} x - 10 & = 0 \\ x & = 8 & x & = 10 \\ \\ \text{Substitute } & \text{into (1),} & \text{Substitute } & \text{into (1),}\\ y & = 18 - 8 & y & = 18 - 10 \\ & = 10 & & = 8 \\ \\ \therefore \text{Dimensions: } & \text{8 m by 10 m} \end{align}

\begin{align} 2x^2 + x - 5 & = 0 \\ \\ [a = 2, b & = 1, c = -5] \\ \\ \text{Sum of roots} & = -{b \over a} \\ \alpha + \beta & = -{1 \over 2} \\ \\ \text{Product of roots} & = {c \over a} \\ (\alpha)(\beta) & = {-5 \over 2} \\ \alpha \beta & = - {5 \over 2} \end{align}

(i)

\begin{align} \alpha^2 + \beta^2 & = (\alpha + \beta)^2 - 2\alpha \beta \\ & = \left(-{1 \over 2}\right)^2 - 2\left(-{5 \over 2}\right) \\ & = {1 \over 4} + 5 \\ & = 5{1 \over 4} \end{align}

(ii)

\begin{align} (\alpha - \beta)^2 & = \alpha^2 - 2\alpha\beta + \beta^2 \\ & = \alpha^2 + \beta^2 - 2 \alpha \beta \\ & = 5{1 \over 4} - 2 \left(-{5 \over 2}\right) \\ & = 5{1 \over 4} + 5 \\ & = 10{1 \over 4} \end{align}

(iii)

\begin{align} \text{Since } \alpha < \beta, & \phantom{.} \alpha - \beta < 0 \\ \\ (\alpha - \beta)^2 & = 10{1 \over 4} \\ & = {41 \over 4} \\ \\ \alpha - \beta & = \pm \sqrt{41 \over 4} \\ & = \pm {\sqrt{41} \over \sqrt{4}} \\ & = \pm {\sqrt{41} \over 2} \\ & = {\sqrt{41} \over 2} \text{ (Reject) or } -{\sqrt{41} \over 2} \end{align}

(iv)

\begin{align} \text{Sum of roots} & = 2\alpha + (-2\beta) \\ & = 2\alpha - 2\beta \\ & = 2(\alpha - \beta) \\ & = 2 \left(-{\sqrt{41} \over 2}\right) \\ & = -\sqrt{41} \\ \\ \text{Product of roots} & = (2\alpha)(-2\beta) \\ & = -4\alpha \beta \\ & = -4\left(-{5 \over 2}\right) \\ & = 10 \\ \\ \\ x^2 - \text{(Sum of roots)}x & + \text{(Product of roots)} = 0 \\ \\ x^2 - (-\sqrt{41})x & + 10 = 0 \\ x^2 + \sqrt{41}x & + 10 = 0 \end{align}

(a)

\begin{align} x^2 + 3 & = 2x + p \\ x^2 - 2x + 3 - p & = 0 \\ \\ [a = 1, b & = -2, c = 3 - p] \\ \\ b^2 - 4ac & \ge 0 \phantom{000} [\text{2 real distinct roots or 2 real equal roots}] \\ (-2)^2 - 4(1)(3 - p) & \ge 0 \\ 4 - 4(3 - p) & \ge 0 \\ 4 - 12 + 4p & \ge 0 \\ -8 + 4p & \ge 0 \\ 4p & \ge 8 \\ p & \ge {8 \over 4} \\ p & \ge 2 \end{align}

(b)

\begin{align} 2x^2 + 2\sqrt{3}x + p & = p(x^2 + 2) \\ 2x^2 + 2\sqrt{3}x + p & = px^2 + 2p \\ 2x^2 - px^2 + 2\sqrt{3}x + p - 2p & = 0 \\ (2 - p)x^2 + 2\sqrt{3}x - p & = 0 \\ \\ [a = 2 - p, b & = 2\sqrt{3}, c = - p] \\ \\ b^2 - 4ac & > 0 \\ (2\sqrt{3})^2 - 4(2 - p)(-p) & > 0 \\ 12 + 4p(2 - p) & > 0 \\ 12 + 8p - 4p^2 & > 0 \\ -4p^2 + 8p + 12 & > 0 \\ \\ 4p^2 - 8p - 12 & < 0 \\ p^2 - 2p - 3 & < 0 \\ (p + 1)(p - 3) & < 0 \end{align}

$$ -1 < p < 3, \phantom{.} p \ne 2 \text{ (Since coefficent of } x^2 \text{ cannot be 0)} $$

