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Punjab Board Class 11 Mathematics

Exercise 5.1 Solutions — Partial Fractions

Complete worked solutions for Exercise 5.1 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 5.1 Solutions

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

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

Denominator \(= (x-1)(x+1)\) — distinct linear factors. \(\frac{A}{x-1} + \frac{B}{x+1}\) likho aur cover-up method se \(x = 1, -1\) rakho.

\[ 2 = A(x+1) + B(x-1) \]

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

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

Question 2: Resolve \(\frac{a-b}{(x-a)(x-b)}\) into partial fractions.

Yahan constants \(a, b\) hain lekin method wohi hai: \(\frac{A}{x-a} + \frac{B}{x-b}\).

\[ a - b = A(x-b) + B(x-a) \]

Put \(x = a\): \(a - b = A(a-b) \implies A = 1\). Put \(x = b\): \(a - b = B(b-a) \implies B = -1\).

Answer: \(\frac{1}{x-a} - \frac{1}{x-b}\)

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

Numerator aur denominator dono ki degree 2 hai — improper fraction! Pehle divide karo: quotient 1, remainder 2.

\[ \frac{x^2+1}{x^2-1} = 1 + \frac{2}{x^2-1} = 1 + \frac{2}{(x-1)(x+1)} \]

From Question 1, \(\frac{2}{(x-1)(x+1)} = \frac{1}{x-1} - \frac{1}{x+1}\).

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

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

Teen distinct linear factors — \(\frac{A}{x+1} + \frac{B}{x+2} + \frac{C}{x+3}\). Har factor ko zero karne wali value rakho.

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

Put \(x = -1\): \(1 = 2A \implies A = \frac{1}{2}\). Put \(x = -2\): \(-1 = -B \implies B = 1\). Put \(x = -3\): \(-3 = 2C \implies C = -\frac{3}{2}\).

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

Question 5: Resolve \(\frac{x^2+4x+5}{(x+1)(x^2+5x+6)}\) into partial fractions.

Pehle denominator factorize karo: \(x^2+5x+6 = (x+2)(x+3)\). Phir teen distinct linear factors wala case.

\[ x^2 + 4x + 5 = A(x+2)(x+3) + B(x+1)(x+3) + C(x+1)(x+2) \]

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

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

Question 6: Resolve \(\frac{4x^3+5x^2-3x-2}{x^2-1}\) into partial fractions.

Improper fraction (degree 3/2). Long division se quotient \(4x+5\) aur remainder \(x+3\) milta hai.

\[ 4x^3 + 5x^2 - 3x - 2 = (4x+5)(x^2-1) + (x+3) \]
\[ \frac{4x^3+5x^2-3x-2}{x^2-1} = 4x + 5 + \frac{x+3}{(x-1)(x+1)} \]
\[ x + 3 = A(x+1) + B(x-1); \quad x = 1 \implies A = 2, \quad x = -1 \implies B = -1 \]
Answer: \(4x + 5 + \frac{2}{x-1} - \frac{1}{x+1}\)

Common mistake: Division ka quotient (\(4x+5\)) jawab mein likhna na bhoolo — sirf remainder ko resolve karke ruk jana ghalat hai.

Question 7: Resolve \(\frac{3x^2-12x+11}{(x-1)(x-2)(x-3)}\) into partial fractions.

Teen distinct linear factors. Identity banao aur \(x = 1, 2, 3\) rakho.

\[ 3x^2 - 12x + 11 = A(x-2)(x-3) + B(x-1)(x-3) + C(x-1)(x-2) \]

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

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

Question 8: Resolve \(\frac{(x-1)(x-2)(x-3)}{(x-4)(x-5)(x-6)}\) into partial fractions.

Dono ki degree 3 hai — improper. Numerator \(= x^3-6x^2+11x-6\), denominator \(= x^3-15x^2+74x-120\). Subtract karne par remainder \(9x^2-63x+114\) milta hai.

\[ \frac{(x-1)(x-2)(x-3)}{(x-4)(x-5)(x-6)} = 1 + \frac{9x^2-63x+114}{(x-4)(x-5)(x-6)} \]
\[ 9x^2 - 63x + 114 = A(x-5)(x-6) + B(x-4)(x-6) + C(x-4)(x-5) \]

Put \(x = 4\): \(6 = 2A \implies A = 3\). Put \(x = 5\): \(24 = -B \implies B = -24\). Put \(x = 6\): \(60 = 2C \implies C = 30\).

Answer: \(1 + \frac{3}{x-4} - \frac{24}{x-5} + \frac{30}{x-6}\)

Question 9: Resolve \(\frac{x^2}{(x^2+a^2)(x^2+b^2)(x^2+c^2)}\) into partial fractions.

