(a)

$$\begin{align} {d \over dx}(3x^{-7}) & = 3(-7)(x^{-8}) \\ & = -21x^{-8} \end{align}$$

(b)

$$\begin{align} {d \over dx}\left( {5 \over 2}\sqrt[5]{x^2} \right) & = {d \over dx} \left( {5 \over 2} x^{2 \over 5} \right) \\ & = \left({5 \over 2}\right) \left({2 \over 5}\right) (x^{{2 \over 5} - 1}) \\ & = x^{-{3 \over 5}} \end{align}$$

(a)

$$\begin{align} {d \over dx} (3x^2 + 4x - 1) & = 3(2)x + 4 - 0 \\ & = 6x + 4 \end{align}$$

(b)

$$\begin{align} {d \over dx} (-4x^3 + 5x^2 - 12) & = -4(3)x^2 + 5(2)x - 0 \\ & = -12x^2 + 10x \end{align}$$

(c)

$$\begin{align} {d \over dx} \left( {\pi^2 x^2 \over 3} - {2x^4 \over 5} \right) & = {d \over dx} \left( {\pi^2 \over 3}x^2 - {2 \over 5}x^4 \right) \\ & = {\pi^2 \over 3}(2)x - {2 \over 5}(4)x^3 \\ & = {2\pi^2 \over 3}x - {8 \over 5}x^3 \end{align}$$

(d)

$$\begin{align} {d \over dx} \left( {3x^8 \over 20} - {\pi x^9 \over 11} \right) & = {d \over dx} \left( {3 \over 20}x^8 - {\pi \over 11}x^9 \right) \\ & = {3 \over 20}(8)x^7 - {\pi \over 11}(9)x^8 \\ & = {6 \over 5}x^7 - {9\pi \over 11}x^8 \end{align}$$

(e) Note a is a constant.

$$\begin{align} {d \over dx} (3a + bx^2) & = (0) + b(2)x \\ & = 2bx \end{align}$$

(f) Note a and b are constants.

$$\begin{align} {d \over dx} \left( {a \over 2} - {b \over 2}x^3 \right) & = (0) - {b \over 2}(3)x^2 \\ & = - {3 \over 2}bx^2 \end{align}$$

(a)

$$\begin{align} {d \over dx} \left( 4x + {2 \over x} \right) & = {d \over dx} [4x + 2(x)^{-1}] \\ & = 4 + 2(-1)x^{-2} \\ & = 4 - 2 \left(1 \over x^2\right) \\ & = 4 - {2 \over x^2} \end{align}$$

(b)

$$\begin{align} {d \over dx} \left( 100x^2 + {100 \over x} \right) & = {d \over dx} (100x^2 + 100x^{-1}) \\ & = 100(2)x + 100(-1)x^{-2} \\ & = 200x - 100 \left(1 \over x^2\right) \\ & = 200x - {100 \over x^2} \end{align}$$

(c)

$$\begin{align} {d \over dx} \left( 9x^2 - {3 \over x^2} \right) & = {d \over dx} (9x^2 - 3x^{-2}) \\ & = 9(2)x - 3(-2)x^{-3} \\ & = 18x + 6 \left(1 \over x^3\right) \\ & = 18x + {6 \over x^3} \end{align}$$

(d)

$$\begin{align} {d \over dx} \left( {6 \over x^3} - {1 \over x} + 3 \right) & = {d \over dx} ( 6x^{-3} - x^{-1} + 3 ) \\ & = 6(-3)x^{-4} - (-1)x^{-2} + 0 \\ & = -18 \left(1 \over x^4\right) + x^{-2} \\ & = -{18 \over x^4} + {1 \over x^2} \end{align}$$

(e)

$$\begin{align} {d \over dx} (3x + 2\sqrt{x} - 3) & = {d \over dx} (3x + 2x^{1 \over 2} - 3) \\ & = 3 + 2\left({1 \over 2}\right)x^{-{1 \over 2}} - 0 \\ & = 3 + x^{-{1 \over 2}} \\ & = 3 + {1 \over \sqrt{x}} \end{align}$$

