Inequalities: ACJC 2025 Promo Q4

(a)

Since $y = |a - bx|$ meets the $x$-axis at $\left(\tfrac{9}{4}, 0\right)$: $$\begin{aligned} \left|a - \frac{9b}{4}\right| &= 0\\ 4a - 9b &= 0. \quad\text{(1)} \end{aligned}$$ The boundaries of the solution set are $x = 1$ and $x = 3$, where $|mx - 6| = |a - bx|$: $$\begin{aligned} &\big((mx-6)+(a-bx)\big)\big((mx-6)-(a-bx)\big) < 0, \end{aligned}$$ with roots $$\begin{aligned} x = \frac{6-a}{m-b} = 1 &\;\Rightarrow\; a - b + m = 6, \quad\text{(2)}\\ x = \frac{6+a}{m+b} = 3 &\;\Rightarrow\; -a + 3b + 3m = 6. \quad\text{(3)} \end{aligned}$$ Solving (1), (2) and (3) simultaneously: $$\begin{aligned} a = 9, \qquad b = 4, \qquad m = 1. \end{aligned}$$

(b)

$$\begin{aligned} \frac{kx^2+1}{x^2+(k+1)x+k} &> -1\\ \frac{kx^2+1}{x^2+(k+1)x+k} + 1 &> 0\\ \frac{(k+1)x^2 + (k+1)x + (k+1)}{x^2+(k+1)x+k} &> 0\\ \frac{(k+1)(x^2 + x + 1)}{x^2+(k+1)x+k} &> 0. \end{aligned}$$ Since $k + 1 > 0$ and $x^2 + x + 1 > 0$ for all real $x$ (discriminant $1 - 4 = -3 < 0$, positive leading coefficient): $$\begin{aligned} x^2 + (k+1)x + k &> 0\\ (x+1)(x+k) &> 0. \end{aligned}$$ As $k > 1$, we have $-k < -1$, so $$\begin{aligned} x < -k \quad\text{or}\quad x > -1. \end{aligned}$$