1. **Problem 17:** Find the negation of \((\forall x)(Q(x) \Rightarrow P(x))\) where \(P(x): x \text{ is positive}\) and \(Q(x): x > 5\). 2. The statement \((\forall x)(Q(x) \Rightarrow P(x))\) means "For all \(x\), if \(x > 5\), then \(x\) is positive." 3. The negation of a universal statement \((\forall x)\) is an existential statement \((\exists x)\) with the negation of the predicate. 4. Negate \(Q(x) \Rightarrow P(x)\). Recall \(A \Rightarrow B\) is logically equivalent to \(\neg A \lor B\), so its negation is \(A \land \neg B\). 5. Thus, negation is \((\exists x)(Q(x) \land \neg P(x))\), meaning "There exists \(x\) such that \(x > 5\) and \(x\) is not positive." 6. Among the options, (B) "There is \(x\) such that \(x > 5\) and \(x\) is not positive" matches this. --- 7. **Problem 18:** Given arithmetic sequence \(\{a_n\}\) with \(a_5 = 21\) and \(a_{10} = 66\), find \(a_2\). 8. Arithmetic sequence formula: \(a_n = a_1 + (n-1)d\). 9. From \(a_5 = a_1 + 4d = 21\) and \(a_{10} = a_1 + 9d = 66\). 10. Subtract: \(a_{10} - a_5 = 5d = 45 \Rightarrow d = 9\). 11. Substitute \(d=9\) into \(a_5 = a_1 + 4 \times 9 = 21 \Rightarrow a_1 = 21 - 36 = -15\). 12. Find \(a_2 = a_1 + d = -15 + 9 = -6\). --- 13. **Problem 19:** UPS revenue in 1998: 24.8, in 2004: 29.8. Estimate 2001 revenue assuming linear growth. 14. Years between 1998 and 2004: 6 years. Revenue increase: 29.8 - 24.8 = 5. 15. Annual increase: \(5 / 6 \approx 0.8333\). 16. Years from 1998 to 2001: 3 years. 17. Estimated 2001 revenue: \(24.8 + 3 \times 0.8333 = 24.8 + 2.5 = 27.3\). --- 18. **Problem 20:** Sum of series \(\sum_{n=1}^\infty \left((-\frac{1}{3})^n + 4^{1-n}\right)\). 19. Split sum: \(\sum (-\frac{1}{3})^n + \sum 4^{1-n}\). 20. First sum is geometric with \(a = -\frac{1}{3}, r = -\frac{1}{3}\), sum = \(\frac{a}{1-r} = \frac{-1/3}{1 + 1/3} = \frac{-1/3}{4/3} = -\frac{1}{4}\). 21. Second sum: \(\sum 4^{1-n} = 4 \sum (\frac{1}{4})^n\), geometric with \(a=4 \times \frac{1}{4} = 1\), ratio \(r=\frac{1}{4}\). 22. Sum = \(\frac{1}{1 - 1/4} = \frac{1}{3/4} = \frac{4}{3}\). 23. Total sum = \(-\frac{1}{4} + \frac{4}{3} = \frac{-3 + 16}{12} = \frac{13}{12}\). --- 24. **Problem 21:** For what \(k\) does \(x^2 + y^2 + 6x - 4y + k = 0\) represent a circle? 25. Complete the square: $$x^2 + 6x + y^2 - 4y + k = 0$$ $$ (x^2 + 6x + 9) + (y^2 - 4y + 4) + k - 9 - 4 = 0$$ $$ (x + 3)^2 + (y - 2)^2 = 13 - k$$ 26. For a circle, radius squared must be positive: \(13 - k > 0 \Rightarrow k < 13\). --- 27. **Problem 22:** "I eat lemons if they are dipped in honey" means \(h \Rightarrow l\). 