(a)

\begin{align} x^2 - 5x + 3 & > 5 - 4x \\ x^2 - 5x + 4x + 3 - 5 & > 0 \\ 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 \\ \\ [a = 3, b & = -6, c = c - 4] \\ \\ \text{For } 3x^2 - 6x + c - 4 & \text{ to be always positive,} \\ \\ b^2 - 4ac & < 0 \\ (-6)^2 - 4(3)(c - 4) & < 0 \\ 36 - 12(c - 4) & < 0 \\ 36 - 12c + 48 & < 0 \\ -12c + 84 & < 0 \\ -12c & < - 84 \\ \\ c & > {-84 \over -12} \\ c & > 7 \end{align}

\begin{align} x^2 + px + 8 & = p \\ x^2 + px + 8 - p & = 0 \\ \\ [a = 1, b & = p, c = 8 - p] \\ \\ b^2 - 4ac & = (p)^2 - 4(1)(8 - p) \\ & = p^2 - 4(8 - p) \\ & = p^2 - 32 + 4p \\ & = p^2 + 4p - 32 \\ & = (p + 8)(p - 4) \end{align}

(i)

\begin{align} b^2 - 4ac & = 0 \\ (p + 8)(p - 4) & = 0 \\ \\ p + 8 = 0 \phantom{00}&\text{or}\phantom{00} p - 4 = 0 \\ p = -8 \phantom{00}&\phantom{or00-4} p = 4 \end{align}

(ii)

\begin{align} b^2 - 4ac & > 0 \\ (p + 8)(p - 4) & > 0 \end{align}

$$ p < -8 \text{ or } p > 4 $$

(iii)

\begin{align} \text{Let } \alpha \text{ and } \beta \text{ denote the } & \text{roots of the equation.} \\ \\ \text{Since one root is positive while} & \text{ the other root is negative,} \\ \alpha \beta & < 0 \\ \text{Product of roots} & < 0 \\ {c \over a} & < 0 \\ {8 - p \over 1} & < 0 \\ 8 - p & < 0 \\ -p & < - 8 \\ p & > 8 \end{align}

(a)

\begin{align} 2x^2 - 4x + 5 & = 2(x^2 - 2x) + 5 \\ & = 2\left[ x^2 - 2x + \left(2 \over 2\right)^2 - \left(2 \over 2\right)^2 \right] + 5 \\ & = 2(x^2 - 2x + 1^2 - 1^2) + 5 \\ & = 2[ (x - 1)^2 - 1] + 5 \\ & = 2(x - 1)^2 - 2 + 5 \\ & = 2(x - 1)^2 + 3 \end{align}
\begin{align} y & = 2x^2 - 4x + 5 \\ & = 2(x - 1)^2 + 3 \\ \\ \text{Comparing with } y & = a(x - h)^2 + k, \phantom{0} a = 2, h = 1 \text{ and } k = 3,\\ \\ \implies & \text{Minimum curve } (\cup) \text{ since } a > 0 \\ \\ \implies & \text{Turning point is } (1, 3) \\ \\ \text{When } & x = 0, \\ y & = 2(0)^2 - 4(0) + 5 \\ & = 5 \\ \implies & y\text{-intercept is } 5 \end{align}

