(a)

\begin{align} {dy \over dx} & = (2)(x) - 5(1) + 0 \\ & = 2x - 5 \\ \\ \text{Let } & {dy \over dx} = 0, \\ 0 & = 2x - 5 \\ 5 & = 2x \\ {5 \over 2} & = x \\ 2{1 \over 2} & = x \\ \\ \text{Substitute } & x = 2{1 \over 2} \text{ into eqn of curve}, \\ y & = \left(2{1 \over 2}\right)^2 - 5\left(2 {1 \over 2}\right) + 1 \\ & = -5{1 \over 4} \\ \\ \text{Coordinates} & \text{ is } \left( 2{1 \over 2}, -5{1 \over 4} \right) \end{align}

(b)

\begin{align} {dy \over dx} & = 0 - 6(1) + (2)(x) \\ & = -6 + 2x \\ \\ \text{Let } & {dy \over dx} = 0, \\ 0 & = -6 + 2x \\ 6 & = 2x \\ {6 \over 2} & = x \\ 3 & = x \\ \\ \text{Substitute } & x = 3 \text{ into eqn of curve}, \\ y & = 5 - 6(3) + (3)^2 \\ & = -4 \\ \\ \text{Coordinates} & \text{ is } (3, - 4) \end{align}

(i)

\begin{align} u & = 2x + 1 &&& v & = x - 1 \\ {du \over dx} & = 2 &&& {dv \over dx} & = 1 \end{align} \begin{align} {dy \over dx} & = {(x - 1)(2) - (2x + 1)(1) \over (x - 1)^2} \phantom{00000} [\text{Quotient rule}] \\ & = {2(x - 1) - (2x + 1) \over (x - 1)^2 } \\ & = {2x - 2 - 2x - 1 \over (x - 1)^2} \\ & = {-3 \over (x - 1)^2} \end{align}

(ii)

\begin{align} \text{For all real values of } & x \text{ where } x \ne 1, \\ (x - 1)^2 & > 0 \\ \implies {1 \over (x - 1)^2} & > 0 \\ {-3 \over (x - 1)^2} & < 0 \\ {dy \over dx} & < 0 \\ \\ \therefore \text{Since } {dy \over dx} \ne 0, \text{ the } & \text{curve has no stationary point} \end{align}

(i)

\begin{align} u & = 2x - 5 &&& v & = x^2 - 4 \\ {du \over dx} & = 2 &&& {dv \over dx} & = 2x \end{align} \begin{align} {dy \over dx} & = {(x^2 - 4)(2) - (2x - 5)(2x) \over (x^2 - 4)^2} \phantom{00000} [\text{Quotient rule}] \\ & = {2(x^2 - 4) - 2x(2x - 5) \over (x^2 - 4)^2} \\ & = {2x^2 - 8 - 4x^2 + 10x \over (x^2 -4)^2} \\ & = {-2x^2 + 10 x - 8 \over (x^2 - 4)^2} \\ & = {-2(x^2 - 5x + 4) \over (x^2 - 4)^2} \\ & = {-2(x - 4)(x - 1) \over (x^2 - 4)^2} \end{align}

(ii)

\begin{align} {dy \over dx} & = {-2(x - 4)(x - 1) \over (x^2 - 4)^2} \\ \\ \text{Let } & x = 0, \\ 0 & = {-2(x - 4)(x - 1) \over (x^2 - 4)^2} \\ 0 & = -2(x - 4)(x - 1) \\ 0 & = (x - 4)(x - 1) \\ \\ x - 4 = 0 & \phantom{00} \text{or} \phantom{000} x - 1 = 0 \\ x = 4 & \phantom{00} \phantom{or000-1} x = 1 \end{align}

(i)

\begin{align} {dy \over dx} & = 2(3)(x^2) - 9(2)(x) + 12(1) - 0 \\ & = 6x^2 - 18x + 12 \\ & = 6(x^2 - 3x + 2) \\ & = 6(x - 1)(x - 2) \phantom{00} \text{ (Shown)} \end{align}

(ii)

\begin{align} {dy \over dx} & = 6(x - 1)(x - 2) \\ \\ \text{Let } & {dy \over dx} = 0, \\ 0 & = 6(x - 1)(x - 2) \\ 0 & = (x - 1)(x - 2) \end{align} \begin{align} x - 1 & = 0 && \text{ or } & x - 2 & = 0 \\ x & = 1 &&& x & = 2 \\ \\ \text{Substitute } & \text{into eqn of curve,} &&& \text{Substitute } & \text{into eqn of curve,} \\ y & = 2(1)^3 - 9(1)^2 + 12(1) - 4 &&& y & = 2(2)^3 - 9(2)^2 + 12(2) - 4 \\ & = 1 &&& & = 0 \\ \\ \therefore & \phantom{.} (1, 1) &&& \therefore & \phantom{.} (2, 0) \end{align}

(iii)

\begin{align} {dy \over dx} & = 6x^2 - 18x + 12 \\ \\ {d^2y \over dx^2} & = 6(2)(x) - 18(1) + 0 \\ & = 12x - 18 \end{align} \begin{align} \text{When } & x = 1, &&& \text{When } & x = 2, \\ {d^2y \over dx^2} & = 12(1) - 18 &&& {d^2y \over dx^2} & = 12(2) - 18 \\ & = -6 &&& & = 6 \\ \\ \therefore (1,1) & \text{ is a maximum point} &&& \therefore (2, 0) & \text{ is a minimum point} \end{align}

