1. Evaluate $ \lim_{x \to 0} \frac{e^{2x} - 1}{x} $. The problem asks for the limit of the expression as $ x $ approaches 0. 2. Use the formula for the derivative of $ e^{2x} $ at 0 or apply L'Hôpital's Rule since direct substitution gives $ \frac{0}{0} $ indeterminate form. 3. Applying L'Hôpital's Rule: $$\lim_{x \to 0} \frac{e^{2x} - 1}{x} = \lim_{x \to 0} \frac{\frac{d}{dx}(e^{2x} - 1)}{\frac{d}{dx}x} = \lim_{x \to 0} \frac{2e^{2x}}{1} = 2e^0 = 2.$$ Answer: e) 2 --- 4. Evaluate $ \lim_{x \to 0} \frac{\ln(1+x) - x}{x^2} $ using L'Hôpital's Rule. 5. Direct substitution gives $ \frac{0}{0} $, so apply L'Hôpital's Rule twice. 6. First derivative: $$\lim_{x \to 0} \frac{\frac{1}{1+x} - 1}{2x} = \lim_{x \to 0} \frac{\frac{1 - (1+x)}{1+x}}{2x} = \lim_{x \to 0} \frac{\frac{-x}{1+x}}{2x} = \lim_{x \to 0} \frac{-1}{2(1+x)} = -\frac{1}{2}.$$ Answer: b) -1/2 --- 5. Given $ y = \ln(\sin(2x^3)) $, find $ \frac{dy}{dx} $. 6. Use chain rule: $$ \frac{dy}{dx} = \frac{1}{\sin(2x^3)} \cdot \cos(2x^3) \cdot \frac{d}{dx}(2x^3) = \frac{\cos(2x^3)}{\sin(2x^3)} \cdot 6x^2 = 6x^2 \cot(2x^3).$$ Answer: c) 6x^2 cot(2x^3) --- 6. For $ y = x^x $, find $ \frac{dy}{dx} $. 7. Use logarithmic differentiation: $$ \ln y = x \ln x \implies \frac{1}{y} \frac{dy}{dx} = \ln x + 1 \implies \frac{dy}{dx} = y(\ln x + 1) = x^x (\ln x + 1).$$ Answer: a) x^x (ln x + 1) --- 7. Find tangent line to $ y = \ln(x^2 + 1) $ at $ x = 1 $. 8. Compute $ y(1) = \ln(1^2 + 1) = \ln 2 $. 9. Derivative: $$ y' = \frac{2x}{x^2 + 1} \implies y'(1) = \frac{2}{2} = 1.$$ 10. Equation of tangent line: $$ y = y(1) + y'(1)(x - 1) = \ln 2 + 1(x - 1).$$ Answer: e) y = ln 2 + (x - 1) --- 8. Evaluate $ \int_1^\infty \frac{1}{x^3} dx $. 9. Integral: $$ \int_1^\infty x^{-3} dx = \lim_{t \to \infty} \left[ \frac{x^{-2}}{-2} \right]_1^t = \lim_{t \to \infty} \left( -\frac{1}{2t^2} + \frac{1}{2} \right) = \frac{1}{2}.$$ Convergent. Answer: b) 1/2, convergent --- 9. For $ \int_1^\infty \frac{1}{x^p} dx $, convergence occurs if $ p > 1 $. Answer: c) Converges for p > 1 --- 10. Evaluate $ \int \frac{1}{x^2 - 7x + 10} dx $ using partial fractions. 11. Factor denominator: $$ x^2 - 7x + 10 = (x - 5)(x - 2).$$ 12. Partial fractions: $$ \frac{1}{(x - 5)(x - 2)} = \frac{A}{x - 5} + \frac{B}{x - 2}.$$ 13. Solve for A and B: $$ 1 = A(x - 2) + B(x - 5).$$ 14. Set $ x = 5 $: $$ 1 = A(3) + B(0) \implies A = \frac{1}{3}.$$ 15. Set $ x = 2 $: $$ 1 = A(0) + B(-3) \implies B = -\frac{1}{3}.$$ 16. Integral: $$ \int \frac{1}{x^2 - 7x + 10} dx = \int \left( \frac{1/3}{x - 5} - \frac{1/3}{x - 2} \right) dx = \frac{1}{3} \ln|x - 5| - \frac{1}{3} \ln|x - 2| + C = \frac{1}{3} \ln \left| \frac{x - 5}{x - 2} \right| + C.$$ Answer: b) (1/3) ln| (x - 5) / (x - 2) | + C --- 11. Evaluate $ \int x e^{2x} dx $. 12. Use integration by parts: Let $ u = x $, $ dv = e^{2x} dx $. 13. Then $ du = dx $, $ v = \frac{1}{2} e^{2x} $. 14. Integral: $$ \int x e^{2x} dx = uv - \int v du = x \cdot \frac{e^{2x}}{2} - \int \frac{e^{2x}}{2} dx = \frac{x e^{2x}}{2} - \frac{1}{2} \cdot \frac{e^{2x}}{2} + C = e^{2x} \left( \frac{x}{2} - \frac{1}{4} \right) + C.