28. So answer is (D) \(h \Rightarrow l\). --- 29. **Problem 23:** Lines \(L_1: 3y - 9 = 4x\) and \(L_2: 2y - 3x + 2 = 0\). 30. Find slopes: \(L_1: 3y = 4x + 9 \Rightarrow y = \frac{4}{3}x + 3\), slope \(m_1 = \frac{4}{3}\). \(L_2: 2y = 3x - 2 \Rightarrow y = \frac{3}{2}x - 1\), slope \(m_2 = \frac{3}{2}\). 31. Angle between lines \(\alpha\) satisfies: $$\tan \alpha = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| = \left| \frac{\frac{4}{3} - \frac{3}{2}}{1 + \frac{4}{3} \times \frac{3}{2}} \right| = \left| \frac{\frac{8}{6} - \frac{9}{6}}{1 + 2} \right| = \frac{1/6}{3} = \frac{1}{18}$$ 32. So answer is (D) \(\frac{1}{18}\). --- 33. **Problem 24:** Find polar form of \(\left(\frac{2}{1 + \sqrt{3}i}\right)^2\). 34. Denominator: \(1 + \sqrt{3}i\) has magnitude \(\sqrt{1^2 + (\sqrt{3})^2} = 2\) and argument \(\theta = \arctan(\sqrt{3}/1) = 60^\circ\). 35. So \(\frac{2}{1 + \sqrt{3}i} = \frac{2}{2 (\cos 60^\circ + i \sin 60^\circ)} = \frac{1}{\cos 60^\circ + i \sin 60^\circ} = \cos(-60^\circ) + i \sin(-60^\circ)\). 36. Square it: $$\left(\cos(-60^\circ) + i \sin(-60^\circ)\right)^2 = \cos(-120^\circ) + i \sin(-120^\circ) = \cos 240^\circ + i \sin 240^\circ$$ 37. Answer is (B). --- 38. **Problem 25:** Multiplying complex number \(z\) by \(i\) rotates it counterclockwise by 90°. 39. So answer is (D). --- 40. **Problem 26:** Line through \((a,b)\) perpendicular to line through \((a,2b)\) and \((b,2a)\). 41. Slope of line through \((a,2b)\) and \((b,2a)\): $$m = \frac{2a - 2b}{b - a} = \frac{2(a - b)}{b - a} = -2$$ 42. Perpendicular slope: \(m_\perp = \frac{1}{2}\). 43. Equation of perpendicular line through \((a,b)\): $$y - b = \frac{1}{2}(x - a) \Rightarrow 2y - 2b = x - a \Rightarrow 2y - x = 2b - a$$ 44. Answer is (C). --- 45. **Problem 27:** Circle with diameter endpoints \((-2,0)\) and \((4,2)\). 46. Center is midpoint: $$\left(\frac{-2 + 4}{2}, \frac{0 + 2}{2}\right) = (1,1)$$ 47. Radius squared is distance squared between center and one endpoint: $$r^2 = (1 + 2)^2 + (1 - 0)^2 = 3^2 + 1^2 = 9 + 1 = 10$$ 48. Equation: $$(x - 1)^2 + (y - 1)^2 = 10$$ 49. Answer is (A). --- 50. **Problem 28:** Asymptotes of hyperbola \(4x^2 - y^2 + 2y = 5\). 51. Rewrite: $$4x^2 - (y^2 - 2y) = 5$$ Complete square: $$4x^2 - (y^2 - 2y + 1) = 5 - 1$$ $$4x^2 - (y - 1)^2 = 4$$ Divide by 4: $$x^2 - \frac{(y - 1)^2}{4} = 1$$ 52. Standard form hyperbola with horizontal transverse axis. 53. Asymptotes: $$\frac{y - 1}{2} = \pm x \Rightarrow y = 1 \pm 2x$$ 54. So asymptotes are \(y = 2x + 1\) and \(y = -2x + 1\). 