1. **Problem (i) Part 1:** Find the range of values of $x$ for which $$\frac{2x+1}{x-2} < 1.$$ **Step 1:** Start with the inequality: $$\frac{2x+1}{x-2} < 1.$$ **Step 2:** Bring all terms to one side: $$\frac{2x+1}{x-2} - 1 < 0.$$ **Step 3:** Write with common denominator: $$\frac{2x+1 - (x-2)}{x-2} < 0.$$ Simplify numerator: $$\frac{2x+1 - x + 2}{x-2} = \frac{x + 3}{x-2} < 0.$$ **Step 4:** The inequality $$\frac{x+3}{x-2} < 0$$ holds when numerator and denominator have opposite signs. - Numerator $x+3 < 0 \Rightarrow x < -3$ (M1) - Denominator $x-2 > 0 \Rightarrow x > 2$ or $x-2 < 0 \Rightarrow x < 2$ (A1) **Step 5:** Combine signs: - If numerator negative ($x < -3$) and denominator positive ($x > 2$) no overlap. - If numerator positive ($x > -3$) and denominator negative ($x < 2$), then $-3 < x < 2$. **Step 6:** Check intervals: - For $x < -3$, numerator negative, denominator negative ($x-2 < 0$), so fraction positive, not less than zero. - For $-3 < x < 2$, numerator positive, denominator negative, fraction negative, satisfies inequality. - For $x > 2$, numerator positive, denominator positive, fraction positive, no. **Step 7:** Exclude $x=2$ (denominator zero). **Answer:** $$\boxed{-3 < x < 2}.$$ 2. **Problem (i) Part 2:** Solve $$\left|\frac{x+1}{2x-3}\right| \leq \frac{x+1}{2x-3}.$$ **Step 1:** Recall that for any real number $a$, $|a| \leq a$ implies $a \geq 0$. **Step 2:** Set $$a = \frac{x+1}{2x-3}.$$ Inequality becomes $$|a| \leq a \Rightarrow a \geq 0.$$ **Step 3:** Solve $$\frac{x+1}{2x-3} \geq 0.$$ **Step 4:** Find critical points where numerator or denominator is zero: - Numerator zero at $x = -1$. - Denominator zero at $x = \frac{3}{2}$. **Step 5:** Test intervals determined by $-1$ and $\frac{3}{2}$: - For $x < -1$, numerator negative, denominator negative (since $2x-3 < 0$), fraction positive. - For $-1 < x < \frac{3}{2}$, numerator positive, denominator negative, fraction negative. - For $x > \frac{3}{2}$, numerator positive, denominator positive, fraction positive. **Step 6:** Inequality holds where fraction $\geq 0$: - $x < -1$ or $x > \frac{3}{2}$. **Step 7:** Exclude $x = \frac{3}{2}$ (denominator zero). **Answer:** $$\boxed{x < -1 \text{ or } x > \frac{3}{2}}.$$ 3. **Problem (ii):** Find real $x$ such that $$e^{3x} - 3e^x - 4e^{-x} = 0,$$ leaving answer in terms of natural logarithms. **Step 1:** Let $$y = e^x > 0.$$ Rewrite equation: $$y^3 - 3y - 4y^{-1} = 0.$$ Multiply both sides by $y$ to clear denominator: $$y^4 - 3y^2 - 4 = 0.$$ **Step 2:** Let $$z = y^2 > 0.$$ Equation becomes: $$z^2 - 3z - 4 = 0.$$ **Step 3:** Solve quadratic: $$z = \frac{3 \pm \sqrt{9 + 16}}{2} = \frac{3 \pm 5}{2}.$$ **Step 4:** Possible values: - $$z = \frac{3 + 5}{2} = 4,$$ - $$z = \frac{3 - 5}{2} = -1$$ (discard negative since $z = y^2 > 0$). **Step 5:** So $$y^2 = 4 \Rightarrow y = 2$$ (since $y = e^x > 0$). **Step 6:** Solve for $x$: $$e^x = 2 \Rightarrow x = \ln 2.$$ **Answer:** $$\boxed{x = \ln 2}.$$