(f)

$$\begin{align} {d \over dx} (8x^2 + 3x - \sqrt{x}) & = {d \over dx} (8x^2 + 3x - x^{1 \over 2}) \\ & = 8(2)x + 3 - \left({1 \over 2}\right)x^{-{1 \over 2}} \\ & = 16x + 3 - {1 \over 2} \left(1 \over \sqrt{x}\right) \\ & = 16x + 3 - {1 \over 2\sqrt{x}} \end{align}$$

(a)

$$\begin{align} {d \over dx} \left( {2x^2 + 4x \over x} \right) & = {d \over dx} \left( {2x^2 \over x} + {4x \over x} \right) \\ & = {d \over dx} (2x + 4) \\ & = 2(1) + 0 \\ & = 2 \end{align}$$

(b)

$$\begin{align} {d \over dx} \left( {x^2 - 6x + 4 \over x} \right) & = {d \over dx} \left( {x^2 \over x} - {6x \over x} + {4 \over x} \right) \\ & = {d \over dx} (x - 6 + 4x^{-1}) \\ & = 1 - 0 + 4(-1)x^{-2} \\ & = 1 - 4 \left(1 \over x^2\right) \\ & = 1 - {4 \over x^2} \end{align}$$

(c)

$$\begin{align} {d \over dx} \left( {4x^3 - 5x - 3 \over 2x} \right) & = {d \over dx} \left( {4x^3 \over 2x} - {5x \over 2x} - {3 \over 2x} \right) \\ & = {d \over dx} \left[ 2x^2 - {5 \over 2} - {3 \over 2}\left(1 \over x\right) \right] \\ & = {d \over dx} \left(2x^2 - {5 \over 2} - {3 \over 2}x^{-1} \right) \\ & = 2(2)x - 0 - {3 \over 2}(-1)x^{-2} \\ & = 4x + {3 \over 2} \left(1 \over x^2\right) \\ & = 4x + {3 \over 2x^2} \end{align}$$

(d)

$$\begin{align} {d \over dx} \left( {x^2 + 4 \over 2x^2} \right) & = {d \over dx} \left( {x^2 \over 2x^2} + {4 \over 2x^2} \right) \\ & = {d \over dx} \left( {1 \over 2} + {2 \over x^2} \right) \\ & = {d \over dx} \left( {1 \over 2} + 2x^{-2} \right) \\ & = 0 + 2(-2)x^{-3} \\ & = - 4 \left(1 \over x^3 \right) \\ & = -{4 \over x^3} \end{align}$$

(a)

$$\begin{align} {d \over dx} \left({3x^2 + x - 1 \over \sqrt{x}}\right) & = {d \over dx} \left({3x^2 \over x^{1 \over 2}} + {x \over x^{1 \over 2}} - {1 \over x^{1 \over 2}} \right) \\ & = {d \over dx} \left( 3x^{3 \over 2} + x^{1 \over 2} - x^{- {1 \over 2}} \right) \phantom{000000000000000000} \left[ {a^m \over a^n} = a^{m - n} \right] \\ & = 3\left({3 \over 2}\right)x^{1 \over 2} + \left({1 \over 2}\right)x^{-{1 \over 2}} - \left(-{1 \over 2}\right)x^{-{3 \over 2}} \\ & = {9 \over 2}x^{1 \over 2} + {1 \over 2}x^{-{1 \over 2}} + {1 \over 2}x^{-{3 \over 2}} \end{align}$$