\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 - 4x - 2x + 5 - c & = 0 \\ 2x^2 - 6x + 5 - c & = 0 \\ \\ [a = 2, b & = -6, c = 5 - c] \\ \\ b^2 - 4ac & = 0 \\ (-6)^2 - 4(2)(5 - c) & = 0 \\ 36 - 8(5 - c) & = 0 \\ 36 - 40 + 8c & = 0 \\ -4 + 8c & = 0 \\ 8c & = 4 \\ c & = {4 \over 8} \\ & = {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 + 5 + 3 \\ 0 & = 4x^2 - px + 8 \\ \\ [a = 4, b & = -p, c = 8] \\ \\ b^2 - 4ac & \ge 0 \phantom{00000} [\text{1 or 2 points of intersection}] \\ (-p)^2 - 4(4)(8) & \ge 0 \\ p^2 - 128 & \ge 0 \\ p^2 - (\sqrt{128})^2 & \ge 0 \\ (p + \sqrt{128})(p - \sqrt{128}) & \ge 0 \end{align}

\begin{align} x \le - \sqrt{128} \phantom{00}&\text{or}\phantom{00} x \ge \sqrt{128} \\ x \le - 11.31 \phantom{00}&\phantom{or00} x \ge 11.31 \\ \\ \therefore \text{Largest negative integer} & = -12 \end{align}

(i)

\begin{align} \text{L.H.S} & = \alpha^2 - \alpha \beta + \beta^2 \\ & = (\alpha^2 + \beta^2) - \alpha \beta \\ & = (\alpha + \beta)^2 - 2\alpha \beta - \alpha \beta \\ & = (\alpha + \beta)^2 - 3\alpha \beta \\ & = \text{R.H.S} \end{align}

(ii)

\begin{align} x^2 - 4x & + 6 = 0 \\ \\ [a = 1, b & = -4, c = 6] \\ \\ \text{Sum of roots} & = -{b \over a} \\ \alpha + \beta & = -{-4 \over 1} \\ \alpha + \beta & = 4 \\ \\ \text{Product of roots} & = {c \over a} \\ (\alpha)(\beta) & = {6 \over 1} \\ \alpha \beta & = 6 \\ \\ \\ \alpha^3 + \beta^3 & = (\alpha + \beta)(\alpha^2 - \alpha\beta + \beta^2) \\ & = (\alpha + \beta)[(\alpha + \beta)^2 - 3\alpha \beta] \phantom{00000} [\text{From part (i)]} \\ & = (4)[(4)^2 - 3(6)] \\ & = -8 \end{align}

(iii)

\begin{align} \text{Sum of roots} & = \alpha^3 + \beta^3 \\ & = -8 \\ \\ \text{Product of roots} & = (\alpha^3)(\beta^3) \\ & = \alpha^3 \beta^3 \\ & = (\alpha \beta)^3 \\ & = (6)^3 \\ & = 216 \\ \\ \\ x^2 - \text{(Sum of roots)}x & + \text{(Product of roots)} = 0 \\ x^2 - (-8)x & + 216 = 0 \\ x^2 + 8x & + 216 = 0 \end{align}

\begin{align} x + 2y & = 10 \\ x & = 10 - 2y \phantom{000} \text{ --- (1)} \\ \\ 2y^2 - 7y + x & = 1 \phantom{000} \text{ --- (2)} \\ \\ \text{Substitute} & \text{ (1) into (2),} \\ 2y^2 - 7y + (10 - 2y) & = 1\\ 2y^2 - 7y + 10 - 2y & = 1\\ 2y^2 - 9y + 9 & = 0 \\ (2y - 3)(y - 3) & = 0 \end{align} \begin{align} 2y - 3 & = 0 & \text{or}\phantom{0000} y - 3 & = 0 \\ 2y & = 3 & y & = 3 \\ y & = {3 \over 2} \\ \\ \text{Substitute } & \text{into (1),} & \text{Substitute } & \text{into (1),} \\ x & = 10 - 2\left(3 \over 2\right) & x & = 10 - 2(3) \\ & = 7 & & = 4 \\ \\ \text{Coordinates are } & \left(7, {3 \over 2}\right) \text{ and } (4, 3). \end{align}