(i)

\begin{align} u & = x - 5 &&& v & = \sqrt{7 + x} \\ & &&& & = (7 + x)^{1 \over 2} \\ {du \over dx} & = 1 &&& {dv \over dx} & = {1 \over 2} (7 + x)^{-{1 \over 2}} . (1) \phantom{00000} [\text{Chain rule}] \\ & &&& & = {1 \over 2} \left(1 \over \sqrt{7 + x}\right) \\ & &&& & = {1 \over 2\sqrt{7 + x}} \end{align} \begin{align} {dy \over dx} & = (x - 5) \left(1 \over 2\sqrt{7 + x}\right) + (\sqrt{7 + x})(1) \phantom{00000} [\text{Product rule}] \\ & = {x - 5 \over 2\sqrt{7 + x}} + \sqrt{7 + x} \\ & = {x - 5 \over 2\sqrt{7 + x}} + {\sqrt{7 + x} (2\sqrt{7 + x}) \over 2\sqrt{7 + x}} \\ & = {x - 5 \over 2\sqrt{7 + x}} + {2(7 + x) \over 2\sqrt{7 + x}} \\ & = {x - 5 + 14 + 2x \over 2\sqrt{7 + x}} \\ & = {3x + 9 \over 2\sqrt{7 + x}} \\ & = {3(x + 3) \over 2\sqrt{7 + x}} \end{align}

(ii)

\begin{align} \text{Let } & {dy \over dx} = 0, \\ 0 & = {3(x + 3) \over 2\sqrt{7 + x}} \\ 0 & = 3(x + 3) \\ 0 & = x + 3 \\ -3 & = x \\ \\ \text{Substitute } & x = -3 \text{ into eqn of curve,} \\ y & = [(-3) - 5] \sqrt{7 + (-3)} \\ & = -16 \\ \\ \text{Stationary} & \text{ point is } (-3, -16) \end{align}

(iii) Note: We are required to use the first derivative test to test for the nature of the stationary point

(a)

\begin{align} {dy \over dx} & = (3)(x^2) - 12(1) \\ & = 3x^2 - 12 \\ \\ {d^2 y \over dx^2} & = 3(2)x \\ & = 6x \\ \\ \text{Let } & {dy \over dx} = 0, \\ 0 & = 3x^2 - 12 \\ 12 & = 3x^2 \\ {12 \over 3} & = x^2 \\ 4 & = x^2 \\ \pm \sqrt{4} & = x \\ \pm 2 & = x \end{align} \begin{align} x & = 2 && \text{ or } & x & = -2 \\ \\ \text{Substitute } & \text{into eqn of curve,} &&& \text{Substitute } & \text{into eqn of curve,} \\ y & = (2)^3 - 12(2) &&& y & = (-2)^3 - 12(-2) \\ & = -16 &&& & = 16 \\ \\ \therefore & \phantom{.} (2, -16) &&& \therefore & \phantom{.} (-2, 16) \\ \\ \text{Substitute } & x = 2 \text{ into } {d^2 y \over dx^2}, &&& \text{Substitute } & x = -2 \text{ into } {d^2 y \over dx^2}, \\ {d^2y \over dx^2} & = 6(2) &&& {d^2y \over dx^2} & = 6(-2) \\ & = 12 &&& & = -12 \\ \\ \therefore (2, -16) & \text{ is a minimum point} &&& \therefore (-2, 16) & \text{ is a maximum point} \end{align}

(b)

\begin{align} {dy \over dx} & = 4x^3 - 8(2)(x) + 0 \\\ & = 4x^3 - 16x \\ \\ {d^2 y \over dx^2} & = 4(3)x^2 - 16(1) \\ & = 12x^2 - 16 \\ \\ \text{Let } & {dy \over dx} = 0, \\ 0 & = 4x^3 - 16x \\ 0 & = x^3 - 4x \\ 0 & = x(x^2 - 4) \\ 0 & = x(x - 2)(x + 2) \\ \\ x = 0 \phantom{00} \text{ or } \phantom{00} x - 2 & = 0 \phantom{00} \text{ or } \phantom{00} x + 2 = 0 \\ x & = 2 \phantom{00 or 00 + 20} x = - 2 \\ \\ \\ \text{Substitute } & x = 0 \text{ into eqn of curve,} \\ y & = (0)^4 - 8(0)^2 + 2 \\ & = 2 \\ \\ \therefore & \phantom{.} (0, 2) \\ \\ \text{Substitute } & x = 0 \text{ into } {d^2 y \over dx^2}, \\ {d^2y \over dx^2} & = 12(0)^2 - 16 \\ & = -16 \\ \\ \therefore (0,2) & \text{ is a maximum point} \\ \\ \\ \text{Substitute } & x = 2 \text{ into eqn of curve,} \\ y & = (2)^4 - 8(2)^2 + 2 \\ & = -14 \\ \\ \therefore & \phantom{.} (2, -14) \\ \\ \text{Substitute } & x = 2 \text{ into } {d^2 y \over dx^2}, \\ {d^2y \over dx^2} & = 12(2)^2 - 16 \\ & = 32 \\ \\ \therefore (2, -14) & \text{ is a minimum point} \\ \\ \\ \text{Substitute } & x = -2 \text{ into eqn of curve,} \\ y & = (-2)^4 - 8(-2)^2 + 2 \\ & = -14 \\ \\ \therefore & \phantom{.} (-2, -14) \\ \\ \text{Substitute } & x = -2 \text{ into } {d^2 y \over dx^2}, \\ {d^2y \over dx^2} & = 12(-2)^2 - 16 \\ & = 32 \\ \\ \therefore (-2, -14) & \text{ is a minimum point} \end{align}