(b)

$$\begin{align} {d \over dx} \left({6x^2 - \sqrt{x} + 2 \over 2x}\right) & = {d \over dx} \left( {6x^2 \over 2x} - {x^{1 \over 2} \over 2x} + {2 \over 2x} \right) \\ & = {d \over dx} \left( 3x - {1 \over 2}x^{-{1 \over 2}} + {1 \over x} \right) \phantom{0000000000} \left[ {a^m \over a^n} = a^{m - n} \right] \\ & = {d \over dx} \left( 3x - {1 \over 2}x^{-{1 \over 2}} + x^{-1} \right) \\ & = 3 -{1 \over 2}\left(-{1 \over 2}\right)x^{-{3 \over 2}} + (-1)x^{-2} \\ & = 3 + {1 \over 4}x^{-{3 \over 2}} - x^{-2} \end{align}$$

(a)

$$\begin{align} {d \over dt} (t + 1)(2t - 1) & = {d \over dt} (2t^2 - t + 2t - 1) \\ & = {d \over dt} (2t^2 + t - 1) \\ & = 2(2)t + 1 - 0 \\ & = 4t + 1 \end{align}$$

(b)

$$\begin{align} {d \over dt} \left[t(\sqrt{t} - 2) \right] & = {d \over dt} [t(t^{1 \over 2} - 2] \\ & = {d \over dt} (t^{3 \over 2} - 2t) \phantom{000000} [ a^m \times a^n = a^{m + n}] \\ & = {3 \over 2}t^{1 \over 2} - 2 \\ & = {3 \over 2}\sqrt{t} - 2 \end{align}$$

(c)

$$\begin{align} {d \over dt} (1 + \sqrt{t})(1 - \sqrt{t}) & = {d \over dt} \left[ (1)^2 - (\sqrt{t})^2 \right] \phantom{000000} [ (a + b)(a - b) = a^2 - b^2] \\ & = {d \over dt} (1 - t) \\ & = 0 - 1 \\ & = -1 \end{align}$$

(d)

$$\begin{align} {d \over dt} \left[ 4t^2(3 - \sqrt{t}) \right] & = {d \over dt} \left[4t^2(3 - t^{1 \over 2}) \right] \\ & = {d \over dt} \left(12t^2 - 4t^{5 \over 2} \right) \phantom{000000} [a^m \times a^n = a^{m + n}] \\ & = 12(2)t - 4\left({5 \over 2}\right)t^{3 \over 2} \\ & = 24t - 10t^{3 \over 2} \end{align}$$

(e)

$$\begin{align} {d \over dt} (2t + 3)^2 & = {d \over dt} [(2t)^2 + 2(2t)(3) + (3)^2] \phantom{000000} [(a + b)^2 = a^2 + 2ab + b^2] \\ & = {d \over dt} (4t^2 + 12t + 9) \\ & = 4(2)t + 12 + 0 \\ & = 8t + 12 \end{align}$$

(f)

$$\begin{align} {d \over dt} \left[ {(2t + 1)(t - 2) \over t} \right] & = {d \over dt} {2t^2 - 3t - 2 \over t} \\ & = {d \over dt} \left( {2t^2 \over t} - {3t \over t} - {2 \over t} \right) \\ & = {d \over dt} \left( 2t - 3 - 2t^{-1} \right) \\ & = 2 - 0 - 2(-1)t^{-2} \\ & = 2 + 2 \left(1 \over t^2\right) \\ & = 2 + {2 \over t^2} \end{align}$$

(a)

$$\begin{align} {dy \over dx} & = {d \over dx} (4x^2 - 6x + 1) \\ & = 4(2)x - 6 + 0 \\ & = 8x - 6 \\ \\ \text{When } & x = 2, \\ {dy \over dx} & = 8(2) - 6 \\ & = 10 \end{align}$$