55. Among options, (A) \(y = -2x + 1\) is an asymptote. --- 56. **Problem 29:** Distance from line to center of sphere. 57. Sphere center: \((-1,0,0)\), radius \(r = 2\). 58. Line: \(x = -1\), \(z = 4 - \sqrt{3} y\). 59. Distance from center to line is shortest distance between point \((-1,0,0)\) and line. 60. Parametrize line: \(x = -1\), \(y = t\), \(z = 4 - \sqrt{3} t\). 61. Vector from center to point on line: $$\vec{v} = (0, t, 4 - \sqrt{3} t)$$ 62. Direction vector of line: $$\vec{d} = (0, 1, -\sqrt{3})$$ 63. Distance formula: $$d = \frac{|\vec{v} \times \vec{d}|}{|\vec{d}|}$$ 64. Compute cross product: $$\vec{v} \times \vec{d} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & t & 4 - \sqrt{3} t \\ 0 & 1 & -\sqrt{3} \end{vmatrix} = \mathbf{i}(t \times (-\sqrt{3}) - (4 - \sqrt{3} t) \times 1) - \mathbf{j}(0 - 0) + \mathbf{k}(0 - 0)$$ $$= \mathbf{i}(-\sqrt{3} t - 4 + \sqrt{3} t) = -4 \mathbf{i}$$ 65. Magnitude \(|\vec{v} \times \vec{d}| = 4\). 66. Magnitude \(|\vec{d}| = \sqrt{0^2 + 1^2 + (-\sqrt{3})^2} = \sqrt{1 + 3} = 2\). 67. Distance \(d = \frac{4}{2} = 2\). 68. Answer is (A). --- 69. **Problem 30:** Complex number \(2i - \pi i = i(2 - \pi)\) corresponds to point \((0, 2 - \pi)\). 70. Answer is (C). --- 71. **Problem 31:** Tangent line to \(f(x) = \frac{3x}{x-1}\) at \(x=2\). 72. Compute \(f(2) = \frac{6}{1} = 6\). 73. Derivative: $$f'(x) = \frac{(3)(x-1) - 3x(1)}{(x-1)^2} = \frac{3x - 3 - 3x}{(x-1)^2} = \frac{-3}{(x-1)^2}$$ 74. At \(x=2\), $$f'(2) = \frac{-3}{(2-1)^2} = -3$$ 75. Equation of tangent line: $$y - 6 = -3(x - 2) \Rightarrow y = -3x + 6 + 6 = -3x + 12$$ 76. Rearranged: $$y + 3x = 12$$ 77. Answer is (D). --- 78. **Problem 32:** \(\lim_{x \to +\infty} \left(\frac{x+4}{x-1}\right)^x\). 79. Rewrite inside limit: $$\frac{x+4}{x-1} = \frac{x-1 + 5}{x-1} = 1 + \frac{5}{x-1}$$ 80. So limit is: $$\lim_{x \to \infty} \left(1 + \frac{5}{x-1}\right)^x = e^5$$ 81. Answer is (D). --- 82. **Problem 33:** \(a_n = \frac{n}{1-n} \left(\frac{3}{\pi}\right)^n\), find \(\lim_{n \to \infty} a_n\). 83. Since \(\left|\frac{3}{\pi}\right| < 1\), \(\left(\frac{3}{\pi}\right)^n \to 0\). 84. Also, \(\frac{n}{1-n} \to -1\) as \(n \to \infty\). 85. So \(a_n \to -1 \times 0 = 0\). 86. Answer is (C). --- 87. **Problem 34:** Divergent sequence among options. 88. (A) \(5^{1/n} \to 1\), convergent. 89. (B) \(\cos(3/n) \to \cos 0 = 1\), convergent. 90. (C) \(\left(-\frac{3}{4}\right)^n \to 0\), convergent. 91. (D) \(\frac{(-1)^n}{10}\) oscillates between \(\pm \frac{1}{10}\), convergent. 