$$ Answer: a) e^(2x) (x/2 - 1/4) + C --- 12. Evaluate $ \int x (x^2 + 1)^{1/3} dx $ using substitution. 13. Let $ u = x^2 + 1 $, then $ du = 2x dx $ or $ x dx = \frac{du}{2} $. 14. Integral becomes: $$ \int x (x^2 + 1)^{1/3} dx = \int u^{1/3} \cdot \frac{du}{2} = \frac{1}{2} \int u^{1/3} du = \frac{1}{2} \cdot \frac{u^{4/3}}{4/3} + C = \frac{3}{8} (x^2 + 1)^{4/3} + C.$$ Answer: c) (3/8) (x² + 1)^(4/3) + C --- 13. Evaluate $ \int_0^1 (3x^2 - 2x + 1) dx $. 14. Integrate term-wise: $$ \int_0^1 3x^2 dx = [x^3]_0^1 = 1,$$ $$ \int_0^1 -2x dx = [-x^2]_0^1 = -1,$$ $$ \int_0^1 1 dx = [x]_0^1 = 1.$$ 15. Sum: $$ 1 - 1 + 1 = 1.$$ Answer: b) 1 --- 14. Find area bounded by $ y = 1 + x^2 $ and $ y = 3 + x $. 15. Find intersection points: $$ 1 + x^2 = 3 + x \implies x^2 - x - 2 = 0 \implies (x - 2)(x + 1) = 0 \implies x = 2, -1.$$ 16. Area: $$ \int_{-1}^2 [(3 + x) - (1 + x^2)] dx = \int_{-1}^2 (2 + x - x^2) dx.$$ 17. Integrate: $$ \left[ 2x + \frac{x^2}{2} - \frac{x^3}{3} \right]_{-1}^2 = \left(4 + 2 - \frac{8}{3}\right) - \left(-2 + \frac{1}{2} + \frac{1}{3}\right) = \left(6 - \frac{8}{3}\right) - \left(-\frac{7}{6}\right) = \frac{10}{3} + \frac{7}{6} = \frac{20}{6} + \frac{7}{6} = \frac{27}{6} = \frac{9}{2}.$$ Answer: c) 9/2 --- 15. Evaluate $ \int x \ln(x^2) dx $. 16. Use integration by parts: Let $ u = \ln(x^2) = 2 \ln x $, $ dv = x dx $. 17. Then $ du = \frac{2}{x} dx $, $ v = \frac{x^2}{2} $. 18. Integral: $$ \int x \ln(x^2) dx = uv - \int v du = \frac{x^2}{2} \cdot 2 \ln x - \int \frac{x^2}{2} \cdot \frac{2}{x} dx = x^2 \ln x - \int x dx = x^2 \ln x - \frac{x^2}{2} + C.$$ 19. Since $ \ln(x^2) = 2 \ln x $, rewrite: $$ \int x \ln(x^2) dx = \frac{x^2}{2} \ln(x^2) - \frac{x^2}{2} + C.$$ Answer: e) (x²/2) ln(x²) - (x²/2) + C --- 10 (second). Evaluate $ \int \frac{1}{(x-1)(x+1)} dx $ using partial fractions. 11. Partial fractions: $$ \frac{1}{(x-1)(x+1)} = \frac{A}{x-1} + \frac{B}{x+1}.$$ 12. Multiply both sides: $$ 1 = A(x+1) + B(x-1).$$ 13. Set $ x=1 $: $$ 1 = A(2) + B(0) \implies A = \frac{1}{2}.$$ 14. Set $ x=-1 $: $$ 1 = A(0) + B(-2) \implies B = -\frac{1}{2}.$$ 15. Integral: $$ \int \left( \frac{1/2}{x-1} - \frac{1/2}{x+1} \right) dx = \frac{1}{2} \ln|x-1| - \frac{1}{2} \ln|x+1| + C.$$ Answer: a) ½ ln |x - 1| - ½ ln |x + 1| - 1/(x^2 - 1) + C --- 17. Evaluate $ \int (5x^4 - \frac{2}{x}) dx $. 18. Integrate term-wise: $$ \int 5x^4 dx = x^5,$$ $$ \int -\frac{2}{x} dx = -2 \ln|x|.$$ 19. Sum: $$ x^5 - 2 \ln|x| + C.$$ Answer: e) x^5 - 2 ln |x| + C --- 18. Evaluate $ \int \frac{2x}{x^2 + 9} dx $. 19. Use substitution: Let $ u = x^2 + 9 $, then $ du = 2x dx $. 20. Integral becomes: $$ \int \frac{du}{u} = \ln|u| + C = \ln|x^2 + 9| + C.$$ Answer: c) ln |x^2 + 9| + C --- 19. Evaluate $ \int \frac{x}{\sqrt{x^2 + 1}} dx $. 20. Use substitution: Let $ u = x^2 + 1 $, then $ du = 2x dx $, so $ x dx = \frac{du}{2} $. 21. 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 --- 20. Evaluate $ \int e^{3x} dx $. 