PSEB 12th Class Maths Book Solutions Guide in Punjabi English Medium

Punjab State Board Syllabus PSEB 12th Class Maths Book Solutions Guide Pdf in English Medium and Punjabi Medium are part of PSEB Solutions for Class 12.

PSEB 12th Class Maths Guide | Maths Guide for Class 12 PSEB in English Medium

Maths Guide for Class 12 PSEB | PSEB 12th Class Maths Book Solutions

PSEB 12th Math Book Pdf Chapter 1 Relations and Functions

PSEB 12th Maths Solution Book Pdf Download Chapter 2 Inverse Trigonometric Functions

PSEB Maths Book 12th Chapter 3 Matrices

PSEB 12th Class Math Solutions Chapter 4 Determinants

PSEB Class 12 Math Syllabus Chapter 5 Continuity and Differentiability

PSEB Class 12th Maths Chapter 6 Application of Derivatives

PSEB 12th Maths Chapter 7 Integrals

PSEB Class 12 Maths Syllabus Chapter 8 Application of Integrals

  • Chapter 8 Application of Integrals Ex 8.1
  • Chapter 8 Application of Integrals Ex 8.2
  • Chapter 8 Application of Integrals Miscellaneous Exercise

PSEB 12th Class Math Solutions Chapter 9 Differential Equations

  • Chapter 9 Differential Equations Ex 9.1
  • Chapter 9 Differential Equations Ex 9.2
  • Chapter 9 Differential Equations Ex 9.3
  • Chapter 9 Differential Equations Ex 9.4
  • Chapter 9 Differential Equations Ex 9.5
  • Chapter 9 Differential Equations Ex 9.6
  • Chapter 9 Differential Equations Miscellaneous Exercise

PSEB Maths Book 12th Chapter 10 Vector Algebra

  • Chapter 10 Vector Algebra Ex 10.1
  • Chapter 10 Vector Algebra Ex 10.2
  • Chapter 10 Vector Algebra Ex 10.3
  • Chapter 10 Vector Algebra Ex 10.4
  • Chapter 10 Vector Algebra Miscellaneous Exercise

Punjab Board Class 12th Maths Chapter 11 Three Dimensional Geometry

  • Chapter 11 Three Dimensional Geometry Ex 11.1
  • Chapter 11 Three Dimensional Geometry Ex 11.2
  • Chapter 11 Three Dimensional Geometry Ex 11.3
  • Chapter 11 Three Dimensional Geometry Miscellaneous Exercise

PSEB 12th Maths Solution Book Pdf Download Chapter 12 Linear Programming

  • Chapter 12 Linear Programming Ex 12.1
  • Chapter 12 Linear Programming Ex 12.2
  • Chapter 12 Linear Programming Miscellaneous Exercise

PSEB 12th Math Book Pdf Chapter 13 Probability

  • Chapter 13 Probability Ex 13.1
  • Chapter 13 Probability Ex 13.2
  • Chapter 13 Probability Ex 13.3
  • Chapter 13 Probability Ex 13.4
  • Chapter 13 Probability Ex 13.5
  • Chapter 13 Probability Miscellaneous Exercise

PSEB 7th Class Hindi Book Solutions | PSEB 7th Class Hindi Guide

Punjab State Board Syllabus PSEB 7th Class Hindi Book Solutions Guide Pdf is part of PSEB Solutions for Class 7.

PSEB 7th Class Hindi Guide | Hindi Guide for Class 7 PSEB

Hindi Guide for Class 7 PSEB | PSEB 7th Class Hindi Book Solutions

PSEB 7th Class Hindi Guide First Language

PSEB 7th Class Hindi Book Vyakaran व्याकरण

पारिभाषिक व्याकरण

व्यावहारिक व्याकरण

PSEB 7th Class Hindi Book Rachana रचना-भाग

PSEB 7th Class Hindi Guide Second Language

PSEB 7th Class Hindi Grammar व्याकरण 2nd Language

PSEB 7th Class Hindi Book Rachana रचना-भाग 2nd Language

PSEB 7th Class Punjabi Book Solutions | PSEB 7th Class Punjabi Guide

Punjab State Board Syllabus PSEB 7th Class Punjabi Book Solutions Guide Pdf is part of PSEB Solutions for Class 7.

