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Punjab Board Class 11 Mathematics

Chapter 5: Partial Fractions — Class 11 Maths Notes

Partial fractions: templates for every denominator type so the method stays consistent and fast. This page has the theory, solved textbook examples, every exercise solution, MCQs and a formula sheet for Chapter 5.

Exercise solutions — Chapter 5

Every exercise of Partial Fractions solved step by step.

Exercise 5.1 Exercise 5.2

Mathematics 11 (Punjab Board)

Chapter 5: Partial Fractions — Theory

Definitions, key results and theory for Punjab Board Class 11 Mathematics, Chapter 5: Partial Fractions.

Rational fractions: proper and improper

A rational fraction \(\frac{P(x)}{Q(x)}\) is proper if the degree of the numerator \(P(x)\) is less than the degree of the denominator \(Q(x)\); otherwise it is improper. Sirf proper fractions ko directly partial fractions mein tora ja sakta hai.

\[ \frac{P(x)}{Q(x)} \text{ is proper } \iff \deg P < \deg Q \]

Key point: Agar fraction improper ho to pehle long division karo: quotient alag likho, phir bache hue proper remainder ko resolve karo.

What are partial fractions?

Partial fractions ka matlab hai aik complicated rational fraction ko chhoti-chhoti simple fractions ke sum mein torna. Har simple fraction ka denominator, original denominator ke aik factor par mushtamil hota hai. Yeh technique integration aur inverse Laplace transforms mein bohat kaam aati hai.

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

Key point: Pehle denominator ko completely factorize karo — partial fractions ki shape isi par depend karti hai.

Case 1: Distinct linear factors

Denominator ke har distinct linear factor \((x - a)\) ke liye \(\frac{A}{x-a}\) jaisi aik term likho, jahan \(A\) aik constant hai.

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

Constants nikalne ka sab se tez tareeqa: dono sides ko denominator se multiply karke identity banao, phir \(x = a, b, c\) rakho — har dafa sirf aik constant bachta hai (cover-up method).

Key point: \(x\) ki woh value rakho jo aik factor ko zero kare — baqi terms khud khatam ho jayengi.

Case 2: Repeated linear factors

Agar \((x - a)\) factor \(n\) dafa repeat ho raha ho, to \(n\) terms likho — powers 1 se \(n\) tak:

\[ \frac{P(x)}{(x-a)^n} = \frac{A_1}{x-a} + \frac{A_2}{(x-a)^2} + \cdots + \frac{A_n}{(x-a)^n} \]

Yahan cover-up method sirf sab se unchi power wale constant ke liye kaam karta hai; baqi constants ke liye coefficients compare karo ya munasib \(x\) values substitute karo.

Key point: Repeated factor ki har power ke liye alag term lazmi hai — koi power skip nahi kar sakte.

Case 3: Irreducible quadratic factors

Agar denominator mein aisa quadratic factor ho jo mazeed factorize na ho sake (discriminant negative), to uske liye \(\frac{Ax+B}{\text{quadratic}}\) jaisi term likho — numerator linear hota hai, constant nahi.

\[ \frac{P(x)}{(x-a)(x^2+bx+c)} = \frac{A}{x-a} + \frac{Bx+C}{x^2+bx+c} \]

Agar quadratic factor repeat ho, to uski har power ke liye alag linear numerator wali term likho.

Key point: Quadratic factor ke liye numerator mein \(Ax + B\) lazmi hai — sirf \(A\) rakhna ghalat hai.

Finding the constants: three methods

1. Substitution (cover-up): \(x\) ki khaas values rakho jo factors ko zero karein. 2. Comparing coefficients: dono sides par \(x^n, x^{n-1}, \ldots\) ke coefficients barabar rakho aur equations solve karo. 3. Mixed: pehle cover-up se jo mil sake nikalo, phir baqi ke liye coefficients compare karo.

\[ P(x) \equiv A(x-b)(x-c) + B(x-a)(x-c) + C(x-a)(x-b) \]

Key point: Aakhir mein apna jawab verify karo: partial fractions ko wapas jama karke dekho ke original fraction banta hai ya nahi.

