1. **Problem 5a:** Express $$\frac{2x^2 + 7x - 6}{(x + 5)(x - 4)}$$ in partial fractions. Step 1: Divide numerator by denominator if possible. Here, degree numerator = degree denominator = 2, so divide. Divide $$2x^2 + 7x - 6$$ by $$x^2 + (5 - 4)x - 20 = x^2 + x - 20$$ (expand denominator): The divisor is $$x^2 + x - 20$$. Step 2: Perform polynomial division: - Leading term division: $$2x^2 \div x^2 = 2$$ - Multiply divisor by 2: $$2(x^2 + x - 20) = 2x^2 + 2x - 40$$ - Subtract from numerator: $$ (2x^2 + 7x - 6) - (2x^2 + 2x - 40) = 0x^2 + 5x + 34$$ Step 3: Write as $$\frac{2x^2 + 7x - 6}{(x + 5)(x - 4)} = 2 + \frac{5x + 34}{(x + 5)(x - 4)}$$ Step 4: Express $$\frac{5x + 34}{(x + 5)(x - 4)} = \frac{A}{x + 5} + \frac{B}{x - 4}$$ Multiply both sides by denominator: $$5x + 34 = A(x - 4) + B(x + 5) = (A + B)x + (-4A + 5B)$$ Equate coefficients: - For $$x$$: $$5 = A + B$$ - Constant: $$34 = -4A + 5B$$ Step 5: Solve system: From first: $$B = 5 - A$$ Substitute into second: $$34 = -4A + 5(5 - A) = -4A + 25 - 5A = 25 - 9A$$ $$9A = 25 - 34 = -9$$ $$A = -1$$ $$B = 5 - (-1) = 6$$ Final partial fractions: $$\frac{2x^2 + 7x - 6}{(x + 5)(x - 4)} = 2 + \frac{-1}{x + 5} + \frac{6}{x - 4}$$ --- 2. **Problem 5b:** Expand $$\frac{2x^2 + 7x - 6}{(x + 5)(x - 4)}$$ in ascending powers of $$x$$ up to $$x^2$$. Step 1: Use the partial fraction form: $$2 + \frac{-1}{x + 5} + \frac{6}{x - 4}$$ Step 2: Rewrite denominators expanding around $$x=0$$: - $$\frac{1}{x + 5} = \frac{1}{5} \cdot \frac{1}{1 + \frac{x}{5}} = \frac{1}{5} \sum_{n=0}^{\infty} (-1)^n \left(\frac{x}{5}\right)^n = \frac{1}{5} - \frac{x}{25} + \frac{x^2}{125} + \cdots$$ - $$\frac{1}{x - 4} = \frac{1}{-4} \cdot \frac{1}{1 - \frac{x}{4}} = -\frac{1}{4} \sum_{n=0}^{\infty} \left(\frac{x}{4}\right)^n = -\frac{1}{4} - \frac{x}{16} - \frac{x^2}{64} + \cdots$$ Step 3: Substitute into partial fractions: $$2 + (-1)\left(\frac{1}{5} - \frac{x}{25} + \frac{x^2}{125}\right) + 6\left(-\frac{1}{4} - \frac{x}{16} - \frac{x^2}{64}\right)$$ Step 4: Simplify terms: - Constants: $$2 - \frac{1}{5} - \frac{6}{4} = 2 - 0.2 - 1.5 = 0.3$$ - Coefficient of $$x$$: $$0 + \frac{1}{25} - \frac{6}{16} = 0.04 - 0.375 = -0.335$$ - Coefficient of $$x^2$$: $$0 - \frac{1}{125} - \frac{6}{64} = -0.008 - 0.09375 = -0.10175$$ Step 5: Write final expansion: $$\frac{2x^2 + 7x - 6}{(x + 5)(x - 4)} = 0.3 - 0.335x - 0.10175 x^2 + \cdots$$ --- 3. **Problem 5c:** State the set of values of $$x$$ for which the expansion is valid. Step 1: The expansions of $$\frac{1}{x+5}$$ and $$\frac{1}{x-4}$$ as geometric series converge if $$\left|\frac{x}{5}\right| < 1 \Rightarrow |x| < 5$$ and $$\left|\frac{x}{4}\right| < 1 \Rightarrow |x| < 4$$ Step 2: The combined expansion is valid on the intersection: $$|x| < 4$$ --- 4. **Problem 6a:** Given $$\frac{3x^2 + 4x - 5}{(x + 3)(x - 2)} = A + \frac{B}{x + 3} + \frac{C}{x - 2}$$ Find $$A, B, C$$. Step 1: Rewrite denominator: $$ (x + 3)(x - 2) = x^2 + x - 6$$ Step 2: Multiply both sides by denominator: $$3x^2 + 4x - 5 = A(x^2 + x - 6) + B(x - 2) + C(x + 3)$$ Step 3: Expand right: $$Ax^2 + Ax - 6A + Bx - 2B + Cx + 3C = Ax^2 + (A + B + C)x + (-6A - 2B + 3C)$$ Step 4: Equate coefficients: - $$x^2: 3 = A$$ - $$x: 4 = A + B + C$$ - Constant: $$-5 = -6A -2B + 3C$$ Step 5: Substitute $$A=3$$ into other equations: - $$4 = 3 + B + C \Rightarrow B + C = 1$$ - $$-5 = -18 - 2B + 3C$$ - Rearrange constant: $$-5 + 18 = -2B + 3C \Rightarrow 13 = -2B + 3C$$ Step 6: From $$B + C = 1$$, express $$C = 1 - B$$. Substitute in constant: $$13 = -2B + 3(1-B) = -2B + 3 - 3B = 3 - 5B$$ $$5B = 3 - 13 = -10$$ $$B = -2$$ Then $$C = 1 - (-2) = 3$$ Final values: $$A=3, B=-2, C=3$$ --- 5. **Problem 6b:** Expand $$\frac{3x^2 + 4x - 5}{(x + 3)(x - 2)}$$ in ascending powers of $$x$$ up to $$x^2$$. Step 1: Use the decomposition: $$3 + \frac{-2}{x + 3} + \frac{3}{x - 2}$$ Step 2: Expand denominators at $$x=0$$: - $$\frac{1}{x + 3} = \frac{1}{3} \frac{1}{1 + \frac{x}{3}} = \frac{1}{3} \sum_{n=0}^\infty (-1)^n \left(\frac{x}{3}\right)^n = \frac{1}{3} - \frac{x}{9} + \frac{x^2}{27} + \cdots$$ - $$\frac{1}{x - 2} = \frac{1}{-2} \frac{1}{1 - \frac{x}{2}} = -\frac{1}{2} \sum_{n=0}^\infty \left(\frac{x}{2}\right)^n = -\frac{1}{2} - \frac{x}{4} - \frac{x^2}{8} + \cdots$$ Step 3: Substitute into partial fractions: $$3 + (-2)\left(\frac{1}{3} - \frac{x}{9} + \frac{x^2}{27}\right) + 3\left(-\frac{1}{2} - \frac{x}{4} - \frac{x^2}{8}\right)$$ Step 4: Simplify: - Constants: $$3 - \frac{2}{3} - \frac{3}{2} = 3 - 0.6667 - 1.5 = 0.8333$$ - Coefficient of $$x$$: $$0 + \frac{2}{9} - \frac{3}{4} = 0.2222 - 0.75 = -0.5278$$ - Coefficient of $$x^2$$: $$0 - \frac{2}{27} - \frac{3}{8} = -0.0741 - 0.375 = -0.4491$$ Step 5: Write final expansion: $$\frac{3x^2 + 4x - 5}{(x + 3)(x - 2)} = \frac{5}{6} - \frac{19}{36} x - \frac{161}{360} x^2 + \cdots$$ (Where coefficients are simplified fractions: $$0.8333 = \frac{5}{6}$$, $$-0.5278 = -\frac{19}{36}$$, $$-0.4491 \approx -\frac{161}{360}$$) --- 6. **Problem 7a:** Given $$f(x) = \frac{2x^2 + 5x + 11}{(2x - 1)^2 (x + 1)}, |x| < \frac{1}{2}$$ with decomposition: $$f(x) = \frac{A}{2x - 1} + \frac{B}{(2x - 1)^2} + \frac{C}{x + 1}$$ Find $$A, B, C$$. Step 1: Multiply both sides by denominator: $$2x^2 + 5x + 11 = A(2x - 1)(x + 1) + B(x + 1) + C(2x - 1)^2$$ Step 2: Expand partial terms: - $$A(2x - 1)(x + 1) = A(2x^2 + 2x - x - 1) = A(2x^2 + x -1)$$ - $$B(x + 1) = Bx + B$$ - $$C(2x -1)^2 = C(4x^2 - 4x + 1)$$ Sum: $$2x^2 + 5x + 11 = A(2x^2 + x -1) + Bx + B + C(4x^2 - 4x + 1)$$ Step 3: Collect like terms: $$2x^2 + 5x + 11 = (2A + 4C) x^2 + (A + B - 4C) x + (-A + B + C)$$ Step 4: Equate coefficients: - $$x^2: 2 = 2A + 4C$$ - $$x: 5 = A + B - 4C$$ - Constant: $$11 = -A + B + C$$ Step 5: From $$2 = 2A + 4C$$, divide by 2: $$1 = A + 2C \Rightarrow A = 1 - 2C$$ Step 6: Substitute $$A = 1 - 2C$$ in other equations: - For $$x$$: $$5 = (1 - 2C) + B - 4C = 1 + B - 6C$$ $$B = 5 - 1 + 6C = 4 + 6C$$ - For constant: $$11 = - (1 - 2C) + B + C = -1 + 2C + B + C = -1 + B + 3C$$ Substitute $$B$$: $$11 = -1 + (4 + 6C) + 3C = 3 + 9C$$ $$9C = 11 - 3 = 8 \Rightarrow C = \frac{8}{9}$$ Step 7: Find $$A$$ and $$B$$: $$A = 1 - 2 \times \frac{8}{9} = 1 - \frac{16}{9} = -\frac{7}{9}$$ $$B = 4 + 6 \times \frac{8}{9} = 4 + \frac{48}{9} = 4 + \frac{16}{3} = \frac{12}{3} + \frac{16}{3} = \frac{28}{3}$$ Final constants: $$A = -\frac{7}{9}, B = \frac{28}{3}, C = \frac{8}{9}$$ --- 7. **Problem 7b:** Find series expansion of $$f(x)$$ up to $$x^2$$. Step 1: Write in partial fractions: $$f(x) = \frac{A}{2x - 1} + \frac{B}{(2x - 1)^2} + \frac{C}{x + 1}$$ Step 2: Expand denominators about $$x=0$$: - $$\frac{1}{2x - 1} = -\frac{1}{1 - 2x} = -\sum_{n=0}^\infty (2x)^n = -1 - 2x - 4x^2 + \cdots$$ - $$\frac{1}{(2x - 1)^2} = \frac{d}{dx} \left(\frac{1}{2x - 1}\right) \times (-1/2) = \text{Alternatively use binomial expansion:}$$ Rewrite: $$\frac{1}{(2x-1)^2} = \frac{d}{d(2x-1)} \left(\frac{1}{2x-1}\right) = -\frac{2}{(2x-1)^3}$$ Better to expand as: $$\frac{1}{(1-2x)^2} = \sum_{n=0}^\infty (n+1)(2x)^n = 1 + 4x + 12x^2 + \cdots$$ Multiply by $$B$$: $$B \times \frac{1}{(2x -1)^2} = \frac{28}{3} (1 + 4x + 12x^2) = \frac{28}{3} + \frac{112}{3} x + 112 x^2$$ Step 3: Expand $$\frac{C}{x+1}$$ as geometric series: $$\frac{1}{x+1} = \sum_{n=0}^\infty (-x)^n = 1 - x + x^2 + \cdots$$ Multiply by $$C = \frac{8}{9}$$: $$\frac{8}{9} (1 - x + x^2) = \frac{8}{9} - \frac{8}{9} x + \frac{8}{9} x^2$$ Step 4: Summing up all parts: - $$A / (2x -1) = -\frac{7}{9}(-1 - 2x - 4x^2) = \frac{7}{9} + \frac{14}{9} x + \frac{28}{9} x^2$$ Total sum: $$f(x) = \left(\frac{7}{9} + \frac{28}{3} + \frac{8}{9}\right) + \left(\frac{14}{9} + \frac{112}{3} - \frac{8}{9}\right) x + \left(\frac{28}{9} + 112 + \frac{8}{9}\right) x^2$$ Simplify constants: - Constant: $$\frac{7}{9} + \frac{28}{3} + \frac{8}{9} = \frac{7+8}{9} + \frac{28}{3} = \frac{15}{9} + \frac{28}{3} = \frac{5}{3} + \frac{28}{3} = \frac{33}{3} = 11$$ - Coefficient of $$x$$: $$\frac{14}{9} + \frac{112}{3} - \frac{8}{9} = \frac{14 - 8}{9} + \frac{112}{3} = \frac{6}{9} + \frac{112}{3} = \frac{2}{3} + \frac{112}{3} = \frac{114}{3} = 38$$ - Coefficient of $$x^2$$: $$\frac{28}{9} + 112 + \frac{8}{9} = \frac{36}{9} + 112 = 4 + 112 = 116$$ Step 5: Final series expansion: $$f(x) = 11 + 38x + 116x^2 + \cdots$$ --- 8. **Problem 7c:** Percentage error in using the series expansion at $$x=0.05$$ Step 1: Calculate exact value: $$f(0.05) = \frac{2(0.05)^2 + 5(0.05) + 11}{(2 \times 0.05 - 1)^2 (0.05 + 1)}$$ Calculate numerator: $$2(0.0025) + 0.25 + 11 = 0.005 + 0.25 + 11 = 11.255$$ Calculate denominator: $$(0.1 - 1)^2 (1.05) = (-0.9)^2 \times 1.05 = 0.81 \times 1.05 = 0.8505$$ Exact value: $$\frac{11.255}{0.8505} \approx 13.23$$ Step 2: Approximate value from series: $$f(0.05) \approx 11 + 38(0.05) + 116(0.05)^2 = 11 + 1.9 + 0.29 = 13.19$$ Step 3: Compute percentage error: $$\frac{|13.23 - 13.19|}{13.23} \times 100 = \frac{0.04}{13.23} \times 100 \approx 0.30\%$$ Answer rounded to 2 significant figures: $$0.30\%$$