PSEB 7th Class Punjabi Guide | Punjabi Guide for Class 7 PSEB

Punjabi Guide for Class 7 PSEB | PSEB 7th Class Punjabi Book Solutions

PSEB 7th Class Punjabi Guide Second Language

PSEB 7th Class Punjabi Grammar ਵਿਆਕਰਨ

PSEB 7th Class Punjabi Rachana ਰਚਨਾ

PSEB 7th Class Punjabi Guide First Language

  • Chapter 1 ਵਣਜਾਰਾ (ਗੀਤ)
  • Chapter 2 ਮੋਤੀ (ਕਹਾਣੀ)
  • Chapter 3 ਪੰਜਾਬ ਦੀ ਲੋਕ-ਗਾਇਕਾ: ਸੁਰਿੰਦਰ ਕੌਰ (ਜੀਵਨੀ)
  • Chapter 4 ਘੜੇ ਦਾ ਪਾਣੀ (ਲੇਖ)
  • Chapter 5 ਮਾਂ ਦਾ ਪਿਆਰ (ਕਵਿਤਾ)
  • Chapter 6 ਬਲਦਾਂ ਵਾਲਾ ਪਿਆਰਾ ਸਿੰਘ (ਕਹਾਣੀ)
  • Chapter 7 ਬਾਲ-ਖੇਡਾਂ (ਕਵਿਤਾ)
  • Chapter 8 ਬਚਿੱਤਰ ਸਿੰਘ ਦੀ ਬਹਾਦਰੀ (ਕਾਵਿ-ਕਹਾਣੀ)
  • Chapter 9 ਮੇਰੇ ਦਾਦੀ (ਲੇਖ)
  • Chapter 10 ਸ਼ੇਰਨੀਆਂ (ਕਹਾਣੀ)
  • Chapter 11 ਬਾਬਾ ਬੰਦਾ ਸਿੰਘ ਬਹਾਦਰ (ਜੀਵਨੀ)
  • Chapter 12 ਸ਼ਾਬਾਸ਼ !ਸੁਮਨ ! (ਕਹਾਣੀ)
  • Chapter 13 ਸਾਉਣ (ਕਵਿਤਾ)
  • Chapter 14 ਕਰਤਾਰ ਸਿੰਘ ਸਰਾਭਾ (ਜੀਵਨੀ)
  • Chapter 15 ਜੀਅ ਕਰੇ (ਕਵਿਤਾ)
  • Chapter 16 ਤ੍ਰਿਲੋਚਨ ਦਾ ਕੱਦ (ਕਹਾਣੀ)
  • Chapter 17 ਕਹਾਣੀ ਲਾਲਾ ਲਾਜਪਤ ਰਾਏ (ਜੀਵਨੀ)
  • Chapter 18 ਗਿਠਮੁਠੀਆਂ ਵਾਲਾ ਖੂਹ (ਕਹਾਣੀ)
  • Chapter 19 ਅਦਭੁਤ ਸੰਸਾਰ (ਕਵਿਤਾ)
  • Chapter 20 ਸੱਤ ਡਾਕਟਰ (ਕਹਾਣੀ)
  • Chapter 21 ਪੁਲਾੜ – ਪਰੀ : ਸੁਨੀਤਾ ਵਿਲੀਅਮਜ਼ (ਜੀਵਨੀ)
  • Chapter 22 ਵਿਰਾਸਤ-ਏ-ਖ਼ਾਲਸਾ
  • Chapter 23 ਮਿਲਖੀ ਦਾ ਵਿਆਹ (ਗੀਤ)
  • Chapter 24 ਜਾਗੋ (ਲੇਖ)
  • Chapter 25 ਕਿਰਤ ਦਾ ਸਤਿਕਾਰ (ਇਕਾਂਗੀ)

PSEB 7th Class Punjabi Grammar ਵਿਆਕਰਨ First Language

  • ਬੋਲੀ, ਵਿਆਕਰਨ ਤੇ ਵਰਨਮਾਲਾ
  • ਸ਼ਬਦ-ਭੇਦ-ਨਾਂਵ
  • ਲਿੰਗ
  • ਵਚਨ
  • ਪੜਨਾਂਵ
  • ਵਿਸ਼ੇਸ਼ਣ
  • ਕਿਰਿਆ, ਕਾਲ, ਕਿਰਿਆ ਵਿਸ਼ੇਸ਼ਣ, ਸੰਬੰਧਕ, ਯੋਜਕ ਤੇ ਵਿਸਮਿਕ
  • ਸੁੰਦਰ ਲਿਖਾਈ ਤੇ ਸ਼ੁੱਧ ਸ਼ਬਦ-ਜੋੜ
  • ਸਮਾਨਾਰਥਕ ਸ਼ਬਦ
  • ਵਿਰੋਧਾਰਥਕ ਸ਼ਬਦ
  • ਬਹੁਤੇ ਸ਼ਬਦਾਂ ਦੀ ਥਾਂ ਇਕ-ਸ਼ਬਦ
  • ਬਹੁ-ਅਰਥਕ ਸ਼ਬਦ
  • ਵਿਸਰਾਮ ਚਿੰਨ੍ਹ
  • ਮੁਹਾਵਰੇ ਤੇ ਵਾਕ-ਅੰਸ਼
  • ਅਖਾਣ

PSEB 7th Class Punjabi Rachana ਰਚਨਾ First Language

  • ਲੇਖ-ਰਚਨਾ
  • ਚਿੱਠੀ-ਪੱਤਰ
  • ਕਹਾਣੀ-ਰਚਨਾ

PSEB 6th Class Punjabi Book Solutions | PSEB 6th Class Punjabi Guide

Punjab State Board Syllabus PSEB 6th Class Punjabi Book Solutions Guide Pdf is part of PSEB Solutions for Class 6.