Denominator \(x^2\) mein quadratic hai, is liye \(x^2 = -a^2, -b^2, -c^2\) substitute karo — har dafa sirf aik constant bachega.

\[ \frac{x^2}{(x^2+a^2)(x^2+b^2)(x^2+c^2)} = \frac{A}{x^2+a^2} + \frac{B}{x^2+b^2} + \frac{C}{x^2+c^2} \]
\[ x^2 = A(x^2+b^2)(x^2+c^2) + B(x^2+a^2)(x^2+c^2) + C(x^2+a^2)(x^2+b^2) \]

Put \(x^2 = -a^2\): \(-a^2 = A(b^2-a^2)(c^2-a^2) \implies A = \frac{-a^2}{(a^2-b^2)(a^2-c^2)}\).

Similarly \(B = \frac{-b^2}{(b^2-a^2)(b^2-c^2)}\), \(C = \frac{-c^2}{(c^2-a^2)(c^2-b^2)}\).

Answer: \(\frac{-a^2}{(a^2-b^2)(a^2-c^2)(x^2+a^2)} + \frac{-b^2}{(b^2-a^2)(b^2-c^2)(x^2+b^2)} + \frac{-c^2}{(c^2-a^2)(c^2-b^2)(x^2+c^2)}\)

Question 10: Resolve \(\frac{x+1}{(x-1)^2}\) into partial fractions.

Repeated linear factor \((x-1)^2\): \(\frac{A}{x-1} + \frac{B}{(x-1)^2}\) likho.

\[ x + 1 = A(x-1) + B \]

Put \(x = 1\): \(B = 2\). Comparing coefficients of \(x\): \(A = 1\).

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

Question 11: Resolve \(\frac{x^2+x}{(x^2-1)^2}\) into partial fractions.

Pehle simplify karo: \(\frac{x(x+1)}{(x-1)^2(x+1)^2} = \frac{x}{(x-1)^2(x+1)}\). Ab aik repeated aur aik distinct linear factor hai.

\[ \frac{x}{(x-1)^2(x+1)} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{x+1} \]
\[ x = A(x-1)(x+1) + B(x+1) + C(x-1)^2 \]

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

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

Question 12: Resolve \(\frac{3x^2+4x-5}{(x-1)^3}\) into partial fractions.

\((x-1)^3\) repeated hai — teen terms: \(\frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{(x-1)^3}\).

\[ 3x^2 + 4x - 5 = A(x-1)^2 + B(x-1) + C \]

Put \(x = 1\): \(C = 2\). Comparing \(x^2\): \(A = 3\). Comparing \(x\): \(-2A + B = 4 \implies B = 10\).

Answer: \(\frac{3}{x-1} + \frac{10}{(x-1)^2} + \frac{2}{(x-1)^3}\)

Question 13: Resolve \(\frac{1}{x(x+1)^3}\) into partial fractions.

Factors: \(x\) (distinct) aur \((x+1)^3\) (repeated). Chaar terms likho.

\[ \frac{1}{x(x+1)^3} = \frac{A}{x} + \frac{B}{x+1} + \frac{C}{(x+1)^2} + \frac{D}{(x+1)^3} \]
\[ 1 = A(x+1)^3 + Bx(x+1)^2 + Cx(x+1) + Dx \]

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

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

Question 14: Resolve \(\frac{4x^2-3x+1}{(x+1)(x-1)^2}\) into partial fractions.

\((x+1)\) distinct aur \((x-1)^2\) repeated hai.

\[ 4x^2 - 3x + 1 = A(x-1)^2 + B(x+1)(x-1) + C(x+1) \]

Put \(x = 1\): \(2 = 2C \implies C = 1\). Put \(x = -1\): \(8 = 4A \implies A = 2\). Comparing \(x^2\): \(A + B = 4 \implies B = 2\).

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

Question 15: Resolve \(\frac{12x^2-48}{(x-2)^2(x+2)^2}\) into partial fractions.

Dono factors repeated hain (squared). Chaar terms likho — lekin note karo ke numerator \(= 12(x^2-4)\) hai, is liye \(B\) aur \(D\) zero niklenge.

\[ \frac{12x^2-48}{(x-2)^2(x+2)^2} = \frac{A}{x-2} + \frac{B}{(x-2)^2} + \frac{C}{x+2} + \frac{D}{(x+2)^2} \]
\[ 12x^2 - 48 = A(x-2)(x+2)^2 + B(x+2)^2 + C(x-2)^2(x+2) + D(x-2)^2 \]

Put \(x = 2\): \(0 = 16B \implies B = 0\). Put \(x = -2\): \(0 = 16D \implies D = 0\). Comparing \(x^3\): \(A + C = 0\). Comparing \(x^2\): \(2A - 2C = 12 \implies A = 3, C = -3\).

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

Common mistake: Jawab ko verify karna na bhoolo: \(\frac{3}{x-2} - \frac{3}{x+2} = \frac{12}{(x-2)(x+2)} = \frac{12x^2-48}{(x-2)^2(x+2)^2}\) — wapas jama karke check karo.