(i)

\begin{align*} d & = 0.002m^2 + 0.02m + 0.005 \\ \\ \text{When } & m = 200, \\ d & = 0.002(200)^2 + 0.02(200) + 0.005 \\ d & = 84.005 \\ \\ \text{Deflection of plastic rod} & = 84.005 \text{ mm} \end{align*}

(ii)

\begin{align} d & = 0.002m^2 + 0.02m + 0.005 \\ \\ 100 \text{ kg} & = (100 \times 1000) \text{ g} \\ & = 100,000 \text{ g} \\ \\ \text{When } & m = 100,000, \\ d & = 0.002(100,000)^2 + 0.02(100,000) + 0.005 \\ d & = 20 \phantom{.} 002 \phantom{.} 000.005 \\ \\ \text{Since deflection is} & \text{ too large, the model is not suitable for a load of 100 kg} \end{align}

(i) I think there's a typo in the question. it should be 'the highest point on the curve' (instead of lowest point) since this curve is a maximum curve (∩)

\begin{align*} y & = -2x^2 +4x + 3 \\ y & = -2(x^2 - 2x) + 3 \\ y & = -2 \left[ \left( x - {2 \over 2} \right)^2 - \left( 2 \over 2 \right)^2 \right] + 3 \\ y & = -2 [ (x - 1)^2 - 1 ] + 3 \\ y & = -2(x - 1)^2 + 2 + 3 \\ y & = -2(x - 1)^2 + 5 \\ \\ \implies & \text{Coordinates of maximum point is } (1, 5) \end{align*}

(ii)

\begin{align*} y & = -2(x - 1)^2 + 5 \\ \\ \text{Let } & y = 0, \\ 0 & = -2(x - 1)^2 + 5 \\ 2(x - 1)^2 & = 5 \\ (x - 1)^2 & = {5 \over 2} \\ (x - 1)^2 & = 2.5 \\ x - 1 & = \pm \sqrt{2.5} \\ x & = \sqrt{2.5} + 1 \text{ or } -\sqrt{2.5} + 1 \\ x & = 2.5811 \text{ or } -0.58113 \\ x & \approx 2.58 \text{ or } -0.581 \end{align*}

(i)

\begin{align*} y & = x^2 + 2x + (8 - p) \\ \\ [a & = 1, b = 2, c = 8 - p] \\ \\ \text{Since } & a > 0, \text{ curve is a minimum curve } (\cup) \\ \\ b^2 - 4ac & = (2)^2 - 4(1)(8 - p) \\ & = 4 - 4(8 - p) \\ & = 4 - 32 + 4p \\ & = 4p - 28 \\ \\ b^2 - 4ac & < 0 \phantom{00000} [\text{No real roots}] \\ 4p - 28 & < 0 \\ 4p & < 28 \\ p & < {28 \over 4} \\ p & < 7 \end{align*}

(ii)

\begin{align*} y & = x^2 + (2)x + 8 - (2) \\ y & = x^2 + 2x + 6 \\ y & = \left(x + {2 \over 2} \right)^2 - \left(2 \over 2\right)^2 + 6 \\ y & = (x + 1)^2 - 1 + 6 \\ y & = (x + 1)^2 + 5 \\ \\ \implies & \text{Coordinates of turning point is } (-1, 5) \\ \\ \text{Let } & x = 0, \\ y & = (0 + 1)^2 + 5 \\ y & = 6 \\ \implies & y\text{-intercept is } (0, 6) \end{align*}

(i)

\begin{align*} \text{Condition 1: } \phantom{0} a & > 0 \phantom{00000} [\text{Minimum curve } (\cup)] \\ \\ \text{Discriminant} & = (-4)^2 - 4(a)(-c) \\ & = 16 + 4ac \\ \\ \text{Discriminant} & < 0 \phantom{00000} [\text{No real roots}] \\ 16 + 4ac & < 0 \\ 4ac & < -16 \\ ac & < {-16 \over 4} \\ \text{Condition 2: } \phantom{0} ac & < - 4 \end{align*}

(ii) Note that there are many possible answers - ensure that the value of ac is smaller -4

\begin{align*} a & = 2, b = -3 \\ \\ a & = 10, b = -1 \\ \\ & \text{(and so on...)} \end{align*}

(i)

\begin{align*} h & = t^2 - 12t + 40 \\ h & = \left( t - {12 \over 2} \right)^2 - \left(12 \over 2\right)^2 + 40 \\ h & = (t - 6)^2 - 36 + 40 \\ h & = (t - 6)^2 + 4 \\ \\ \implies \text{Coordinates }& \text{of minimum point is } (6, 4) \\ \\ \text{Minimum height} & = 4 \text{ m} \\ \\ \therefore \text{Capsule } & \text{cannot reach height of 2m} \end{align*}

(ii) The coordinates of the minimum point is (6, 4). In the context of the question, the roller coaster reaches a height of 4 m when t = 6 seconds.