PSEB 6th Class Punjabi Guide | Punjabi Guide for Class 6 PSEB

Punjabi Guide for Class 6 PSEB | PSEB 6th Class Punjabi Book Solutions

PSEB 6th Class Punjabi Guide Second Language

PSEB 6th Class Punjabi Grammar ਵਿਆਕਰਨ

PSEB 6th Class Punjabi Rachana ਰਚਨਾ

PSEB 6th Class Punjabi Guide First Language

  • Chapter 1 ਤਿਰੰਗਾ
  • Chapter 2 ਆਪਣੇ-ਆਪਣੇ ਥਾਂ ਸਾਰੇ ਚੰਗੇ
  • Chapter 3 ਮਹਾਤਮਾ ਗਾਂਧੀ
  • Chapter 4 ਦੇਸ ਪੰਜਾਬ
  • Chapter 5 ਲਿਫ਼ਾਫ਼ੇ
  • Chapter 6 ਬਾਬਾ ਬੁੱਢਾ ਜੀ
  • Chapter 7 ਬਸੰਤ
  • Chapter 8 ਸਾਰਾ ਜੱਗ ਨਹੀਂ ਜਿੱਤਿਆ ਜਾਂਦਾ
  • Chapter 9 ਥਾਲ
  • Chapter 10 ਕੀੜੀ
  • Chapter 11 ਦਾਤੇ
  • Chapter 12 ਪਹਿਲ
  • Chapter 13 ਭਗਤ ਕਬੀਰ ਜੀ
  • Chapter 14 ਆਲੋਕ ਮੁਖੀ, ਗੁਆਂਢੀ ਦੁਖੀ ! ਨਾ ਬਈ ਨਾ!
  • Chapter 15 ਤਿੰਨ ਇਨਕਲਾਬੀ-ਸ਼ਹੀਦ ਭਗਤ ਸਿੰਘ, ਰਾਜਗੁਰੂ ਤੇ ਸੁਖਦੇਵ
  • Chapter 16 ਵਿਸਾਖੀ ਦਾ ਮੇਲਾ
  • Chapter 17 ਝੀਲ, ਪਸ਼ੂ-ਪੰਛੀ ਅਤੇ ਬੱਚੇ
  • Chapter 18 ਸੜਕੀ ਦੁਰਘਟਨਾਵਾਂ ਤੋਂ ਬਚਾਅ
  • Chapter 19 ਤਿੰਨ ਸਵਾਲ
  • Chapter 20 ਧਰਤੀ ਦਾ ਗੀਤ
  • Chapter 21 ਪਿੰਡ ਇਉਂ ਬੋਲਦੈ
  • Chapter 22 ਲੋਕ-ਨਾਇਕ ਦਾ ਚਲਾਣਾ
  • Chapter 23 ਹਾਕੀ ਖਿਡਾਰਨ ਅਜਿੰਦਰ ਕੌਰ
  • Chapter 24 ਵੱਡੇ ਕੰਮ ਦੀ ਭਾਲ
  • Chapter 25 ਭਾਰਤ ਰਤਨ : ਡਾ. ਭੀਮ ਰਾਓ ਅੰਬੇਦਕਰ
  • Chapter 26 ਫੁੱਲਾਂ ਦਾ ਸੁਨੇਹਾ

ਵਿਆਕਰਨ
ਲੇਖ-ਰਚਨਾ
ਚਿੱਠੀ-ਪੱਤਰ
ਕਹਾਣੀ-ਰਚਨਾ

PSEB 8th Class Punjabi Book Solutions | PSEB 8th Class Punjabi Guide

Punjab State Board Syllabus PSEB 8th Class Punjabi Book Solutions Guide Pdf is part of PSEB Solutions for Class 8.