92. All converge, but (D) is oscillating but bounded, so convergent. 93. None diverge, but (D) is not divergent. 94. So none diverge, but (D) is oscillatory. 95. Possibly (D) is considered divergent due to oscillation. --- 96. **Problem 35:** Derivative of \(f(x) = \frac{\sin x}{1 + \cos x}\). 97. Use quotient rule: $$f'(x) = \frac{(\cos x)(1 + \cos x) - \sin x(-\sin x)}{(1 + \cos x)^2} = \frac{\cos x + \cos^2 x + \sin^2 x}{(1 + \cos x)^2}$$ 98. Since \(\sin^2 x + \cos^2 x = 1\), numerator: $$\cos x + \cos^2 x + 1 = 1 + \cos x + \cos^2 x$$ 99. Simplify numerator: $$1 + \cos x + \cos^2 x = (1 + \cos x)^2$$ 100. So derivative: $$f'(x) = \frac{(1 + \cos x)^2}{(1 + \cos x)^2} = 1$$ 101. None of the options is 1, so check carefully. 102. Re-express numerator: $$\cos x + \cos^2 x + \sin^2 x = \cos x + \cos^2 x + (1 - \cos^2 x) = \cos x + 1$$ 103. So numerator is \(1 + \cos x\), derivative: $$f'(x) = \frac{1 + \cos x}{(1 + \cos x)^2} = \frac{1}{1 + \cos x}$$ 104. Answer is (B). --- 105. **Problem 36:** \(f(x) = \cos x + 5^x\), find \(f'(x)\). 106. Derivative: $$f'(x) = -\sin x + 5^x \ln 5$$ 107. Answer is (C). --- 108. **Problem 37:** Function $$f(x) = \begin{cases} \frac{\tan kx}{x}, & x < 0 \\ 3x + 2k^2, & x \geq 0 \end{cases}$$ 109. For continuity at \(x=0\), left and right limits must be equal. 110. \(\lim_{x \to 0^-} \frac{\tan kx}{x} = \lim_{x \to 0^-} \frac{kx}{x} = k\). 111. \(f(0) = 3 \times 0 + 2k^2 = 2k^2\). 112. Set equal: $$k = 2k^2 \Rightarrow 2k^2 - k = 0 \Rightarrow k(2k - 1) = 0$$ 113. So \(k = 0\) or \(k = \frac{1}{2}\). 114. Since \(k \neq 0\), answer is (C) \(\frac{1}{2}\). --- 115. **Problem 38:** \(g(x) = x f(3x + 1)\), given \(f(4) = 1\), \(f'(4) = 2\), find \(g'(1)\). 116. Use product and chain rule: $$g'(x) = f(3x + 1) + x \cdot f'(3x + 1) \cdot 3$$ 117. At \(x=1\): $$g'(1) = f(4) + 1 \times f'(4) \times 3 = 1 + 3 \times 2 = 7$$ 118. Answer is (A). --- 119. **Problem 39:** \(a_n = \frac{1}{n^2} + \frac{2}{n^2} + \cdots + \frac{n}{n^2} = \frac{1 + 2 + \cdots + n}{n^2}\). 120. Sum numerator: $$1 + 2 + \cdots + n = \frac{n(n+1)}{2}$$ 121. So $$a_n = \frac{n(n+1)/2}{n^2} = \frac{n+1}{2n} = \frac{1}{2} + \frac{1}{2n}$$ 122. Limit as \(n \to \infty\) is \(\frac{1}{2}\). 123. Answer is (B). --- 124. **Problem 40:** \(\lim_{x \to 4} \frac{\sqrt{x} - 2}{x^2 - 4x}\). 