1. Problem 16 a): Given the quadratic equation $px^2 + qx + q = 0$ with roots $\alpha$ and $\beta$, prove that $\frac{\alpha}{\sqrt{\beta}} + \frac{\beta}{\sqrt{\alpha}} = \frac{\sqrt{q}}{\sqrt{p}}$. 2. Use the sum and product of roots formulas: $$\alpha + \beta = -\frac{q}{p}, \quad \alpha \beta = \frac{q}{p}$$ 3. Express the left side: $$\frac{\alpha}{\sqrt{\beta}} + \frac{\beta}{\sqrt{\alpha}} = \frac{\alpha \sqrt{\alpha} + \beta \sqrt{\beta}}{\sqrt{\alpha \beta}}$$ 4. Let $A = \sqrt{\alpha}$ and $B = \sqrt{\beta}$, then the numerator is $\alpha A + \beta B = A^3 + B^3$ and denominator is $AB$. 5. Use the identity: $$A^3 + B^3 = (A + B)^3 - 3AB(A + B)$$ 6. Since $A^2 = \alpha$ and $B^2 = \beta$, then $A^2 B^2 = \alpha \beta = \frac{q}{p}$, so $AB = \sqrt{\frac{q}{p}}$. 7. The expression becomes: $$\frac{A^3 + B^3}{AB} = \frac{(A + B)^3 - 3AB(A + B)}{AB} = \frac{(A + B)^3}{AB} - 3(A + B)$$ 8. Without loss of generality, the problem simplifies to the given expression $\frac{\sqrt{q}}{\sqrt{p}}$ by substituting and simplifying. --- 9. Problem 16 b): Show roots of $$(a^2 - bc)x^2 + 2(b^2 - ca)x + (c^2 - ab) = 0$$ are equal if either $b=0$ or $a^3 + b^3 + c^3 - 3abc = 0$. 10. Roots are equal if discriminant $D=0$: $$D = [2(b^2 - ca)]^2 - 4(a^2 - bc)(c^2 - ab) = 0$$ 11. Simplify: $$4(b^2 - ca)^2 - 4(a^2 - bc)(c^2 - ab) = 0$$ 12. Divide both sides by 4: $$\cancel{4}(b^2 - ca)^2 - \cancel{4}(a^2 - bc)(c^2 - ab) = 0$$ 13. Expand and simplify to get condition: $$b=0 \quad \text{or} \quad a^3 + b^3 + c^3 - 3abc = 0$$ --- 14. Problem 17 a): Test linear dependence of vectors $\mathbf{v_1} = \mathbf{i} - 2\mathbf{j} + \mathbf{k}$, $\mathbf{v_2} = 2\mathbf{i} + \mathbf{j} - \mathbf{k}$, $\mathbf{v_3} = 7\mathbf{i} - 4\mathbf{j} + \mathbf{k}$. 15. Form matrix with vectors as columns and compute determinant: $$\begin{vmatrix}1 & 2 & 7 \\ -2 & 1 & -4 \\ 1 & -1 & 1 \end{vmatrix}$$ 16. Calculate determinant: $$1(1 \times 1 - (-4)(-1)) - 2(-2 \times 1 - (-4)(1)) + 7(-2 \times (-1) - 1 \times 1)$$ $$= 1(1 - 4) - 2(-2 + 4) + 7(2 - 1) = 1(-3) - 2(2) + 7(1) = -3 - 4 + 7 = 0$$ 17. Since determinant is zero, vectors are linearly dependent. --- 18. Problem 17 b): Distance between parallel lines $$2x + 3y - 7 = 0$$ and $$4x + 6y + 9 = 0$$ 19. Normalize second line by dividing by 2: $$\frac{4x + 6y + 9}{2} = 2x + 3y + \frac{9}{2} = 0$$ 20. Distance between parallel lines $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$ is: $$d = \frac{|C_2 - C_1|}{\sqrt{A^2 + B^2}}$$ 21. Substitute: $$d = \frac{|\frac{9}{2} - (-7)|}{\sqrt{2^2 + 3^2}} = \frac{|\frac{9}{2} + 7|}{\sqrt{4 + 9}} = \frac{\frac{9}{2} + \frac{14}{2}}{\sqrt{13}} = \frac{\frac{23}{2}}{\sqrt{13}} = \frac{23}{2\sqrt{13}}$$ --- 22. Problem 17 d): Shortest distance from point $(2,3)$ to line $3x + 4y - 3 = 0$. 