1. **Stating the problem:** We need to complete the sign table for $N_1(x)$, $N_2(x)$, $D_1(x)$, and $D_2(x)$, then solve the inequality $$\frac{N_1(x) \cdot N_2(x)}{D_1(x) \cdot D_2(x)} \leq 0$$ and deduce the solutions for $$\frac{4 - x^2}{2x - 3x^2} \leq 0.$$ 2. **Given sign information:** - $D_1(x)$ is negative on $(-\infty,0)$, zero at $0$, positive on $(0,2)$, zero at $2$, negative on $(2,+\infty)$. - $D_2(x)$ is positive on $(-\infty,2)$, zero at $2$, negative on $(2,+\infty)$. - Their product $D_1(x) \cdot D_2(x)$ is negative on $(-\infty,0)$, positive on $(0,2)$, zero at $0$ and $2$, negative on $(2,+\infty)$. 3. **(i) Complete the sign table for $N_1(x)$ and $N_2(x)$:** Since $N_1(x) \geq 0$ is given, and the problem states $N_1(x) \geq 0$ on $[-2,0]$ (from the graph description), we can infer: - $N_1(x) \geq 0$ on $[-2,0]$ and negative elsewhere. For $N_2(x)$, no explicit sign info is given, so we assume it is positive on the domain where the product is considered. 4. **(ii) Solve the inequality $$\frac{N_1(x) \cdot N_2(x)}{D_1(x) \cdot D_2(x)} \leq 0.$$** - The fraction is less than or equal to zero when the numerator and denominator have opposite signs or numerator is zero. - Since $N_1(x) \geq 0$ on $[-2,0]$ and $N_2(x)$ is positive, numerator $\geq 0$ on $[-2,0]$. - Denominator $D_1(x) \cdot D_2(x)$ is negative on $(-\infty,0)$, positive on $(0,2)$, zero at $0$ and $2$, negative on $(2,+\infty)$. Check intervals: - For $x \in (-\infty,-2)$: numerator negative, denominator negative $\Rightarrow$ fraction positive $>0$ (not solution). - For $x \in [-2,0)$: numerator $\geq 0$, denominator negative $\Rightarrow$ fraction $\leq 0$ (solution). - At $x=0$: denominator zero, fraction undefined. - For $x \in (0,2)$: numerator positive, denominator positive $\Rightarrow$ fraction positive $>0$ (not solution). - At $x=2$: denominator zero, fraction undefined. - For $x > 2$: numerator positive, denominator negative $\Rightarrow$ fraction negative $<0$ (solution). So solution set is $$[-2,0) \cup (2,+\infty).$$ 5. **(iii) Solve $$\frac{4 - x^2}{2x - 3x^2} \leq 0.$$** Rewrite numerator and denominator: - Numerator: $4 - x^2 = (2 - x)(2 + x)$ - Denominator: $2x - 3x^2 = x(2 - 3x)$ Sign analysis: - Numerator zero at $x = \pm 2$. - Denominator zero at $x=0$ and $x=\frac{2}{3}$. Intervals to test: $(-\infty, -2)$, $(-2,0)$, $(0, \frac{2}{3})$, $(\frac{2}{3}, 2)$, $(2, +\infty)$. Check signs: - $x < -2$: numerator positive (since $4 - x^2 > 0$ is false, actually negative), denominator negative (since $x$ negative, $2-3x$ positive), so numerator negative, denominator negative, fraction positive $>0$ (not solution). - $-2 < x < 0$: numerator positive, denominator negative, fraction negative $<0$ (solution). - $0 < x < \frac{2}{3}$: numerator positive, denominator positive, fraction positive $>0$ (not solution). - $\frac{2}{3} < x < 2$: numerator positive, denominator negative, fraction negative $<0$ (solution). - $x > 2$: numerator negative, denominator negative, fraction positive $>0$ (not solution). Include points where fraction equals zero: - Numerator zero at $x=\pm 2$, fraction zero. - Denominator zero at $x=0, \frac{2}{3}$, fraction undefined. Final solution: $$[-2,0) \cup \left(\frac{2}{3}, 2\right].$$