(c)

\begin{align} y & = x(x - 6)^2 \\ & = x[x^2 - 2(x)(6) + 6^2] \\ & = x(x^2 - 12x + 36) \\ & = x^3 - 12x^2 + 36x \\ \\ {dy \over dx} & = 3x^2 - 12(2)x + 36(1) \\ & = 3x^2 - 24x + 36 \\ \\ {d^2 y \over dx^2} & = 3(2)x - 24(1) \\ & = 6x - 24 \\ \\ \text{Let } & {dy \over dx} = 0, \\ 0 & = 3x^2 - 24x + 36 \\ 0 & = x^2 - 8x + 12 \\ 0 & = (x - 2)(x - 6) \end{align} \begin{align} x - 2 & = 0 && \text{ or } & x - 6 & = 0 \\ x & = 2 &&& x & = 6 \\ \\ \text{Substitute } & x = 2 \text{ into the eqn of curve,} &&& \text{Substitute } & x = 6 \text{ into the eqn of curve,} \\ y & = (2)[(2) - 6]^2 &&& y & = (6)[(6) - 6)]^2 \\ & = 32 &&& & = 0 \\ \\ \therefore & \phantom{.} (2, 32) &&& \therefore & \phantom{.} (6, 0) \\ \\ \text{Substitute } & x = 2 \text{ into } {d^2 y \over dx^2}, &&& \text{Substitute } & x = 6 \text{ into } {d^2 y \over dx^2}, \\ {d^2 y \over dx^2} & = 6(2) - 24 &&& {d^2 y \over dx^2} & = 6(6) - 24 \\ & = -12 &&& & = 12 \\ \\ \therefore (2, 32) & \text{ is a maximum point} &&& \therefore (6, 0) & \text{ is a minimum point} \end{align}

(a)

\begin{align} y & = 2x + {18 \over x} \\ & = 2x + 18 x^{-1} \\ \\ {dy \over dx} & = 2(1) + 18(-1)x^{-2} \\ & = 2 - 18 x^{-2} \\ & = 2 - {18 \over x^2} \\ \\ {d^2 y \over dx^2} & = - 18 (-2) x^{-3} \\ & = 36 x^{-3} \\ & = {36 \over x^3} \\ \\ \text{Let } & {dy \over dx} = 0, \\ 0 & = 2 - {18 \over x^2} \\ {18 \over x^2} & = 2 \\ 18 & = 2x^2 \\ {18 \over 2} & = x^2 \\ 9 & = x^2 \\ \pm \sqrt{9} & = x \\ \pm 3 & = x \end{align} \begin{align} x & = 3 && \text{ or } & x & = -3 \\ \\ \text{Substitute } & x = 3 \text{ into eqn of curve,} &&& \text{Substitute } & x = -3 \text{ into eqn of curve,} \\ y & = 2(3) + {18 \over 3} &&& y & = 2(-3) + {18 \over -3} \\ & = 12 &&& & = -12 \\ \\ \therefore & \phantom{.} (3, 12) &&& \therefore & \phantom{.} (-3, -12) \\ \\ \text{Substitute } & x = 3 \text{ into } {d^2 y \over dx^2}, &&& \text{Substitute } & x = -3 \text{ into } {d^2 y \over dx^2}, \\ {d^2y \over dx^2} & = {36 \over (3)^3} &&& {d^2y \over dx^2} & = {36 \over (-3)^3} \\ & = {4 \over 3} &&& & = -{4 \over 3} \\ \\ \therefore (3, 12) & \text{ is a minimum point} &&& \therefore (-3, -12) & \text{ is a maximum point} \end{align}

(b)

\begin{align} y & = x^2 + {16 \over x} \\ & = x^2 + 16x^{-1} \\ \\ {dy \over dx} & = 2x + 16(-1) x^{-2} \\ & = 2x - 16 x^{-2} \\ & = 2x - {16 \over x^2} \\ \\ {d^2 y \over dx^2} & = 2(1) - 16(-2) x^{-3} \\ & = 2 + 32x^{-3} \\ & = 2 + {32 \over x^3} \\ \\ \text{Let } & {dy \over dx} = 0, \\ 0 & = 2x - {16 \over x^2} \\ {16 \over x^2} & = 2x \\ 16 & = 2x^3 \\ {16 \over 2} & = x^3 \\ 8 & = x^3 \\ \sqrt[3]{8} & = x \\ 2 & = x \\ \\ \text{Substitute } & x = 2 \text{ into eqn of curve,} \\ y & = (2)^2 + {16 \over (2)} \\ & = 12 \\ \\ \therefore & \phantom{.} (2, 12) \\ \\ \text{Substitute } & x = 2 \text{ into } {d^2 y \over dx^2}, \\ {d^2y \over dx^2} & = 2 + {32 \over (2)^3} \\ & = 6 \\ \\ \therefore (2, 12) & \text{ is a minimum point} \end{align}