PSEB 8th Class Punjabi Guide | Punjabi Guide for Class 8 PSEB

Punjabi Guide for Class 8 PSEB | PSEB 8th Class Punjabi Book Solutions

PSEB 8th Class Punjabi Guide Second Language

PSEB 8th Class Punjabi Grammar ਵਿਆਕਰਨ

PSEB 8th Class Punjabi Rachana ਰਚਨਾ

PSEB 8th Class Punjabi Guide First Language

  • Chapter 1 ਜੈ ਭਾਰਤ ਮਾਤਾ (ਕਵਿਤਾ)
  • Chapter 2 ਪੇਮੀ ਦੇ ਨਿਆਣੇ (ਕਹਾਣੀ)
  • Chapter 3 ਛਿੰਝ ਛਰਾਹਾਂ ਦੀ (ਲੇਖ)
  • Chapter 4 ਸ੍ਰੀ ਗੁਰੂ ਅਰਜਨ ਦੇਵ ਜੀ (ਜੀਵਨੀ)
  • Chapter 5 ਉੱਦਮ ਕਰੀਂ ਜ਼ਰੂਰ (ਕਵਿਤਾ)
  • Chapter 6 ਦਲੇਰੀ (ਕਹਾਣੀ)
  • Chapter 7 ਰੂਪਨਗਰ (ਲੇਖ)
  • Chapter 8 ਬਾਬਾ ਫ਼ਰੀਦ (ਜੀਵਨੀ)
  • Chapter 9 ਪੰਜਾਬ (ਕਵਿਤਾ)
  • Chapter 10 ਹਰਿਆਵਲ ਦੇ ਬੀਜ (ਕਹਾਣੀ)
  • Chapter 11 ਪੰਜਾਬੀ ਲੋਕ-ਨਾਚ : ਗਿੱਧਾ (ਲੇਖ)
  • Chapter 12 ਪੰਜਾਬ ਦਾ ਸੁਪਨਸਾਜ਼ ਡਾ. ਮਹਿੰਦਰ ਸਿੰਘ ਰੰਧਾਵਾ (ਜੀਵਨੀ)
  • Chapter 13 ਧਰਤੀ (ਕਵਿਤਾ)
  • Chapter 14 ਸਾਂਝੀ ਮਾਂ (ਕਹਾਣੀ)
  • Chapter 15 ਰਾਬਿੰਦਰ ਨਾਥ ਟੈਗੋਰ (ਜੀਵਨੀ)
  • Chapter 16 ਗੁਲਾਬ ਦੀ ਫ਼ਸਲ (ਕਵਿਤਾ)
  • Chapter 17 ਲੋਹੜੀ (ਲੇਖ)
  • Chapter 18 ਹਾਕੀ ਦਾ ਜਾਦੂਗਰ : ਧਿਆਨ ਚੰਦ (ਜੀਵਨੀ)
  • Chapter 19 ਅੰਮੜੀ ਦਾ ਵਿਹੜਾ (ਕਵਿਤਾ)
  • Chapter 20 ਛੱਲੀਆਂ ਦੇ ਰਾਖੇ (ਕਹਾਣੀ)
  • Chapter 21 ਗਿਆਨ, ਵਿਗਿਆਨ ਤੇ ਮਨੋਰੰਜਨ ਦਾ ਅਨੋਖਾ ਸੰਗਮ : ਸਾਇੰਸ-ਸਿਟੀ (ਲੇਖ)
  • Chapter 22 ਸ਼ਿਵ ਸਿੰਘ : ਬੁੱਤ – ਘਾੜਾ (ਜੀਵਨੀ)
  • Chapter 23 ਪਿੰਡ ਦੀ ਘੁਲਾੜੀ (ਕਵਿਤਾ)
  • Chapter 24 ਭੂਆ (ਕਹਾਣੀ)
  • Chapter 25 ਰੱਬ ਦੀ ਪੌੜੀ (ਸਫ਼ਰਨਾਮਾ)
  • Chapter 26 ਗੱਗੂ (ਕਹਾਣੀ)
  • Chapter 27 ਵੱਡੇ ਭੈਣ ਜੀ (ਇਕਾਂਗੀ)

PSEB 8th Class Punjabi Grammar ਵਿਆਕਰਨ First Language

  • ਬੋਲੀ, ਭਾਵਾਂਸ਼, ਸ਼ਬਦ, ਵਿਆਕਰਨ ਤੇ ਵਰਨ
  • ਸ਼ਬਦ-ਜੋੜ
  • ਨਾਂਵ
  • ਪੜਨਾਂਵ
  • ਲਿੰਗ
  • ਵਚਨ
  • ਵਿਸ਼ੇਸ਼ਣ
  • ਕਿਰਿਆ
  • ਕਾਲ
  • ਕਿਰਿਆ ਵਿਸ਼ੇਸ਼ਣ
  • ਸੰਬੰਧਕ
  • ਯੋਜਕ
  • ਵਿਸਮਿਕ ਝ ਵਾਕ-ਬੋਧ, ਵਾਕ-ਜੋੜ, ਵਾਕ-ਤੋੜ ਅਤੇ ਵਾਕ-ਵੰਡ ਤੇ ਵਾਕ-ਵਟਾਂਦਰਾ
  • ਅਰਥ-ਬੋਧ
    • ਵਿਰੋਧਾਰਥਕ (ਉਲਟ-ਭਾਵੀ) ਸ਼ਬਦ
    • ਬਹੁਤੇ ਸ਼ਬਦਾਂ ਦੀ ਥਾਂ ਇਕ-ਸ਼ਬਦ
    • ਸਮਾਨਾਰਥਕ ਸ਼ਬਦ
    • ਬਹੁ-ਅਰਥਕ ਸ਼ਬਦ
  • ਵਿਸਰਾਮ ਚਿੰਨ੍ਹ
  • ਮੁਹਾਵਰਿਆਂ ਦੀ ਵਾਕਾਂ ਵਿਚ ਵਰਤੋਂ
  • ਅਖਾਣ
  • ਸ਼ਬਦ-ਜਾਲ ਵਿਚੋਂ ਸਾਰਥਕ ਸ਼ਬਦ ਲੱਭਣਾ

PSEB 8th Class Punjabi Rachana ਰਚਨਾ First Language

  • ਲੇਖ-ਰਚਨਾ
  • ਚਿੱਠੀ-ਪੱਤਰ

PSEB 8th Class Hindi Book Solutions | PSEB 8th Class Hindi Guide

Punjab State Board Syllabus PSEB 8th Class Hindi Book Solutions Guide Pdf is part of PSEB Solutions for Class 8.

PSEB 8th Class Hindi Guide | Hindi Guide for Class 8 PSEB

Hindi Guide for Class 8 PSEB | PSEB 8th Class Hindi Book Solutions

PSEB 8th Class Hindi Guide First Language

PSEB 8th Class Hindi Book Vyakaran व्याकरण

पारिभाषिक व्याकरण

व्यावहारिक व्याकरण

PSEB 8th Class Hindi Book Rachana रचना-भाग

PSEB 8th Class Hindi Guide Second Language

PSEB 8th Class Hindi Book Vyakaran व्याकरण Second Language

पारिभाषिक व्याकरण

मुहावरे और लोकोक्तियाँ

व्यावहारिक व्याकरण

PSEB 8th Class Hindi Book Rachana रचना-भाग Second Language

PSEB 6th Class Hindi Book Solutions | PSEB 6th Class Hindi Guide

Punjab State Board Syllabus PSEB 6th Class Hindi Book Solutions Guide Pdf is part of PSEB Solutions for Class 6.