125. Factor denominator: $$x^2 - 4x = x(x - 4)$$ 126. Substitute \(x = 4\) gives \(0/0\), use L'Hôpital's rule. 127. Derivative numerator: $$\frac{1}{2\sqrt{x}}$$ 128. Derivative denominator: $$2x - 4$$ 129. Limit: $$\lim_{x \to 4} \frac{1/(2\sqrt{x})}{2x - 4} = \frac{1/(2 \times 2)}{8 - 4} = \frac{1/4}{4} = \frac{1}{16}$$ 130. Answer is (D). --- 131. **Problem 41:** Volume of solid formed by rotating \(y = \sin x\) on \([0, \pi]\) about x-axis. 132. Volume formula: $$V = \pi \int_0^\pi (\sin x)^2 dx$$ 133. Use identity: $$\sin^2 x = \frac{1 - \cos 2x}{2}$$ 134. So $$V = \pi \int_0^\pi \frac{1 - \cos 2x}{2} dx = \frac{\pi}{2} \left[ x - \frac{\sin 2x}{2} \right]_0^\pi = \frac{\pi}{2} (\pi - 0) = \frac{\pi^2}{2}$$ 135. Answer is (A). --- 136. **Problem 42:** Area enclosed by \(y^2 = x\) on \([1,4]\). 137. Express \(y = \sqrt{x}\), area: $$A = \int_1^4 y dx = \int_1^4 \sqrt{x} dx = \int_1^4 x^{1/2} dx$$ 138. Compute integral: $$\left[ \frac{2}{3} x^{3/2} \right]_1^4 = \frac{2}{3} (4^{3/2} - 1) = \frac{2}{3} (8 - 1) = \frac{2}{3} \times 7 = \frac{14}{3}$$ 139. Answer is (A). --- 140. **Problem 43:** Increasing intervals of \(f(x) = \frac{x}{x^2 + 9}\). 141. Derivative: $$f'(x) = \frac{(x^2 + 9) - x(2x)}{(x^2 + 9)^2} = \frac{x^2 + 9 - 2x^2}{(x^2 + 9)^2} = \frac{9 - x^2}{(x^2 + 9)^2}$$ 142. Numerator positive when \(9 - x^2 > 0 \Rightarrow |x| < 3\). 143. So \(f\) increasing on \((-3, 3)\). 144. Answer is (B). --- 145. **Problem 44:** Max value of \(f(x) = x^2 - x\) on \([-2, 2]\). 146. Critical points: $$f'(x) = 2x - 1 = 0 \Rightarrow x = \frac{1}{2}$$ 147. Evaluate \(f\) at \(-2, 2, 1/2\): \(f(-2) = 4 + 2 = 6\) \(f(2) = 4 - 2 = 2\) \(f(1/2) = (1/4) - (1/2) = -1/4\) 148. Maximum is 6 at \(x = -2\). 149. Answer is (C). --- 150. **Problem 45:** \(\int_0^2 x^2 \ln(1 + x^3) dx\). 151. Use substitution \(t = 1 + x^3\), \(dt = 3x^2 dx\). 152. Rewrite integral: $$\int_0^2 x^2 \ln(1 + x^3) dx = \frac{1}{3} \int_1^9 \ln t dt$$ 153. Integral of \(\ln t\): $$\int \ln t dt = t \ln t - t + C$$ 154. Evaluate: $$\frac{1}{3} [t \ln t - t]_1^9 = \frac{1}{3} (9 \ln 9 - 9 - (1 \times 0 - 1)) = \frac{1}{3} (9 \ln 9 - 9 + 1) = 3 \ln 9 - \frac{8}{3}$$ 155. Since \(\ln 9 = 2 \ln 3\), $$3 \times 2 \ln 3 - \frac{8}{3} = 6 \ln 3 - \frac{8}{3}$$ 156. Answer is (C). --- 157. **Problem 46:** \(f(x) = 2x|x|\). 158. For \(x > 0\), \(f(x) = 2x^2\), derivative \(4x\). 159. For \(x < 0\), \(f(x) = -2x^2\), derivative \(-4x\). 160. At \(x=0\), left derivative \(\lim_{x \to 0^-} f'(x) = 0\), right derivative \(\lim_{x \to 0^+} f'(x) = 0\). 161. So \(f\) is differentiable at 0 with \(f'(0) = 0\). 162. Answer is (D). --- 163. **Problem 47:** Hyperbola \(4(y-1)^2 - 16(x+3)^2 = 1\). 