(a)

\begin{align} |5 - 9| & = |-4| \\ & = 4 \end{align}

(b)

\begin{align} |3 - \pi - 1| & = |3 - 1 - \pi| \\ & = |2 - \pi | \\ & = -(2 - \pi) \\ & = -2 + \pi \\ & = \pi - 2 \end{align}

(c)

\begin{align} 2|-5| - |8| & = 2(5) - 8 \\ & = 2 \end{align}

(i)

\begin{align} \text{When } x & = 1, \\ y & = |(1)^2 - 8(1) + 2| \\ & = |-5| \\ & = 5 \end{align}

(ii)

\begin{align} \text{When } x & = -1, \\ y & = |(-1)^2 - 8(-1) + 2| \\ & = |11| \\ & = 11 \end{align}

(iii)

\begin{align} \text{When } x & = 2, \\ y & = |(2)^2 - 8(2) + 2| \\ & = |-10| \\ & = 10 \end{align}

(a)

\begin{align} |2 - \sqrt{2}| + |-1| & = (2 - \sqrt{2}) + 1 \\ & = 2 + 1 - \sqrt{2} \\ & = 3 - \sqrt{2} \end{align}

(b)

\begin{align} |2 - \sqrt{5}| + |3 - \sqrt{5}| & = -(2 - \sqrt{5}) + (3 - \sqrt{5}) \\ & = -2 + \sqrt{5} + 3 - \sqrt{5} \\ & = -2 + 3 + \sqrt{5} - \sqrt{5} \\ & = 1 \end{align}

(c)

\begin{align} 2|\sqrt{6} - \sqrt{3}| + |\sqrt{3} - \sqrt{6}| & = 2(\sqrt{6} - \sqrt{3}) - (\sqrt{3} - \sqrt{6}) \\ & = 2\sqrt{6} - 2\sqrt{3} - \sqrt{3} + \sqrt{6} \\ & = 2\sqrt{6} + \sqrt{6} - 2\sqrt{3} - \sqrt{3} \\ & = 3\sqrt{6} - 3 \sqrt{3} \end{align}

(i)

\begin{align} \text{When } x & = 8, \\ \\ y & = 2|3 - 8| + 4 \\ & = 2|-5| + 4 \\ & = 2(5) + 4 \\ & = 10 + 4 \\ & = 14 \end{align}

(ii)

\begin{align} \text{When } y & = 18, \\ \\ 18 & = 2|3 - x| + 4 \\ 18 - 4 & = 2|3 - x| \\ 14 & = 2|3 - x| \\ {14 \over 2} & = |3 - x| \\ 7 & = |3 - x| \end{align} \begin{align} 3 - x & = 7 \phantom{000}&\text{or}\phantom{000} 3 - x & = - 7 \\ -x & = 7 - 3 & -x & = -7 - 3 \\ -x & = 4 & -x & = -10 \\ x & = -4 & x & = 10 \end{align}
$$ 18 = 2|3 - x| + 4 $$ \begin{align} \text{For } & x = -4, \phantom{0} & \phantom{00000} \text{For } & x = 10, \\ \\ \text{R.H.S} & = 2|3 - (-4)| + 4 & \text{R.H.S} & = 2|3 - 10| + 4 \\ & = 2| 7| + 4 & & = 2|-7| + 4 \\ & = 2(7) + 4 & & = 2(7) + 4 \\ & = 18 & & = 18 \\ & = \text{L.H.S} & & = \text{L.H.S} \\ \\ \\ \therefore x & = -4, 10 \end{align}

(a)

$$ |x - 4| = 6 $$ \begin{align} x - 4 & = 6 \phantom{000}&\text{or}\phantom{000} x - 4 & = - 6 \\ x & = 6 + 4 & x & = -6 + 4 \\ & = 10 & & = -2 \end{align}
$$ |x - 4| = 6 $$ \begin{align} \text{For } & x = 10, \phantom{0} & \phantom{00000} \text{For } & x = -2, \\ \\ \text{L.H.S} & = |10 - 4| & \text{L.H.S} & = |-2 - 4| \\ & = |6| & & = |-6| \\ & = 6 & & = 6 \\ & = \text{R.H.S} & & = \text{R.H.S} \\ \\ \\ \therefore x & = 10, -2 \end{align}

(b)

$$ |2x + 3| = 5 $$ \begin{align} 2x + 3 & = 5 \phantom{000}&\text{or}\phantom{000} 2x + 3 & = -5 \\ 2x & = 5 - 3 & 2x & = -5 - 3 \\ & = 2 & & = - 8 \\ \\ x & = {2 \over 2} & x & = {-8 \over 2} \\ & = 1 & & = -4 \end{align}
$$ |2x + 3| = 5 $$ \begin{align} \text{For } & x = 1, \phantom{0} & \phantom{00000} \text{For } & x = -4, \\ \\ \text{L.H.S} & = |2(1) + 3| & \text{L.H.S} & = |2(-4) + 3| \\ & = |5| & & = |-5| \\ & = 5 & & = 5 \\ & = \text{R.H.S} & & = \text{R.H.S} \\ \\ \\ \therefore x & = 1, -4 \end{align}