PSEB 6th Class Hindi Guide | Hindi Guide for Class 6 PSEB

Hindi Guide for Class 6 PSEB | PSEB 6th Class Hindi Book Solutions

PSEB 6th Class Hindi Book Solutions First Language

PSEB 6th Class Hindi Book Vyakaran व्याकरण

PSEB 6th Class Hindi Book Rachana रचना-भाग

PSEB 6th Class Hindi Guide Second Language

PSEB 6th Class Hindi Grammar व्याकरण Second Language

PSEB 6th Class Hindi Book Rachana रचना-भाग Second Language

PSEB 8th Class Welcome Life Book Solutions Guide in Punjabi English Medium

Punjab State Board Syllabus PSEB 8th Class Welcome Life Book Solutions Guide Pdf in English Medium and Punjabi Medium are part of PSEB Solutions for Class 8.

PSEB 8th Class Welcome Life Guide | Welcome Life Guide for Class 8 PSEB in English Medium

Welcome Life Guide for Class 8 PSEB | PSEB 8th Class Welcome Life Book Solutions

PSEB 8th Class Welcome Life Book Solutions Guide in Hindi Medium

PSEB 8th Class Welcome Life Structure of Question Paper

कक्षा – आठवीं (पंजाब)
स्वागत ज़िन्दगी

समय : 2 घंटे

लिखित : 50
सी.सी.ई. : 10
प्रयोगी : 40
कुल अंक : 100

प्रश्न-पत्र की रूप-रेखा

1. सभी प्रश्न करने अनिवार्य होंगे।
2. प्रश्न-पत्र चार भागों में बंटा होगा। प्रत्येक भाग सारे पाठ्यक्रम पर आधारित होगा।
भाग – क में प्रश्न नं. 1 (i – xv) में 15 बहु-विकल्पीय प्रश्न होंगे। प्रत्येक प्रश्न एक अंक का होगा। (15 × 1 = 15)

भाग – ख में प्रश्न नं. 2 – 11 तक 10 वस्तुनिष्ठ प्रश्न (जैसे कि एक शब्द से एक वाक्य वाले प्रश्न/रिक्त स्थान भरना/सही-गलत बताना) होंगे। प्रत्येक प्रश्न एक अंक का होगा। (10 × 1 = 10)

भाग – ग में प्रश्न नं. 12 – 16 तक 5 छोटे उत्तरों वाले प्रश्न होंगे। प्रत्येक प्रश्न दो अंकों का होगा। (5 × 2 = 10)

भाग – घ में प्रश्न नं. 17 के दो भाग होंगे। भाग (i) में 2 काल्पनिक (Hypothetical) स्थिति पर आधारित प्रश्न होंगे तथा भाग (ii) में 1 वास्तविक जीवन से जुड़ी समस्या को हल करने से सम्बन्धित प्रश्न होगा। प्रत्येक प्रश्न पांच अंकों का होगा। सभी प्रश्नों में आन्तरिक छूट होगी। (3 × 5 = 15)

प्रश्न-पत्र की बनावट

भाग प्रश्नों की किस्म अंकों का विभाजन
भाग – क 15 बहुविकल्पी प्रश्न 15 × 1 = 15 अंक
भाग – ख 10 वस्तुनिष्ठ प्रश्न (एक शब्द से एक वाक्य वाले प्रश्न/रिक्त स्थान भरें/सही-ग़लत बताना) 10 × 1 = 10 अंक
भाग – ग 5 छोटे उत्तरों वाले प्रश्न (20-25 शब्दों में) (5 × 2 = 10 अंक)
भाग – घ (i) 2 (तीन में से) काल्पनिक परिस्थिति के आधार पर प्रश्न 5 × 2 = 10 अंक
(ii) 1 (दो में से) वास्तविक जीवन से जुड़ी समस्या को हल करने से सम्बन्धित प्रश्न 1 × 5 = 5 अंक
कुल 50 अंक

प्रयोगी परीक्षा का अंक विभाजन : 40 अंक

  • प्रोजेक्ट/एक्टीविटी वर्क : 10 अंक
  • विद्यार्थी का सामाजिक योगदान : 10 अंक
  • स्थिति आधारित मौखिक परीक्षा : 10 अंक
  • पाठ्य-पुस्तक पर आधारित क्रिया : 10 अंक

पाठ्यक्रम

  • पाठ 1 शरीर की सफ़ाई
  • पाठ 2 स्व: नियंत्रण
  • पाठ 3 बुजुर्गों से प्रेम तथा धन्यवाद
  • पाठ 4 हमारी कीमती ज़िन्दगी की सुरक्षा
  • पाठ 5 पशुओं के प्रति नैतिकता
  • पाठ 6 निर्णय लेना
  • पाठ 7 खेल भावना
  • पाठ 8 स्कूल तथा सार्वजनिक सम्पत्तियों का सम्मान
  • पाठ 9 लिंग समानता

PSEB 11th Class Maths Book Solutions Guide in Punjabi English Medium

Punjab State Board Syllabus PSEB 11th Class Maths Book Solutions Guide Pdf in English Medium and Punjabi Medium are part of PSEB Solutions for Class 11.