164. Standard form: $$\frac{(y-1)^2}{\frac{1}{4}} - \frac{(x+3)^2}{\frac{1}{16}} = 1$$ 165. For hyperbola \(\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1\), distance between foci: $$2c, \quad c^2 = a^2 + b^2$$ 166. Here \(a^2 = \frac{1}{4}, b^2 = \frac{1}{16}\). 167. So $$c = \sqrt{\frac{1}{4} + \frac{1}{16}} = \sqrt{\frac{4}{16} + \frac{1}{16}} = \sqrt{\frac{5}{16}} = \frac{\sqrt{5}}{4}$$ 168. Distance between foci: $$2c = \frac{\sqrt{5}}{2}$$ 169. Answer is (B). --- 170. **Problem 48:** \(\int \frac{1}{\sqrt{x}} \cos \sqrt{x} dx\). 171. Substitute \(t = \sqrt{x}\), \(x = t^2\), \(dx = 2t dt\). 172. Integral becomes: $$\int \frac{1}{t} \cos t \times 2t dt = 2 \int \cos t dt = 2 \sin t + C = 2 \sin \sqrt{x} + C$$ 173. Answer is (C). --- 174. **Problem 49:** Local minimum of \(f(x) = -3x^4 + 4x^3 + 12x^2 + 1\). 175. Find critical points by \(f'(x) = 0\). 176. \(f'(x) = -12x^3 + 12x^2 + 24x\). 177. Factor: $$f'(x) = 12x(-x^2 + x + 2) = 12x(-(x-2)(x+1))$$ 178. Critical points: \(x = 0, 2, -1\). 179. Second derivative: $$f''(x) = -36x^2 + 24x + 24$$ 180. Evaluate at critical points: - At \(x = -2\): Not critical point. - At \(x = -1\): $$f''(-1) = -36 + (-24) + 24 = -36$$ (negative, local max) - At \(x = 0\): $$f''(0) = 24 > 0$$ (local min) - At \(x = 2\): $$f''(2) = -36(4) + 48 + 24 = -144 + 72 = -72 < 0$$ (local max) 181. Local minimum at \(x=0\), \(f(0) = 1\). 182. Answer is (C). --- 183. **Problem 50:** \(\int_4^5 \frac{dx}{(x-3)^2}\). 184. Integral: $$\int \frac{dx}{(x-3)^2} = -\frac{1}{x-3} + C$$ 185. Evaluate: $$\left[-\frac{1}{x-3}\right]_4^5 = -\frac{1}{2} + \frac{1}{1} = \frac{1}{2}$$ 186. Answer is (C). --- 187. **Problem 51:** Vector \(\mathbf{u}\) magnitude 15, direction 120°. 188. Components: $$u_x = 15 \cos 120^\circ = 15 \times (-\frac{1}{2}) = -7.5$$ $$u_y = 15 \sin 120^\circ = 15 \times \frac{\sqrt{3}}{2} = \frac{15 \sqrt{3}}{2}$$ 189. Answer closest to (A) \(-15/2, 15\sqrt{3}/2\). --- 190. **Problem 52:** Solve \(\sin 2x = \cos x\) on \([0, 2\pi]\). 191. Use identity \(\sin 2x = 2 \sin x \cos x\). 192. Equation: $$2 \sin x \cos x = \cos x$$ 193. Cases: - \(\cos x = 0 \Rightarrow x = \frac{\pi}{2}, \frac{3\pi}{2}\) - Else: $$2 \sin x = 1 \Rightarrow \sin x = \frac{1}{2} \Rightarrow x = \frac{\pi}{6}, \frac{5\pi}{6}$$ 194. Solution set: $$\left\{ \frac{\pi}{2}, \frac{3\pi}{2}, \frac{\pi}{6}, \frac{5\pi}{6} \right\}$$ 195. Answer is (C). --- 196. **Problem 53:** Function \(f(x) = -2 \sin (x/3)\). 197. Period of \(\sin (bx)\) is \(\frac{2\pi}{b}\). 198. Here \(b = \frac{1}{3}\), so period: $$2\pi \times 3 = 6\pi$$ 199. So (A) is false. 200. Check (B): \(f(x) < 0\) for \(0 < x < 3\). 