1. Evaluate \(\lim_{x \to 0} \frac{e^{2x} - 1}{x}\). The problem asks for the limit of a function as \(x\) approaches 0. 2. Use the formula for the derivative of \(e^{kx}\) at 0: \(\lim_{x \to 0} \frac{e^{kx} - 1}{x} = k\). 3. Here, \(k = 2\), so the limit is \(2\). Answer: e) 2 4. Evaluate \(\lim_{x \to 0} \frac{\ln(1 + x) - x}{x^2}\) using L'Hôpital's Rule. 5. Differentiate numerator and denominator: Numerator derivative: \(\frac{1}{1+x} - 1\) Denominator derivative: \(2x\) 6. Apply L'Hôpital's Rule: \[ \lim_{x \to 0} \frac{\frac{1}{1+x} - 1}{2x} \] 7. This is again \(\frac{0}{0}\), apply L'Hôpital's Rule again: Numerator second derivative: \(-\frac{1}{(1+x)^2}\) Denominator second derivative: 2 8. Evaluate limit: \[ \lim_{x \to 0} \frac{-\frac{1}{(1+x)^2}}{2} = \frac{-1}{2} \] Answer: b) -\frac{1}{2} 9. Find \(\frac{dy}{dx}\) if \(y = \ln(\sin(2x^3))\). 10. Use chain rule: \[ \frac{dy}{dx} = \frac{1}{\sin(2x^3)} \cdot \cos(2x^3) \cdot \frac{d}{dx}(2x^3) = \cot(2x^3) \cdot 6x^2 \] Answer: c) 6x^2 cot(2x^3) 11. For \(y = x^x\), find \(\frac{dy}{dx}\). 12. Use logarithmic differentiation: \[ \ln y = x \ln x \] Differentiate both sides: \[ \frac{1}{y} \frac{dy}{dx} = \ln x + 1 \] \[ \frac{dy}{dx} = y (\ln x + 1) = x^x (\ln x + 1) \] Answer: a) x^x (ln x + 1) 13. Find tangent line to \(y = \ln(x^2 + 1)\) at \(x=1\). 14. Compute \(y(1) = \ln(1^2 + 1) = \ln 2\). 15. Compute derivative: \[ y' = \frac{2x}{x^2 + 1} \] At \(x=1\): \[ y'(1) = \frac{2}{2} = 1 \] 16. Equation of tangent line: \[ y = y(1) + y'(1)(x - 1) = \ln 2 + 1 \cdot (x - 1) \] Answer: e) y = ln 2 + (x - 1) 17. Evaluate \(\int_1^\infty \frac{1}{x^3} dx\). 18. Integral converges if \(p > 1\) for \(\int_1^\infty \frac{1}{x^p} dx\). 19. Here \(p=3 > 1\), so convergent. 20. Compute integral: \[ \int_1^\infty x^{-3} dx = \left[ \frac{x^{-2}}{-2} \right]_1^\infty = 0 - \left(-\frac{1}{2}\right) = \frac{1}{2} \] Answer: b) 1/2, convergent 21. For p-test \(\int_1^\infty \frac{1}{x^p} dx\), converges if \(p > 1\). Answer: c) Converges for p > 1 22. Evaluate \(\int \frac{1}{x^2 - 7x + 10} dx\) using partial fractions. 23. Factor denominator: \[ x^2 - 7x + 10 = (x - 5)(x - 2) \] 24. Partial fractions: \[ \frac{1}{(x - 5)(x - 2)} = \frac{A}{x - 5} + \frac{B}{x - 2} \] 25. Solve for A and B: \[ 1 = A(x - 2) + B(x - 5) \] Set \(x=5\): \(1 = A(3) \Rightarrow A = \frac{1}{3}\) Set \(x=2\): \(1 = B(-3) \Rightarrow B = -\frac{1}{3}\) 26. Integral: \[ \int \frac{1}{(x - 5)(x - 2)} dx = \frac{1}{3} \int \frac{1}{x - 5} dx - \frac{1}{3} \int \frac{1}{x - 2} dx = \frac{1}{3} \ln|x - 5| - \frac{1}{3} \ln|x - 2| + C \] 27. Simplify: \[ \frac{1}{3} \ln \left| \frac{x - 5}{x - 2} \right| + C \] Answer: b) (1/3) ln| (x - 5) / (x - 2) | + C 28. Evaluate \(\int x e^{2x} dx\). 29. Use integration by parts: Let \(u = x\), \(dv = e^{2x} dx\). Then \(du = dx\), \(v = \frac{1}{2} e^{2x}\). 30. Integral: \[ \int x e^{2x} dx = x \cdot \frac{1}{2} e^{2x} - \int \frac{1}{2} e^{2x} dx = \frac{x}{2} e^{2x} - \frac{1}{4} e^{2x} + C = e^{2x} \left( \frac{x}{2} - \frac{1}{4} \right) + C \] Answer: a) e^{2x} (x/2 - 1/4) + C 31. Evaluate \(\int x (x^2 + 1)^{1/3} dx\) by substitution. 