1. **Problem 1:** Find constants $a$, $b$, and $c$ in the expansion of $(2 + bx)^8 = a + 256x + cx^2$. 2. The binomial expansion formula is $$(p + q)^n = \sum_{k=0}^n \binom{n}{k} p^{n-k} q^k.$$ 3. Here, $p=2$, $q=bx$, and $n=8$. The first three terms are: $$T_0 = \binom{8}{0} 2^8 (bx)^0 = 2^8 = 256,$$ $$T_1 = \binom{8}{1} 2^7 (bx)^1 = 8 \times 128 \times b x = 1024 b x,$$ $$T_2 = \binom{8}{2} 2^6 (bx)^2 = 28 \times 64 \times b^2 x^2 = 1792 b^2 x^2.$$ 4. Given the expansion is $a + 256x + cx^2$, matching terms: - Constant term $a = 256$ - Coefficient of $x$ is $256$, so $1024 b = 256 \Rightarrow b = \frac{256}{1024} = \frac{1}{4}$ - Coefficient of $x^2$ is $c$, so $1792 b^2 = c$. Substitute $b=\frac{1}{4}$: $$c = 1792 \times \left(\frac{1}{4}\right)^2 = 1792 \times \frac{1}{16} = 112.$$ 5. **Answer:** $a=256$, $b=\frac{1}{4}$, $c=112$. --- 6. **Problem 2:** Interpret $x \propto y$ and give two examples. 7. The notation $x \propto y$ means $x$ is directly proportional to $y$, i.e., there exists a constant $k$ such that: $$x = k y.$$ 8. **Example 1:** If $x$ is the distance traveled and $y$ is time at constant speed, then $x \propto y$. 9. **Example 2:** If $x$ is the cost of apples and $y$ is the weight bought, then $x \propto y$. --- 10. **Problem 3:** Given points $(5,12)$ and $(3,20)$ on the graph of $\log y^2$ vs $x$, find $y$ in the form $y = 10^{ax + b}$. 11. Since $\log y^2$ is plotted against $x$, the equation of the line is: $$\log y^2 = m x + c,$$ where $m$ is slope and $c$ is intercept. 12. Calculate slope: $$m = \frac{20 - 12}{3 - 5} = \frac{8}{-2} = -4.$$ 13. Use point $(5,12)$ to find $c$: $$12 = -4 \times 5 + c \Rightarrow c = 12 + 20 = 32.$$ 14. So, $$\log y^2 = -4x + 32.$$ 15. Rewrite: $$2 \log y = -4x + 32 \Rightarrow \log y = -2x + 16.$$ 16. Convert from log form: $$y = 10^{-2x + 16} = 10^{16} \times 10^{-2x} = 10^{16} \cdot 10^{-2x}.$$ 17. Hence, $$y = 10^{ax + b}$$ with $a = -2$ and $b = 16$. --- 18. **Problem 4:** Find $\int (3x - 14 + \frac{32}{x^2}) dx$ and determine the shaded area under $y = 3x - 14 + \frac{32}{x^2}$ between points $P$ and $Q$ where the curve cuts the x-axis. 19. Integrate term-by-term: $$\int 3x dx = \frac{3x^2}{2},$$ $$\int -14 dx = -14x,$$ $$\int \frac{32}{x^2} dx = 32 \int x^{-2} dx = 32 \times (-x^{-1}) = -\frac{32}{x}.$$ 20. So, $$\int (3x - 14 + \frac{32}{x^2}) dx = \frac{3x^2}{2} - 14x - \frac{32}{x} + C.$$ 21. To find $P$ and $Q$, solve $y=0$: $$3x - 14 + \frac{32}{x^2} = 0 \Rightarrow 3x^3 - 14x^2 + 32 = 0.$$ 22. Test possible roots: try $x=2$: $$3(8) - 14(4) + 32 = 24 - 56 + 32 = 0,$$ so $x=2$ is a root. 23. Divide polynomial by $(x-2)$: $$3x^3 - 14x^2 + 32 = (x-2)(3x^2 - 8x - 16).$$ 24. Solve quadratic: $$3x^2 - 8x -16 = 0,$$ $$x = \frac{8 \pm \sqrt{64 + 192}}{6} = \frac{8 \pm \sqrt{256}}{6} = \frac{8 \pm 16}{6}.$$ 25. Roots: $$x = \frac{8 + 16}{6} = 4, \quad x = \frac{8 - 16}{6} = -\frac{4}{3}.$$ 26. Points where curve cuts x-axis are $P = -\frac{4}{3}$ and $Q = 2$ (assuming positive interval for shaded area). 27. The shaded area between $P$ and $Q$ is: $$A = \left| \int_{-\frac{4}{3}}^{2} (3x - 14 + \frac{32}{x^2}) dx \right|.$$ 28. Evaluate: $$F(x) = \frac{3x^2}{2} - 14x - \frac{32}{x}.$$ 29. Calculate: $$F(2) = \frac{3(4)}{2} - 14(2) - \frac{32}{2} = 6 - 28 - 16 = -38,$$ $$F\left(-\frac{4}{3}\right) = \frac{3(\frac{16}{9})}{2} - 14 \left(-\frac{4}{3}\right) - \frac{32}{-\frac{4}{3}} = \frac{48}{18} + \frac{56}{3} + 24 = \frac{8}{3} + \frac{56}{3} + 24 = \frac{64}{3} + 24 = \frac{64}{3} + \frac{72}{3} = \frac{136}{3}.$$ 30. Area: $$A = \left| -38 - \frac{136}{3} \right| = \left| -\frac{114}{3} - \frac{136}{3} \right| = \left| -\frac{250}{3} \right| = \frac{250}{3}.$$ **Final answers:** - $a=256$, $b=\frac{1}{4}$, $c=112$ - $x \propto y$ means $x = ky$ - $y = 10^{-2x + 16}$ - $P = -\frac{4}{3}$, $Q=2$ - $\int (3x - 14 + \frac{32}{x^2}) dx = \frac{3x^2}{2} - 14x - \frac{32}{x} + C$ - Area under curve between $P$ and $Q$ is $\frac{250}{3}$.