201. Since \(\sin (x/3)\) is positive for \(0 < x/3 < \pi\), i.e., \(0 < x < 3\pi\), \(f(x) = -2 \sin (x/3) < 0\) in \(0 < x < 3\pi\). 202. So (B) is true. --- 203. **Problem 54:** Simplify \(\sin (x - \frac{\pi}{2}) \tan x\). 204. Use identity: $$\sin (x - \frac{\pi}{2}) = -\cos x$$ 205. So expression: $$-\cos x \times \tan x = -\cos x \times \frac{\sin x}{\cos x} = -\sin x$$ 206. Answer is (A). --- 207. **Problem 55:** \(\arctan (\tan (\frac{2}{3} \pi))\). 208. \(\tan\) is periodic with period \(\pi\), \(\arctan\) returns values in \((-\frac{\pi}{2}, \frac{\pi}{2})\). 209. \(\frac{2}{3} \pi = 120^\circ\) is outside principal range. 210. Equivalent angle in principal range: $$\frac{2}{3} \pi - \pi = -\frac{\pi}{3}$$ 211. So answer is (B). --- 212. **Problem 56:** Vectors \(u = (6, x + y)\), \(v = (y - 2x, -1)\), and \(2v - u = (0,1)\). 213. Compute: $$2v - u = (2(y - 2x) - 6, 2(-1) - (x + y)) = (0,1)$$ 214. Equate components: $$2(y - 2x) - 6 = 0 \Rightarrow 2y - 4x = 6$$ $$-2 - x - y = 1 \Rightarrow -x - y = 3$$ 215. From second: $$x + y = -3$$ 216. From first: $$2y - 4x = 6$$ 217. Substitute \(y = -3 - x\) into first: $$2(-3 - x) - 4x = 6 \Rightarrow -6 - 2x - 4x = 6 \Rightarrow -6 - 6x = 6 \Rightarrow -6x = 12 \Rightarrow x = -2$$ 218. Then \(y = -3 - (-2) = -1\). 219. Answer is (D). --- 220. **Problem 57:** Solve \(2 \sin^2 x - \cos x - 1 = 0\) on \([0, \pi]\). 221. Use \(\sin^2 x = 1 - \cos^2 x\): $$2(1 - \cos^2 x) - \cos x - 1 = 0 \Rightarrow 2 - 2 \cos^2 x - \cos x - 1 = 0$$ $$1 - 2 \cos^2 x - \cos x = 0$$ 222. Rearrange: $$2 \cos^2 x + \cos x - 1 = 0$$ 223. Let \(t = \cos x\), quadratic: $$2t^2 + t - 1 = 0$$ 224. Solve: $$t = \frac{-1 \pm \sqrt{1 + 8}}{4} = \frac{-1 \pm 3}{4}$$ 225. Roots: $$t = \frac{2}{4} = \frac{1}{2}, \quad t = -1$$ 226. \(\cos x = \frac{1}{2} \Rightarrow x = \frac{\pi}{3}\) 227. \(\cos x = -1 \Rightarrow x = \pi\) 228. Solution set: \(\{ \frac{\pi}{3}, \pi \}\). 229. Answer is (C). --- 230. **Problem 58:** Period of \(f(x) = \tan \left( \frac{2x - \pi}{3} \right)\). 231. Period of \(\tan bx\) is \(\frac{\pi}{|b|}\). 232. Here, argument is \(\frac{2x - \pi}{3} = \frac{2}{3} x - \frac{\pi}{3}\). 233. So \(b = \frac{2}{3}\), period: $$\frac{\pi}{2/3} = \frac{3\pi}{2}$$ 234. Answer is (B). --- 235. **Problem 59:** Identify incorrect identity. 236. (A) \(\sin x \cos^2 x - \sin x = \sin x (\cos^2 x - 1) = -\sin^3 x\) correct. 237. (B) \(\frac{\cos x \tan x - \sin x}{\cot x} = \frac{\cos x \frac{\sin x}{\cos x} - \sin x}{\frac{\cos x}{\sin x}} = \frac{\sin x - \sin x}{\frac{\cos x}{\sin x}} = 0\) correct. 