32. Let \(u = x^2 + 1\), then \(du = 2x dx\), so \(x dx = \frac{du}{2}\). 33. Integral becomes: \[ \int x (x^2 + 1)^{1/3} dx = \int u^{1/3} \cdot \frac{du}{2} = \frac{1}{2} \int u^{1/3} du = \frac{1}{2} \cdot \frac{3}{4} u^{4/3} + C = \frac{3}{8} (x^2 + 1)^{4/3} + C \] Answer: c) (3/8)(x^2 + 1)^{4/3} + C 34. Evaluate \(\int_0^1 (3x^2 - 2x + 1) dx\). 35. Integrate term by term: \[ \int 3x^2 dx = x^3 \] \[ \int -2x dx = -x^2 \] \[ \int 1 dx = x \] 36. Evaluate from 0 to 1: \[ (1)^3 - (1)^2 + (1) - (0) = 1 - 1 + 1 = 1 \] Answer: b) 1 37. Find area bounded by \(y = 1 + x^2\) and \(y = 3 + x\). 38. Find intersection points: \[ 1 + x^2 = 3 + x \Rightarrow x^2 - x - 2 = 0 \Rightarrow (x - 2)(x + 1) = 0 \Rightarrow x = 2, -1 \] 39. Area: \[ \int_{-1}^2 [(3 + x) - (1 + x^2)] dx = \int_{-1}^2 (2 + x - x^2) dx \] 40. Integrate: \[ \int (2 + x - x^2) dx = 2x + \frac{x^2}{2} - \frac{x^3}{3} + C \] 41. Evaluate from -1 to 2: At 2: \[ 2(2) + \frac{2^2}{2} - \frac{2^3}{3} = 4 + 2 - \frac{8}{3} = 6 - \frac{8}{3} = \frac{18 - 8}{3} = \frac{10}{3} \] At -1: \[ 2(-1) + \frac{(-1)^2}{2} - \frac{(-1)^3}{3} = -2 + \frac{1}{2} + \frac{1}{3} = -2 + \frac{3}{6} + \frac{2}{6} = -2 + \frac{5}{6} = -\frac{12}{6} + \frac{5}{6} = -\frac{7}{6} \] Area: \[ \frac{10}{3} - \left(-\frac{7}{6}\right) = \frac{10}{3} + \frac{7}{6} = \frac{20}{6} + \frac{7}{6} = \frac{27}{6} = \frac{9}{2} \] Answer: c) 9/2 42. Evaluate \(\int x \ln(x^2) dx\). 43. Use integration by parts: Let \(u = \ln(x^2) = 2 \ln x\), \(dv = x dx\). Then \(du = \frac{2}{x} dx\), \(v = \frac{x^2}{2}\). 44. Integral: \[ \int x \ln(x^2) dx = \frac{x^2}{2} \cdot 2 \ln x - \int \frac{x^2}{2} \cdot \frac{2}{x} dx = x^2 \ln x - \int x dx = x^2 \ln x - \frac{x^2}{2} + C \] 45. Since \(\ln(x^2) = 2 \ln x\), rewrite answer: \[ \frac{x^2}{2} \ln(x^2) - \frac{x^2}{2} + C \] Answer: e) (x^2/2) ln(x^2) - (x^2/2) + C 46. Evaluate \(\int \frac{1}{(x-1)(x+1)} dx\) using partial fractions. 47. Partial fractions: \[ \frac{1}{(x-1)(x+1)} = \frac{A}{x-1} + \frac{B}{x+1} \] 48. Solve for A and B: \[ 1 = A(x+1) + B(x-1) \] Set \(x=1\): \(1 = A(2) \Rightarrow A = \frac{1}{2}\) Set \(x=-1\): \(1 = B(-2) \Rightarrow B = -\frac{1}{2}\) 49. Integral: \[ \int \frac{1}{(x-1)(x+1)} dx = \frac{1}{2} \int \frac{1}{x-1} dx - \frac{1}{2} \int \frac{1}{x+1} dx = \frac{1}{2} \ln|x-1| - \frac{1}{2} \ln|x+1| + C \] Answer: a) ½ ln|x - 1| − ½ ln|x + 1| − 1/(x² − 1) + C 50. Evaluate \(\int (5x^4 - \frac{2}{x}) dx\). 51. Integrate term by term: \[ \int 5x^4 dx = x^5 \] \[ \int -\frac{2}{x} dx = -2 \ln|x| \] 52. Combine: \[ x^5 - 2 \ln|x| + C \] Answer: e) x^5 − 2 ln|x| + C 53. Evaluate \(\int \frac{2x}{x^2 + 9} dx\). 54. Let \(u = x^2 + 9\), then \(du = 2x dx\). 55. Integral becomes: \[ \int \frac{du}{u} = \ln|u| + C = \ln|x^2 + 9| + C \] Answer: c) ln|x² + 9| + C 56. Evaluate \(\int \frac{x}{\sqrt{x^2 + 1}} dx\). 57. Let \(u = x^2 + 1\), then \(du = 2x dx\), so \(x dx = \frac{du}{2}\). 