(c)

\begin{align} y & = {4x^2 + 9 \over x} \\ & = {4x^2 \over x} + {9 \over x} \\ & = 4x + 9x^{-1} \\ \\ {dy \over dx} & = 4(1) + 9(-1) x^{-2} \\ & = 4 - 9 x^{-2} \\ & = 4 - {9 \over x^2} \\ \\ {d^2 y \over dx^2} & = - 9(-2) x^{-3} \\ & = 18x^{-3} \\ & = {18 \over x^3} \\ \\ \text{Let } & {dy \over dx} = 0, \\ 0 & = 4 - {9 \over x^2} \\ {9 \over x^2} & = 4 \\ 9 & = 4x^2 \\ {9 \over 4} & = x^2 \\ \pm \sqrt{9 \over 4} & = x \\ \pm {3 \over 2} & = x \end{align} \begin{align} x & = {3 \over 2} &&& x & = -{3 \over 2} \\ \\ \text{Substitute } & x = {3 \over 2} \text{ into eqn of curve,} &&& \text{Substitute } & x = -{3 \over 2} \text{ into eqn of curve,} \\ y & = {4\left({3 \over 2}\right)^2 + 9 \over ({3 \over 2})} &&& y & = {4(-{3 \over 2})^2 + 9 \over (-{3 \over 2})} \\ & = 12 &&& & = -12 \\ \\ \therefore & \phantom{.} \left( {3 \over 2}, 12 \right) &&& \therefore & \phantom{.} \left(-{3 \over 2}, -12 \right) \\ \\ \text{Substitute } & x = {3 \over 2} \text{ into } {d^2 y \over dx^2}, &&& \text{Substitute } & x = -{3 \over 2} \text{ into } {d^2 y \over dx^2}, \\ {d^2y \over dx^2} & = {18 \over ({3 \over 2})^3} &&& {d^2y \over dx^2} & = {18 \over (-{3 \over 2})^3} \\ & = {16 \over 3} &&& & = -{16 \over 3} \\ \\ \therefore \left({3 \over 2}, 12 \right) & \text{ is a minimum point} &&& \therefore \left(-{3 \over 2}, -12 \right) & \text{ is a maximum point} \end{align}

(d)

(i)

\begin{align} y & = 8x + {1 \over 2x^2} \\ & = 8x + {1 \over 2} \left(1 \over x^2\right) \\ & = 8x + {1 \over 2} x^{-2} \\ \\ {dy \over dx} & = 8(1) + {1 \over 2}(-2)x^{-3} \\ & = 8 - x^{-3} \\ & = 8 - {1 \over x^3} \\ \\ \text{Let } & {dy \over dx} = 0, \\ 0 & = 8 - {1 \over x^3} \\ {1 \over x^3} & = 8 \\ {1 \over 8} & = x^3 \\ \sqrt[3]{1 \over 8} & = x \\ {1 \over 2} & = x \\ \\ \text{Substitute } & x = {1 \over 2} \text{ into eqn of curve,} \\ y & = 8\left(1 \over 2\right) + {1 \over 2({1 \over 2})^2} \\ & = 6 \\ \\ \text{Turning} & \text{ point is } \left( {1 \over 2}, 6 \right) \end{align}

(ii)

\begin{align} {dy \over dx} & = 8(1) + {1 \over 2}(-2)x^{-3} \\ & = 8 - x^{-3} \\ \\ {d^2 y \over dx^2} & = - (-3) x^{-4} \\ & = 3 x^{-4} \\ & = {3 \over x^4} \\ \\ \text{When } & x = {1 \over 2}, \\ {d^2y \over dx^2} & = {3 \over ({1 \over 2})^4} \\ & = 48 \\ \\ \therefore \left({1 \over 2}, 6\right) & \text{ is a minimum point} \end{align}

(i)

\begin{align} {dy \over dx} & = (3)x^2 - 6(2)x + 0 \\ & = 3x^2 - 12x \end{align}

(ii)

\begin{align} \text{Let } & {dy \over dx} = 0, \\ 0 & = 3x^2 - 12x \\ 0 & = x^2 - 4x \\ 0 & = (x)(x - 4) \\ \\ x = 0 \phantom{00} & \text{or} \phantom{000} x - 4 = 0 \\ & \phantom{or000-4} x = 4 \end{align}

(iii) We need the consider the value of ${dy \over dx}$ between $x = 0$ and $x = 4$

\begin{align} {dy \over dx} & = 3x^2 - 12x \\ & = 3x(x - 4) \\ \\ \text{For } 0 < x < 4, & \phantom{.} 3x > 0 \text{ and } x - 4 < 0 \\ \\ \implies 3x(x - 4) & < 0 \\ \implies {dy \over dx} & < 0 \\ \\ \therefore \text{Gradient of the curve} & \text{ between the stationary points is always negative} \end{align}

(i)

\begin{align} u & = 2x - 1 &&& v & = x^2 + 2 \\ {du \over dx} & = 2 &&& {dv \over dx} & = 2x \end{align} \begin{align} {dy \over dx} & = {(x^2 + 2) (2) - (2x - 1)(2x) \over (x^2 + 2)^2} \phantom{00000} [\text{Quotient rule}] \\ & = {2(x^2 + 2) - 2x(2x - 1) \over (x^2 + 2)^2} \\ & = {2x^2 + 4 - 4x^2 + 2x \over (x^2 + 2)^2} \\ & = {-2x^2 + 2x + 4 \over (x^2 + 2)^2} \\ & = {-2(x^2 - x - 2) \over (x^2 + 2)^2} \\ & = {-2(x - 2)(x + 1) \over (x^2 + 2)^2} \end{align}

(ii)

\begin{align} \text{Let } & {dy \over dx} = 0, \\ 0 & = {-2(x - 2)(x + 1) \over (x^2 + 2)^2} \\ 0 & = -2(x - 2)(x + 1) \\ 0 & = (x - 2)(x + 1) \\ \\ x - 2 = 0 \phantom{00} & \text{or} \phantom{000} x + 1 = 0 \\ x = 2 \phantom{00} & \phantom{or000+1} x = - 1 \end{align}