(b)

$$\begin{align} y & = \sqrt{x}(2 - x) \\ & = x^{1 \over 2} (2 - x) \\ & = 2x^{1 \over 2} - x^{3 \over 2} \phantom{00000000} [a^m \times a^n = a^{m + n}] \\ \\ {dy \over dx} & = {d \over dx} (2x^{1 \over 2} - x^{3 \over 2}) \\ & = 2\left( {1 \over 2} \right)x^{-{1 \over 2}} - {3 \over 2}x^{1 \over 2} \\ & = x^{-{1 \over 2}} - {3 \over 2}x^{1 \over 2} \\ \\ \text{When } & x = 9, \\ {dy \over dx} & = (9)^{-{1 \over 2}} - {3 \over 2}(9)^{1 \over 2} \\ & = -{25 \over 6} \end{align}$$

(c)

$$\begin{align} y & = {(x + 1)(2x - 3) \over x} \\ & = {2x^2 - x - 3 \over x} \\ & = {2x^2 \over x} - {x \over x} - {3 \over x} \\ & = 2x - 1 - 3x^{-1} \\ \\ {dy \over dx} & = {d \over dx} ( 2x - 1 - 3x^{-1}) \\ & = 2(1) - 0 - 3(-1)x^{-2} \\ & = 2 + 3 \left(1 \over x^2\right) \\ & = 2 + {3 \over x^2} \\ \\ \text{When } & x = -1, \\ {dy \over dx} & = 2 + {3 \over (-1)^2} \\ & = 5 \end{align}$$

For parts (a)-(c), the x-coordinates can be found by substituting the provided y-coordinates into the equation of the curve

(a)

$$\begin{align} {dy \over dx} & = {d \over dx} (x^2 - 2x) \\ & = (2)x - 2 \\ & = 2x - 2 \\ \\ \text{Substitute } & y = -1 \text{ into eqn of curve,}\\ -1 & = x^2 - 2x \\ 0 & = x^2 - 2x + 1 \\ 0 & = (x - 1)^2 \\ \pm \sqrt{0} & = x - 1 \\ 0 & = x - 1 \\ 1 & = x \\ \\ \text{Substitute } & x = 1 \text{ into } {dy \over dx}, \\ {dy \over dx} & = 2(1) - 2 \\ & = 0 \end{align}$$

(b)

$$\begin{align} {dy \over dx} & = {d \over dx} (2x^2 + 3x) \\ & = 2(2)(x) + 3(1) \\ & = 4x + 3 \\ \\ \text{Substitute } & y = 2 \text{ into eqn of curve,} \\ 2 & = 2x^2 + 3x \\ 0 & = 2x^2 + 3x - 2 \\ 0 & = (2x - 1)(x + 2) \\ \end{align}$$ $$\begin{align} 2x - 1 & = 0 && \text{ or } & x + 2 & = 0 \\ 2x & = 1 &&& x & = -2 \\ x & = {1 \over 2} \\ \\ \text{Substitute } & \text{into } {dy \over dx}, &&& \text{Substitute } & \text{into } {dy \over dx}, \\ {dy \over dx} & = 4\left(1 \over 2\right) + 3 &&& {dy \over dx} & = 4(-2) + 3 \\ & = 5 &&& & = -5 \end{align}$$

(c)

$$\begin{align} y & = {x^2 + 4 \over x} \\ & = {x^2 \over x} + {4 \over x} \\ & = x + 4x^{-1} \\ \\ {dy \over dx} & = {d \over dx} (x + 4x^{-1}) \\ & = 1 + 4(-1)x^{-2} \\ & = 1 - 4 \left(1 \over x^2\right) \\ & = 1 - {4 \over x^2} \\ \\ \text{Substitute } & y = 5 \text{ into eqn of curve,} \\ 5 & = {x^2 + 4 \over x} \\ 5x & = x^2 + 4 \\ 0 & = x^2 - 5x + 4 \\ 0 & = (x - 4)(x - 1) \end{align}$$ $$\begin{align} x - 4 & = 0 && \text{ or } & x - 1 & = 0 \\ x & = 4 &&& x & = 1 \\ \\ \text{Substitute } & \text{into } {dy \over dx}, &&& \text{Substitute } & \text{into } {dy \over dx}, \\ {dy \over dx} & = 1 - {4 \over (4)^2} &&& {dy \over dx} & = 1 - {4 \over (1)^2} \\ & = {3 \over 4} &&& & = -3 \end{align}$$