Mathematics 11 (Punjab Board)

Solved Examples: Chapter 5 – Partial Fractions

Textbook solved examples for Punjab Board Class 11 Mathematics, Chapter 5: Partial Fractions, worked out step by step.

Example 1: Resolve \(\frac{7x+25}{(x+3)(x+4)}\) into partial fractions.

Denominator ke factors distinct linear hain, is liye \(\frac{A}{x+3} + \frac{B}{x+4}\) likho. Cover-up method se \(x = -3\) aur \(x = -4\) rakho.

\[ 7x + 25 = A(x+4) + B(x+3) \]

Put \(x = -3\): \(7(-3) + 25 = 4 = A(1) \implies A = 4\).

Put \(x = -4\): \(7(-4) + 25 = -3 = B(-1) \implies B = 3\).

Answer: \(\frac{4}{x+3} + \frac{3}{x+4}\)

Example 2: Resolve \(\frac{x^2-10x+13}{(x-1)(x^2-5x+6)}\) into partial fractions.

Pehle denominator factorize karo: \(x^2-5x+6 = (x-2)(x-3)\). Phir teen distinct linear factors ke liye \(A, B, C\) nikalo.

\[ \frac{x^2-10x+13}{(x-1)(x-2)(x-3)} = \frac{A}{x-1} + \frac{B}{x-2} + \frac{C}{x-3} \]
\[ x^2 - 10x + 13 = A(x-2)(x-3) + B(x-1)(x-3) + C(x-1)(x-2) \]

Put \(x = 1\): \(4 = A(2) \implies A = 2\). Put \(x = 2\): \(-3 = B(-1) \implies B = 3\). Put \(x = 3\): \(-8 = C(2) \implies C = -4\).

Answer: \(\frac{2}{x-1} + \frac{3}{x-2} - \frac{4}{x-3}\)

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

Yeh improper fraction hai (degree 3/3). Pehle divide karo: denominator \(= 2x^3+x^2-3x\), numerator se \(2x\) zyada hai, to quotient 1 aur remainder \(2x\) milega.

\[ \frac{2x^3+x^2-x-3}{x(2x+3)(x-1)} = 1 + \frac{2x}{x(2x+3)(x-1)} = 1 + \frac{2}{(2x+3)(x-1)} \]
\[ \frac{2}{(2x+3)(x-1)} = \frac{A}{2x+3} + \frac{B}{x-1}, \quad 2 = A(x-1) + B(2x+3) \]

Put \(x = 1\): \(2 = 5B \implies B = \frac{2}{5}\). Put \(x = -\frac{3}{2}\): \(2 = -\frac{5}{2}A \implies A = -\frac{4}{5}\).

Answer: \(1 - \frac{4}{5(2x+3)} + \frac{2}{5(x-1)}\)

Common mistake: Improper fraction ko baghair divide kiye resolve karna — hamesha pehle check karo ke numerator ki degree denominator se kam hai ya nahi.

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

Repeated linear factor \((x+2)^3\) hai, is liye teen terms likho: \(\frac{A}{x+2} + \frac{B}{(x+2)^2} + \frac{C}{(x+2)^3}\).

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

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

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

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

Denominator \(= (x+1)^2(x-1)(x+1) = (x-1)(x+1)^3\). Aik distinct aur aik repeated (cubed) linear factor hai.

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

Put \(x = 1\): \(1 = 8A \implies A = \frac{1}{8}\). Put \(x = -1\): \(1 = -2D \implies D = -\frac{1}{2}\).

Comparing \(x^3\): \(A + B = 0 \implies B = -\frac{1}{8}\). Comparing \(x^2\): \(3A + B + C = 0 \implies C = -\frac{1}{4}\).

Answer: \(\frac{1}{8(x-1)} - \frac{1}{8(x+1)} - \frac{1}{4(x+1)^2} - \frac{1}{2(x+1)^3}\)

Example 6: Resolve \(\frac{3x-11}{(x^2+1)(x+3)}\) into partial fractions.

