1. **Solve the equation** $|2x - 1| - 3x + 4 = 0$. We consider two cases based on the absolute value definition: **Case 1:** $2x - 1 \geq 0 \Rightarrow x \geq \frac{1}{2}$. Then $|2x - 1| = 2x - 1$, so the equation becomes: $$2x - 1 - 3x + 4 = 0 \Rightarrow -x + 3 = 0 \Rightarrow x = 3.$$ Check if $x=3$ satisfies $x \geq \frac{1}{2}$: yes. **Case 2:** $2x - 1 < 0 \Rightarrow x < \frac{1}{2}$. Then $|2x - 1| = -(2x - 1) = -2x + 1$, so the equation becomes: $$-2x + 1 - 3x + 4 = 0 \Rightarrow -5x + 5 = 0 \Rightarrow x = 1.$$ Check if $x=1$ satisfies $x < \frac{1}{2}$: no. So the only solution is $x = 3$. 2. **Solve the inequality** $|4x + 12| > 8$. Rewrite as: $$|4x + 12| > 8 \Rightarrow |4(x + 3)| > 8 \Rightarrow 4|x + 3| > 8 \Rightarrow |x + 3| > 2.$$ This splits into two inequalities: $$x + 3 > 2 \Rightarrow x > -1,$$ $$x + 3 < -2 \Rightarrow x < -5.$$ So the solution set is: $$(-\infty, -5) \cup (-1, \infty).$$ 3. **Solve the inequality** $|x + 1| \leq 3$. This means: $$-3 \leq x + 1 \leq 3,$$ which gives: $$-4 \leq x \leq 2.$$ 4. **Find the limit** $$\lim_{x \to -1} \frac{x^9 + 1}{x^5 + x^2}.$$ Factor numerator and denominator: $$x^9 + 1 = (x + 1)(x^8 - x^7 + x^6 - x^5 + x^4 - x^3 + x^2 - x + 1),$$ $$x^5 + x^2 = x^2(x^3 + 1) = x^2(x + 1)(x^2 - x + 1).$$ Cancel $(x + 1)$: $$\frac{x^8 - x^7 + x^6 - x^5 + x^4 - x^3 + x^2 - x + 1}{x^2(x^2 - x + 1)}.$$ Evaluate at $x = -1$: Numerator: $$(-1)^8 - (-1)^7 + (-1)^6 - (-1)^5 + (-1)^4 - (-1)^3 + (-1)^2 - (-1) + 1 = 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 9.$$ Denominator: $$(-1)^2((-1)^2 - (-1) + 1) = 1(1 + 1 + 1) = 3.$$ Limit is: $$\frac{9}{3} = 3.$$ 5. **Find the limit** $$\lim_{x \to \infty} \frac{4x^5 - 7x - 1}{8x^5 + 5}.$$ Divide numerator and denominator by $x^5$: $$\lim_{x \to \infty} \frac{4 - \frac{7}{x^4} - \frac{1}{x^5}}{8 + \frac{5}{x^5}} = \frac{4}{8} = \frac{1}{2}.$$ 6. **Inclined plane problem:** Given weight $W = 800$ gm.wt, $\sin \theta = \frac{6}{10} = 0.6$. The body is in equilibrium under three forces: weight $W$ vertically down, horizontal force $F$, and reaction $R$ from the plane. Resolve forces parallel and perpendicular to the plane: - Component of weight down the plane: $$W \sin \theta = 800 \times 0.6 = 480.$$ - Component of weight perpendicular to the plane: $$W \cos \theta = 800 \times 0.8 = 640,$$ since $\cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{1 - 0.36} = 0.8.$ - Horizontal force $F$ acts horizontally, so its component along the plane is: $$F \cos \theta,$$ and perpendicular to the plane is: $$F \sin \theta.$$ For equilibrium along the plane: $$F \cos \theta = W \sin \theta = 480 \Rightarrow F = \frac{480}{0.8} = 600.$$ For equilibrium perpendicular to the plane: $$R = W \cos \theta + F \sin \theta = 640 + 600 \times 0.6 = 640 + 360 = 1000.$$ So, $$F = 600,$$ $$R = 1000.$$ 7. **Ball on wall problem:** Weight $W = 90\sqrt{2}$ N, string length $= 2r$ (twice radius), ball rests on smooth wall and is suspended by string fixed at ball surface and wall point vertically above contact. The string makes an angle $45^\circ$ with the vertical because the length is twice the radius and the ball touches the wall at the bottom. For equilibrium: - Tension $T$ in the string balances components of weight and wall reaction $R$. - Resolve forces: Vertical: $$T \cos 45^\circ = W = 90\sqrt{2}.$$ Since $\cos 45^\circ = \frac{\sqrt{2}}{2}$, $$T \times \frac{\sqrt{2}}{2} = 90\sqrt{2} \Rightarrow T = 90\sqrt{2} \times \frac{2}{\sqrt{2}} = 180.$$ Horizontal: $$R = T \sin 45^\circ = 180 \times \frac{\sqrt{2}}{2} = 90\sqrt{2}.$$ So, $$T = 180,$$ $$R = 90\sqrt{2}.$$