(c)

\begin{align} |x - 2| + 3 & = 5 \\ |x - 2| & = 5 - 3 \\ |x - 2| & = 2 \end{align} \begin{align} x - 2 & = 2 \phantom{000}&\text{or}\phantom{000} x - 2 & = -2 \\ x & = 2 + 2 & x & = -2 + 2 \\ & = 4 & & = 0 \end{align}
$$ |x - 2| + 3 = 5 $$ \begin{align} \text{For } & x = 4, \phantom{0} & \phantom{00000} \text{For } & x = 0, \\ \\ \text{L.H.S} & = |4 - 2| + 3 & \text{L.H.S} & = |0 - 2| + 3 \\ & = |2| + 3 & & = |-2| + 3 \\ & = 2 + 3 & & = 2 + 3 \\ & = 5 & & = 5 \\ & = \text{R.H.S} & & = \text{R.H.S} \\ \\ \\ \therefore x & = 4, 0 \end{align}

(a)

(b)

(c)

(d)

(a)

\begin{align} y & = x - 3 \phantom{00000} [\text{Straight line } y = mx + c] \\ \\ \text{When } & x = 0, \\ y & = 0 - 3 \\ & = -3 \\ \\ \implies & y\text{-intercept is } -3 \\ \\ \text{When } & y = 0, \\ 0 & = x - 3 \\ 3 & = x \\ \\ \implies & x\text{-intercept is } 3 \\ \\ \text{When } x & = 4, \\ y & = 4 - 3 \\ & = 1 \\ \\ \text{Right } & \text{limit is } (4, 1) \end{align}

Reflect the portion of the graph below the $y$-axis to obtain the graph of $y = |x - 3|$.

(b)

\begin{align} y & = x + 2 \\ \\ \text{When } & x = 0, \\ y & = 0 + 2 \\ & = 2 \\ \\ \implies & y\text{-intercept is } 2 \\ \\ \text{When } & y = 0, \\ 0 & = x + 2 \\ -2 & = x \\ \\ \implies & x\text{-intercept is } -2 \\ \\ \text{When } & x = -3, \\ y & = -3 + 2 \\ & = -1 \\ \text{Left} & \text{ limit is } (-3, -1) \\ \\ \text{When } & x = 3, \\ y & = 3 + 2 \\ & = 5 \\ \\ \text{Right} & \text{ limit is } (3, 5) \end{align}

Reflect the portion of the graph below the $y$-axis to obtain the graph of $y = |x + 2|$.

\begin{align} f(a) & = (a) + 2 - |3 - 2(a)| \\ & = a + 2 - |3 - 2a| \\ & = a + 2 - [-(3 - 2a)] \phantom{00000} [\text{Since } a > 2 \implies 3 - 2a < 0] \\ & = a + 2 - (-3 + 2a) \\ & = a + 2 + 3 - 2a \\ & = 2 + 3 - 2a + a \\ & = 5 - a \text{ (Shown)} \end{align}

\begin{align} f(c) & = |6 - c| - (c)^2 \\ & = |6 - c| - c^2 \\ & = (6 - c) - c^2 \phantom{00000} [\text{Since } c < 0 \implies 6 - c > 0] \\ & = 6 - c - c^2 \\ & = (3 + c)(2 - c) \\ \\ \text{For the} & \text{ negative root, } x < 0, \\ f(x) & = |6 - x| - x^2 \\ & = (3 + x)(2 - x) \\ \\ \text{When } & f(x) = 0, \\ 0 & = (3 + x)(2 - x) \\ \\ 3 + x = 0 \phantom{000}&\text{or}\phantom{000} 2 - x = 0 \\ x = -3 \phantom{0.}&\phantom{or0002} -x = -2 \\ & \phantom{or00002-} x = 2 \text{ (Reject)} \end{align}

(a)

\begin{align} 2|x + 1| + 3 & = 9 \\ 2|x + 1| & = 9 - 3 \\ 2|x + 1| & = 6 \\ |x + 1| & = {6 \over 2} \\ |x + 1| & = 3 \end{align}
$$ 2|x + 1| + 3 = 9 $$ \begin{align} \text{For } & x = 2, \phantom{0} & \phantom{00000} \text{For } & x = -4, \\ \\ \text{L.H.S} & = 2|2 + 1| + 3 & \text{L.H.S} & = 2|-4 + 1| + 3 \\ & = 2|3| + 3 & & = 2|-3| + 3 \\ & = 2(3) + 3 & & = 2(3) + 3 \\ & = 9 & & = 9 \\ & = \text{R.H.S} & & = \text{R.H.S} \\ \\ \therefore x & = 2, - 4 \end{align}