\(x^2+1\) irreducible quadratic hai, is liye uske liye \(\frac{Ax+B}{x^2+1}\) (linear numerator) likho, aur \(\frac{C}{x+3}\) alag.

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

Put \(x = -3\): \(-20 = 10C \implies C = -2\). Comparing \(x^2\): \(A + C = 0 \implies A = 2\). Comparing \(x\): \(3A + B = 3 \implies B = -3\).

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

Example 7: Resolve \(\frac{4x^2+8x}{x^4+2x^2+9}\) into partial fractions.

Denominator factorize karo: \(x^4+2x^2+9 = (x^2+2x+3)(x^2-2x+3)\) — dono irreducible quadratics hain (discriminant \(4-12 < 0\)). Har aik ke liye linear numerator wali term.

\[ \frac{4x^2+8x}{(x^2+2x+3)(x^2-2x+3)} = \frac{Ax+B}{x^2+2x+3} + \frac{Cx+D}{x^2-2x+3} \]
\[ 4x^2+8x = (Ax+B)(x^2-2x+3) + (Cx+D)(x^2+2x+3) \]

Comparing \(x^3\): \(A + C = 0\). Comparing \(x\): \(-2B + 2D = 8\). Comparing constants: \(3B + 3D = 0\).

From these: \(B = -2\), \(D = 2\); comparing \(x^2\): \(-4A + B + D = 4 \implies A = -1\), \(C = 1\).

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

Example 8: Resolve \(\frac{4x^2}{(x^2+1)^2(x-1)}\) into partial fractions.

Repeated irreducible quadratic \((x^2+1)^2\) plus aik distinct linear factor. Teen terms: \(\frac{Ax+B}{x^2+1} + \frac{Cx+D}{(x^2+1)^2} + \frac{E}{x-1}\).

\[ 4x^2 = (Ax+B)(x^2+1)(x-1) + (Cx+D)(x-1) + E(x^2+1)^2 \]

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

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

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

Mathematics 11 (Punjab Board)

MCQs: Chapter 5 – Partial Fractions

Choose the correct option. The answer key is checked against the worked solutions.

  1. \(\frac{P(x)}{Q(x)}\) is a proper fraction if:

    1. \(\deg P > \deg Q\)
    2. \(\deg P < \deg Q\)
    3. \(\deg P = \deg Q\)
    4. \(Q(x) = 0\)
    Proper fraction ki definition hi yahi hai: numerator ki degree denominator se kam ho.

  2. To resolve \(\frac{2}{x^2-1}\) into partial fractions, the correct form is:

    1. \(\frac{A}{x-1} + \frac{B}{x+1}\)
    2. \(\frac{A}{x-1} + \frac{B}{(x-1)^2}\)
    3. \(\frac{Ax+B}{x^2-1}\)
    4. \(\frac{A}{x^2} + \frac{B}{1}\)
    \(x^2-1 = (x-1)(x+1)\) distinct linear factors hain, to har aik ke liye constant numerator wali term.

  3. In \(\frac{3x^2+4x-5}{(x-1)^3} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{(x-1)^3}\), the value of \(C\) is:

    1. \(3\)
    2. \(10\)
    3. \(2\)
    4. \(-5\)
    \(x = 1\) rakho: \(3 + 4 - 5 = 2 = C\) — cover-up method seedha jawab deta hai.

  4. For an irreducible quadratic factor \(x^2+1\), the partial fraction term is:

    1. \(\frac{A}{x^2+1}\)
    2. \(\frac{A}{x} + \frac{B}{1}\)
    3. \(\frac{Ax}{x^2+1}\)
    4. \(\frac{Ax+B}{x^2+1}\)
    Quadratic factor ke liye numerator linear (\(Ax+B\)) hota hai — sirf constant rakhna ghalat hai.

  5. \(\frac{x^2+1}{x^2-1}\) is:

    1. A proper fraction
    2. An improper fraction
    3. Not a rational fraction
    4. Equal to \(1\)
    Dono ki degree 2 hai (barabar), is liye improper — pehle divide karna parega.