(iii) We need the consider the value of ${dy \over dx}$ between $x = -1$ and $x = 2$

\begin{align} {dy \over dx} & = {-2(x - 2)(x + 1) \over (x^2 + 2)^2} \\ \\ \text{For } -1 < x < 2, & \phantom{.} (x - 2) < 0 \text{ and } (x + 1) > 0 \\ \text{For } -1 < x < 2, & \phantom{.} (x^2 + 2)^2 > 0 \\ \\ \implies {-2(x - 2)(x + 1) \over (x^2 + 2)^2} & > 0 \\ \implies {dy \over dx} & > 0 \\ \\ \therefore \text{Function is increas} & \text{ing between the stationary points} \end{align}

(i)

\begin{align} {dy \over dx} & = 3(4)x^3 + k(1) \\ & = 12x^3 + k \\ \\ \text{Let } & {dy \over dx} = 0, \\ 0 & = 12x^3 + k \\ -k & = 12x^3 \\ -{k \over 12} & = x^3 \\ \sqrt[3]{-{k \over 12}} & = x \\ \\ \therefore \text{Curve } & \text{only has 1 stationary point} \end{align}

(ii)

\begin{align} \text{When } & k = 12, \\ x & = -\sqrt[3]{(12) \over 12} \\ & = -\sqrt[3]{1} \\ & = -1 \\ \\ \text{Substitute } x = -1 & \text{ and } k = 12 \text{ into eqn of curve,} \\ y & = 3(-1)^4 + (12)(-1) \\ & = -9 \\ \\ \text{Turning} & \text{ point is } (-1, -9) \\ \\ {dy \over dx} & = 12x^3 + k \\ & = 12x^3 + 12 \\ \\ {d^2 y \over dx^2} & = 12(3)x^2 \\ & = 36x^2 \\ \\ \text{When } & x = -1, \\ {d^2y \over dx^2} & = 36(-1)^2 \\ & = 36 > 0 \\ \\ \therefore (-1, -9) & \text{ is a minimum point} \end{align}

\begin{align} y & = ax + {b \over x^2} \\ \\ \text{Using } & (3, 5), \\ (5) & = a(3) + {b \over (3)^2} \\ 5 & = 3a + {b \over 9} \\ 45 & = 27a + b \\ 45 - 27a & = b \\ \\ b & = 45 - 27a \phantom{0} \text{ --- (1)} \\ \\ \\ y & = ax + {b \over x^2} \\ & = ax + bx^{-2} \\ \\ {dy \over dx} & = a(1) + b(-2)x^{-3} \\ & = a - 2b x^{-3} \\ & = a - {2b \over x^3} \\ \\ \text{When } & {dy \over dx} = 0 \text{ and } x = 3, \phantom{00000} [(3, 5) \text{ is a stationary pt}] \\ 0 & = a - {2b \over (3)^3} \\ 0 & = a - {2b \over 27} \\ {2b \over 27} & = a \\ \\ a & = {2b \over 27} \phantom{0} \text{ --- (2)} \\ \\ \text{Substitute} & \text{ (2) into (1),} \\ b & = 45 - 27\left( 2b \over 27 \right) \\ b & = 45 - 2b \\ b + 2b & = 45 \\ 3b & = 45 \\ b & = {45 \over 3} \\ & = 15 \\ \\ \text{Substitute } & b = 15 \text{ into (2),} \\ a & = {2(15) \over 27} \\ & = {30 \over 27} \\ & = {10 \over 9} \\ \\ \therefore a & = {10 \over 9}, b = 15 \end{align}

\begin{align} \text{Curve 1: } y & = 2x^2 - 4x + 5 \\ \\ {dy \over dx} & = 2(2)(x) - 4(1) + 0 \\ & = 4x - 4 \\ \\ \text{Let } & {dy \over dx} = 0, \\ 0 & = 4x - 4 \\ 4 & = 4x \\ {4 \over 4} & = x \\ 1 & = x \\ \\ \text{Substitute } & x = 1 \text{ into eqn of curve,} \\ y & = 2(1)^2 - 4(1) + 5 \\ & = 3 \\ \\ \text{Common} & \text{ turning point is } (1, 3) \\ \\ \\ \text{Curve 2: } y & = x^3 + ax^2 + x + b \\ \\ \text{Using } & (1, 3), \\ (3) & = (1)^3 + a(1)^2 + (1) + b \\ 3 & = 1 + a + 1 + b \\ 1 & = a + b \\ 1 - a & = b \\ \\ b & = 1 - a \phantom{0} \text{ --- (1)} \\ \\ y & = x^3 + ax^2 + x + b \\ \\ {dy \over dx} & = (3)(x^2) + a(2)(x) + (1) + 0 \\ & = 3x^2 + 2ax + 1 \\ \\ \text{When } & {dy \over dx} = 0 \text{ and } x = 1, \\ 0 & = 3(1)^2 + 2a(1) + 1 \\ 0 & = 3 + 2a + 1 \\ 0 & = 4 + 2a \\ -4 & = 2a \\ {-4 \over 2} & = a\\ -2 & = a \\ \\ \text{Substitute } & a = -2 \text{ into (1),} \\ b & = 1 - (-2) \\ & = 3 \\ \\ \therefore a & = -2, b = 3 \end{align}

