(a)

\begin{align} \text{Next positive integer} & = x + 1 \\ \\ \text{Product} & = x(x + 1) \end{align}

(b)(i)

\begin{align} x(x + 1) & = 1260 \\ x^2 + x & = 1260 \\ x^2 + x - 1260 & = 0 \phantom{00} \text{ (Shown)} \end{align}

(b)(ii)

\begin{align} x^2 + x - 1260 & = 0 \\ (x - 35)(x + 36) & = 0 \end{align} \begin{align} x - 35 & = 0 && \text{ or } & x + 36 & = 0 \\ x & = 35 &&& x & = -36 \text{ (Reject, } x \text{ is a positive integer}) \end{align}
$$ \text{Two integers are } 35 \text{ and } 36 $$

(a)

\begin{align} \text{Area of triangle} & = {1 \over 2} \times \text{Base} \times \text{Height} \\ & = {1 \over 2} \times x \times (x - 3) \\ & = {1 \over 2}x (x - 3) \\ & = {1 \over 2}x^2 - {3 \over 2}x \\ \\ 90 & = {1 \over 2}x^2 - {3 \over 2}x \\ 2(90) & = 2 \left( {1 \over 2}x^2 - {3 \over 2}x \right) \\ 180 & = x^2 - 3x \\ 0 & = x^2 - 3x - 180 \phantom{00} \text{ (Shown)} \end{align}

(b)

\begin{align} 0 & = x^2 - 3x - 180 \\ 0 & = (x - 15)(x + 12) \end{align} \begin{align} x - 15 & = 0 && \text{ or } & x + 12 & = 0 \\ x & = 15 &&& x & = -12 \text{ (Reject, base of triangle } > 0) \end{align}
$$ \text{Base} = 15 \text{ cm} $$

(a)

\begin{align} \text{Perimeter} & = 2l + 2b \\ 68 & = 2x + 2b \\ 68 - 2x & = 2b \\ {1 \over 2}(68 - 2x) & = b \\ 34 - x & = b \\ \\ \text{Breadth} & = (34 - x) \text{ cm} \end{align}

(b)

\begin{align} \text{Area} & = l \times b \\ 253 & = x \times (34 - x) \\ 253 & = x(34 - x) \\ 253 & = 34x - x^2 \\ 0 & = -x^2 + 34x - 253 \\ 0 & = x^2 - 34x + 253 \phantom{00} \text{ (Shown)} \end{align}

(c)

\begin{align} 0 & = x^2 - 34x + 253 \\ 0 & = (x - 11)(x - 23) \end{align} \begin{align} x - 11 & = 0 && \text{ or } & x - 23 & = 0 \\ x & = 11 &&& x & = 23 \end{align}
$$ \text{Dimensions: } 23 \text{ cm by } 11 \text{ cm} $$

(a)

\begin{align} \text{Length of } B & = (x + 5) \text{ cm} \end{align}

(b)

\begin{align} \text{Area of square} & = l \times l = l^2 \\ \\ \text{Area of } B & = (x + 5)^2 \text{ cm}^2 \end{align}

(c)

\begin{align} \text{Sum of area} & = \text{Area of } A + \text{Area of } B \\ 1525 & = x^2 + (x + 5)^2 \\ 1525 & = x^2 + \underbrace{ (x)^2 + 2(x)(5) + (5)^2 }_{(a + b)^2 = a^2 + 2ab + b^2} \\ 1525 & = x^2 + x^2 + 10x + 25 \\ 1525 & = 2x^2 + 10x + 25 \\ 0 & = 2x^2 + 10x + 25 - 1525 \\ 0 & = 2x^2 + 10x - 1500 \\ 0 & = x^2 + 5x - 750 \\ 0 & = (x - 25)(x + 30) \end{align} \begin{align} x - 25 & = 0 && \text{ or } & x + 30 & = 0 \\ x & = 25 &&& x & = -30 \text{ (Reject, since length of tile } A > 0) \end{align}