PSEB 11th Class Maths Guide | Maths Guide for Class 11 PSEB in English Medium

PSEB 11 Class Math Book Pdf Chapter 1 Sets

  • Chapter 1 Sets Ex 1.1
  • Chapter 1 Sets Ex 1.2
  • Chapter 1 Sets Ex 1.3
  • Chapter 1 Sets Ex 1.4
  • Chapter 1 Sets Ex 1.5
  • Chapter 1 Sets Ex 1.6

Punjab Board Maths Book Class 11 Solutions Chapter 2 Relations and Function

  • Chapter 2 Relations and Functions Ex 2.1
  • Chapter 2 Relations and Functions Ex 2.2
  • Chapter 2 Relations and Functions Ex 2.3
  • Chapter 2 Relations and Functions Miscellaneous Exercise

11th Class Math Book PSEB Chapter 3 Trigonometric Functions

  • Chapter 3 Trigonometric Functions Ex 3.1
  • Chapter 3 Trigonometric Functions Ex 3.2
  • Chapter 3 Trigonometric Functions Ex 3.3
  • Chapter 3 Trigonometric Functions Ex 3.4
  • Chapter 3 Trigonometric Functions Miscellaneous Exercise

PSEB Class 11 Maths Solutions Chapter 4 Principle of Mathematical Induction

  • Chapter 4 Principle of Mathematical Induction Ex 4.1

PSEB Class 11 Maths Book Pdf Download Chapter 5 Complex Numbers and Quadratic Equations

  • Chapter 5 Complex Numbers and Quadratic Equations Ex 5.1
  • Chapter 5 Complex Numbers and Quadratic Equations Ex 5.2
  • Chapter 5 Complex Numbers and Quadratic Equations Ex 5.3
  • Chapter 5 Complex Numbers and Quadratic Equations Miscellaneous Exercise

PSEB Class 11 Maths Syllabus Chapter 6 Linear Inequalities

  • Chapter 6 Linear Inequalities Ex 6.1
  • Chapter 6 Linear Inequalities Ex 6.2
  • Chapter 6 Linear Inequalities Ex 6.3
  • Chapter 6 Linear Inequalities Miscellaneous Exercise

PSEB 11th Class Math Book Pdf Chapter 7 Permutations and Combinations

  • Chapter 7 Permutations and Combinations Ex 7.1
  • Chapter 7 Permutations and Combinations Ex 7.2
  • Chapter 7 Permutations and Combinations Ex 7.3
  • Chapter 7 Permutations and Combinations Ex 7.4
  • Chapter 7 Permutations and Combinations Miscellaneous Exercise

11th Class Maths Solutions PSEB Chapter 8 Binomial Theorem

  • Chapter 8 Binomial Theorem Ex 8.1
  • Chapter 8 Binomial Theorem Ex 8.2
  • Chapter 8 Binomial Theorem Miscellaneous Exercise

PSEB 11th Class Maths Solutions Chapter 9 Sequences and Series

  • Chapter 9 Sequences and Series Ex 9.1
  • Chapter 9 Sequences and Series Ex 9.2
  • Chapter 9 Sequences and Series Ex 9.3
  • Chapter 9 Sequences and Series Ex 9.4
  • Chapter 9 Sequences and Series Miscellaneous Exercise

PSEB 11th Class Math Book Pdf Chapter 10 Straight Lines

  • Chapter 10 Straight Lines 10.1
  • Chapter 10 Straight Lines 10.2
  • Chapter 10 Straight Lines 10.3
  • Chapter 10 Straight Lines Miscellaneous Exercise

PSEB Class 11 Maths Syllabus Chapter 11 Conic Sections

  • Chapter 11 Conic Sections 11.1
  • Chapter 11 Conic Sections 11.2
  • Chapter 11 Conic Sections 11.3
  • Chapter 11 Conic Sections 11.4
  • Chapter 11 Conic Sections Miscellaneous Exercise

PSEB Class 11 Maths Book Pdf Download Chapter 12 Introduction to three Dimensional Geometry

  • Chapter 12 Introduction to three Dimensional Geometry Ex 12.1
  • Chapter 12 Introduction to three Dimensional Geometry Ex 12.2
  • Chapter 12 Introduction to three Dimensional Geometry Ex 12.3
  • Chapter 12 Introduction to three Dimensional Geometry Miscellaneous Exercise

PSEB Class 11 Maths Solutions Chapter 13 Limits and Derivatives

  • Chapter 13 Limits and Derivatives Ex 13.1
  • Chapter 13 Limits and Derivatives Ex 13.2
  • Chapter 13 Limits and Derivatives Miscellaneous Exercise

11th Class Math Book PSEB Chapter 14 Mathematical Reasoning

  • Chapter 14 Mathematical Reasoning Ex 14.1
  • Chapter 14 Mathematical Reasoning Ex 14.2
  • Chapter 14 Mathematical Reasoning Ex 14.3
  • Chapter 14 Mathematical Reasoning Ex 14.4
  • Chapter 14 Mathematical Reasoning Ex 14.5
  • Chapter 14 Mathematical Reasoning Miscellaneous Exercise

Punjab Board Maths Book Class 11 Solutions Chapter 15 Statistics

  • Chapter 15 Statistics Ex 15.1
  • Chapter 15 Statistics Ex 15.2
  • Chapter 15 Statistics Ex 15.3
  • Chapter 15 Statistics Miscellaneous Exercise