(a) The point where the curve crosses the y-axis is the y-intercept, where x = 0

$$\begin{align} {dy \over dx} & = {d \over dx} (2x^2 - 5x + 1) \\ & = 2(2)x - 5 + 0 \\ & = 4x - 5 \\ \\ \text{When } & x = 0, \\ {dy \over dx} & = 4(0) - 5 \\ & = - 5 \end{align}$$

(b) The point where the curve crosses the x-axis is the x-intercept, where y = 0

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

(c)

(a)

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

(b)

$$\begin{align} {dy \over dx} & = {d \over dx} \left( ax^2 + {b \over x} \right) \\ & = {d \over dx} ( ax^2 + bx^{-1}) \\ & = a(2)x + b(-1)x^{-2} \\ & = 2ax - {b \over x^2} \\ \\ \text{When } & x = 1 \text{ and } {dy \over dx} = 2, \\ 2 & = 2a(1) - {b \over (1)^2} \\ 2 & = 2a - b \\ \\ b & = 2a - 2 \phantom{000} \text{ --- (1)} \\ \\ \text{When } & x = 4 \text{ and } {dy \over dx} = -1, \\ -1 & = 2a(4) - {b \over (4)^2} \\ -1 & = 8a - {b \over 16} \\ -16 & = 128a - b \\ \\ b & = 128a + 16 \phantom{000} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ 2a - 2 & = 128a + 16 \\ -2 - 16 & = 128a - 2a \\ -18 & = 126a \\ {-18 \over 126} & = a \\ -{1 \over 7} & = a \\ \\ \text{Substitute } & a = -{1 \over 7} \text{ into (1)}, \\ b & = 2\left(-{1 \over 7}\right) - 2 \\ & = -{16 \over 7} \\ \\ \therefore a & = -{1 \over 7}, b = -{16 \over 7} \end{align}$$

The x-coordinates of P and Q can be found by solving simultaneous equations using the equations of the curve and the line.

$$\begin{align} 3x^2 - 8y & = -5 \\ -8y & = -3x^2 - 5 \\ 8y & = 3x^2 + 5 \\ y & = {3 \over 8}x^2 + {5 \over 8} \\ \\ {dy \over dx} & = {3 \over 8}(2)(x) + 0 \\ & = {3 \over 4}x \\ \\ \text{Eqn of line: } \phantom{00} 4y - 3x & = 7 \\ 4y & = 3x + 7 \\ 8y & = 6x + 14 \phantom{000} \text{ --- (1)} \\ \\ \text{Eqn of curve: } \phantom{00} 3x^2 - 8y & = -5 \phantom{000} \text{ --- (2)} \\ \\ \text{Substitute (1)} & \text{ into (2)}, \\ 3x^2 - (6x + 14) & = -5 \\ 3x^2 - 6x - 14 & = -5 \\ 3x^2 - 6x - 9 & = 0 \\ x^2 - 2x - 3 & = 0 \\ (x - 3)(x + 1) & = 0 \end{align}$$ $$\begin{align} x - 3 & = 0 && \text{ or } & x + 1 & =0 \\ x & = 3 &&& x & = -1 \\ \\ \text{Substitute } & \text{into } {dy \over dx}, &&& \text{Substitute } & \text{into } {dy \over dx}, \\ {dy \over dx} & = {3 \over 4}(3) &&& {dy \over dx} & = {3 \over 4}(-1) \\ & = {9 \over 4} &&& & = -{3 \over 4} \end{align}$$