(b)

\begin{align} 11 - 2|x + 3| & = 1 \\ 11 - 1 & = 2|x + 3| \\ 10 & = 2|x + 3| \\ {10 \over 2} & = |x + 3| \\ 5 & = |x + 3| \end{align} \begin{align} x + 3 = 5 \phantom{0000}&\text{or}\phantom{000} x + 3 = -5 \\ x = 5 - 3 \phantom{(} &\phantom{or000+3} x = -5 - 3 \\ x = 2 \phantom{0000} &\phantom{or000+3} x = -8 \end{align}
$$ 11 - 2|x + 3| = 1 $$ \begin{align} \text{For } & x = 2, \phantom{0} & \phantom{00000} \text{For } & x = -8, \\ \\ \text{L.H.S} & = 11 - 2|2 + 3| & \text{L.H.S} & = 11 - 2|-8 + 3| \\ & = 11 - 2|5| & & = 11 - 2|-5| \\ & = 11 - 2(5) & & = 11 - 2(5) \\ & = 1 & & = 1 \\ & = \text{R.H.S} & & = \text{R.H.S} \\ \\ \therefore x & = 2, - 8 \end{align}

(a)

$$ |2x - 3| = x $$ \begin{align} 2x - 3 = x \phantom{0000} & \text{or} \phantom{000} 2x - 3 = -x \\ 2x - x = 3 \phantom{0000} & \phantom{or000} 2x + x = 3 \\ x = 3 \phantom{0000} & \phantom{or000+x} 3x = 3 \\ & \phantom{or000+x3} x = {3 \over 3} \\ & \phantom{or000+x3} x = 1 \end{align}
$$ |2x - 3| = x $$ \begin{align} \text{For } x & = 3, \phantom{0} & \phantom{00000} \text{For } x & = 1, \\ \\ \text{L.H.S} & = |2(3) -3| & \text{L.H.S} & = |2(1) - 3| \\ & = |3| & & = |-1| \\ & = 3 & & = 1 \\ \\ \text{R.H.S} & = 3 & \text{R.H.S} & = 1 \\ & = \text{L.H.S} & & = \text{L.H.S} \\ \\ \therefore x & = 3, 1 \end{align}

(b)

$$ |5x - 8| = 3x $$ \begin{align} 5x - 8 = 3x \phantom{0000} &\text{or} \phantom{000} 5x - 8 = -3x \\ 5x - 3x = 8 \phantom{00000} & \phantom{or00} 5x + 3x = 8 \\ 2x = 8 \phantom{00000} & \phantom{or00+3x} 8x = 8 \\ x = {8 \over 2} \phantom{000()} & \phantom{or00+3x8} x = {8 \over 8} \\ x = 4 \phantom{00000} & \phantom{or00+3x8} x = 1 \end{align}
$$ |5x - 8| = 3x $$ \begin{align} \text{For } x & = 4, \phantom{0} & \phantom{00000} \text{For } x & = 1, \\ \\ \text{L.H.S} & = |5(4) - 8| & \text{L.H.S} & = |5(1) - 8| \\ & = |12| & & = |-3| \\ & = 12 & & = 3 \\ \\ \text{R.H.S} & = 3(4) & \text{R.H.S} & = 3(1) \\ & = 12 & & = 3 \\ & = \text{L.H.S} & & = \text{L.H.S} \\ \\ \therefore x & = 4, 1 \end{align}

(c)

\begin{align} 4|x - 6| - 5x & = 0 \\ 4|x - 6| & = 5x \\ |x - 6| & = {5 \over 4}x \end{align} \begin{align} x - 6 & = {5 \over 4}x \phantom{0} &\text{or} \phantom{00000} x - 6 & = -{5 \over 4}x \\ -6 & = {5 \over 4}x - x &\phantom{or00000} x + {5 \over 4}x & = 6 \\ -6 & = {1 \over 4}x & {9 \over 4}x & = 6 \\ -24 & = x & x & = 6 \div {9 \over 4} \\ & & x & = {8 \over 3} \end{align}
$$ 4|x - 6| - 5x = 0 $$ \begin{align} \text{For } x & = -24, \phantom{0} & \phantom{00000} \text{For } x & = {8 \over 3}, \\ \\ \text{L.H.S} & = 4|-24 - 6| - 5(-24) & \text{L.H.S} & = 4\left| {8 \over 3} - 6 \right| - 5\left(8 \over 3\right) \\ & = 4|-30| + 120 & & = 4 \left|-{10 \over 3}\right| - {40 \over 3} \\ & = 4(30) + 120 & & = 4\left(10 \over 3\right) - {40 \over 3} \\ & = 240 & & = 0 \\ & \ne \text{R.H.S} & & = \text{R.H.S} \\ \\ \therefore x & = {8 \over 3} \end{align}