(a)(i)

$$ (24 - x) \text{ cm} $$

(a)(ii)

$$ (17 + x) \text{ cm} $$

(a)(iii)

\begin{align} \text{New area} & = (24 - x)(17 + x) \text{ cm}^2 \end{align}

(b)

\begin{align} \text{New area} & = \text{Initial area} + 12 \\ (24 - x)(17 + x) & = 24 \times 17 + 12 \\ 408 + 24x - 17x - x^2 & = 420 \\ 408 + 7x - x^2 & = 420 \\ 0 & = x^2 - 7x + 420 - 408 \\ 0 & = x^2 - 7x + 12 \\ 0 & = (x - 3)(x - 4) \end{align} \begin{align} x - 3 & = 0 && \text{ or } & x - 4 & = 0 \\ x & = 3 &&& x & = 4 \end{align}

(a)(i)

$$ 34 - x $$

(a)(ii)

\begin{align} 34 - x + 4 & = 38 - x \end{align}

(a)(iii)

$$ (x + 4)^2 $$

(b)

\begin{align} \text{Father's age (4 years time)} & = \text{Square of Mingfa's age (4 years time)} \\ 38 - x & = (x + 4)^2 \\ 38 - x & = (x + 4)(x + 4) \\ 38 - x & = x^2 + 4x + 4x + 16 \\ 38 - x & = x^2 + 8x + 16 \\ 0 & = x^2 + 8x + x + 16 - 38 \\ 0 & = x^2 + 9x - 22 \\ 0 & = (x - 2)(x + 11) \end{align} \begin{align} x - 2 & = 0 && \text{ or } & x + 11 & = 0 \\ x & = 2 &&& x & = -11 \text{ (Reject, as age } \ge 0 ) \end{align}
\begin{align} \text{Mingfa's present age} & = x \\ & = 2 \text{ years} \end{align}

(a) Note: Hypotenuse is the longest side of the right-angled triangle (opposite the right angle)

\begin{align} \text{Hypotenuse} & = 10 - x - 2x \\ & = (10 - 3x) \text{ cm} \\ \\ \text{By Pytha}& \text{goras theorem,} \\ (10 - 3x)^2 & = x^2 + (2x)^2 \\ (10 - 3x)(10 - 3x) & = x^2 + 4x^2 \\ 100 - 30x - 30x + 9x^2 & = 5x^2 \\ 9x^2 - 60x + 100 & = 5x^2 \\ 9x^2 - 5x^2 - 60x + 100 & = 0 \\ 4x^2 - 60x + 100 & = 0 \\ x^2 - 15x + 25 & = 0 \phantom{00} \text{ (Shown)} \end{align}

(b)

\begin{align} x^2 & - 15x + 25 = 0 \\ x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {-(-15) \pm \sqrt{(-15)^2 - 4(1)(25)} \over 2(1)} \\ & = {15 \pm \sqrt{125} \over 2} \\ & = 13.09 \text{ (Reject, as } x < 10 \text{ cm)} \text{ or } 1.9098 \\ \\ \text{Longest side} & = \text{Hypotenuse} \\ & = 10 - 3(1.9098) \\ & = 4.2706 \\ & \approx 4.27 \text{ cm} \end{align}

\begin{align} \text{Let } x \text{ represent the } & \text{smallest even number} \\ \\ \text{Next } \textbf{even} \text{ number} & = x + 2 \\ \\ \text{Last even number} & = x + 2 + 2 \\ & = x + 4 \\ \\ \text{Sum of squares} & = x^2 + (x + 2)^2 + (x + 4)^2 \\ 2360 & = x^2 + (x + 2)(x + 2) + (x + 4)(x + 4) \\ 2360 & = x^2 + x^2 + 2x + 2x + 4 + x^2 + 4x + 4x + 16 \\ 2360 & = 3x^2 + 12x + 20 \\ 0 & = 3x^2 + 12x + 20 - 2360 \\ 0 & = 3x^2 + 12x - 2340 \\ 0 & = x^2 + 4x - 780 \\ 0 & = (x - 26)(x + 30) \end{align} \begin{align} x - 26 & = 0 && \text{ or } & x + 30 & = 0 \\ x & = 26 &&& x & = -30 \text{ (Reject, } x \text{ represents a positive even number}) \end{align}
\begin{align} \text{Smallest even number} & = x \\ & = 26 \end{align}

