\begin{align*} \text{Let price of item} & = P \\ \\ \text{Initially, } P & = 300 \\ \\ \text{After 1 month, } P & = 300 \times 0.95 \phantom{00000} [100\% - 5\% = 95\% = {95 \over 100} = 0.95] \\ P & = 300(0.95) \\ \\ \text{After 2 months, } P & = 300(0.95) \times 0.95 \\ P & = 300(0.95)^2 \\ \\ \text{After 3 months, } P & = 300(0.95)^2 \times 0.95 \\ P & = 300(0.95)^3 \\ \\ \implies \text{After } n \text{ months, } P & = 300(0.95)^n \\ \\ \text{Let } & P = 120, \\ 120 & = 300 (0.95)^n \\ {120 \over 300} & = (0.95)^n \\ 0.4 & = (0.95)^n \\ \lg 0.4 & = \lg (0.95)^n \\ \lg 0.4 & = n \lg 0.95 \phantom{000000} [\text{Power law (logarithms)}] \\ \\ n & = {\lg 0.4 \over \lg 0.95} \\ n & = 17.86 \\ \\ \text{When } & n = 17, \\ P & = 300(0.95)^{17} \approx \$ 125.44 \\ \\ \text{When } & n =18, \\ P & = 300(0.95)^{18} \approx \$ 119.16 < \$ 120 \\ \\ \therefore 18 \text{ complete months} & \text{ have elapsed before item is sold.} \end{align*}

(a)(i)

\begin{align*} a^3 - b^3 & = (a - b)(a^2 + ab + b^2) \phantom{000000} [\text{Need to memorise}] \end{align*}

(a)(ii)

\begin{align*} 347^3 - 286^3 & = (347 - 286)[(347)^2 + (347)(286) + (286)^2] \\ & = (61)(301 \phantom{.} 447) \\ \\ \text{Factor} & = 61 \text{ or } 301 \phantom{.} 447 \end{align*}

(b)

\begin{align*} {(x + n)^3 \over 2} - {(x - n)^3 \over 2} & = { (x + n)^3 - (x - n)^3 \over 2 } \\ & = { [ (x + n) - (x - n) ] [ (x + n)^2 + (x + n)(x - n) + (x - n)^2 ] \over 2 } \phantom{000000} [a = x + n, b = x - n] \\ & = { ( x + n - x + n ) [ (x + n)^2 + (x + n)(x - n) + (x - n)^2 ] \over 2 } \\ & = { 2n [(x + n)^2 + (x + n)(x - n) + (x - n)^2] \over 2 } \\ & = n [(x + n)^2 + (x + n)(x - n) + (x - n)^2] \\ \\ \therefore \text{Since expression is a } & \text{multiple of } n, \text{ it is divisible by } n \end{align*}

(a)

\begin{align*} u & = x^{n + 1} &&& v &= \ln x \\ {du \over dx} & =(n + 1)x^n &&& {dv \over dx} & = {1 \over x} \end{align*} \begin{align*} {d \over dx} (x^{n + 1} \ln x) & = (x^{n + 1})\left(1 \over x\right) + (\ln x)(n + 1)(x^n) \phantom{000900} [\text{Product rule}] \\ & = (x^{n + 1}) (x^{-1}) + (\ln x)(n + 1)(x^n) \\ & = x^{n + 1 + (-1)} + (\ln x)(n + 1)(x^n) \phantom{000000000} [a^m \times a^n = a^{m + n} ] \\ & = x^n + (\ln x)(n + 1)(x^n) \end{align*}

(b)

\begin{align*} {d \over dx} (x^{n + 1} \ln x) & = x^n + (\ln x)(n + 1)(x^n) \\ \\ \implies \int x^n \phantom{.} dx + \int (\ln x)(n + 1)(x^n) \phantom{.} dx & = x^{n + 1} \ln x \\ \int (\ln x)(n + 1)(x^n) \phantom{.} dx & = x^{n + 1} \ln x - \int x^n \phantom{.} dx \\ \underbrace{ (n + 1) \int (\ln x)(x^n) \phantom{.} dx }_{\text{Can factorise } (n + 1) \text{ since it is a constant}} & = x^{n + 1} \ln x - {x^{n + 1} \over n + 1} \\ \int (\ln x)(x^n) \phantom{.} dx & = {1 \over n + 1} \left( {x^{n + 1} \ln x \over 1} - {x^{n + 1} \over n + 1} \right) \\ \int (\ln x)(x^n) \phantom{.} dx & = { x^{n + 1} \ln x \over n + 1} - {x^{n + 1} \over (n + 1)^2} + C \end{align*}

(a)

\begin{align*} \text{Eqn of chord: } & 2y - 3x =1 7 \\ 2y & = 3x + 17 \\ y & = {3 \over 2}x + {17 \over 2} \phantom{000000} [y = mx + c] \\ \\ \text{Gradient of chord} & = {3 \over 2} \\ \text{Gradient of diameter // } AB & = {3 \over 2} \\ \\ y & = mx + c \\ y & = {3 \over 2}x + c \\ \\ \text{Using } & D\left(-{1 \over 2}, {1 \over 4}\right), \\ {1 \over 4} & = {3 \over 2}\left(-{1 \over 2}\right) + c \\ {1 \over 4} & = -{3 \over 4} + c \\ 1 & = c \\ \\ \text{Eqn of diameter // } AB: & y = {3 \over 2}x + 1 \\ \\ \text{Let } & y = -2, \phantom{000000} [y \text{-coordinate of centre} = -2] \\ -2 & = {3 \over 2}x + 1 \\ -{3 \over 2}x & = 3 \\ x & = 3 \div -{3 \over 2} \\ x & = -2 \\ \\ \text{Centre, }C & (-2, -2) \\ \\ \text{Radius} & = AC \\ & = \sqrt{ [-2 -(-1)]^2 + (-2 - 7)^2 } \\ & = \sqrt{82} \\ \\ \text{Eqn of circle: } & [x - (-2)]^2 + [y - (-2)]^2 = (\sqrt{82})^2 \\ & \phantom{000-.} (x + 2)^2 + (y + 2)^2 = 82 \end{align*}

(b)

\begin{align*} \text{Eqn of circle: } & (x + 2)^2 + (y + 2)^2 = 82 \\ \\ \text{Let } & x = 7, \\ (7 + 2)^2 + (y + 2)^2 & = 82 \\ 81 + (y + 2)^2 & = 82 \\ (y + 2)^2 & = 1 \\ y + 2 & = \pm \sqrt{1} \\ y + 2 & = 1 \text{ or } - 1 \\ y & = -1 \text{ or } - 3 \end{align*}

\begin{align*} [\text{Any point on the } \text{blue line will have } & x \text{-coordinate }7 \text{ AND be inside the circle}] \\ \\ \text{Set of values: } \{ \alpha: & \alpha \in \mathbb{R}, -3 < \alpha < -1 \} \end{align*}