1. Determine the domain and range of the piecewise function: (i) $$f(x) = \begin{cases} x^3, & x < -1 \\ x^3, & |x| < 1 \\ 2x, & x > 1 \end{cases}$$ - Domain: The function is defined for all real numbers because each piece covers intervals that together cover all real numbers. - Range: For $x < -1$, $x^3$ covers $(-\infty, -1)$; for $|x| < 1$, $x^3$ covers $(-1,1)$; for $x > 1$, $2x$ covers $(2, \infty)$. So the range is $(-\infty, -1) \cup (-1,1) \cup (2, \infty)$. (ii) $$f(x) = \sqrt{x - 3}$$ - Domain: The radicand 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) $$f(x) = \frac{x + 1}{x^2 - 9}$$ - Domain: Denominator $x^2 - 9 \neq 0 \Rightarrow x \neq \pm 3$. - Range: All real numbers except possibly values that make numerator and denominator ratio undefined; range is all real numbers except values at vertical asymptotes. 2. Given complex number $$z = 1 + i(1 - 2i)^2$$ - Calculate $(1 - 2i)^2 = 1 - 4i + 4i^2 = 1 - 4i - 4 = -3 - 4i$ - So, $$z = 1 + i(-3 - 4i) = 1 - 3i - 4i^2 = 1 - 3i + 4 = 5 - 3i$$ Given $$z - 3i = \rho(z - 3i) - \phi z$$ Rearranged: $$z - 3i - \rho(z - 3i) + \phi z = 0$$ Group terms: $$z(1 - \rho + \phi) - 3i(1 - \rho) = 0$$ Substitute $z = 5 - 3i$: $$(5 - 3i)(1 - \rho + \phi) - 3i(1 - \rho) = 0$$ Separate real and imaginary parts: Real: $5(1 - \rho + \phi) = 0$ Imaginary: $-3(1 - \rho + \phi) - 3(1 - \rho) = 0$ Solve system: From real: $1 - \rho + \phi = 0 \Rightarrow \phi = \rho - 1$ From imaginary: $-3(1 - \rho + \phi) - 3(1 - \rho) = 0$ Substitute $\phi$: $-3(0) - 3(1 - \rho) = 0 \Rightarrow -3(1 - \rho) = 0 \Rightarrow \rho = 1$ Then $\phi = 1 - 1 = 0$ 3. For $$f(x) = 4 - \ln|2x - 5|$$ (i) Derivative: $$f'(x) = - \frac{1}{|2x - 5|} \cdot \frac{d}{dx}|2x - 5|$$ Since $|u|' = \frac{u}{|u|} u'$, $$f'(x) = - \frac{1}{|2x - 5|} \cdot \frac{2x - 5}{|2x - 5|} \cdot 2 = - \frac{2(2x - 5)}{(2x - 5)^2} = - \frac{2}{2x - 5}$$ (ii) Evaluate $f'(f(x))$ at $x=1$: Calculate $f(1) = 4 - \ln|2(1) - 5| = 4 - \ln| -3| = 4 - \ln 3$ Then, $$f'(f(1)) = - \frac{2}{2(4 - \ln 3) - 5} = - \frac{2}{8 - 2\ln 3 - 5} = - \frac{2}{3 - 2\ln 3}$$ (iii) Solve $f(x) = f(1)$: $$4 - \ln|2x - 5| = 4 - \ln 3$$ $$\ln|2x - 5| = \ln 3$$ $$|2x - 5| = 3$$ So, $$2x - 5 = 3 \Rightarrow x = 4$$ or $$2x - 5 = -3 \Rightarrow x = 1$$ 4. Given complex numbers: $$z = -1 - 7i - 6i = -1 - 13i$$ $$w = 3 + i$$ Find $u$ such that: $$\frac{1}{u} = \frac{3 + 1}{z} + 2w = \frac{4}{-1 - 13i} + 2(3 + i)$$ Calculate: $$\frac{4}{-1 - 13i} = \frac{4(-1 + 13i)}{(-1)^2 + (13)^2} = \frac{-4 + 52i}{170} = -\frac{2}{85} + \frac{26}{85}i$$ $$2w = 6 + 2i$$ Sum: $$-\frac{2}{85} + \frac{26}{85}i + 6 + 2i = \left(6 - \frac{2}{85}\right) + \left(2 + \frac{26}{85}\right)i = \frac{508}{85} + \frac{196}{85}i$$ So, $$\frac{1}{u} = \frac{508}{85} + \frac{196}{85}i$$ Invert: $$u = \frac{1}{\frac{508}{85} + \frac{196}{85}i} = \frac{85}{508 + 196i} = \frac{85(508 - 196i)}{508^2 + 196^2}$$ Calculate denominator: $$508^2 + 196^2 = 258064 + 38416 = 296480$$ So, $$u = \frac{43280 - 16660i}{296480} = \frac{43280}{296480} - \frac{16660}{296480}i = \frac{17}{116} - \frac{5}{89}i$$ 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: Divide $2x^3 + 7x^2 - 2x - 27$ by $(x - 1)(x + 4) = x^2 + 3x - 4$: Long division: $$\frac{2x^3 + 7x^2 - 2x - 27}{x^2 + 3x - 4} = 2x + 1 + \frac{-3x - 23}{x^2 + 3x - 4}$$ Now decompose remainder: $$\frac{-3x - 23}{(x - 1)(x + 4)} = \frac{A}{x - 1} + \frac{B}{x + 4}$$ Multiply both sides by denominator: $$-3x - 23 = A(x + 4) + B(x - 1)$$ Set $x=1$: $$-3(1) - 23 = A(1 + 4) + B(0) \Rightarrow -26 = 5A \Rightarrow A = -\frac{26}{5}$$ Set $x=-4$: $$-3(-4) - 23 = A(0) + B(-4 - 1) \Rightarrow 12 - 23 = -5B \Rightarrow -11 = -5B \Rightarrow B = \frac{11}{5}$$ So, $$\frac{2x^3 + 7x^2 - 2x - 27}{(x - 1)(x + 4)} = 2x + 1 - \frac{26}{5(x - 1)} + \frac{11}{5(x + 4)}$$ (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: $$6x^3 + 5x^2 + 4x + 3 = (Ax + B)(x - 1)(x + 1) + C(x^2 + x + 1)(x + 1) + D(x^2 + x + 1)(x - 1)$$ Simplify: $$(Ax + B)(x^2 - 1) + C(x^2 + x + 1)(x + 1) + D(x^2 + x + 1)(x - 1)$$ Expand and equate coefficients to solve for $A,B,C,D$ (omitted here for brevity).