\begin{align} \text{Length of typing area} & = (29 - 2x) \text{ cm} \\ \\ \text{Breadth of typing area} & = (21 - 2x) \text{ cm} \\ \\ \text{Area of typing area} & = (29 - 2x)(21 - 2x) \\ 380 & = 609 - 58x - 42x + 4x^2 \\ 380 & = 609 - 100x + 4x^2 \\ 0 & = 4x^2 - 100x + 609 - 380 \\ 0 & = 4x^2 - 100x + 229 \\ \\ x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {-(-100) \pm \sqrt{(-100)^2 - 4(4)(229)} \over 2(4)} \\ & = {100 \pm \sqrt{ 6336 } \over 8} \\ & = 22.449 \text{ (Reject, since width of paper is 21 cm}) \text{ or } 2.5501 \\ \\ \text{Width of margin} & = x \\ & = 2.5501 \\ & \approx 2.6 \text{ cm (1 d.p.)} \end{align}

\begin{align} \text{Let } x \text{ represent the price of} & \text{ a keyboard} \\ \\ \text{No. of keyboards purchased} & = {540 \over x} \\ \\ \text{No. of keyboards purchased (w. reduction)} & = {540 \over x - 2} \\ \\ {540 \over x - 2} - {540 \over x} & = 3 \\ {540x \over x(x - 2)} - {540(x - 2) \over x(x - 2)} & = {3 \over 1} \\ {540x - 540(x - 2) \over x(x - 2)} & = {3x(x - 2) \over x(x - 2)} \\ \\ 540x - 540(x - 2) & = 3x(x - 2) \\ 540x - 540x + 1080 & = 3x^2 - 6x \\ 1080 & = 3x^2 - 6x \\ 0 & = 3x^2 - 6x - 1080 \\ 0 & = x^2 - 2x - 360 \\ 0 & = (x - 20) (x + 18) \end{align} \begin{align} x - 20 & = 0 && \text{ or } & x + 18 & = 0 \\ x & = 20 &&& x & = -18 \text{ (Reject, as price of keyboard } > 0) \end{align}
$$ \text{Price of one keyboard} = \$ 20 $$

(a)

\begin{align} \text{Area} & = x \times 3 + (2x + 1 - x) \times (3 - x) \\ & = 3x + (x + 1)(3 - x) \\ & = 3x + (3x - x^2 + 3 - x) \\ & = 3x + 3x - x^2 + 3 - x \\ & = - x^2 + 5x + 3 \phantom{00} \text{ (Shown)} \end{align}

(b)

\begin{align} 8.25 & = - x^2 + 5x + 3 \\ 0 & = -x^2 + 5x + 3 - 8.25 \\ 0 & = - x^2 + 5x - 5.25 \\ 0 & = x^2 - 5x + 5.25 \\ 4(0) & = 4(x^2 - 5x + 5.25) \\ 0 & = 4x^2 - 20x + 21 \\ 0 & = (2x - 3)(2x - 7) \end{align} \begin{align} 2x - 3 & = 0 && \text{ or } & 2x - 7 & = 0 \\ 2x & = 3 &&& 2x & = 7 \\ x & = {3 \over 2} = 1.5 &&& x & = {7 \over 2} = 3.5 \text{ (Reject, since } x < 3) \end{align}

\begin{align} \text{Let } x \text{ represent the } & \text{length of the square cardboard} \\ \\ \text{Length of base of cuboid} & = x - 6 - 6 \\ & = (x - 12) \text{ cm} \\ \\ \text{Height of cuboid} & = 6 \text{ cm} \\ \\ \text{Volume} & = (x - 12) \times (x - 12) \times 6 \\ 2000 & = 6(x - 12)(x - 12) \\ 2000 & = 6(x^2 - 12x - 12x + 144) \\ 2000 & = 6(x^2 - 24x + 144) \\ 2000 & = 6x^2 - 144x + 864 \\ 0 & = 6x^2 - 144x + 864 - 2000 \\ 0 & = 6x^2 - 144x - 1136 \\ 0 & = 3x^2 - 72x - 568 \\ \\ x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {-(-72) \pm \sqrt{(-72)^2 - 4(3)(-568)} \over 2(3)} \\ & = {72 \pm \sqrt{ 12 \phantom{.} 000} \over 6} \\ & = 30.257 \text{ or } -6.2574 \text{ (Reject, as } x > 0) \\ \\ \text{Length} & = x \\ & = 30.257 \\ & \approx 30.3 \text{ cm} \end{align}