\begin{align} \text{Firm A's rate} - \text{Firm B's rate} & = {1 \over 2} \times \text{Difference in rates between firm B and firm C} \\ \\ x - y & = {1 \over 2}|y - 5| \\ 2(x - y) & = |y - 5| \\ \\ 2x - 2y & = |y - 5| \end{align} \begin{align} 2x - 2y & = y - 5 \phantom{0} &\text{or} \phantom{00000} -(2x - 2y) & = y - 5 \\ 2x + 5 & = y + 2y & -2x + 2y & = y - 5 \\ 2x + 5 & = 3y & 2y - y & = 2x - 5 \\ {2 \over 3}x + {5 \over 3} & = y & y & = 2x - 5 \end{align}

(a)

$$ |x^2 - 4| = 2 $$ \begin{align} x^2 - 4 & = 2 \phantom{0} &\text{or} \phantom{00000} x^2 - 4 & = -2 \\ x^2 & = 2 + 4 & x^2 & = -2 + 4 \\ x^2 & = 6 & x^2 & = 2 \\ x & = \pm \sqrt{6} & x & = \pm \sqrt{2} \end{align}
$$ |x^2 - 4| = 2 $$ \begin{align} \text{For } x & = \pm\sqrt{6}, \phantom{0} & \phantom{00000} \text{For } x & = \pm\sqrt{2}, \\ \\ \text{L.H.S} & = |(\pm\sqrt{6})^2 - 4| & \text{L.H.S} & = |(\pm \sqrt{2})^2 - 4| \\ & = |6 - 4| & & = |2 - 4| \\ & = |2| & & = |-2| \\ & = 2 & & = 2 \\ & = \text{R.H.S} & & = \text{R.H.S} \\ \\ \therefore x & = \pm\sqrt{6}, \pm \sqrt{2} \end{align}

(b)

\begin{align} |x|^2 + x^2 & = 4 \\ x^2 + x^2 & = 4 \\ 2x^2 & = 4 \\ x^2 & = {4 \over 2} \\ & = 2 \\ x & = \pm \sqrt{2} \end{align}
$$ |x|^2 + x^2 = 4 $$ \begin{align} \text{For } x & = \sqrt{2}, \phantom{0} & \phantom{00000} \text{For } x & = -\sqrt{2}, \\ \\ \text{L.H.S} & = |\sqrt{2}|^2 + (\sqrt{2})^2 & \text{L.H.S} & = |-\sqrt{2}|^2 + (-\sqrt{2})^2 \\ & = (\sqrt{2})^2 + 2 & & = (\sqrt{2})^2 + 2 \\ & = 2 + 2 & & = 2 + 2 \\ & = 4 & & = 4 \\ & = \text{R.H.S} & & = \text{R.H.S} \\ \\ \therefore x & = \pm \sqrt{2} \end{align}

(c)

\begin{align} |x(x - 2)| & = 2x - 3 \\ |x^2 - 2x| & = 2x - 3 \end{align} \begin{align} x^2 - 2x & = 2x - 3 \phantom{0} &\text{or} \phantom{00000} x^2 - 2x & = -(2x - 3) \\ x^2 - 2x - 2x + 3 & = 0 & x^2 -2x & = -2x + 3 \\ x^2 - 4x + 3 & = 0 & x^2 - 2x + 2x & = 3 \\ (x - 1)(x - 3) & = 0 & x^2 & = 3 \\ & & x & = \pm\sqrt{3} \\ x = 1 \text{ or } x & = 3 \end{align}
$$ |x(x - 2)| = 2x - 3 $$ \begin{align} \text{For } x & = 1, \phantom{0} & \phantom{00000} \text{For } x & = 3, \\ \\ \text{L.H.S} & = |1(1 - 2)| & \text{L.H.S} & = |3(3 - 2)| \\ & = |-1| & & = |3| \\ & = 1 & & = 3 \\ \\ \text{R.H.S} & = 2(1) - 3 & \text{R.H.S} & = 2(3) - 3 \\ & = -1 & & = 3 \\ & \ne \text{L.H.S} & & = \text{R.H.S} \end{align} \begin{align} \text{For } x & = \sqrt{3}, \phantom{0} & \phantom{00000} \text{For } x & = -\sqrt{3}, \\ \\ \text{L.H.S} & = |\sqrt{3}(\sqrt{3} - 2)| & \text{L.H.S} & = |-\sqrt{3}(-\sqrt{3} - 2)| \\ & = |3 - 2\sqrt{3}| & & = |\sqrt{3} + 2\sqrt{3}| \\ & = -(3 - 2\sqrt{3}) & & = |3 \sqrt{3}| \\ & = -3 + 2\sqrt{3} & & = 3\sqrt{3} \\ & = 2\sqrt{3} - 3 \\ \\ \text{R.H.S} & = 2(\sqrt{3}) - 3 & \text{R.H.S} & = 2(-\sqrt{3}) - 3 \\ & = 2\sqrt{3} - 3 & & = -2\sqrt{3} - 3 \\ & = \text{L.H.S} & & \ne \text{L.H.S} \\ \\ \\ \therefore x & = 3, \sqrt{3} \end{align}