$$\begin{align} y & = x(x - 2)^2 - 3 \\ & = x(x^2 - 4x + 4) - 3 \\ & = x^3 - 4x^2 + 4x - 3 \\ \\ {dy \over dx} & = 3x^2 - 4(2)x + 4 \\ & = 3x^2 - 8x + 4 \\ \\ \text{When } & {dy \over dx} = 7, \\ 7 & = 3x^2 - 8x + 4 \\ 0 & = 3x^2 - 8x - 3 \\ 0 & = (3x + 1)(x - 3) \end{align}$$ $$\begin{align} 3x + 1 & = 0 && \text{ or } & x - 3 & =0 \\ 3x & = -1 &&& x & = 3 \\ x & = -{1 \over 3} \\ \\ \text{Substitute } & \text{into eqn of curve}, &&& \text{Substitute } & \text{into eqn of curve}, \\ y & = \left(-{1 \over 3}\right) \left(-{1 \over 3} - 2\right)^2 - 3 &&& y & = (3)(3 - 2)^2 - 3 \\ & = -4{22 \over 27} &&& & = 0 \\ \\ \therefore & \phantom{.} \left(-{1 \over 3}, -4{22 \over 27}\right) &&& \therefore & \phantom{.} (3, 0) \end{align}$$

The x-axis is a horizontal line with gradient = 0. Since the tangent to the curve is parallel to the x-axis, the gradient of the tangent is 0 as well.

$$\begin{align} y & = {9x^2 + 1 \over x} \\ & = {9x^2 \over x} + {1 \over x} \\ & = 9x + x^{-1} \\ \\ {dy \over dx} & = 9 + (-1)x^{-2} \\ & = 9 - {1 \over x^2} \\ \\ \text{When } & {dy \over dx} = 0, \\ 0 & = 9 - {1 \over x^2} \\ -9 & = -{1 \over x^2} \\ 9 & = {1 \over x^2} \\ 9x^2 & = 1 \\ x^2 & = {1 \over 9} \\ x & = \pm\sqrt{1 \over 9} \\ & = \pm{1 \over 3} \end{align}$$ $$\begin{align} \text{Substitute } & x = {1 \over 3} \text{ into eqn of curve}, &&& \text{Substitute } & x = -{1 \over 3} \text{ into eqn of curve}, \\ y & = {9({1 \over 3})^2 + 1 \over {1 \over 3}} &&& y & = {9(-{1 \over 3})^2 + 1 \over -{1 \over 3}} \\ & = 6 &&& & = -6 \\ \\ \therefore & \phantom{.} \left({1 \over 3}, 6 \right) &&& \therefore & \phantom{.} \left( -{1 \over 3}, -6 \right) \end{align}$$

$$\begin{align} y & = {\pi\sqrt{x} \over 2} + {3 \over \sqrt{x}} \\ & = {\pi \over 2}\sqrt{x} + {3 \over \sqrt{x}} \\ & = {\pi \over 2}x^{1 \over 2} + {3 \over x^{1 \over 2}} \\ & = {\pi \over 2}x^{1 \over 2} + 3x^{-{1 \over 2}} \\ \\ {dy \over dx} & = {\pi \over 2}\left({1 \over 2}\right)x^{-{1 \over 2}} + 3\left(-{1 \over 2}\right)x^{-{3 \over 2}} \\ & = {\pi \over 4}x^{-{1 \over 2}} - {3 \over 2}x^{-{3 \over 2}} \\ \\ \text{When } & x = 1, \\ {dy \over dx} & = {\pi \over 4}(1)^{-{1 \over 2}} - {3 \over 2}(1)^{-{3 \over 2}} \\ & = {\pi \over 4}(1) - {3 \over 2}(1) \\ & = {\pi \over 4} - {3 \over 2} \end{align}$$

(i) To find the values of two unknown constants, we need to form two equations linking the constants.