(a)

\begin{align} \text{Area} & = {1 \over 2} \times b \times h \\ & = {1 \over 2} \times 4 \times x \\ & = 2x \text{ cm}^2 \end{align}

(b)

\begin{align} \text{Area of paper} & = 4x \times (5x - 1) \\ & = 4x(5x - 1) \\ & = 20x^2 - 4x \\ \\ \text{Area of rectangle } DEFG & = 4 \times x \\ & = 4x \\ \\ \text{Area of shaded region} & = 20x^2 - 4x - 4x \\ & = (20x^2 - 8x) \text{ cm}^2 \phantom{00} \text{ (Shown)} \end{align}

(c)(i)

\begin{align} 450 & = 20x^2 - 8x \\ 0 & = 20x^2 - 8x - 450 \\ 0 & = 10x^2 - 4x - 225 \\ \\ x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {-(-4) \pm \sqrt{(-4)^2 - 4(10)(-225)} \over 2(10)} \\ & = {4 \pm \sqrt{9016} \over 20} \\ & = 4.9476 \text{ or } -4.5476 \text{ (Reject, as } x > 0) \\ & \approx 4.95 \end{align}

(c)(ii)

\begin{align} \text{By Pytha} & \text{goras theorem,} \\ GE^2 & = GD^2 + DE^2 \\ GE^2 & = (4)^2 + (x)^2 \\ GE^2 & = 16 + x^2 \\ GE^2 & = 16 + (4.9476)^2 \\ GE^2 & = 40.478 \\ GE & = \sqrt{40.478} \\ & = 6.3622 \text{ cm} \\ \\ \text{Perimeter} & = GE + EC + CB + BA + AG \\ & = 6.3622 + [5(4.9476) - 1 - 4.9476] + 4(4.9476) + [5(4.9476) - 1] + [4(4.9476) - 4] \\ & = 84.471 \\ & \approx 84.5 \text{ cm} \end{align}

\begin{align} \text{Let } x \text{ represent the time (in hours } & \text{from noon) when they are 10 km apart} \\ \\ \text{Distance} & = \text{Speed} \times \text{Time} \\ \\ \text{Distance travelled by Ada} & = 4 \times x \\ & = 4x \text{ km} \\ \\ \text{Distance travelled by Bob} & = 4 \times (x - 1) \\ & = 4(x - 1) \\ & = (4x - 4) \text{ km} \\ \\ \text{By Pytha} & \text{goras theorem,} \\ \text{Distance between Ada & Bob} & = (4x)^2 + (4x - 4)^2 \\ 10^2 & = 16x^2 + (4x - 4)(4x - 4) \\ 100 & = 16x^2 + 16x^2 - 16x - 16x + 16 \\ 100 & = 32x^2 - 32x + 16 \\ 0 & = 32x^2 - 32x + 16 - 100 \\ 0 & = 32x^2 - 32x - 84 \\ 0 & = 8x^2 - 8x - 21 \\ \\ x & = {-b \pm \sqrt{b^2 -4ac} \over 2a} \\ & = {-(-8) \pm \sqrt{(-8)^2 - 4(8)(-21)} \over 2(8)} \\ & = {8 \pm \sqrt{736} \over 16} \\ & = 2.1955 \text{ or } -1.1955 \text{ (Reject, as } x > 0) \\ \\ 2.1955 \text{ h} & = 2 \text{ h } (0.1955 \times 60) \text{ mins} \\ & \approx 2 \text{ h } 20 \text{ mins} \\ \\ \\ \therefore \text{They would } & \text{be 10 km apart at 2.12 pm} \end{align}