(d)

\begin{align} 2|x| & = x^2 - 4x \\ |x| & = {1 \over 2}x^2 - 2x \end{align} \begin{align} x & = {1 \over 2}x^2 - 2x \phantom{0} &\text{or} \phantom{00000} x & = -\left({1 \over 2}x^2 - 2x\right) \\ 0 & = {1 \over 2}x^2 - 2x - x & x & = -{1 \over 2}x^2 + 2x \\ 0 & = {1 \over 2}x^2 - 3x & 0 & = -{1 \over 2}x^2 + 2x - x \\ 0 & = x^2 - 6x & 0 & = -{1 \over 2}x^2 + x \\ 0 & = x(x - 6) & 0 & = x^2 - 2x \\ & & 0 & = x(x - 2) \\ \\ x & = 0 \text{ or } x = 6 & x & = 0 \text{ or } x = 2 \end{align}
$$ 2|x| = x^2 - 4x $$ \begin{align} \text{For } x & = 0, \phantom{0} & \phantom{00000} \text{For } x & = 6, \phantom{0} & \phantom{00000} \text{For } x & = 2, \\ \\ \text{L.H.S} & = 2|0| & \text{L.H.S} & = 2|6| & \text{L.H.S} & = 2|2| \\ & = 2(0) & & = 2(6) & & = 2(2) \\ & = 0 & & = 12 & & = 4 \\ \\ \text{R.H.S} & = (0)^2 - 4(0) & \text{R.H.S} & = (6)^2 - 4(6) & \text{R.H.S} & = (2)^2 - 4(2) \\ & = 0 & & = 12 & & = - 4 \\ & = \text{L.H.S} & & = \text{L.H.S} & & \ne \text{L.H.S} \\ \\ \\ \therefore x & = 0, 6 \end{align}

\begin{align} y & = 2x - 6 \phantom{00000} [\text{Straight line } y = mx + c] \\ \\ \text{When } & x = 0, \\ y & = 2(0) - 6 \\ & = -6 \\ \\ \implies & y\text{-intercept is } -6 \\ \\ \text{When } & y = 0, \\ 0 & = 2x - 6 \\ 6 & = 2x \\ {6 \over 2} & = x \\ 3 & = x \\ \\ \implies & x\text{-intercept is } 3 \end{align}

Reflect the portion of the graph below the $y$-axis to obtain $y = |2x - 6|$.

(i)

\begin{align} y & = 2|x - 3| \\ & = |2||x - 3| \\ & = |2(x - 3)| \\ & = |2x - 6| \end{align}

(ii)

\begin{align} y & = |6 - 2x| \\ & = |(-1)(-6 + 2x)| \\ & = |(-1)(2x - 6)| \\ & = |-1| |2x - 6| \\ & = (1)|2x - 6| \\ & = |2x - 6| \end{align}

(a)

\begin{align} y & = 1 - x^2 \\ \\ \text{Since coefficient of } x^2 & \text{ is negative} \implies \text{Maximum curve } (\cap) \\ \\ \text{When } & y = 0, \\ 0 & = 1 - x^2 \\ 0 & = (1 + x)(1 - x) \\ \\ x & = - 1 \text{ or } x = 1 \\ \\ \implies x &\text{-intercepts are } -1 \text{ and } 1 \\ \\ \text{When } & x = 0, \\ y & = (1 + 0)(1 - 0) \\ & = 1 \\ \\ \implies y & \text{-intercept is } 1 \\ \\ x\text{-coordinate of turning point} & = {-1 + 1 \over 2} \\ & = 0 \\ \\ \text{When } & x = 1, \\ y & = 1 - (1)^2 \\ & = 0 \\ \\ \implies & \text{ Turning point is } (1, 0) \\ \\ \text{When } & x = -2, \\ y & = 1 - (-2)^2 \\ & = -3 \\ \\ & \text{Left limit is } (-2, -3) \end{align}

Reflect the portion of the graph below the $y$-axis to obtain $y = |1 - x^2|$.