$$\begin{align} y & = x^3 + px + q \\ \\ \text{Using } & (3, 16), \\ 16 & = (3)^3 + p(3) + q \\ 16 & = 27 + 3p + q \\ q & = -3p - 11 \phantom{00} \text{---(1)} \\ \\ {dy \over dx} & = (3)x^2 + p \\ & = 3x^2 + p \\ \\ \text{When } & x = 3 \text{ and } {dy \over dx} = 20, \\ 20 & = 3(3)^2 + p \\ 20 & = 27 + p \\ 20 - 27 & = p \\ -7 & = p \\ \\ \text{Substitute } & p = -7 \text{ into (1),} \\ q & = -3(-7) - 11 \\ & = 21 - 11 \\ & = 10 \\ \\ \therefore p & = -7, q = 10 \end{align}$$

(ii)

$$\begin{align} {dy \over dx} & = 3x^2 + p \\ \\ \text{When } & {dy \over dx} = 20 \text{ and } p = -7, \\ 20 & = 3x^2 - 7 \\ 20 + 7 & = 3x^2 \\ 27 & = 3x^2 \\ {27 \over 3} & = x^2 \\ 9 & = x^2 \\ \pm \sqrt{9} & = x \\ \pm 3 & = x \\ \\ \therefore x\text{-coordinate} & \text{ of other point is } -3 \\ \\ \text{Substitute } & x = -3 \text{ into eqn of curve,} \\ y & = (-3)^3 + (-7)(-3) + (10) \\ & = -27 + 21 + 10 \\ & = 4 \\ \\ \therefore & \phantom{.} (-3, 4) \end{align}$$

I don't think this is necessary for O levels.

$$\begin{align} f(x) & = \sqrt{x} \\ \\ f'(x) & = \lim_{h \rightarrow 0} { f(x + h) - f(x) \over h} \\ & = \lim_{h \rightarrow 0} {\sqrt{x + h} - \sqrt{x} \over h} \\ & = \lim_{h \rightarrow 0} {(\sqrt{x + h} - \sqrt{x})(\sqrt{x + h} + \sqrt{x}) \over h(\sqrt{x + h} + \sqrt{x})} \\ & = \lim_{h \rightarrow 0} { (\sqrt{x + h})^2 - (\sqrt{x})^2 \over h(\sqrt{x + h} + \sqrt{x})} \\ & = \lim_{h \rightarrow 0} { x + h - x \over h (\sqrt{x + h} + \sqrt{x})} \\ & = \lim_{h \rightarrow 0} { h \over h(\sqrt{x + h} + \sqrt{x})} \\ & = \lim_{h \rightarrow 0} {1 \over \sqrt{x + h} + \sqrt{x}} \\ & = {1 \over \sqrt{x} + \sqrt{x}} \\ & = {1 \over 2\sqrt{x}} \end{align}$$

$$\begin{align} y & = (x^3 + 1)(x^2 + 1)(x + 1) \\ & = (x^5 + x^3 + x^2 + 1)(x + 1) \\ & = x^6 + x^5 + x^4 + x^3 + x^3 + x^2 + x + 1 \\ & = x^6 + x^5 + x^4 + 2x^3 + x^2 + x + 1 \\ \\ {dy \over dx} & = 6x^5 + 5x^4 + 4x^3 + 6x^2 + 2x + 1 \end{align}$$

Note the powers of $2$ (i.e. $2^n$) are $2$, $4$, $8$, $16$, $32$, …, $1024$, $2048$ and so on…

$$\begin{align} f(x) & = (x - 1)(x + 1)(x^2 + 1)(x^4 + 1)...(x^{1024} + 1) \\ & = (x^2 - 1)(x^2 + 1)(x^4 + 1)...(x^{1024} + 1) \\ & = (x^4 - 1) (x^4 + 1) ... (x^{1024} + 1) \\ & = (x^{8} - 1) ... (x^{1024} + 1) \\ & \phantom{= }. \\ & \phantom{= }. \\ & \phantom{= }. \\ & = (x^{1024} - 1)(x^{1024} + 1) \\ & = x^{2048} - 1 \\ \\ f'(x) & = 2048x^{2047} \end{align}$$