21. Integral: $$ \frac{1}{3} e^{3x} + C.$$ Answer: d) 1/3 e^(3x) + C --- 21. Evaluate $ \int_0^{\pi/2} \sin x \cos x dx $. 22. Use substitution: Let $ u = \sin x $, then $ du = \cos x dx $. 23. Integral becomes: $$ \int_0^{\pi/2} u du = \left[ \frac{u^2}{2} \right]_0^{\pi/2} = \frac{\sin^2(\pi/2)}{2} - 0 = \frac{1}{2}.$$ Answer: e) 1/2 --- 22. Evaluate $ \int \frac{1}{x} dx $. 23. Integral: $$ \ln|x| + C.$$ Answer: none of the options exactly match, but the closest is e) ln e^2 (which is 2), so the integral is $ \ln|x| + C $. --- 23. Evaluate and decide convergence: $ \int_1^\infty \frac{\ln x}{x^2} dx $. 24. Use integration by parts or known result; integral converges. 25. The value is 1, convergent. Answer: b) 1, convergent --- 24. Evaluate and decide convergence: $ \int_0^1 \frac{1}{\sqrt{x}} dx $. 25. Integral: $$ \int_0^1 x^{-1/2} dx = \left[ 2 \sqrt{x} \right]_0^1 = 2.$$ Convergent. Answer: c) 2, convergent --- 25. Use logarithmic differentiation for $ y = (x^2 - 2x)^{1/x} $, $ x > \sqrt{2} $. 26. Take natural log: $$ \ln y = \frac{1}{x} \ln(x^2 - 2x).$$ 27. Differentiate both sides: $$ \frac{1}{y} \frac{dy}{dx} = -\frac{1}{x^2} \ln(x^2 - 2x) + \frac{1}{x} \cdot \frac{2x - 2}{x^2 - 2x} = -\frac{1}{x^2} \ln(x^2 - 2x) + \frac{2(x - 1)}{x(x^2 - 2x)}.$$ 28. Multiply both sides by $ y $: $$ \frac{dy}{dx} = y \left( -\frac{1}{x^2} \ln(x^2 - 2x) + \frac{2(x - 1)}{x(x^2 - 2x)} \right) = (x^2 - 2x)^{1/x} \left( -\frac{1}{x^2} \ln(x^2 - 2x) + \frac{2(x - 1)}{x(x^2 - 2x)} \right).$$ --- 26. Find local max and min of $ f(x) = 1 - 9x - 6x^2 - x^3 $. 27. Find critical points by $ f'(x) = -9 - 12x - 3x^2 = 0 $. 28. Solve: $$ -3x^2 - 12x - 9 = 0 \implies x^2 + 4x + 3 = 0 \implies (x+1)(x+3) = 0 \implies x = -1, -3.$$ 29. Use second derivative: $$ f''(x) = -12 - 6x.$$ 30. Evaluate at critical points: $$ f''(-1) = -12 + 6 = -6 < 0 \Rightarrow \text{local max at } x = -1,$$ $$ f''(-3) = -12 + 18 = 6 > 0 \Rightarrow \text{local min at } x = -3.$$ Answer: d) Local max at x = -1; local min at x = -3 --- 27. For $ y = \frac{2x^2 - 3}{x^2 - 1} $, find asymptotes. 28. Vertical asymptotes where denominator zero: $$ x^2 - 1 = 0 \implies x = \pm 1.$$ 29. Horizontal asymptote by degrees: Degrees numerator and denominator both 2, leading coefficients ratio: $$ \frac{2}{1} = 2.$$ Answer: d) Vertical: x = ±1, horizontal: y = 2 --- 28. For $ f(x) = x^3 + 5x + 2 $, find concavity. 29. Second derivative: $$ f''(x) = 6x.$$ 30. Concave up where $ f''(x) > 0 $, i.e., $ x > 0 $. Answer: b) (0, ∞) --- 29. From graph description, increasing/decreasing and local extrema: 30. Increasing on (-∞, -2) and (1, ∞), decreasing on (-2, 1), local max at -2, local min at 1. Answer: b) Increasing on (-∞, -2) ∪ (1, ∞); decreasing on (-2, 1); local max at x = -2, local min at x = 1 --- 30. Concavity and inflection point: 31. Concave down on (-∞, -0.5), concave up on (-0.5, ∞), inflection at x = -0.5. Answer: d) Concave down on (-∞, -0.5); concave up on (-0.5, ∞); inflection at x = -0.5