(i)

\begin{align} y & = 2x^3 + ax^2 + b \\ \\ \text{Using } & (-3, 19), \\ (19) & = 2(-3)^3 + a(-3)^2 + b \\ 19 & = -54 + 9a + b \\ 73 & = 9a + b \\ 73 - 9a & = b \\ \\ b & = 73 - 9a \phantom{0} \text{ --- (1)} \\ \\ y & = 2x^3 + ax^2 + b \\ \\ {dy \over dx} & = 2(3)(x^2) + a(2)(x) + 0 \\ & = 6x^2 + 2ax \\ \\ \text{When } & {dy \over dx} = 0 \text{ and } x = -3, \phantom{00000} [(-3, 19) \text{ is a stationary pt}] \\ 0 & = 6(-3)^2 + 2a(-3) \\ 0 & = 54 - 6a \\ 6a & = 54 \\ a & = {54 \over 6} \\ & = 9 \\ \\ \text{Substitute } & a = 9 \text{ into (1),} \\ b & = 73 - 9(9) \\ & = -8 \\ \\ \therefore a & = 9, b = -8 \end{align}

(ii)

\begin{align} {dy \over dx} & = 6x^2 + 2ax \\ & = 6x^2 + 2(9)x \\ & = 6x^2 + 18x \\ \\ {d^2 y \over dx^2} & = 6(2)(x) + 18(1) \\ & = 12x + 18 \\ \\ \text{When } & x = -3, \\ {d^2y \over dx^2} & = 12(-3) + 18 \\ & = -18 \\ \\ \therefore (-3, 19) & \text{ is a maximum point} \end{align}

(i)

\begin{align} y & = ax + {b \over 2x - 1} \\ \\ \text{Using } & A(2, 7), \\ (7) & = a(2) + {b \over 2(2) - 1} \\ 7 & = 2a + {b \over 3} \\ 21 & = 6a + b \\ 21 - 6a & = b \\ \\ b & = 21 - 6a \phantom{0} \text{ --- (1)} \\ \\ y & = ax + {b \over 2x - 1} \\ & = ax + b(2x - 1)^{-1} \\ \\ {dy \over dx} & = a(1) + b(-1)(2x - 1)^{-2}.(2) \\ & = a - 2b(2x - 1)^{-2} \\ & = a - 2b \left[ {1 \over (2x - 1)^2} \right] \\ & = a - {2b \over (2x - 1)^2} \\ \\ \text{When } & {dy \over dx} = 0 \text{ and } x = 2, \phantom{00000} [A(2, 7) \text{ is a stationary point}] \\ 0 & = a - {2b \over (2(2) - 1)^2} \\ 0 & = a - {2b \over 9} \\ {2b \over 9} & = a \\ \\ a & = {2b \over 9} \phantom{0} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(2) into (1),} \\ b & = 21 - 6\left( 2b \over 9 \right) \\ b & = 21 - {4b \over 3} \\ 3b & = 63 - 4b \\ 7b & = 63 \\ b & = {63 \over 7} \\ & = 9 \\ \\ \text{Substitute } & b = 9 \text{ into (2),} \\ a & = {2(9) \over 9} \\ & = 2 \\ \\ \therefore a & = 2, b = 9 \end{align}

(ii)

\begin{align} {dy \over dx} & = a - {2b \over (2x - 1)^2} \\ & = 2 - {18 \over (2x - 1)^2} \\ \\ \text{Let } & {dy \over dx} = 0, \\ 0 & = 2 - {18 \over (2x - 1)^2} \\ {18 \over (2x - 1)^2} & = 2 \\ 18 & = 2(2x - 1)^2 \\ 9 & = (2x - 1)^2 \\ \pm \sqrt{9} & = 2x - 1 \\ \pm 3 & = 2x - 1 \end{align} \begin{align} 2x - 1 & = 3 && \text{ or } & 2x - 1 & = -3 \\ 2x & = 4 &&& 2x & = -2 \\ x & = {4 \over 2} &&& x & = {-2 \over 2} \\ & = 2 \phantom{000} [\text{Point } A] &&& & = -1 \\ \\ & &&& \text{Substitute } & \text{into eqn of curve,} \\ & &&& y & = a(-1) + {b \over 2(-1) - 1} \\ & &&& & = 2(-1) + {9 \over 2(-1) - 1} \\ & &&& & = -5 \\ \\ & &&& \therefore & \phantom{.} (-1, -5) \end{align}

(iii)

\begin{align} {dy \over dx} & = 2 - {18 \over (2x - 1)^2} \\ & = 2 - 18(2x - 1)^{-2} \\ \\ {d^2 y \over dx^2} & = -18(-2)(2x - 1)^{-3} . (2) \\ & = 72(2x - 1)^{-3} \\ & = {72 \over (2x - 1)^3} \\ \\ \text{When } & x = 2, \\ {d^2 y \over dx^2} & = {72 \over [2(2) - 1]^3} \\ & = {8 \over 3} \\ \\ \therefore A(2, 7) & \text{ is a minimum point} \end{align}

(i)

\begin{align} u & = x^2 &&& v & = (x - k)^2 \\ {du \over dx} & = 2x &&& {dv \over dx} & = 2(x - k). (1) \\ & &&& & = 2(x - k) \end{align} \begin{align} {dy \over dx} & = (x^2)[2(x - k)] + (x - k)^2 (2x) \phantom{00000} [\text{Product rule}] \\ & = 2x^2 (x - k) + 2x(x - k)^2 \\ & = 2x(x - k) [ x + (x - k) ] \\ & = 2x (x - k) (x + x - k) \\ & = 2x (x - k) (2x - k) \end{align}

