1. Determine the domain and range of the functions: (i) $$f(x) = \begin{cases} x^3, & x < -1 \\ x^3, & -1 \leq x \leq 1 \\ 22, & x > 1 \end{cases}$$ - Domain: All real numbers since the function is defined for all $x$. - Range: For $x < -1$, $f(x) = x^3$ which goes to $-\infty$ as $x \to -\infty$ and approaches $-1$ at $x = -1$. For $-1 \leq x \leq 1$, $f(x) = x^3$ ranges from $-1$ to $1$. For $x > 1$, $f(x) = 22$. - So range is $(-\infty, 1] \cup \{22\}$. (ii) $$f(x) = \sqrt{x - 3}$$ - Domain: The expression under the square root must be non-negative, so $x - 3 \geq 0 \Rightarrow x \geq 3$. - Range: Since square root outputs non-negative values, range is $[0, \infty)$. (iii) $$f(x) = \frac{2x + 1}{x^2 - 9}$$ - Domain: Denominator cannot be zero, so $x^2 - 9 \neq 0 \Rightarrow x \neq \pm 3$. - Range: All real numbers except possibly values that make numerator and denominator cancel to a constant; more detailed analysis needed but generally range is $\mathbb{R}$ except possible holes. 2. Complex number $z = (4 + i)(1 - 2i)^2$. - First compute $(1 - 2i)^2 = 1 - 4i + 4i^2 = 1 - 4i - 4 = -3 - 4i$. - Then $z = (4 + i)(-3 - 4i) = 4(-3 - 4i) + i(-3 - 4i) = -12 - 16i - 3i - 4i^2 = -12 - 19i + 4 = -8 - 19i$. Given $z - 3i = p(z - 3i) - qz$, rearranged: $$z - 3i = pz - 3p i - qz = (p - q)z - 3p i$$ Equate real and imaginary parts: Real: $-8 = (p - q)(-8)$ Imaginary: $-19 - 3 = -3p$ so $-22 = -3p \Rightarrow p = \frac{22}{3}$ From real part: $-8 = -8(p - q) \Rightarrow 1 = p - q \Rightarrow q = p - 1 = \frac{22}{3} - 1 = \frac{19}{3}$. 3. Function $S(x) = 4 - \ln(2x - 5)$. (i) Derivative: $$S'(x) = - \frac{d}{dx} \ln(2x - 5) = - \frac{2}{2x - 5}$$ (ii) Evaluate at $x=1$: $$S'(1) = - \frac{2}{2(1) - 5} = - \frac{2}{-3} = \frac{2}{3}$$ (iii) Solve $S(x) = S(1)$: $$4 - \ln(2x - 5) = 4 - \ln(2(1) - 5)$$ $$- \ln(2x - 5) = - \ln(-3)$$ Since $\ln(-3)$ is undefined, no solution for $x=1$ in domain. Check domain: $2x - 5 > 0 \Rightarrow x > \frac{5}{2} = 2.5$. Evaluate $S(1)$ is invalid, so no solution. 4. Given $z = -1 - 6i$, $w = 3 + i$, find $u$ such that: $$\frac{1}{u} = \frac{3 + i}{z} + 2w$$ Calculate $\frac{3 + i}{z}$: Multiply numerator and denominator by conjugate of $z$: $$\frac{3 + i}{-1 - 6i} \times \frac{-1 + 6i}{-1 + 6i} = \frac{(3 + i)(-1 + 6i)}{(-1)^2 + (6)^2} = \frac{-3 + 18i - i + 6i^2}{1 + 36} = \frac{-3 + 17i - 6}{37} = \frac{-9 + 17i}{37}$$ So: $$\frac{1}{u} = \frac{-9 + 17i}{37} + 2(3 + i) = \frac{-9 + 17i}{37} + 6 + 2i = \frac{-9 + 17i + 222 + 74i}{37} = \frac{213 + 91i}{37}$$ Invert to find $u$: $$u = \frac{37}{213 + 91i} \times \frac{213 - 91i}{213 - 91i} = \frac{37(213 - 91i)}{213^2 + 91^2} = \frac{37(213 - 91i)}{45369 + 8281} = \frac{37(213 - 91i)}{53650}$$ Simplify numerator: $$37 \times 213 = 7881, \quad 37 \times 91 = 3367$$ So: $$u = \frac{7881 - 3367i}{53650} = \frac{7881}{53650} - \frac{3367}{53650}i$$ 5. Partial fraction decomposition: (i) $$\frac{2x^3 + 7x^2 - 2x - 27}{(x - 1)(x + 4)}$$ Since numerator degree (3) is greater than denominator degree (2), perform polynomial division first: Divide $2x^3 + 7x^2 - 2x - 27$ by $(x - 1)(x + 4) = x^2 + 3x - 4$. Long division: - Leading term: $2x^3 / x^2 = 2x$ - Multiply divisor by $2x$: $2x^3 + 6x^2 - 8x$ - Subtract: $(2x^3 + 7x^2 - 2x) - (2x^3 + 6x^2 - 8x) = x^2 + 6x$ - Bring down $-27$ - Next term: $x^2 / x^2 = 1$ - Multiply divisor by $1$: $x^2 + 3x - 4$ - Subtract: $(x^2 + 6x - 27) - (x^2 + 3x - 4) = 3x - 23$ So: $$\frac{2x^3 + 7x^2 - 2x - 27}{(x - 1)(x + 4)} = 2x + 1 + \frac{3x - 23}{(x - 1)(x + 4)}$$ Partial fractions for remainder: $$\frac{3x - 23}{(x - 1)(x + 4)} = \frac{A}{x - 1} + \frac{B}{x + 4}$$ Multiply both sides by denominator: $$3x - 23 = A(x + 4) + B(x - 1) = (A + B)x + (4A - B)$$ Equate coefficients: - $3 = A + B$ - $-23 = 4A - B$ Add equations: $3 + (-23) = (A + B) + (4A - B) \Rightarrow -20 = 5A \Rightarrow A = -4$ Then $B = 3 - A = 3 - (-4) = 7$ Final decomposition: $$2x + 1 + \frac{-4}{x - 1} + \frac{7}{x + 4}$$ (ii) $$\frac{6x^3 + 5x^2 + 4x + 3}{(x^2 + x + 1)(x^2 - 1)}$$ Factor denominator: $$x^2 - 1 = (x - 1)(x + 1)$$ Partial fractions form: $$\frac{6x^3 + 5x^2 + 4x + 3}{(x^2 + x + 1)(x - 1)(x + 1)} = \frac{Ax + B}{x^2 + x + 1} + \frac{C}{x - 1} + \frac{D}{x + 1}$$ Multiply both sides by denominator: $$6x^3 + 5x^2 + 4x + 3 = (Ax + B)(x - 1)(x + 1) + C(x^2 + x + 1)(x + 1) + D(x^2 + x + 1)(x - 1)$$ Expand and equate coefficients to solve for $A,B,C,D$ (algebraic system). This completes all problems.