PSEB 11 Class Math Book Pdf Chapter 16 Probability

  • Chapter 16 Probability Ex 16.1
  • Chapter 16 Probability Ex 16.2
  • Chapter 16 Probability Ex 16.3
  • Chapter 16 Probability Miscellaneous Exercise

PSEB 12th Class Maths Solutions Chapter 3 Matrices Ex 3.1

Punjab State Board PSEB 12th Class Maths Book Solutions Chapter 3Matrices Ex 3.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.1

Question 1.
In the matrix A = \(=\left[\begin{array}{cccc}
2 & 5 & 19 & -7 \\
35 & -2 & \frac{5}{2} & 12 \\
\sqrt{3} & 1 & -5 & 17
\end{array}\right]\), write:
(i) The order of the matrix
(ii) The number of elements,
(iii) The elements a13, a21, a33, a24, a23.
Solution.
(i) In the given matrix, the number of rows is 3 and the number of columns is 4. Therefore, the order of the matrix is 3 × 4.
(ii) Since, the order of the matrix is 3 × 4, so there are 3 × 4 = 12 elements in it.
(iii) a13 = 19, a21 = 35, a33= – 5, a24 = 12, a23 = \(\frac{5}{2}\).

PSEB 12th Class Maths Solutions Chapter 3 Matrices Ex 3.1

Question 2.
If a matrix has 24 elements, what are the possible orders it can have? What, if it has 13 elements?
Solution.
We know that if a matrix is of the order m × n, it has mn elements. Thus, to find all the possible orders of a matrix having 24 elements, we have to find all the ordered pairs of natural numhers whose product is 24.
The ordered pairs are: (1, 24), (24,1), (2,12), (12, 2), (3, 8), (8, 3), (4, 6) and (6, 4).
Hence, the possible orders of a matrix having 24 elements are 1 × 24,24 × 1, 2 × 12,12 × 2, 3 × 8, 8 × 3, 4 × 6, and 6 × 4 (1, 13) and (13, 1) are the ordered pairs of natural numbers whose product is 13.
Hence, the possible orders of a matrix having 13 elements are 1 x 13 and 13×1.

Question 3.
If a matrix has 18 elements, what are the possible orders it can have? What, if it has 5 elements?
Solution.
We know that if a matrix is of the order m × n, it has mn elements. Thus, to find all the possible orders of a matrix having 18 elements, we have to find all the ordered pairs of natural numbers whose product is 18.
The ordered pairs are: (1, 18), (18, 1), (2, 9), (9, 2), (3, 6), and (6, 3)
Hence, the possible orders of a matrix having 18 elements are 1 × 18, 18 × 1, 2 × 9, 9 × 2, 3 × 6 and 6 × 3.
(1, 5)and (5, 1)are ordered pairs of natural numbers whose product is 5.
Hence, the possible orders of a matrix having 5 elements are 1 × 5 and 5 × 1.

PSEB 12th Class Maths Solutions Chapter 3 Matrices Ex 3.1

Question 4.
Construct 2 × 2 matrix, A = [aij] whose elements are given by
(i) aij = \(\frac{(i+j)^{2}}{2}\)

(ii) aij = \(\frac{i}{j}\)

(iii) aij = \(\frac{(i+2 j)^{2}}{2}\)
Solution.
(i) The order of the given matrix is 2 × 2, so A = \(\left[\begin{array}{ll}
a_{11} & a_{12} \\
a_{21} & a_{22}
\end{array}\right]_{2 \times 2}\)
where aij = \(\frac{(i+2 j)^{2}}{2}\).
To find a11, put i = 1 and j = 1
∴ a11 = \(\frac{(1+1)^{2}}{2}\) = 2 Similarly

a12 = \(\frac{(1+2)^{2}}{2}=\frac{9}{2}\)

a21 = \(\frac{(2+1)^{2}}{2}=\frac{9}{2}\) and

a22 = \(\frac{(2+2)^{2}}{2}\) = 8

(ii) Here, A = \(\left[\begin{array}{ll}
a_{11} & a_{12} \\
a_{21} & a_{22}
\end{array}\right]_{2 \times 2}\), where aij = \(\frac{i}{j}\).
∴ a11 = \(\frac{1}{1}\) = 1,

a12 = \(\frac{1}{2}\),

a21 = \(\frac{2}{1}\) = 2 and

a22 = \(\frac{2}{1}\) = 1

Hence, the required matrix is A = \(\left[\begin{array}{cc}
1 & 1 / 2 \\
2 & 1
\end{array}\right]_{2 \times 2}\)

(iii) Here, A = \(\left[\begin{array}{ll}
a_{11} & a_{12} \\
a_{21} & a_{22}
\end{array}\right]_{2 \times 2}\), where aij = \(\frac{(i+2 j)^{2}}{2}\)
∴ a11 = \(\frac{(1+2)^{2}}{2}=\frac{9}{2}\)

a12 = \(\frac{(1+4)^{2}}{2}=\frac{25}{2}\)

a21 = \(\frac{(2+2)^{2}}{2}\) = 8

a22 = \(\frac{(2+4)^{2}}{2}\) = 18.

PSEB 12th Class Maths Solutions Chapter 3 Matrices Ex 3.1

Hence, the required matrix is A = \(\left[\begin{array}{cc}
9 / 2 & 25 / 2 \\
8 & 18
\end{array}\right]_{2 \times 2}\).