(a)

\begin{align} \text{Let } \alpha \text{ denote } & \text{the smaller root.} \\ \text{Let } \beta \text{ denote } & \text{the larger root.} \\ \\ \therefore \beta & = 2\alpha \\ \\ x^2 - 2(p + 1)x & + p^2 + p = 0 \\ x^2 + (-2p - 2)x & + (p^2 + p) = 0 \\ \\ [a = 1, b & = -2p - 2, c = p^2 + p] \\ \\ \text{Sum of roots} & = -{b \over a} \\ \alpha + 2\alpha & = -{-2p - 2 \over 1} \\ 3\alpha & = -(-2p - 2) \\ 3\alpha & = 2p + 2 \\ \alpha & = {2 \over 3}p + {2 \over 3} \phantom{000} \text{ --- (1)} \\ \\ \text{Product of roots} & = {c \over a} \\ (\alpha)(2\alpha) & = {p^2 + p \over 1} \\ 2\alpha^2 & = p^2 + p \\ \alpha^2 & = {1 \over 2}p^2 + {1 \over 2}p \phantom{000} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ \left({2 \over 3}p + {2 \over 3}\right)^2 & = {1 \over 2}p^2 + {1 \over 2}p \\ \left({2 \over 3}p\right)^2 + 2\left({2 \over 3}p\right)\left(2 \over 3\right) + \left(2 \over 3\right)^2 & = {1 \over 2}p^2 + {1 \over 2}p \\ {4 \over 9}p^2 + {8 \over 9}p + {4 \over 9} & = {1 \over 2}p^2 + {1 \over 2}p \\ 0 & = {1 \over 2}p^2 - {4 \over 9}p^2 + {1 \over 2}p - {8 \over 9}p - {4 \over 9} \\ 0 & = {1 \over 18}p^2 - {7 \over 18}p - {4 \over 9} \\ 0 & = p^2 - 7p - 8 \\ 0 & = (p + 1)(p - 8) \end{align} \begin{align} p + 1 & = 0 \phantom{0}& \text{or} \phantom{000} p - 8 & = 0 \\ p & = -1 & p & = 8 \\ \\ \text{Substitute } & \text{into (1),} & \text{Substitute } & \text{into (1),} \\ \alpha & = {2 \over 3}(-1) + {2 \over 3} & \alpha & = {2 \over 3}(8) + {2 \over 3} \\ & = 0 \implies \text{Reject } p = - 1 & & = 6 \\ \\ \therefore p & = 8 \end{align}

(b)

\begin{align} 3 - 2x - x^2 & \ge 0 \phantom{00000} [\text{Non-negative} \implies \text{Positive or zero}] \\ -x^2 - 2x + 3 & \ge 0 \\ \\ x^2 + 2x - 3 & \le 0 \\ (x + 3)(x - 1) & \le 0 \end{align}

$$ -3 \le x \le 1 $$

(i)

\begin{align} x^2 - \sqrt{12}x & + 1 = 0 \\ \\ \text{Sum of roots} & = -{-\sqrt{12} \over 1} \\ c + d & = \sqrt{12} \\ \\ \text{Product of roots} & = {1 \over 1} \\ (c)(d) & = 1 \\ cd & = 1 \\ \\ {1 \over c} + {1 \over d} & = {d \over cd} + {c \over cd} \\ & = {d + c \over cd} \\ & = {\sqrt{12} \over 1} \\ & = \sqrt{12} \\ & = \sqrt{4}\sqrt{3} \\ & = 2\sqrt{3} \text{ (Shown)} \end{align}

(ii)

\begin{align} \text{Sum of roots} & = {1 \over c} + {1 \over d} \\ & = 2\sqrt{3} \\ \\ \text{Product of roots} & = \left(1 \over c\right)\left(1 \over d\right) \\ & = {1 \over cd} \\ & = {1 \over 1} \\ & = 1 \\ \\ \\ x^2 - \text{(Sum of roots)}x & + \text{(Product of roots)} = 0 \\ x^2 - 2\sqrt{3} x & + 1 = 0 \end{align}