  6. \(\frac{1}{x-1} - \frac{1}{x+1}\) simplifies to:

    1. \(\frac{2}{x^2-1}\)
    2. \(\frac{2x}{x^2-1}\)
    3. \(\frac{0}{x^2-1}\)
    4. \(\frac{-2}{x^2-1}\)
    \(\frac{(x+1)-(x-1)}{(x-1)(x+1)} = \frac{2}{x^2-1}\) — jawab verify karne ka yahi tareeqa hai.

  7. To resolve \(\frac{x+1}{(x-1)^2}\), the number of constants needed is:

    1. \(1\)
    2. \(3\)
    3. \(4\)
    4. \(2\)
    \((x-1)^2\) repeated hai, to do terms \(\frac{A}{x-1} + \frac{B}{(x-1)^2}\) — do constants.

  8. In \(\frac{7x+25}{(x+3)(x+4)} = \frac{A}{x+3} + \frac{B}{x+4}\), the value of \(A\) is:

    1. \(3\)
    2. \(7\)
    3. \(4\)
    4. \(25\)
    \(x = -3\) rakho: \(7(-3)+25 = 4 = A(-3+4) = A\) — to \(A = 4\).

  9. Which of the following quadratics is irreducible over reals?

    1. \(x^2 - 4\)
    2. \(x^2 + 1\)
    3. \(x^2 - 3x + 2\)
    4. \(x^2 + 5x + 6\)
    \(x^2+1\) ka discriminant \(-4 < 0\) hai — real linear factors mein nahi toot-ta.

  10. \(\frac{4x^3+5x^2-3x-2}{x^2-1}\) on division gives quotient:

    1. \(4x - 5\)
    2. \(4x + 1\)
    3. \(x + 3\)
    4. \(4x + 5\)
    Long division se: \(4x^3 \div x^2 = 4x\), phir \(5x^2 \div x^2 = 5\) — quotient \(4x+5\), remainder \(x+3\).

  11. The partial fraction form of \(\frac{1}{x(x+1)^3}\) has how many terms?

    1. \(4\)
    2. \(2\)
    3. \(3\)
    4. \(5\)
    \(x\) ke liye 1 term aur \((x+1)^3\) ke liye 3 terms — kul 4 terms.

  12. If \(\frac{2x+3}{(x+1)(x+2)(x+3)} = \frac{A}{x+1} + \frac{B}{x+2} + \frac{C}{x+3}\), then \(B\) equals:

    1. \(\frac{1}{2}\)
    2. \(-\frac{3}{2}\)
    3. \(1\)
    4. \(-1\)
    \(x = -2\) rakho: \(2(-2)+3 = -1 = B(-1)(1)\) — to \(B = 1\).

  13. \(x^4 + 2x^2 + 9\) factors as:

    1. \((x^2+3)^2\)
    2. \((x^2+2x+3)(x^2-2x+3)\)
    3. \((x^2+1)(x^2+9)\)
    4. \((x^2+2x-3)(x^2-2x-3)\)
    \((x^2+3)^2 - (2x)^2 = x^4+2x^2+9\) — difference of squares se verify karo.

  14. In \(\frac{3x^2-12x+11}{(x-1)(x-2)(x-3)} = \frac{A}{x-1} + \frac{B}{x-2} + \frac{C}{x-3}\), \(A + B + C\) equals:

    1. \(1\)
    2. \(2\)
    3. \(3\)
    4. \(0\)
    \(A = 1, B = 1, C = 1\) (cover-up se), to sum \(3\) hai.

  15. Before resolving an improper fraction, the first step is:

    1. Factorize the numerator
    2. Divide numerator by denominator
    3. Find the conjugate
    4. Substitute \(x = 0\)
    Improper fraction ko pehle division se proper banao — yahi rule hai.

  16. \(\frac{2}{x-1} - \frac{1}{x+1}\) for the fraction \(\frac{x+3}{x^2-1}\) is:

    1. Incorrect
    2. Correct
    3. Partially correct
    4. Not a resolution
    Wapas jama karo: \(\frac{2(x+1)-(x-1)}{x^2-1} = \frac{x+3}{x^2-1}\) — bilkul sahi hai.