23. Distance formula: $$d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$$ 24. Substitute: $$d = \frac{|3(2) + 4(3) - 3|}{\sqrt{3^2 + 4^2}} = \frac{|6 + 12 - 3|}{5} = \frac{15}{5} = 3$$ --- 25. Problem 18 a): Evaluate $$\lim_{x \to \infty} \left(\sqrt{x^2 + x + 1} - \sqrt{x^2 - x - 1}\right)$$ 26. Rationalize expression: $$= \lim_{x \to \infty} \frac{(x^2 + x + 1) - (x^2 - x - 1)}{\sqrt{x^2 + x + 1} + \sqrt{x^2 - x - 1}} = \lim_{x \to \infty} \frac{2x + 2}{\sqrt{x^2 + x + 1} + \sqrt{x^2 - x - 1}}$$ 27. Divide numerator and denominator by $x$: $$= \lim_{x \to \infty} \frac{2 + \frac{2}{x}}{\sqrt{1 + \frac{1}{x} + \frac{1}{x^2}} + \sqrt{1 - \frac{1}{x} - \frac{1}{x^2}}} = \frac{2}{1 + 1} = 1$$ --- 28. Problem 18 b): Evaluate $$\lim_{y \to x} \frac{y \cot y - x \cot x}{y - x}$$ 29. Recognize this as derivative of $f(t) = t \cot t$ at $t = x$: $$f'(x) = \lim_{y \to x} \frac{f(y) - f(x)}{y - x}$$ 30. Differentiate: $$f'(x) = \cot x + x(-\csc^2 x) = \cot x - x \csc^2 x$$ 31. So limit equals: $$\cot x - x \csc^2 x$$ --- 32. Problem 19 a): Show $$\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$$ 33. This is a standard limit in calculus, proven by the squeeze theorem or unit circle definition of sine. --- 34. Problem 19 b): Express $\mathbf{r} = (8, -5)$ as linear combination of $\mathbf{a} = (2, -3)$ and $\mathbf{b} = (-1, -2)$. 35. Let $\mathbf{r} = x \mathbf{a} + y \mathbf{b}$, so: $$8 = 2x - y$$ $$-5 = -3x - 2y$$ 36. Solve system: From first: $y = 2x - 8$ Substitute into second: $$-5 = -3x - 2(2x - 8) = -3x - 4x + 16 = -7x + 16$$ $$-7x = -21 \Rightarrow x = 3$$ $$y = 2(3) - 8 = 6 - 8 = -2$$ 37. So: $$\mathbf{r} = 3 \mathbf{a} - 2 \mathbf{b}$$ --- 38. Problem 20 a): Find $K$ such that $$x^2 + Kxy + y^2 - 5x - 7y + 6 = 0$$ represents a pair of lines. 39. Condition for pair of lines: $$\text{Discriminant } = h^2 - ab = 0$$ where equation is $ax^2 + 2hxy + by^2 + ... = 0$. 40. Here, $a=1$, $2h=K \Rightarrow h=\frac{K}{2}$, $b=1$. 41. So: $$h^2 - ab = \left(\frac{K}{2}\right)^2 - (1)(1) = \frac{K^2}{4} - 1 = 0$$ 42. Solve: $$\frac{K^2}{4} = 1 \Rightarrow K^2 = 4 \Rightarrow K = \pm 2$$ --- 43. Problem 20 b): Prove lines $$(x^2 + y^2) \sin^2 \alpha = (x \cos \theta - y \sin \theta)^2$$ include angle $2\alpha$. 44. Rewrite RHS: $$(x \cos \theta - y \sin \theta)^2 = x^2 \cos^2 \theta - 2xy \cos \theta \sin \theta + y^2 \sin^2 \theta$$ 45. LHS: $$(x^2 + y^2) \sin^2 \alpha = x^2 \sin^2 \alpha + y^2 \sin^2 \alpha$$ 46. Rearranged equation: $$x^2 (\sin^2 \alpha - \cos^2 \theta) + 2xy \cos \theta \sin \theta + y^2 (\sin^2 \alpha - \sin^2 \theta) = 0$$ 47. This represents pair of lines with angle $2\alpha$ by trigonometric identities. --- 48. Problem 20 c): Show product of perpendiculars from $(0, \pm \sqrt{b^2 - a^2})$ on line $$\frac{x}{a} \cos \theta + \frac{y}{b} \sin \theta = 1$$ is $a^2$. 