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Punjab Board Class 11 Mathematics

Exercise 5.2 Solutions — Partial Fractions

Complete worked solutions for Exercise 5.2 from Chapter 5 (Partial Fractions), Punjab Board Class 11 Mathematics. Each question shows every step and names the formula used.

Mathematics 11 (Punjab Board)

Exercise 5.2 Solutions

Worked solutions for Punjab Board Class 11 Mathematics, Chapter 5: Partial Fractions.

Question 1: Resolve \(\frac{2x^2+3x+3}{(x+1)(x^2+1)}\) into partial fractions.

Aik distinct linear factor aur aik irreducible quadratic — is liye \(\frac{A}{x+1} + \frac{Bx+C}{x^2+1}\) likho (quadratic ke liye numerator linear!).

\[ 2x^2 + 3x + 3 = A(x^2+1) + (Bx+C)(x+1) \]

Put \(x = -1\): \(2 = 2A \implies A = 1\). Comparing \(x^2\): \(A + B = 2 \implies B = 1\). Comparing \(x\): \(B + C = 3 \implies C = 2\).

Answer: \(\frac{1}{x+1} + \frac{x+2}{x^2+1}\)

Question 2: Resolve \(\frac{2x+1}{(x-2)(x^2+3x+5)}\) into partial fractions.

\(x^2+3x+5\) ka discriminant \(9-20 < 0\) hai — irreducible quadratic. Wohi template: \(\frac{A}{x-2} + \frac{Bx+C}{x^2+3x+5}\).

\[ 2x + 1 = A(x^2+3x+5) + (Bx+C)(x-2) \]

Put \(x = 2\): \(5 = 15A \implies A = \frac{1}{3}\). Comparing \(x^2\): \(A + B = 0 \implies B = -\frac{1}{3}\). Comparing \(x\): \(3A - 2B + C = 2 \implies 1 + \frac{2}{3} + C = 2 \implies C = \frac{1}{3}\).

Answer: \(\frac{1}{3(x-2)} + \frac{1-x}{3(x^2+3x+5)}\)

Question 3: Resolve \(\frac{2x+32}{(x-2)(x^2+2)^2}\) into partial fractions.

\((x^2+2)^2\) repeated irreducible quadratic hai — uski dono powers ke liye alag linear numerators: \(\frac{A}{x-2} + \frac{Bx+C}{x^2+2} + \frac{Dx+E}{(x^2+2)^2}\).

\[ 2x + 32 = A(x^2+2)^2 + (Bx+C)(x-2)(x^2+2) + (Dx+E)(x-2) \]

Put \(x = 2\): \(36 = 36A \implies A = 1\). Comparing \(x^4\): \(A + B = 0 \implies B = -1\). Comparing \(x^3\): \(-2B + C = 0 \implies C = -2\).

Comparing \(x^2\): \(4A + 2B - 2C + D = 0 \implies 4 - 2 + 4 + D = 0 \implies D = -6\). Comparing \(x\): \(-4B + 2C - 2D + E = 2 \implies 4 - 4 + 12 + E = 2 \implies E = -10\).

Answer: \(\frac{1}{x-2} + \frac{-x-2}{x^2+2} + \frac{-6x-10}{(x^2+2)^2}\)

Common mistake: Repeated quadratic ki dono powers ke liye terms likhna lazmi hai — sirf \(\frac{Bx+C}{x^2+2}\) rakh kar \(\frac{Dx+E}{(x^2+2)^2}\) bhool jana ghalat hai.

Question 4: Resolve \(\frac{3x^2+3}{(x+1)(x^2-x+1)}\) into partial fractions.

\(x^2-x+1\) irreducible hai (discriminant \(1-4 < 0\)). Template: \(\frac{A}{x+1} + \frac{Bx+C}{x^2-x+1}\).

\[ 3x^2 + 3 = A(x^2-x+1) + (Bx+C)(x+1) \]

Put \(x = -1\): \(6 = 3A \implies A = 2\). Comparing \(x^2\): \(A + B = 3 \implies B = 1\). Comparing \(x\): \(-A + B + C = 0 \implies C = 1\).

Answer: \(\frac{2}{x+1} + \frac{x+1}{x^2-x+1}\)

Question 5: Resolve \(\frac{x^4}{(x^2+1)^2}\) into partial fractions.

Improper fraction (degree 4/4). Pehle divide karo: \(x^4 = 1\cdot(x^4+2x^2+1) + (-2x^2-1)\). Phir proper part ko resolve karo.

\[ \frac{x^4}{(x^2+1)^2} = 1 - \frac{2x^2+1}{(x^2+1)^2} \]
\[ \frac{2x^2+1}{(x^2+1)^2} = \frac{Ax+B}{x^2+1} + \frac{Cx+D}{(x^2+1)^2}, \quad 2x^2+1 = (Ax+B)(x^2+1) + Cx + D \]

Comparing \(x^3\): \(A = 0\). Comparing \(x^2\): \(B = 2\). Comparing \(x\): \(C = 0\). Constants: \(B + D = 1 \implies D = -1\).

\[ \frac{x^4}{(x^2+1)^2} = 1 - \left(\frac{2}{x^2+1} - \frac{1}{(x^2+1)^2}\right) = 1 - \frac{2}{x^2+1} + \frac{1}{(x^2+1)^2} \]
Answer: \(1 - \frac{2}{x^2+1} + \frac{1}{(x^2+1)^2}\)

Question 6: Resolve \(\frac{6x^4+40x^2}{(2-x)(2+x)(x^2+4)^2}\) into partial fractions.

Do distinct linear factors \((2-x), (2+x)\) aur aik repeated irreducible quadratic \((x^2+4)^2\). Chh terms: \(\frac{A}{2-x} + \frac{B}{2+x} + \frac{Cx+D}{x^2+4} + \frac{Ex+F}{(x^2+4)^2}\).

\[ 6x^4+40x^2 = A(2+x)(x^2+4)^2 + B(2-x)(x^2+4)^2 + (Cx+D)(4-x^2)(x^2+4) + (Ex+F)(4-x^2) \]

Put \(x = 2\): \(256 = 256A \implies A = 1\). Put \(x = -2\): \(256 = 256B \implies B = 1\).

Comparing \(x^5\): \(A - B - C = 0 \implies C = 0\). Comparing \(x^4\): \(2A + 2B - D = 6 \implies D = -2\).

Comparing \(x^3\): \(8A - 8B - E = 0 \implies E = 0\). Comparing \(x^2\): \(16A + 16B - F = 40 \implies F = -8\).

Answer: \(\frac{1}{2-x} + \frac{1}{2+x} - \frac{2}{x^2+4} - \frac{8}{(x^2+4)^2}\)