\begin{align*} \text{Time taken} & = 6 \text{ seconds} \end{align*}

(i)

\begin{align*} y & = ax^2 + bx + c \\ \\ \text{Using } & (0, 1), \\ 1 & = a(0)^2 + b(0) + c \\ 1 & = c \\ \\ y & = ax^2 + bx + 1 \\ \\ \text{Using } & (2, 7), \\ 7 & = a(2)^2 + b(2) + 1 \\ 7 & = 4a + 2b + 1 \\ 6 & = 4a + 2b \\ 3 & = 2a + b \\ \\ b & = 3 - 2a \phantom{000} \text{ --- (1)} \\ \\ \text{Using } & (4, 17), \\ 17 & = a(4)^2 + b(4) + 1 \\ 17 & = 16a + 4b + 1 \\ 16 & = 16a + 4b \phantom{000} \text{ --- (2)} \\ \\ \text{Substitute } & (1) \text{ into } (2), \\ 16 & = 16a + 4(3 - 2a) \\ 16 & = 16a + 12 - 8a \\ 4 & = 8a \\ {4 \over 8} & = a \\ {1 \over 2} & = a \\ \\ \text{Substitute } & a = {1 \over 2} \text{ into } (1), \\ b & = 3 - 2\left(1 \over 2\right) \\ b & = 2 \\ \\ \therefore y & = {1 \over 2}x^2 + 2x + 1 \end{align*}

(ii)

\begin{align*} y & = {1 \over 2}x^2 + 2x + 1 \\ \\ \text{Let } & y = 0, \\ 0 & = {1 \over 2}x^2 + 2x + 1 \\ 0 & = x^2 + 4x + 2 \\ 0 & = \left(x + {4 \over 2} \right)^2 - \left(4 \over 2\right)^2 + 2 \\ 0 & = (x + 2)^2 - 4 + 2 \\ 0 & = (x + 2)^2 - 2 \\ 2 & = (x + 2)^2 \\ \pm \sqrt{2} & = x + 2 \\ \\ x & = \sqrt{2} - 2 \text{ or } -\sqrt{2} - 2 \\ \\ \text{Since the graph} & \text{ of the function is a minimum curve } (\cup), \\ \\ \text{for } y < 0, & \phantom{.} -\sqrt{2} - 2 < x < \sqrt{2} - 2 \end{align*}

(i)

\begin{align*} y & = 2x^2 - 10x + m \\ y & = 2(x^2 - 5x) + m \\ y & = 2 \left[ \left( x - {5 \over 2} \right)^2 - \left(5 \over 2\right)^2 \right] + m \\ y & = 2 [ (x - 2.5)^2 - 6.25] + m \\ y & = 2(x - 2.5)^2 - 12.5 + m \\ y & = 2(x - 2.5)^2 + (m - 12.5) \end{align*}

(ii)

\begin{align*} y & = 2(x - 2.5)^2 + (m - 12.5) \\ \\ \implies & \text{Coordinates of minimum point is } (2.5, m - 12.5) \\ \\ m - 12.5 & = -{19 \over 2} \\ m & = -{19 \over 2} + 12.5 \\ m & = 3 \end{align*}

(iii)

\begin{align*} y & = 2x^2 - 10x + m \\ \\ b^2 - 4ac & = (-10)^2 - 4(2)(m) \\ & = 100 - 8m \\ \\ b^2 - 4ac & > 0 \phantom{00000} [\text{2 real and distinct roots}] \\ 100 - 8m & > 0 \\ -8m & > -100 \\ 8m & < 100 \\ m & < {100 \over 8} \\ m & < 12.5 \end{align*}

(i)

\begin{align*} y & = ax^2 - x + c \\ \\ \text{Using } & (-2, -3), \\ -3 & = a(-2)^2 - (-2) + c \\ -3 & = a(4) + 2 + c \\ -3 & = 4a + 2 + c \\ -5 & = 4a + c \\ \\ c & = -5 - 4a \phantom{000} \text{ --- (1)} \\ \\ \text{Using } & (1, 3), \\ 3 & = a(1)^2 - (1) + c \\ 3 & = a(1) - 1 + c \\ 3 & = a - 1 + c \\ 4 & = a + c \phantom{000} \text{ --- (2)} \\ \\ \text{Substitute} & \text{ (1) into (2),} \\ \\ 4 & = a + (-5 - 4a) \\ 4 & = a - 5 - 4a \\ 4 + 5 & = a - 4a \\ 9 & = -3a \\ {9 \over -3} & = a \\ -3 & = a \\ \\ \text{Substitute } & a = -3 \text{ into (1),} \\ \\ c & = -5 - 4(-3) \\ c & = 7 \end{align*}