(b)

\begin{align} y & = x^2 - 4 \\ \\ \text{Since coefficient of } x^2 & \text{ is positive} \implies \text{Minimum curve } (\cup) \\ \\ \text{When } & y = 0, \\ 0 & = x^2 - 4 \\ 0 & = (x + 2)(x - 2) \\ x & = - 2 \text{ or } x = 2 \\ \\ \implies x &\text{-intercepts are } -2 \text{ and } 2 \\ \\ \text{When } & x = 0, \\ y & = (0)^2 - 4 \\ & = - 4 \\ \\ \implies y & \text{-intercept is } -4 \\ \\ x\text{-coordinate of turning point} & = {-2 + 2 \over 2} \\ & = 0 \\ \\ \text{When } & x = 0, \\ y & = (0)^2 - 4 \\ & = -4 \\ \\ \implies & \text{ Turning point is } (0, -4) \end{align}

Reflect the portion of the graph below the $y$-axis to obtain $y = |x^2 - 4|$.

(c)

\begin{align} y & = x^2 - 4x + 3 \\ \\ \text{Since coefficient of } x^2 & \text{ is positive} \implies \text{Minimum curve } (\cup) \\ \\ \text{When } & y = 0, \\ 0 & = (x - 1)(x - 3) \\ \\ x & = 1 \text{ or } x = 3 \\ \\ \implies x &\text{-intercepts are } 1 \text{ and } 3 \\ \\ \text{When } & x = 0, \\ y & = (0 - 1)(0 - 3) \\ & = 3 \\ \\ \implies y & \text{-intercept is } 3 \\ \\ x\text{-coordinate of turning point} & = {1 + 3 \over 2} \\ & = 2 \\ \\ \text{When } & x = 2, \\ y & = (2 - 1)(2 - 3) \\ & = -1 \\ \\ \implies & \text{ Turning point is } (2, - 1) \\ \\ \text{When } x & = 5, \\ y & = (5 - 1)(5 - 3) \\ & = 8 \\ \\ & \text{Right limit is } (5, 8) \end{align}

Reflect the portion of the graph below the $y$-axis to obtain $y = |(x - 1)(x - 3)|$.

(d)

\begin{align} y & = x^2 - 4x \\ \\ \text{Since coefficient of } x^2 & \text{ is positive} \implies \text{Minimum curve } (\cup) \\ \\ \text{When } & y = 0, \\ 0 & = x^2 - 4x \\ 0 & = x(x - 4) \\ \\ x & = 0 \text{ or } x = 4 \\ \\ \implies x &\text{-intercepts are } 0 \text{ and } 4 \\ \\ \text{When } & x = 0, \\ y & = (0)^2 - 4(0) \\ & = 0 \\ \\ \implies y & \text{-intercept is } 0 \\ \\ x\text{-coordinate of turning point} & = {0 + 4 \over 2} \\ & = 2 \\ \\ \text{When } & x = 2, \\ y & = (2)^2 - 4(2) \\ & = -8 \\ \implies & \text{ Turning point is } (2, - 8) \end{align}

Reflect the portion of the graph below the $y$-axis to obtain $y = |x^2 - 4x|$.

(i)

\begin{align} y & = |2x + c| \\ \\ \text{When } x = 1 & \text{ and } y = 2, \\ 2 & = |2(1) + c| \\ 2 & = |2 + c| \\ \\ 2 = 2 + c \phantom{000}&\text{or}\phantom{000} -2 = 2 + c \\ 0 = c \phantom{00020()} &\phantom{or000} -4 = c \\ \\ \\ \text{When } x = 0 & \text{ and } y = 4, \\ 4 & = |2(0) + c| \\ 4 & = |c| \\ \\ 4 = c \phantom{000}&\text{or} \phantom{000} -4 = c \\ \\ \\ \therefore c & = - 4 \end{align}

(ii)

\begin{align} y & = |2x + (-4)| \\ & = |2x - 4| \\ \\ \text{Consider: } y & = 2x - 4 \phantom{00000} [\text{Straight line } y = mx + c] \\ \\ \text{When } & x = 0, \\ y & = 2(0) - 4 \\ & = - 4 \\ \\ \implies & y\text{-intercept is } -4 \\ \\ \text{When } & y = 0, \\ 0 & = 2x - 4 \\ 4 & = 2x \\ {4 \over 2} & = x \\ 2 & = x \\ \\ \implies & x\text{-intercept is } 2 \end{align}

Reflect the portion of the graph below the $y$-axis to obtain $y = |2x - 4|$.

