1. (a) Evaluate $3 \log_2 4 + \frac{1}{2} \log_2 16 - \frac{1}{3} \log_2 8$. - Recall that $\log_b a^n = n \log_b a$ and $\log_2 4 = 2$, $\log_2 16 = 4$, $\log_2 8 = 3$. - Substitute values: $3 \times 2 + \frac{1}{2} \times 4 - \frac{1}{3} \times 3 = 6 + 2 - 1 = 7$. (b) Solve for $x$ given $\log x^4 - \log x^3 = \log 5x - \log 2x$. - Use $\log a - \log b = \log \frac{a}{b}$. - Left side: $\log \frac{x^4}{x^3} = \log x$. - Right side: $\log \frac{5x}{2x} = \log \frac{5}{2}$. - So, $\log x = \log \frac{5}{2} \implies x = \frac{5}{2}$. (c) Express $\frac{3\sqrt{5} - \sqrt{3}}{\sqrt{5} - \sqrt{3}}$ in the form $a + b\sqrt{c}$. - Multiply numerator and denominator by conjugate $\sqrt{5} + \sqrt{3}$: $$\frac{(3\sqrt{5} - \sqrt{3})(\sqrt{5} + \sqrt{3})}{(\sqrt{5} - \sqrt{3})(\sqrt{5} + \sqrt{3})} = \frac{3\sqrt{5} \times \sqrt{5} + 3\sqrt{5} \times \sqrt{3} - \sqrt{3} \times \sqrt{5} - \sqrt{3} \times \sqrt{3}}{5 - 3}$$ - Simplify numerator: $$3 \times 5 + 3 \sqrt{15} - \sqrt{15} - 3 = 15 + 2\sqrt{15} - 3 = 12 + 2\sqrt{15}$$ - Denominator: $2$. - Final form: $\frac{12}{2} + \frac{2\sqrt{15}}{2} = 6 + \sqrt{15}$. - So, $a=6$, $b=1$, $c=15$. 2. Consider $f(x) = 5 - \sqrt{x} + 3 = 8 - \sqrt{x}$. (a) Domain: $x \geq 0$ because of $\sqrt{x}$. Range: Since $\sqrt{x} \geq 0$, $f(x) \leq 8$, so range is $(-\infty, 8]$. (b) Find inverse $f^{-1}(x)$: - Set $y = 8 - \sqrt{x}$. - Solve for $x$: $\sqrt{x} = 8 - y$. - Square both sides: $x = (8 - y)^2$. - Swap $x$ and $y$ for inverse: $f^{-1}(x) = (8 - x)^2$. (c) Domain of $f^{-1}$ is range of $f$: $(-\infty, 8]$. Range of $f^{-1}$ is domain of $f$: $[0, \infty)$. 3. (a) Venn diagram for $(A \cap B^c) \cup (A \cap C^c)$ highlights elements in $A$ but not in $B$ or not in $C$. (b) Given $A = \{1,2,3,4\}$, $B = \{4,5,6,7,9\}$, $C = \{7,8,9\}$: (i) $(A \cup B) \cap C$: - $A \cup B = \{1,2,3,4,5,6,7,9\}$. - Intersection with $C$: $\{7,8,9\} \cap \{1,2,3,4,5,6,7,9\} = \{7,9\}$. (ii) $(A \cup B) - C$: - Remove elements of $C$ from $A \cup B$: $\{1,2,3,4,5,6,7,9\} - \{7,8,9\} = \{1,2,3,4,5,6\}$. (iii) Symmetric difference $(A \cap B) \oplus C$: - $A \cap B = \{4\}$. - Symmetric difference is elements in either set but not both. - $\{4\} \oplus \{7,8,9\} = \{4,7,8,9\}$. 4. (a) Simplify $\frac{(x+1)!}{(x-1)!}$: - Expand numerator: $(x+1)! = (x+1) x (x-1)!$. - Cancel $(x-1)!$: $$\frac{(x+1)!}{(x-1)!} = (x+1) x$$. (b) Number of permutations of 6 letters taken 3 at a time: - Formula: $P(6,3) = \frac{6!}{(6-3)!} = \frac{6!}{3!} = \frac{720}{6} = 120$. (c) Show $\binom{n}{k} = \binom{n}{n-k}$: - By definition: $$\binom{n}{k} = \frac{n!}{k! (n-k)!}$$ - Similarly: $$\binom{n}{n-k} = \frac{n!}{(n-k)! k!}$$ - Both expressions are equal. (d) Show $\binom{8}{6} = \binom{8}{2}$: - Using above: $$\binom{8}{6} = \binom{8}{8-6} = \binom{8}{2}$$ 5. (a) Differentiate: (i) $f(x) = 2x^4 + e^x$: - $f'(x) = 8x^3 + e^x$. (ii) $f(x) = 3 e^{3x} \sin x$: - Use product rule: $(uv)' = u'v + uv'$. - $u = 3 e^{3x}$, $u' = 9 e^{3x}$. - $v = \sin x$, $v' = \cos x$. - So, $$f'(x) = 9 e^{3x} \sin x + 3 e^{3x} \cos x = 3 e^{3x} (3 \sin x + \cos x)$$. (b) Integrals: (i) $\int_0^3 3 e^{\frac{5}{6} u} du$: - Let $a = \frac{5}{6}$. - Integral: $$3 \int_0^3 e^{a u} du = 3 \left[ \frac{1}{a} e^{a u} \right]_0^3 = 3 \times \frac{1}{a} (e^{3a} - 1) = \frac{3}{5/6} (e^{\frac{5}{2}} - 1) = \frac{18}{5} (e^{2.5} - 1)$$. (ii) $\int_2^3 (6u^2 - 2u + 3) du$: - Integrate termwise: $$\left[ 2u^3 - u^2 + 3u \right]_2^3 = (2 \times 27 - 9 + 9) - (2 \times 8 - 4 + 6) = (54 - 9 + 9) - (16 - 4 + 6) = 54 - 16 = 38$$. Final answers: 1.(a) 7 1.(b) $x=\frac{5}{2}$ 1.(c) $6 + \sqrt{15}$ with $a=6$, $b=1$, $c=15$ 2.(a) Domain: $[0, \infty)$, Range: $(-\infty, 8]$ 2.(b) $f^{-1}(x) = (8 - x)^2$ 2.(c) Domain of $f^{-1}$: $(-\infty, 8]$, Range: $[0, \infty)$ 3.(b)(i) $\{7,9\}$ 3.(b)(ii) $\{1,2,3,4,5,6\}$ 3.(b)(iii) $\{4,7,8,9\}$ 4.(a) $(x+1) x$ 4.(b) 120 4.(c) Shown equality 4.(d) Shown equality 5.(a)(i) $8x^3 + e^x$ 5.(a)(ii) $3 e^{3x} (3 \sin x + \cos x)$ 5.(b)(i) $\frac{18}{5} (e^{2.5} - 1)$ 5.(b)(ii) 38