(ii)

\begin{align*} y & = -3x^2 - x + 7 \\ y & = -3 \left( x^2 + {1 \over 3}x \right) + 7 \\ y & = -3 \left[ \left(x + {1 \over 6} \right)^2 - \left(1 \over 6\right)^2 \right] + 7 \\ y & = -3 \left[ \left(x + {1 \over 6} \right)^2 - {1 \over 36} \right] + 7 \\ y & = -3 \left( x + {1 \over 6} \right)^2 + {1 \over 12} + 7 \\ y & = -3 \left( x + {1 \over 6} \right)^2 + 7{1 \over 12} \\ \\ \implies & \text{Coordinates of maximum point is } \left(-{1 \over 6}, 7{1 \over 12} \right) \\ \\ \text{Since maximum} & \text{ value of } y = 7{1 \over 12}, \text{ the value of } y \text{ can never exceed } 8 \end{align*}

(i)

\begin{align*} h & = 4t - t^2 + 12 \\ \\ \text{When } & t =3, \\ h & = 4(3) - (3)^2 + 12 \\ h & = 15 \\ \\ \text{Height of ball} & = 15 \text{ m} \end{align*}

(ii)

\begin{align*} h & = 4t - t^2 + 12 \\ h & = -t^2 + 4t + 12 \\ h & = -(t^2 - 4t) + 12 \\ h & = - \left[ \left( t - {4 \over 2} \right)^2 - \left(4 \over 2\right)^2 \right] + 12 \\ h & = - [ (t - 2)^2 - 4] + 12 \\ h & = - (t - 2)^2 + 4 + 12 \\ h & = - (t - 2)^2 + 16 \\ \\ \implies & \text{Coordinates of maximum point is } (2, 16) \\ \\ \therefore \text{Maximum height} & = 16 \text{ m} \end{align*}

(iii)

\begin{align*} h & = - (t - 2)^2 + 16 \\ \\ \text{When } & h = 0, \\ 0 & = -(t - 2)^2 + 16 \\ (t - 2)^2 & = 16 \\ t - 2 & = \pm \sqrt{16} \\ t - 2 & = \pm 4 \\ t & = 4 + 2 \text{ or } - 4 + 2 \\ t & = 6 \text{ or } -2 \end{align*}

(iv)

The ball reaches the ground after 6 seconds

(i)

\begin{align*} 2x + 2y & = 16 \\ 2y & = 16 - 2x \\ y & = 8 - x \end{align*}

(ii)

\begin{align*} A & = xy \\ \\ \text{Since } & y = 8 - x, \\ A & = x(8 - x) \end{align*}

(iii)

\begin{align*} A & = x(8 - x) \\ A & = 8x - x^2 \\ A & = -x^2 + 8x \\ A & = -(x^2 - 8x) \\ A & = - \left[ \left( x - {8 \over 2} \right)^2 - \left(8 \over 2\right)^2 \right] \\ A & = - [ (x - 4)^2 - 16] \\ A & = - (x - 4)^2 + 16 \\ \\ \implies & \text{Coordinates of (maximum) turning point is } (4, 16) \\ \\ \therefore \text{Greatest area} & = 16 \text{ cm}^2 \text{ when } x = 4 \\ \\ \text{When } x = 4, \phantom{.} y & = 8 - (4) \\ y & = 4 \\ \\ \therefore \text{Greatest area} & \text{ occurs when } x = y \end{align*}

\begin{align*} \text{Line of symmetry, } x & = {0 + 180 \over 2} \\ x & = 90 \\ \\ \implies \text{Coordinates of mini} & \text{mum point is } (90, 30) \\ \\ y & = a(x - h)^2 + k \\ y & = a(x - 90)^2 + 30 \\ \\ \text{Using } & (0, 75), \\ 75 & = a(0 - 90)^2 + 30 \\ 75 & = a(8100) + 30 \\ 75 & = 8100a + 30 \\ 75 - 30 & = 8100a \\ 45 & = 8100a \\ {45 \over 8100} & = a \\ {1 \over 180} & = a \\ \\ \therefore y & = {1 \over 180}(x - 90)^2 + 30 \end{align*}

(a)

\begin{align*} \text{Condition 1: } \phantom{0} a & > 0 \phantom{00000} [\text{Minimum curve } (\cup)] \\ \\ b^2 - 4ac & = (4)^2 - 4(a)(6) \\ & = 16 - 24a \\ \\ b^2 - 4ac & < 0 \phantom{00000} [\text{No real roots}] \\ 16 - 24a & < 0 \\ -24a & < -16 \\ 24a & > 16 \\ a & > {16 \over 24} \\ \text{Condition 2: } \phantom{0} a & > {2 \over 3} \\ \\ \\ \therefore \text{Smallest integer value of } a & = 1 \end{align*}

(b)

\begin{align*} \text{Condition 1 met: } \phantom{0} a & = -6 \phantom{00000} [\text{Maximum curve } (\cap)] \\ \\ b^2 - 4ac & = (b)^2 - 4(-6)(-3) \\ & = b^2 - 72 \\ \\ b^2 - 4ac & < 0 \phantom{00000} [\text{No real roots}] \\ b^2 - 72 & < 0 \\ b^2 & < 72 \text{ (Shown)} \end{align*}