(i)

\begin{align} y & = 2|x + c| - 5 \\ \\ \text{When } x = -1 & \text{ and } y = 3, \\ 3 & = 2|(-1) + c| - 5 \\ 3 + 5 & = 2|-1 + c| \\ 8 & = 2|c - 1| \\ {8 \over 2} & = |c - 1| \\ 4 & = |c - 1| \\ \\ 4 = c - 1 \phantom{000}&\text{or}\phantom{000} -4 = c - 1 \\ 5 = c \phantom{00020()} &\phantom{or000} -3 = c \end{align}

(ii)

\begin{align} \text{If } & c = 5, \\ y & = 2|x + (5)| \\ & = 2|x + 5| \\ \\ \\ \text{When } & x = 0, \\ y & = 2|0 + 5| \\ & = 2|5| \\ & = 2(5) \\ & = 10 \\ \\ \implies & y\text{-intercept is } 10 \\ \\ \text{When } & y = 0, \\ 0 & = 2|x + 5| \\ 0 & = |x + 5| \\ \pm 0 & = x + 5 \\ 0 & = x + 5 \\ -5 & = x \\ \\ \implies & x\text{-intercept is } -5 \end{align}

\begin{align} \text{If } & c = -3, \\ y & = 2|x + (-3)| \\ & = 2|x - 3| \\ \\ \text{When } & x = 0, \\ y & = 2|0 - 3| \\ & = 2|-3| \\ & = 2(3) \\ & = 6 \\ \\ \implies & y\text{-intercept is } 6 \\ \\ \text{When } & y = 0, \\ 0 & = 2|x - 3| \\ 0 & = |x - 3| \\ \pm 0 & = x - 3 \\ 0 & = x - 3 \\ 3 & = x \\ \\ \implies & x\text{-intercept is } 3 \end{align}

(i)

\begin{align} y & = |3 - kx| \\ \\ \text{When } x = 5 & \text{ and } y = 12, \\ 12 & = |3 - k(5)| \\ 12 & = |3 - 5k| \\ \\ 12 = 3 - 5k \phantom{000}&\text{or}\phantom{000} -12 = 3 - 5k \\ 5k = 3 - 12 \phantom{2.+} &\phantom{or0000+} 5k = 3 + 12 \\ 5k = -9 \phantom{0000+.} &\phantom{or0000+} 5k = 15 \\ k = -{9 \over 5} \phantom{0000..} &\phantom{or000000.} k = 3 \end{align}

(ii)(a)

\begin{align} \text{If } & k = 3, \\ y & = |3 - 3x| \\ \\ \text{Consider } y & = 3 - 3x \phantom{00000} [\text{Straight line } y = mx + c] \\ \\ \text{When } & x = 0, \\ y & = 3 - 3(0) \\ & = 3 \\ \\ \implies & y\text{-intercept is } 3 \\ \\ \text{When } & y = 0, \\ 0 & = 3 - 3x \\ 3x & = 3 \\ x & = 1 \\ \\ \implies & x\text{-intercept is } 1 \\ \\ \text{When } & x = -1, \\ y & = 3 - 3(-1) \\ & = 6 \\ \\ & \text{Left limit is } (-1, 6) \\ \\ \text{When } & x = 5, \\ y & = 3 - 3(5) \\ & = -12 \\ \\ & \text{Right limit is } (5, -12) \end{align}

Reflect the portion of the graph below the $y$-axis to obtain $y = |3 - 3x|$.

(ii)(b)

\begin{align} |3 - 3x| & = 3 \\ \\ 3 - 3x = 3 \phantom{000+}&\text{or}\phantom{000} 3 - 3x = -3 \\ -3x = 3 - 3 \phantom{0} &\phantom{or00.+} -3x = -3 - 3 \\ -3x = 0 \phantom{0 - 3} &\phantom{or00.+} -3x = -6 \\ x = 0 \phantom{0 - 3} &\phantom{or00.+0-} x = 2 \end{align}
$$ |3 - 3x| = 3 $$ \begin{align} \text{For } x & = 0, \phantom{0} & \phantom{00000} \text{For } x & = 2, \\ \\ \text{L.H.S} & = |3 - 3(0)| & \text{L.H.S} & = |3 - 3(2)| \\ & = |3| & & = |-3| \\ & = 3 & & = 3 \\ & = \text{R.H.S} & & = \text{R.H.S} \\ \\ \\ \therefore x & = 0, 2 \end{align}

(i)

\begin{align} y & = 2x - 5 \phantom{00000} [\text{Straight line } y = mx + c] \\ \\ \text{When } & x = 0, \\ y & = 2(0) - 5 \\ & = -5 \\ \\ \implies & y\text{-intercept is } -5 \\ \\ \text{When } & y = 0, \\ 0 & = 2x - 5 \\ 5 & = 2x \\ {5 \over 2} & = x \\ 2.5 & = x \\ \\ \implies & x\text{-intercept is } 2.5 \end{align}

Reflect the portion of the graph below the $y$-axis to obtain $y = |2x - 5|$.