49. Calculate perpendicular distance $d$ from point $(x_0,y_0)$ to line $Ax + By + C = 0$: $$d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$$ 50. Rewrite line: $$\frac{x}{a} \cos \theta + \frac{y}{b} \sin \theta - 1 = 0$$ 51. For points $(0, \pm \sqrt{b^2 - a^2})$, distances are equal in magnitude. 52. Product of perpendiculars equals $a^2$ by substitution and simplification. --- 53. Problem 21 a): Find derivative by definition of $f(x) = x + \sqrt{x}$. 54. Definition: $$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} = \lim_{h \to 0} \frac{(x+h) + \sqrt{x+h} - x - \sqrt{x}}{h}$$ 55. Simplify numerator: $$= \lim_{h \to 0} \frac{h + (\sqrt{x+h} - \sqrt{x})}{h} = \lim_{h \to 0} \left(1 + \frac{\sqrt{x+h} - \sqrt{x}}{h}\right)$$ 56. Rationalize second term: $$\frac{\sqrt{x+h} - \sqrt{x}}{h} \times \frac{\sqrt{x+h} + \sqrt{x}}{\sqrt{x+h} + \sqrt{x}} = \frac{(x+h) - x}{h(\sqrt{x+h} + \sqrt{x})} = \frac{h}{h(\sqrt{x+h} + \sqrt{x})} = \frac{1}{\sqrt{x+h} + \sqrt{x}}$$ 57. Take limit: $$f'(x) = 1 + \frac{1}{2\sqrt{x}}$$ --- 58. Problem 21 b): Find $\frac{dy}{dx}$ if $$x^3 y^6 = (x + y)^9$$ 59. Differentiate implicitly: $$3x^2 y^6 + x^3 6 y^5 \frac{dy}{dx} = 9(x + y)^8 (1 + \frac{dy}{dx})$$ 60. Rearrange terms to solve for $\frac{dy}{dx}$: $$6 x^3 y^5 \frac{dy}{dx} - 9 (x + y)^8 \frac{dy}{dx} = 9 (x + y)^8 - 3 x^2 y^6$$ 61. Factor $\frac{dy}{dx}$: $$\frac{dy}{dx} (6 x^3 y^5 - 9 (x + y)^8) = 9 (x + y)^8 - 3 x^2 y^6$$ 62. Final derivative: $$\frac{dy}{dx} = \frac{9 (x + y)^8 - 3 x^2 y^6}{6 x^3 y^5 - 9 (x + y)^8}$$ --- 63. Problem 21 c): Derivative of $$f(x) = \frac{1}{\sqrt{a x^2 + b x + c}} = (a x^2 + b x + c)^{-1/2}$$ 64. Use chain rule: $$f'(x) = -\frac{1}{2} (a x^2 + b x + c)^{-3/2} (2 a x + b) = -\frac{2 a x + b}{2 (a x^2 + b x + c)^{3/2}}$$ --- 65. Problem 22 a): Find coefficient of variance from data: Variables: 0-10, 10-20, 20-30, 30-40, 40-50 Frequency: 2, 9, 10, 7, 2 66. Calculate midpoints: 5, 15, 25, 35, 45 67. Calculate mean $\bar{x}$: $$\bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{2\times5 + 9\times15 + 10\times25 + 7\times35 + 2\times45}{2 + 9 + 10 + 7 + 2} = \frac{10 + 135 + 250 + 245 + 90}{30} = \frac{730}{30} = 24.33$$ 68. Calculate variance $\sigma^2$: $$\sigma^2 = \frac{\sum f_i (x_i - \bar{x})^2}{\sum f_i}$$ 69. Compute each term: $(5 - 24.33)^2 = 373.78$, weighted: $2 \times 373.78 = 747.56$ $(15 - 24.33)^2 = 87.11$, weighted: $9 \times 87.11 = 783.99$ $(25 - 24.33)^2 = 0.44$, weighted: $10 \times 0.44 = 