(a)

(b)

\begin{align} \text{No, since the element } 1 \notin B \end{align}

(a)

(b)

\begin{align} \text{Yes, since every element in } C \text{ is in } D \text{ and } C \ne D \end{align}

(a)

(b)

\begin{align} M \text{ and } N \text{ are disjoint sets} \end{align}

(a)

\begin{align} A' & = \{ \text{diamond, jade, platinum, ruby} \} \end{align}

(b)

(a)

\begin{align} \xi & = \{ \text{c, o, m, p, l, e, n, t} \} \\ \\ A & = \{ \text{e, l, m, n, t} \} \end{align}

(b)

(c)

\begin{align} A' & = \{ \text{c, o, p} \} \end{align}

(a)

\begin{align} A & = \{ \text{democracy, justice, peace} \} \\ \\ B & = \{ \text{democracy, justice, progress} \} \end{align}

(b)

(a)(i)

\begin{align} 11 \in A \end{align}

(a)(ii)

\begin{align} \{ 13, 15 \} \not \subset B \end{align}

(b)

(a)

\begin{align} & \text{Yes, since all triangles with three equal sides has at least two equal sides, but} \\ & \text{not all triangles with at least two equal sides have three equal sides} \end{align}

(b)

(a)

\begin{align} 2x^2 - 7x + 3 & = 0 \\ (2x - 1)(x - 3) & = 0 \end{align} \begin{align} 2x - 1 & = 0 && \text{ or } & x - 3 & = 0 \\ 2x & = 1 &&& x & = 3 \\ x & = {1 \over 2} \end{align}
\begin{align} \therefore A & = \{ 3 \} \\ \\ \\ 5^{x^2} & = 125^x \\ 5^{x^2} & = (5^3)^x \\ 5^{x^2} & = 5^{3x} \phantom{000000} [ (a^m)^n = a^{mn}] \\ \\ x^2 & = 3x \\ x^2 - 3x & = 0 \\ x(x - 3) & = 0 \end{align} \begin{align} x & = 0 && \text{ or } & x - 3 & = 0 \\ & &&& x & = 3 \end{align}
$$ \therefore B = \{ 0, 3 \} $$

(b)

(a)(i)

\begin{align} A' & = \{ x : x \text{ is not an odd integer} \} \end{align}

(a)(ii)

\begin{align} A' & = \{ 0, 2, 4, 6, ... \} \end{align}

(b) Note: All multiples of 4 are even numbers (i.e. not odd numbers)

(a)

\begin{align} A' & = \{ \text{sodium, potassium, magnesium, iron} \} \end{align}

(b)

(c)(i)

\begin{align} n(A) + n(A') & = 3 + 4 = 7 \end{align}

(c)(ii)

\begin{align} n(A) + n(A') & = n(\xi) \end{align}

(c)(iii)

\begin{align} \text{Yes. For a finite universal set, every element not in } A \text{ must be in } A' \end{align}

(a)(i)

\begin{align} Q \subset R \end{align}

(a)(ii)

\begin{align} \{1 , 2 \} \subset P \phantom{000000} [\text{Note } \{ 1, 2 \} \text{ is a set} ] \end{align}

(a)(iii)

\begin{align} 2 \in R \end{align}

(a)(iv)

\begin{align} 9 \notin P \end{align}

(b)

\begin{align} P: k & = 3 \times 8 = 24 \\ \\ Q: k & = 1 \times 5 = 5 \\ \\ R: k & = 1 \times 10 = 10 \end{align}

(c)

\begin{align} & \text{Find the product of the common element(s) in the two sets} \\ \\ & \text{For example, the HCF of } P(24) \text{ and } R(10) \text{ is } 1 \times 2 = 2 \end{align}

Note: Only 1 prime number (2) is even and some perfect squares (4, 16, 36, ..) are even

(a)

$$ \text{True} $$

(b)

(a)

\begin{align} \xi & = \{ 1, 2, 3, 6, 8 \} \end{align}

(b)

\begin{align} C & = \{ 3, 8 \} \end{align}

(c)