(a)

\begin{align} \text{Let } \alpha \text{ denote } & \text{the smaller root of the equation.} \\ \text{Let } \beta \text{ denote } & \text{the larger root of the equation.} \\ \\ \therefore \beta & = 3\alpha \\ \\ 6x^2 - 2kx & + k = 0 \\ \\ [a = 6, b & = -2k, c = k] \\ \\ \text{Sum of roots} & = -{b \over a} \\ \alpha + 3\alpha & = -{-2k \over 6} \\ 4\alpha & = {k \over 3} \\ \alpha & = {k \over 12} \phantom{000} \text{ --- (1)} \\ \\ \text{Product of roots} & = {c \over a} \\ (\alpha)(3\alpha) & = {k \over 6} \\ 3\alpha^2 & = {k \over 6} \\ \alpha^2 & = {k \over 18} \phantom{000} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ \left(k \over 12\right)^2 & = {k \over 18} \\ {k^2 \over 144} & = {k \over 18} \\ 18k^2 & = 144k \\ 18k^2 - 144k & = 0 \\ k^2 - 8k & = 0 \\ k(k - 8) & = 0 \\ \\ k = 0 \text{ (Reject)} \phantom{00}&\text{or}\phantom{00} k - 8 = 0 \\ & \phantom{or00-8} k = 8 \end{align}

(b)

\begin{align} x^2 & = 3x + 5 \\ x^2 - 3x - 5 & = 0 \\ \\ [a = 1, b & = -3, c = -5] \\ \\ \text{Sum of roots} & = -{b \over a} \\ \alpha + \beta & = -{-3 \over 1} \\ & = 3 \\ \\ \text{Product of roots} & = {c \over a} \\ (\alpha)(\beta) & = {-5 \over 1} \\ \alpha \beta & = -5 \\ \\ {1 \over \alpha^2} + {1 \over \beta^2} & = {\beta^2 \over \alpha^2 \beta^2} + {\alpha^2 \over \alpha^2 \beta^2} \\ & = {\beta^2 + \alpha^2 \over \alpha^2 \beta^2} \\ & = {(\alpha + \beta)^2 - 2\alpha \beta \over (\alpha \beta)^2} \\ & = {(3)^2 - 2(-5) \over (-5)^2} \\ & = {19 \over 25} \end{align}
\begin{align} x^2 & = 3x + 5 \\ \\ \text{When } x & = \alpha, \\ \alpha^2 & = 3\alpha + 5 \\ \\ \text{Squaring } & \text{both sides,} \\ (\alpha^2)^2 & = (3\alpha + 5)^2 \\ \alpha^4 & = (3\alpha)^2 + 2(3\alpha)(5) + 5^2 \\ \alpha^4 & = 9\alpha^2 + 30\alpha + 25 \\ \\ \text{Since } \alpha^2 & = 3\alpha + 5, \\ \alpha^4 & = 9(3\alpha + 5) + 30\alpha + 25 \\ & = 27\alpha + 45 + 30\alpha + 25 \\ & = 57\alpha + 70 \text{ (Shown)} \end{align}

(a) Since the curve is entirely below the x-axis, it does not touch the x-axis. Thus, there are no real roots.

\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}

$$ -2 < k < 8 $$

(b)(i)

\begin{align} y & = (x - 3)(x + 1) \\ \\ \text{When } & x = 0, \\ y & = (0 - 3)(0 + 1) \\ & = -3 \\ \implies & y\text{-intercept is } -3 \\ \\ \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 & x\text{-intercepts are } 3, - 1 \\ \\ \\ x\text{-coordinate of turning point} & = {3 + (-1) \over 2} \\ & = 1 \\ \\ \text{When } & x = 1, \\ y & = (1 - 3)(1 + 1) \\ & = -4 \\ \implies & \text{Turning point is } (1, -4). \\ \\ \text{Since the coefficient of } x^2 \text{ is positive,} & \text{ this is a minimum curve } (\cup) \end{align}

(b)(ii)

\begin{align} (x - 3)&(x + 1)= p \\ \\ \text{Since } & y = (x - 3)(x + 1), \\ y & = p \end{align}
For the equation to have equal real roots (i.e. only 1 real root), the line y = p must intersect the curve from part (i) only once. Thus, the line must pass through the turning point (1, -4).

$$ \therefore p = - 4 $$ $$ \text{Root, } x = 1 $$

(a) If x is real, the equation has at least one real root, i.e. b² - 4ac ≥ 0

\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)