(ii)

\begin{align} y & = x^2 (x - k)^2 \\ \\ \text{Using } & (1, 1), \\ 1 & = (1)^2 (1 - k)^2 \\ 1 & = (1) [1^2 - 2(1)(k) + k^2] \\ 1 & = (1 - 2k + k^2) \\ 1 & = 1 - 2k + k^2 \\ 0 & = k^2 - 2k \\ 0 & = k(k - 2) \\ \\ k & = 0 \text{ (Reject, since } k \ne 0) \phantom{00} \text{ or } \phantom{00} k = 2 \\ \\ \text{Eqn of curve: } y & = x^2 (x - k)^2 \\ & = x^2 (x - 2)^2 \\ \\ {dy \over dx} & = 2x(x - k)(2x - k) \\ & = 2x(x - 2)(2x - 2) \\ & = 2x(x - 2)(2)(x - 1) \\ & = 4x(x - 2)(x - 1) \\ \\ \text{Let } & {dy \over dx} = 0, \\ 0 & = 4x(x - 2)(x - 1) \end{align} \begin{align} 4x & = 0 &&& x - 2 & = 0 &&& x - 1 & = 0 \\ x & = 0 &&& x & = 2 &&& x & = 1 \\ \\ \text{Substitute } & \text{into eqn of curve,} &&& \text{Substitute } & \text{into eqn of curve,} &&& \text{Substitute } & \text{into eqn of curve,} \\ y & = (0)^2 (0 - 2)^2 &&& y & = (2)(2 - 2)^2 &&& y & = (1)^2 (1 - 2)^2 \\ & = 0 &&& & = 0 &&& & = 1 \\ \\ \therefore & \phantom{.} (0, 0) &&& \therefore & \phantom{.} (2, 0) &&& \therefore & \phantom{.} (1, 1) \end{align}

(iii)

\begin{align} {dy \over dx} & = 4x(x - 2)(x - 1) \\ & = 4x(x^2 - x - 2x + 2) \\ & = 4x(x^2 - 3x + 2) \\ & = 4x^3 - 12x^2 + 8x \\ \\ {d^2 y \over dx^2} & = 4(3)x^2 - 12(2)x + 8(1) \\ & = 12x^2 - 24x + 8 \\ \\ \text{When } & x = 0, \\ {d^2 y \over dx^2} & = 12(0)^2 - 24(0) + 8 \\ & = 8 \\ \\ \therefore (0, 0) & \text{ is a minimum point} \\ \\ \text{When } & x = 2, \\ {d^2 y \over dx^2} & = 12(2)^2 - 24(2) + 8 \\ & = 8 \\ \\ \therefore (2, 0) & \text{ is a minimum point} \\ \\ \text{When } & x = 1, \\ {d^2 y \over dx^2} & = 12(1)^2 - 24(1) + 8 \\ & = -4 \\ \\ \therefore (1, 1) & \text{ is a maximum point} \end{align}

(iv)

\begin{align} y & = x^2 (x - 2)^2 \\ \\ \text{When } & x = -1, \\ y & = (-1)^2 (-1 - 2)^2 \\ & = 9 \\ \\ \text{End-point} & \text{ is } (-1, 9) \\ \\ \text{When } & x = 3, \\ y & = (3)^2 (3 - 2)^2 \\ & = 9 \\ \\ \text{End-point} & \text{ is } (3, 9) \end{align}

(i)

\begin{align} \text{Eqn of curve: } y & = {1 \over 20}x^5 - {7 \over 6}x^3 + 3x^2 \\ \\ \text{Gradient function: } {dy \over dx} & = {1 \over 20}(5)x^4 - {7 \over 6}(3)x^2 + 3(2)x \\ & = {1 \over 4} x^4 - {7 \over 2} x^2 + 6x \\ \\ {d^2 y \over dx^2} & = {1 \over 4}(4)x^3 - {7 \over 2}(2) x + 6(1) \\ & = x^3 - 7x + 6 \\ \\ \text{Let } & {d^2 y \over dx^2} = 0, \\ 0 & = x^3 - 7x + 6 \\ \\ \text{Let } f(x) & = x^3 - 7x + 6 \\ \\ f(1) & = (1)^3 - 7(1) + 6 \\ & = 0 \\ \\ \text{By Factor} & \text{ theorem, } x - 1 \text{ is a factor of } f(x) \\ \\ x^3 - 7x + 6 & = (x - 1)(x^2 + bx + c) \\ & = x^3 + bx^2 + cx - x^2 - bx - c \\ & = x^3 + bx^2 - x^2 + cx - bx - c \\ & = x^3 + (b - 1)x^2 + (c - b)x - c \\ \\ \text{Comparing } & \text{the constant,} \\ 6 & = -c \\ c & = -6 \\ \\ \text{Comparing } & \text{coefficient of } x^2, \\ 0 & = b - 1 \\ 1 & = b \\ \\ x^3 - 7x + 6 & = (x - 1)(x^2 + x - 6) \\ & = (x - 1)(x - 2)(x + 3) \\ \\ \\ 0 & = (x - 1)(x - 2)(x + 3) \end{align} \begin{align} x - 1 & = 0 &&& x - 2 & = 0 &&& x + 3 & = 0 \\ x & = 1 &&& x & = 2 &&& x & = -3 \\ \\ \text{Sub } & \text{into gradient function,} &&& \text{Sub } & \text{into gradient function,} &&& \text{Sub } & \text{into gradient function,} \\ {dy \over dx} & = {1 \over 4}(1)^4 - {7 \over 2}(1)^2 + 6(1) &&& {dy \over dx} & = {1 \over 4}(2)^4 - {7 \over 2}(2)^2 + 6(2) &&& {dy \over dx} & = {1 \over 4}(-3)^4 - {7 \over 2}(-3)^2 + 6(-3) \\ & = 2{3 \over 4} &&& & = 2 &&& & = -29{1 \over 4} \\ \\ \therefore & \phantom{.} \left(1, 2{3 \over 4}\right) &&& \therefore & \phantom{.} (2, 2) &&& \therefore & \phantom{.} \left(-3, -29{1 \over 4}\right) \end{align}