238. (C) \(\sin^4 x - \cos^4 x = (\sin^2 x - \cos^2 x)(\sin^2 x + \cos^2 x) = \sin^2 x - \cos^2 x\) correct. 239. (D) \(\frac{\sin x}{1 + \cos x} \neq \sec x - \tan x\). 240. Check (D): $$\sec x - \tan x = \frac{1}{\cos x} - \frac{\sin x}{\cos x} = \frac{1 - \sin x}{\cos x}$$ 241. \(\frac{\sin x}{1 + \cos x} \neq \frac{1 - \sin x}{\cos x}\) generally. 242. So (D) is incorrect. --- 243. **Problem 60:** Triangle vertices \(A(1,2), B(3,4), C(2,5)\), find interior angle at \(B\). 244. Vectors: $$\vec{BA} = (1 - 3, 2 - 4) = (-2, -2)$$ $$\vec{BC} = (2 - 3, 5 - 4) = (-1, 1)$$ 245. Angle \(\theta\) between vectors: $$\cos \theta = \frac{\vec{BA} \cdot \vec{BC}}{|\vec{BA}| |\vec{BC}|} = \frac{(-2)(-1) + (-2)(1)}{\sqrt{(-2)^2 + (-2)^2} \sqrt{(-1)^2 + 1^2}} = \frac{2 - 2}{\sqrt{8} \sqrt{2}} = 0$$ 246. So \(\theta = 90^\circ\). 247. Answer is (D). --- 248. **Problem 61:** Translation \(T\) takes \((1, -3)\) to \((-1, 0)\). 249. Translation vector: $$\vec{v} = (-1 - 1, 0 - (-3)) = (-2, 3)$$ 250. Image of origin under \(T\) is: $$(0,0) + \vec{v} = (-2, 3)$$ 251. Answer is (C). --- 252. **Problem 62:** Parametric line \((x,y) = (2,5) + t(1,-2)\). 253. Parametric equations: $$x = 2 + t, \quad y = 5 - 2t$$ 254. Express \(t = x - 2\), substitute into \(y\): $$y = 5 - 2(x - 2) = 5 - 2x + 4 = 9 - 2x$$ 255. Rearranged: $$2x + y = 9$$ 256. Answer is (A). --- 257. **Problem 63:** Ellipse \(\frac{x^2}{4} + (y-1)^2 = 1\) rotated 90° anticlockwise about origin. 258. Rotation matrix: $$R = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$$ 259. New coordinates: $$x' = -y, \quad y' = x$$ 260. Substitute: $$\frac{(-y)^2}{4} + (x - 1)^2 = 1 \Rightarrow \frac{y^2}{4} + (x - 1)^2 = 1$$ 261. Answer is (C). --- 262. **Problem 64:** Kiros at 20 m, 35° north of west; Guta at 30 m, 25° south of west. 263. Convert to coordinates (west is negative x, north positive y): - Kiros: $$x_K = -20 \cos 35^\circ, \quad y_K = 20 \sin 35^\circ$$ - Guta: $$x_G = -30 \cos 25^\circ, \quad y_G = -30 \sin 25^\circ$$ 264. Distance between K and G: $$d = \sqrt{(x_K - x_G)^2 + (y_K - y_G)^2}$$ 265. Compute numerically: $$x_K \approx -20 \times 0.8192 = -16.384$$ $$y_K \approx 20 \times 0.574 = 11.48$$ $$x_G \approx -30 \times 0.9063 = -27.189$$ $$y_G \approx -30 \times 0.4226 = -12.678$$ 266. Differences: $$\Delta x = -16.384 + 27.189 = 10.805$$ $$\Delta y = 11.48 + 12.678 = 24.158$$ 267. Distance: $$\sqrt{10.805^2 + 24.158^2} \approx \sqrt{116.8 + 583.6} = \sqrt{700.4} \approx 26.46$$ 268. Closest answer is (D) 25 m. ---