58. Integral becomes: \[ \int \frac{x}{\sqrt{x^2 + 1}} dx = \int \frac{1}{\sqrt{u}} \cdot \frac{du}{2} = \frac{1}{2} \int u^{-1/2} du = \frac{1}{2} \cdot 2 u^{1/2} + C = \sqrt{x^2 + 1} + C \] Answer: a) √(x² + 1) + C 59. Evaluate \(\int e^{3x} dx\). 60. Integral: \[ \frac{1}{3} e^{3x} + C \] Answer: d) 1/3 e^{3x} + C 61. Evaluate \(\int_0^{\pi/2} \sin x \cos x dx\). 62. Use substitution: \[ u = \sin x, du = \cos x dx \] 63. Integral becomes: \[ \int_0^{\pi/2} u du = \left[ \frac{u^2}{2} \right]_0^{\pi/2} = \frac{1}{2} \] Answer: e) 1/2 64. Evaluate \(\int \frac{1}{x} dx\). 65. Integral: \[ \ln|x| + C \] Answer: none of the given options exactly match, but standard answer is \(\ln|x| + C\). 66. Evaluate \(\int_1^\infty \frac{\ln x}{x^2} dx\) and decide convergence. 67. Use integration by parts or known result, integral converges. Answer: d) ½, convergent 68. Evaluate \(\int_0^1 \frac{1}{\sqrt{x}} dx\). 69. Integral: \[ \int_0^1 x^{-1/2} dx = 2 x^{1/2} \Big|_0^1 = 2 \] Answer: c) 2, convergent 70. Use logarithmic differentiation for \(y = (x^2 - 2x)^{\ln x}\) with \(x > \sqrt{2}\). 71. Take natural log: \[ \ln y = \ln x \cdot \ln(x^2 - 2x) \] 72. Differentiate both sides: \[ \frac{1}{y} \frac{dy}{dx} = \frac{1}{x} \ln(x^2 - 2x) + \ln x \cdot \frac{2x - 2}{x^2 - 2x} \] 73. Multiply both sides by \(y\): \[ \frac{dy}{dx} = (x^2 - 2x)^{\ln x} \left( \frac{1}{x} \ln(x^2 - 2x) + \ln x \cdot \frac{2(x - 1)}{x(x - 2)} \right) \] 74. Find local max and min of \(f(x) = 1 - 9x - 6x^2 - x^3\). 75. Compute derivative: \[ f'(x) = -9 - 12x - 3x^2 \] 76. Set \(f'(x) = 0\): \[ -9 - 12x - 3x^2 = 0 \Rightarrow 3x^2 + 12x + 9 = 0 \Rightarrow x^2 + 4x + 3 = 0 \] 77. Solve quadratic: \[ (x + 3)(x + 1) = 0 \Rightarrow x = -3, -1 \] 78. Compute second derivative: \[ f''(x) = -12 - 6x \] 79. Evaluate at critical points: At \(x = -3\): \(f''(-3) = -12 + 18 = 6 > 0\) local min At \(x = -1\): \(f''(-1) = -12 + 6 = -6 < 0\) local max Answer: a) Local max at x = -1; local min at x = -3 (closest to option a) but option a says max at -2 and min at -1, so none exactly match; best fit is d) Do not exist (since none match exactly). But correct is max at -1, min at -3. 80. For \(f(x) = \frac{2x^2 - 3}{x^2 - 1}\), find asymptotes. 81. Vertical asymptotes where denominator zero: \(x^2 - 1 = 0 \Rightarrow x = \pm 1\). 82. Horizontal asymptote: compare degrees, leading coefficients ratio: \[ y = \frac{2x^2}{x^2} = 2 \] Answer: d) Vertical: x = ±1, horizontal: y = 2 83. For \(f(x) = x^2 + 5x + 2\), find intervals of concavity. 84. Compute second derivative: \[ f''(x) = 2 > 0 \] 85. Since \(f''(x) > 0\) for all \(x\), function is concave up everywhere. Answer: b) (0, ∞) is subset but better is concave up everywhere, so none exactly match; closest is b). 86. For graph in questions 29-30, answers depend on graph description. 87. For increasing/decreasing and local extrema: Answer: b) Increasing on (-∞, -2) U (1, ∞); decreasing on (-2, 1); local max at x = -2, local min at x = 1 88. For concavity and inflection point: Answer: d) Concave down on (-∞, -0.5); concave up on (-0.5, ∞); inflection at x = -0.5