  17. In \(\frac{2x^2+3x+3}{(x+1)(x^2+1)} = \frac{A}{x+1} + \frac{Bx+C}{x^2+1}\), the value of \(A\) is:

    1. \(2\)
    2. \(3\)
    3. \(1\)
    4. \(-1\)
    \(x = -1\) rakho: \(2 - 3 + 3 = 2 = 2A\) — to \(A = 1\).

  18. The degree of the numerator in each partial fraction term for a repeated quadratic factor is at most:

    1. \(0\)
    2. \(2\)
    3. \(3\)
    4. \(1\)
    Quadratic denominator ke liye numerator linear (\(Ax+B\)) hota hai — degree at most 1.

  19. \(\frac{3}{x-2} - \frac{3}{x+2}\) simplifies to:

    1. \(\frac{12}{x^2-4}\)
    2. \(\frac{6x}{x^2-4}\)
    3. \(0\)
    4. \(\frac{12x}{x^2-4}\)
    \(\frac{3(x+2)-3(x-2)}{x^2-4} = \frac{12}{x^2-4}\).

  20. A partial fraction decomposition is verified by:

    1. Adding the parts back together
    2. Differentiating the result
    3. Setting \(x = 0\) only
    4. Factoring the numerator
    Parts ko wapas jama karke original fraction milna chahiye — yahi sab se reliable check hai.

Mathematics 11 (Punjab Board)

Chapter 5: Partial Fractions — Formula Sheet

Quick-revision formulas for Punjab Board Class 11 Mathematics, Chapter 5: Partial Fractions.

Rational Fraction Definitions

\[ \frac{P(x)}{Q(x)} \text{ is proper } \iff \deg P < \deg Q \]
\[ \text{Improper: divide first } \implies \text{quotient} + \frac{\text{remainder}}{Q(x)} \]

Resolution Templates

\[ \frac{P(x)}{(x-a)(x-b)} = \frac{A}{x-a} + \frac{B}{x-b} \quad \text{(distinct linear)} \]
\[ \frac{P(x)}{(x-a)^n} = \frac{A_1}{x-a} + \frac{A_2}{(x-a)^2} + \cdots + \frac{A_n}{(x-a)^n} \quad \text{(repeated linear)} \]
\[ \frac{P(x)}{(x-a)(x^2+bx+c)} = \frac{A}{x-a} + \frac{Bx+C}{x^2+bx+c} \quad \text{(irreducible quadratic)} \]
\[ \frac{P(x)}{(x^2+bx+c)^n} = \frac{A_1x+B_1}{x^2+bx+c} + \cdots + \frac{A_nx+B_n}{(x^2+bx+c)^n} \quad \text{(repeated quadratic)} \]

Finding the Constants

\[ \text{Cover-up: put } x = a \text{ to kill every factor except } (x-a) \]
\[ \text{Compare coefficients of } x^n, x^{n-1}, \ldots \text{ on both sides} \]
\[ \text{Check: add the partial fractions back to recover } \frac{P(x)}{Q(x)} \]

Chapter 5 FAQs

What is covered in Chapter 5 (Partial Fractions)?
This chapter covers Proper and improper fractions, repeated factors, irreducible quadratic factors. The page includes theory and definitions, solved textbook examples, complete exercise solutions, MCQs and a formula sheet.
How many marks does Chapter 5 carry in the Punjab Board paper?
Chapter 5 (Partial Fractions) carries approximately 5 marks in the Punjab Board Class 11 Mathematics paper.
Are these notes for the new PCTB syllabus?
Yes. These notes follow the PCTB 2025 new book for Class 11 Mathematics, used by BISE Lahore, Gujranwala, Rawalpindi, Faisalabad, Multan, Sahiwal, Sargodha, Bahawalpur and D.G. Khan.
Where are the exercise solutions for Chapter 5?
Use the exercise index on this page — every exercise of Chapter 5 has its own page with complete step-by-step solutions.