\begin{align} y & = x + 2 \phantom{00000} [\text{Straight line } y = mx + c] \\ \\ \text{When } & x = 0, \\ y & = 0 + 2 \\ & = 2 \\ \\ \implies & y\text{-intercept is } 2 \\ \\ \text{When } & y = 0, \\ 0 & = x + 2 \\ -2 & = x \\ \\ \implies & x\text{-intercept is } -2 \end{align}

(ii)

\begin{align} \underbrace{x + 2}_\text{Straight line} & = \underbrace{|2x - 5|}_\text{Modulus graph} \end{align} $$ \text{From part (i), there are 2 solutions.} $$

(i)

\begin{align} y & = x^2 + 2x - 3 \\ y & = (x + 3)(x - 1) \\ \\ \text{Since the coefficient of } & x^2 \text{ is positive} \implies \text{Minimum curve } (\cup) \\ \\ \text{When } & x = 0, \\ y & = (0 + 3)(0 - 1) \\ & = - 3 \\ \\ \implies & y\text{-intercept is } -3 \\ \\ \text{When } & y = 0, \\ 0 & = (x + 3)(x - 1) \\ \\ x & = - 3 \text{ or } x = 1 \\ \\ \implies & x\text{-intercept are } -3 \text{ and } 1 \\ \\ x-\text{coordinate of turning point} & = {-3 + 1 \over 2} \\ & = -1 \\ \\ \text{When } & x = -1, \\ y & = (-1 + 3)(-1 -1) \\ & = -4 \\ \\ \implies & \text{Turning point is } (-1, -4) \end{align}

Reflect the portion of the graph below the $y$-axis to obtain $y = |x^2 + 2x - 3|$.

(ii)

\begin{align} |x - 1| & = {2 \over |x + 3|} \\ |x - 1||x + 3| & = 2 \\ |(x - 1)(x + 3)| & = 2 \\ |x^2 + 3x - x - 3| & = 2 \\ |x^2 + 2x - 3| & = 2 \\ \\ \text{Since } y = |x^2 + 2x - 3|, & \text{ draw } y = 2 \end{align}

$$ \therefore \text{4 solutions} $$

(i)

\begin{align} y & = x^2 - 6x + 5 \\ y & = (x - 1)(x - 5) \\ \\ \text{Since the coefficient of } & x^2 \text{ is positive} \implies \text{Minimum curve } (\cup) \\ \\ \text{When } & x = 0, \\ y & = (0 - 1)(0 - 5) \\ & = 5 \\ \\ \implies & y\text{-intercept is } 5 \\ \\ \text{When } & y = 0, \\ 0 & = (x - 1)(x - 5) \\ \\ x & = 1 \text{ or } x = 5 \\ \\ \implies & x\text{-intercept are } 1 \text{ and } 5 \\ \\ x-\text{coordinate of turning point} & = {1 + 5 \over 2} \\ & = 3 \\ \\ \text{When } & x = 3, \\ y & = (3 - 1)(3 - 5) \\ & = -4 \\ \\ \implies & \text{Turning point is } (3, -4) \\ \\ \text{When } x & = -1, \\ y & = (-1 - 1)(-1 - 5) \\ & = 12 \\ \\ & \text{Left limit is } (-1, 12) \\ \\ \text{When } x & = 7, \\ y & = (7 - 1)(7 - 5) \\ & = 12 \\ \\ & \text{Right limit is } (7, 12) \end{align}

Reflect the portion of the graph below the $y$-axis to obtain $y = |x^2 - 6x + 5|$.

(ii)

\begin{align} |x^2 - 6x + 5| & = k \\ \\ \text{Since } y = |x^2 - 6x + 5 |, \text{ draw } y & = k \end{align}

(a)

$$ k = 0 \text{ or } k > 4 $$

(b)

$$ k = 4 $$

(c)

$$ 0 < k < 4 $$

(a)

\begin{align} & \text{No, } |x| \text{ is not always positive} \\ \\ & \text{If } x = 0, |x| = |0| = 0 \end{align}

(b)

$$ \text{If } x \text{ is a real number, } |x| \ge 0 $$

(c)

$$ \text{For all real values of } x $$

(i)(a)

\begin{align} & \text{False, as } x^2 \ge 0 \text{ and } \sqrt{x^2} \ge 0 \\ \\ & \text{For example, } \sqrt{(-1)^2} =\sqrt{1} = 1 \phantom{00000} [\text{Instead of } - 1] \end{align}

(i)(b)

\begin{align} & \text{False, as } x^2 \ge 0 \text{ and } \sqrt{x^2} \ge 0 \\ \\ & \text{For example, } \sqrt{(-1)^2} =\sqrt{1} = 1 \phantom{00000} [\text{Instead of } \pm 1] \end{align}

(ii)

\begin{align} & \text{Resembles graph of } y = |x| \\ \\ & \implies \sqrt{x^2} = |x| \end{align}