(ii)

\begin{align} {d^2 y \over dx^2} & = x^3 - 7x + 6 \\ \\ {d^3 y \over dx^3} & = 3x^2 - 7(1) \\ & = 3x^2 - 7 \\ \\ \text{When } & x = 1, \\ {d^3 y \over dx^3} & = 3(1)^2 - 7 \\ & = -4 \\ \\ \therefore \left(1, 2{3 \over 4}\right) & \text{ is a maximum point} \\ \\ \text{When } & x = 2, \\ {d^3 y \over dx^3} & = 3(2)^2 - 7 \\ & = 5 \\ \\ \therefore (2, 2) & \text{ is a minimum point} \\ \\ \text{When } & x = -3, \\ {d^3 y \over dx^3} & = 3(-3)^2 - 7 \\ & = 20 \\ \\ \therefore \left(-3, -29{1 \over 4}\right) & \text{ is a minimum point} \end{align}

(i) Since the curve touches the $x$-axis at $x = 1$, the $x$-axis is tangent to the curve $x = 1$. In other words, when $x = 1$, ${dy \over dx} = 0$

\begin{align} y & = x^3 + ax^2 + bx + c \\ \\ \text{Using } & (1, 0), \\ 0 & = (1)^3 + a(1)^2 + b(1) + c \\ 0 & = 1 + a(1) + b + c \\ -1 & = a + b + c \phantom{000} \text{--- (1)} \\ \\ \text{Using } & (4, 0), \\ 0 & = (4)^3 + a(4)^2 + b(4) + c \\ 0 & = 64 + a(16) + 4b + c \\ -64 & = 16a + 4b + c \phantom{000} \text{--- (2)} \\ \\ \text{(2) } & - \text{(1),} \\ -64 - (-1) & = 16a + 4b + c - (a + b + c) \\ -63 & = 16a + 4b + c - a - b - c \\ -63 & = 15a + 3b \\ -21 & = 5a + b \phantom{000} \text{--- (3)} \\ \\ y & = x^3 + ax^2 + bx + c \\ \\ {dy \over dx} & = 3x^2 + a(2)x + b(1) \\ & = 3x^2 + 2ax + b \\ \\ \text{When } & {dy \over dx} = 0 \text{ and } x = 1, \\ 0 & = 3(1)^2 + 2a(1) + b \\ 0 & = 3 + 2a + b \\ \\ b & = -2a - 3 \phantom{000} \text{--- (4)} \\ \\ \text{Substitute } & \text{(4) into (3),} \\ -21 & = 5a + (-2a - 3) \\ -21 & = 5a - 2a - 3 \\ -21 + 3 & = 5a - 2a \\ -18 & = 3a \\ {-18 \over 3} & = a \\ -6 & = a \\ \\ \text{Substitute } & a = -6 \text{ into (4),} \\ b & = -2(-6) - 3 \\ & = 9 \\ \\ \text{Substitute } & a = -6 \text{ and } b = 9 \text{ into (1),} \\ -1 & = -6 + 9 + c \\ -1 & = 3 + c \\ -4 & = c \\ \\ \therefore a & = -6, b = 9, c = -4 \end{align}

(ii)

\begin{align} y & = x^3 + ax^2 + bx + c \\ & = x^3 - 6x^2 + 9x - 4 \\ \\ {dy \over dx} & = 3x^2 + 2ax + b \\ & = 3x^2 - 12x + 9 \\ \\ {d^2 y \over dx^2} & = 3(2)x - 12(1) \\ & = 6x - 12 \\ \\ \text{Let } & {dy \over dx} = 0, \\ 0 & = 3x^2 - 12x + 9 \\ 0 & = x^2 - 4x + 3 \\ 0 & = (x - 1)(x - 3) \end{align} \begin{align} x - 1 & = 0 && \text{ or } & x - 3 & = 0 \\ x & = 1 &&& x & = 3 \\ \\ & &&& \text{Substitute } & \text{into eqn of curve,} \\ & &&& y & = (3)^3 - 6(3)^2 + 9(3) - 4 \\ & &&& & = -4 \\ \\ \therefore & \phantom{.} (1, 0) &&& \therefore & \phantom{.} (3, -4) \\ \\ \text{Substitute } & x = 1 \text{ into } {d^2 y \over dx^2}, &&& \text{Substitute } & x = 3 \text{ into } {d^2 y \over dx^2}, \\ {d^2 y \over dx^2} & = 6(1) - 12 &&& {d^2 y \over dx^2} & = 6(3) - 12 \\ & = -6 &&& & = 6 \\ \\ \therefore (1, 0) & \text{ is a max. point} &&& \therefore (3, -4) & \text{ is a min. point} \end{align}

(iii)

\begin{align} \text{Substitute } & x = 0 \text{ into eqn of curve,} \\ y & = (0)^3 - 6(0)^2 + 9(0) - 4 \\ & = -4 \\ \\ \text{Coordinates} & \text{ of point is } (0, -4) \\ \\ \text{Substitute } & x = 0 \text{ into } {dy \over dx}, \\ {dy \over dx} & = 3(0)^2 - 12(0) + 9 \\ & = 9 \\ \\ y & = mx + c \\ y & = 9x + c \\ \\ \text{Using } & (0, -4), \\ -4 & = 9(0) + c \\ -4 & = c \\ \\ \text{Eqn of tangent: } & y = 9x - 4 \end{align}