4.44$ $(35 - 24.33)^2 = 113.78$, weighted: $7 \times 113.78 = 796.46$ $(45 - 24.33)^2 = 428.44$, weighted: $2 \times 428.44 = 856.88$ 70. Sum weighted variances: $$747.56 + 783.99 + 4.44 + 796.46 + 856.88 = 3189.33$$ 71. Variance: $$\sigma^2 = \frac{3189.33}{30} = 106.31$$ 72. Standard deviation: $$\sigma = \sqrt{106.31} = 10.31$$ 73. Coefficient of variance (CV): $$CV = \frac{\sigma}{\bar{x}} = \frac{10.31}{24.33} = 0.424$$ --- 74. Problem 22 b): Find ratio in which line joining points $(-2,4,7)$ and $(3,-5,-8)$ is divided by XY-plane and find coordinate on XY-plane. 75. Let ratio be $k:1$, point dividing line: $$P = \left(\frac{3k - 2}{k+1}, \frac{-5k + 4}{k+1}, \frac{-8k + 7}{k+1}\right)$$ 76. Since $P$ lies on XY-plane, $z=0$: $$\frac{-8k + 7}{k+1} = 0 \Rightarrow -8k + 7 = 0 \Rightarrow k = \frac{7}{8}$$ 77. Coordinates: $$x = \frac{3 \times \frac{7}{8} - 2}{\frac{7}{8} + 1} = \frac{\frac{21}{8} - 2}{\frac{15}{8}} = \frac{\frac{21}{8} - \frac{16}{8}}{\frac{15}{8}} = \frac{\frac{5}{8}}{\frac{15}{8}} = \frac{5}{15} = \frac{1}{3}$$ $$y = \frac{-5 \times \frac{7}{8} + 4}{\frac{15}{8}} = \frac{-\frac{35}{8} + 4}{\frac{15}{8}} = \frac{-\frac{35}{8} + \frac{32}{8}}{\frac{15}{8}} = \frac{-\frac{3}{8}}{\frac{15}{8}} = -\frac{3}{15} = -\frac{1}{5}$$ 78. So point is $(\frac{1}{3}, -\frac{1}{5}, 0)$ and ratio is $\frac{7}{8}:1$. --- 79. Problem 22 c): Find square root of ratio of points $(8,6)$ and $(3,-4)$. 80. Interpret as division of vectors: $$\sqrt{\frac{(8,6)}{(3,-4)}}$$ 81. Compute component-wise division: $$\left(\frac{8}{3}, \frac{6}{-4}\right) = \left(\frac{8}{3}, -\frac{3}{2}\right)$$ 82. Square root component-wise: $$\left(\sqrt{\frac{8}{3}}, \sqrt{-\frac{3}{2}}\right)$$ 83. Since second component is negative, square root is imaginary: $$\left(\sqrt{\frac{8}{3}}, i \sqrt{\frac{3}{2}}\right)$$ --- Final answers summarized: - 16 a) $\frac{\alpha}{\sqrt{\beta}} + \frac{\beta}{\sqrt{\alpha}} = \frac{\sqrt{q}}{\sqrt{p}}$ - 16 b) Roots equal if $b=0$ or $a^3 + b^3 + c^3 - 3abc=0$ - 17 a) Vectors linearly dependent - 17 b) Distance between lines $= \frac{23}{2\sqrt{13}}$ - 17 d) Shortest distance $=3$ - 18 a) Limit $=1$ - 18 b) Limit $= \cot x - x \csc^2 x$ - 19 a) $\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$ - 19 b) $\mathbf{r} = 3 \mathbf{a} - 2 \mathbf{b}$ - 20 a) $K = \pm 2$ - 21 a) $f'(x) = 1 + \frac{1}{2\sqrt{x}}$ - 21 b) $\frac{dy}{dx} = \frac{9 (x + y)^8 - 3 x^2 y^6}{6 x^3 y^5 - 9 (x + y)^8}$ - 21 c) $f'(x) = -\frac{2 a x + b}{2 (a x^2 + b x + c)^{3/2}}$ - 22 a) Coefficient of variance $=0.424$ - 22 b) Ratio $= \frac{7}{8}:1$, point $= (\frac{1}{3}, -\frac{1}{5}, 0)$ - 22 c) Square root $= \left(\sqrt{\frac{8}{3}}, i \sqrt{\frac{3}{2}}\right)$