(i)

$$\begin{align} x[x^2 + kx + (2k - 3)] & = 0 \\ \\ x = 0 & \end{align}$$

(ii)

$$\begin{align} x^2 + kx + (2k - 3) & = 0 \\ \\ b^2 - 4ac & = (k)^2 - 4(1)(2k - 3) \\ & = k^2 - 4(2k - 3) \\ & = k^2 - 8k + 12 \\ \\ b^2 - 4ac & < 0 \phantom{000000} [\text{No real roots}] \\ k^2 - 8k + 12 & < 0 \\ (k - 2)(k - 6) & < 0 \end{align}$$

$$2 < k < 6$$

(iii)

$$\begin{align} f(x) & = x[x^2 + kx + (2k - 3)] \\ f(x) & = x^3 + kx^2 + (2k - 3)x \\ \\ f'(x) & = 3x^2 + 2kx + (2k - 3) \\ \\ 0 & = 3x^2 + 2kx + (2k - 3) \\ \\ \alpha + \beta & = -{b \over a} \\ & = -{2k \over 3} \\ & = -{2 \over 3}k \\ \\ \alpha \beta & = {c \over a} \\ & = {2k - 3 \over 3} \\ & = {2k \over 3} - {3 \over 3} \\ & = {2 \over 3}k - 1 \\ \\ (\alpha - \beta)^2 & = \alpha^2 - 2 \alpha \beta + \beta^2 \\ & = (\alpha^2 + \beta^2) - 2\alpha \beta \\ & = (\alpha + \beta)^2 - 2\alpha \beta - 2 \alpha \beta \\ & = (\alpha + \beta)^2 - 4 \alpha \beta \\ & = \left(-{2 \over 3}k\right)^2 - 4 \left({2 \over 3}k - 1\right) \\ & = {4 \over 9}k^2 - {8 \over 3}k + 4 \\ \\ \\ \text{If } & k = 3, \\ (\alpha - \beta)^2 & = {4 \over 9}(3)^2 - {8 \over 3}(3) + 4 \\ & = 0 \\ \implies \alpha = \beta & \text{ and } f'(x) = 0 \text{ has only 1 real root} \\ \\ \\ \therefore \text{If } & f'(x) = 0 \text{ has 2 real roots, } k \ne 3 \end{align}$$

(iv)

$$\begin{align} |\alpha - \beta| & \le {2 \over 3} \\ (\alpha - \beta)^2 & \le \left(2 \over 3\right)^2 \\ {4 \over 9}k^2 - {8 \over 3}k + 4 & \le {4 \over 9} \\ 4k^2 - 24k + 36 & \le 4 \\ 4k^2 - 24k + 32 & \le 0 \\ k^2 - 6k + 8 & \le 0 \\ (k - 2)(k - 4) & \le 0 \end{align}$$

$$\begin{align} 2 \le & \phantom{.} k \le 4 \\ \\ \text{From (ii), } 2 < & \phantom{.} k < 6 \\ \\ \text{From (iii), } k & \ne 3 \\ \\ \\ \therefore k & = 4 \end{align}$$

$$\begin{align} \text{Function 1: } & y = {5 \over 2}x^2 \\ \\ \text{Prove: } & {dy \over dx} = {5 \over 2}(2)x = 5x \\ \\ \\ \text{Function 2: } & y = {5 \over 2}x^2 - 1 \\ \\ \text{Prove: } & {dy \over dx} = {5 \over 2}(2)x - 0 = 5x \end{align}$$

Discussion on this concept from Mathematics Stack Exchange (Thread 1, Thread 2)

$${d \over dx} (x + x + \phantom{.} ... \phantom{.} + x) = 1 + 1 + \phantom{.} ... \phantom{.} + 1$$ The step above is wrong as the number of terms depends on $x$.