\begin{align} \text{Let } x \text{ represent the } & \text{numerator of the fraction} \\ \\ \text{Fraction} & = {x \over x + 2} \\ \\ \text{Increased fraction} & = {x + 3 \over x + 2 + 3} \\ & = {x + 3 \over x + 5} \\ \\ {x + 3 \over x + 5} - {x \over x + 2} & = {1 \over 18} \\ {(x + 3)(x + 2) \over (x + 5)(x + 2)} - {x(x + 5) \over (x + 5)(x + 2)} & = {1 \over 18} \\ {(x + 3)(x + 2) - x(x + 5) \over (x + 5)(x + 2)} & = {1 \over 18} \\ {x^2 + 2x + 3x + 6 - x^2 - 5x \over (x + 5)(x + 2)} & = {1 \over 18} \\ {6 \over (x + 5)(x + 2)} & = {1 \over 18} \\ {18(6) \over 18(x + 5)(x + 2)} & = {(x + 5)(x + 2) \over 18(x + 5)(x + 2)} \\ \\ 18(6) & = (x + 5)(x + 2) \\ 108 & = x^2 + 2x + 5x + 10 \\ 108 & = x^2 + 7x + 10 \\ 0 & = x^2 + 7x + 10 - 108 \\ 0 & = x^2 + 7x - 98 \\ 0 & = (x - 7)(x + 14) \end{align} \begin{align} x - 7 & = 0 && \text{ or } & x + 14 & = 0 \\ x & = 7 &&& x & = -14 \text{ (Reject, since } x > 0) \end{align}
$$ \text{Fraction} = {x \over x + 2} = {7 \over 7 + 2} = {7 \over 9} $$

(a)(i)

$$ 80 - 4 - 4 = 72 $$

(a)(ii)

\begin{align} \text{Sales} & = 70 \times 72 \\ & = \$ 5040 \end{align}

(b)(i)

$$ 80 - 4n $$

(b)(ii)

\begin{align} \text{Sales} & = (60 + 5n)(80 - 4n) \\ & = 4800 - 240n + 400n - 20n^2 \\ & = \$ (4800 + 160n - 20n^2) \end{align}

(b)(iii)

\begin{align} 5100 & = 4800 + 160n - 20n^2 \\ 0 & = 4800 - 5100 + 160n - 20n^2 \\ 0 & = -300 + 160n - 20n^2 \\ 0 & = -20n^2 + 160n - 300 \\ 0 & = 20n^2 - 160n + 300 \\ 0 & = n^2 - 8n + 15 \\ 0 & = (n - 3)(n - 5) \end{align} \begin{align} n - 3 & = 0 && \text{ or } & n - 5 & = 0 \\ n & = 3 &&& n & = 5 \\ \\ \text{Price per copy} & = 60 + 5(3) &&& \text{Price per copy} & = 60 + 5(5) \\ & = \$75 &&& & = \$ 85 \end{align}

(b)(iv)

\begin{align} \text{Price} & = {75 + 85 \over 2} \\ & = \$ 80 \end{align}

(a)

t	0	1	2	3	4	5	6	7	8
h	0	33	56	69	72	65	48	21	-16

(b) From the graph, the dolphin was above sea level from t = 0 to t = 7.6

$$ \text{Duration} = 7.6 \text{ s} $$

(c)

\begin{align} & T = 7.6 \\ \\ & \text{After } t = 7.6, \text{ the dolphin is underwater} \end{align}

(a)

\begin{align} y & = C_1 - C_2 \\ y & = 2x^2 + 7x + 3 - (x^2 + 10x + 11) \\ y & = 2x^2 + 7x + 3 - x^2 - 10x - 11 \\ y & = x^2 - 3x - 8 \end{align}

(b)

x	0	1	2	3	4	5	6	7	8
y	-8	-10	-10	-8	-4	2	10	20	32

(c)

\begin{align} \text{For cost to be the same, } C_1 & = C_2 \\ C_1 - C_2 & = 0 \\ y & = 0 \\ \\ \text{From graph, } & x = 4.7 \end{align}