\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}

$$ -\sqrt{10} < k < \sqrt{10} $$ $$ \text{For line to be tangent to the curve, } b^2 - 4ac = 0 $$ $$ \therefore k = \pm \sqrt{10} $$

(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 + 6 + 8 & \le 0 \\ -x^2 + 5x + 14 & \le 0 \\ x^2 - 5x - 14 & \ge 0 \\ (x + 2)(x - 7) & \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 - x^2 + 4x - 3x - x - 3 & < 0 \\ x^2 - 3 & < 0 \\ x^2 - (\sqrt{3})^2 & < 0 \\ (x + \sqrt{3})(x - \sqrt{3}) & < 0 \end{align}

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

(i)

\begin{align} \text{When } & r = 40, \\ 40 & = (t - 10)^2 - 2(t - 4)^2 \\ 40 & = t^2 - 2(t)(10) + 10^2 - 2[t^2 - 2(t)(4) + 4^2] \\ 40 & = t^2 - 20t + 100 - 2(t^2 - 8t + 16) \\ 40 & = t^2 - 20t + 100 - 2t^2 + 16t - 32 \\ 40 & = -t^2 - 4t + 68 \\ 0 & = -t^2 - 4t + 68 - 40 \\ 0 & = -t^2 -4t + 28 \\ 0 & = t^2 + 4t - 28 \text{ (Shown)} \end{align}

(ii)

\begin{align} 0 & = t^2 + 4t - 28 \\ \\ a = 1, b & = 4, c = - 28 \\ \\ b^2 - 4ac & = (4)^2 - 4(1)(-28) \\ & = 128 \\ \\ \text{Since } b^2 - 4ac > 0, & \text{ the equation has real roots.} \end{align}

(iii)

\begin{align} [a = 1, b & = 4, c = - 28] \\ \\ \text{Sum of roots} & = -{b \over a} \\ & = -{4 \over 1} \\ & = 4 \\ \\ \text{Product of roots} & = {c \over a} \\ & = {-28 \over 1} \\ & = -28 \\ \\ \text{Since product of roots is negative}, & \text{ one root is negative while the other is positive.} \\ \\ \text{Since time is either zero or positive,} & \text{ the rate is 40 mg/s only at one particular time.} \end{align}

(iv)

\begin{align} r & = (t - 10)^2 - 2(t - 4)^2 \\ & = t^2 - 2(t)(10) + 10^2 - 2[t^2 - 2(t)(4) + 4^2] \\ & = t^2 - 20t + 100 - 2(t^2 - 8t + 16) \\ & = t^2 - 20t + 100 - 2t^2 + 16t - 32 \\ & = -t^2 - 4t + 68 \\ \\ \text{When } & t = 0, \\ r & = -(0)^2 - 4(0) + 68 \\ & = 68 \\ \implies & r\text{-intercept is } 68 \\ \\ \text{When } & r = 0, \\ 0 & = -t^2 - 4t + 68 \\ 0 & = t^2 + 4t - 68 \\ \\ t & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {-4 \pm \sqrt{4^2 - 4(1)(-68)} \over 2(1)} \\ & = {-4 \pm \sqrt{288} \over 2} \\ & = {-4 \pm \sqrt{16}\sqrt{9}\sqrt{2} \over 2} \\ & = {-4 \pm 12\sqrt{2} \over 2} \\ & = -2 \pm 6\sqrt{2} \\ & = 6.485 \text{ or } -10.485 \\ \\ \implies & t \text{-intercepts are } 6.485, -10.485 \\ \\ t\text{-coordinate of turning point} & = {(-2 + 6\sqrt{2}) + (-2 - 6\sqrt{2}) \over 2} \\ & = {-2 + 6\sqrt{2} - 2 - 6\sqrt{2} \over 2} \\ & = {-4 \over 2} \\ & = -2 \\ \\ \text{When } t & = -2, \\ r & = -(-2)^2 - 4(-2) + 68 \\ & = 72 \\ \\ \implies & \text{Turning point is } (-2, 72). \\ \\ \text{Since of coefficient of } t^2 \text{ is negative,} & \phantom{.} \text{this is a maximum curve } (\cap) \end{align}