Question 5.
Construct a 3 × 4 matrix, whose elements are given by:
(i) aij = \(\frac{4}{4}\) |- 3i + j|
(ii) aij = 2i – j
Solution.
In general, a 3 × 4 matrix is given by A = \(\left[\begin{array}{llll}
a_{11} & a_{12} & a_{13} & a_{14} \\
a_{21} & a_{22} & a_{23} & a_{24} \\
a_{31} & a_{32} & a_{33} & a_{34}
\end{array}\right]\)

(i) (i) aij = \(\frac{1}{2}\) |- 3i + j|, i = 1, 2, 3, 4 and j =1, 2, 3, 4
a11 = \(\frac{1}{2}\) |- 3 × 1 + 1|
= \(\frac{1}{2}\) |- 3 + 1|
= \(\frac{1}{2}\) |- 2|
= \(\frac{21}{2}\) = 1

a21 = \(\frac{1}{2}\) |- 3 × 2 + 1|
= \(\frac{1}{2}\) |- 6 + 1|
= \(\frac{1}{2}\) |- 5|
= \(\frac{5}{2}\)

a31 = \(\frac{1}{2}\) |- 3 × 3 + 1|
= \(\frac{1}{2}\) |- 9 + 1|
= \(\frac{1}{2}\) |- 8|
= \(\frac{8}{2}\) = 4

a12 = \(\frac{1}{2}\) |- 3 × 1 + 2|
= \(\frac{1}{2}\) |- 3 + 2|
= \(\frac{1}{2}\) |- 1|
= \(\frac{1}{2}\)

a22 = \(\frac{1}{2}\) |- 3 × 2 + 2|
= \(\frac{1}{2}\) |- 6 + 2|
= \(\frac{1}{2}\) |- 4|
= \(\frac{4}{2}\) = 2

a32 = \(\frac{1}{2}\) |- 3 × 3 + 2|
= \(\frac{1}{2}\) |- 9 + 2|
= \(\frac{1}{2}\) |- 7|
= \(\frac{7}{2}\)

a13 = \(\frac{1}{2}\) |- 3 × 1 + 3|
= \(\frac{1}{2}\) |- 3 + 3| = 0

a23 = \(\frac{1}{2}\) |- 3 × 2 + 3|
= \(\frac{1}{2}\) |- 6 + 3|
= \(\frac{1}{2}\) |- 3|
= \(\frac{3}{2}\)

a33 = \(\frac{1}{2}\) |- 3 × 3 + 3|
= \(\frac{1}{2}\) |- 9 + 3|
= \(\frac{1}{2}\) |- 6|
= \(\frac{6}{2}\) = 3

a14 = \(\frac{1}{2}\) |- 3 × 1 + 4|
= \(\frac{1}{2}\) |- 3 + 4|
= \(\frac{1}{2}\) |1|
= \(\frac{1}{2}\)

a24 = \(\frac{1}{2}\) |- 3 × 2 + 4|
= \(\frac{1}{2}\) | – 6 + 4|
= \(\frac{1}{2}\) |- 2|
= \(\frac{2}{2}\) = 1

a34 = \(\frac{1}{2}\) |- 3 × 3 + 4|
= \(\frac{1}{2}\) |- 9 + 4|
= \(\frac{1}{2}\) |- 5|
= \(\frac{5}{2}\)

Therfore, the required matrix is A = \(\left[\begin{array}{cccc}
1 & \frac{1}{2} & 0 & \frac{1}{2} \\
\frac{5}{2} & =2 & \frac{3}{2} & 1 \\
4 & \frac{7}{2} & 3 & \frac{5}{2}
\end{array}\right]_{3 \times 4}\)

PSEB 12th Class Maths Solutions Chapter 3 Matrices Ex 3.1

(ii) aij = 2i – j, i = 1, 2, 3, 4 and j = 1, 2, 3, 4
a11 = 2 × 1 – 1 = 2 – 1 = 1
a12 = 2 × 2 – 1 = 4 – 1 = 3
a13 = 2 × 3 – 1 = 6 – 1 = 5

a21 = 2 × 1 – 2 = 2 – 2 = 0
a22 = 2 × 2 – 2 = 4 – 2 = 2
a23 = 2 × 3 – 2 = 6 – 2 = 4

a31 = 2 × 1 – 3 = 2 – 3 = – 1
a32 = 2 × 2 – 3 = 4 – 3 = 1
a33 = 2 × 3 – 3 = 6 – 3 = 3

a41 = 2 × 1 – 4 = 2 – 4 = – 2
a42 = 2 × 2 – 4 = 4 – 4 = 0
a43 = 2 × 3 – 4 = 6 – 4 = 0

Therfore, the required matrix is A = \(\left[\begin{array}{cccc}
1 & 0 & -1 & -2 \\
3 & 2 & 1 & 0 \\
5 & 4 & 3 & 2
\end{array}\right]_{3 \times 4}\).

PSEB 12th Class Maths Solutions Chapter 3 Matrices Ex 3.1

Question 6.
Find the values of x, y, and z from the following equations:
(i) \(\left[\begin{array}{ll}
\mathbf{4} & \mathbf{3} \\
\boldsymbol{x} & 5
\end{array}\right]=\left[\begin{array}{ll}
\boldsymbol{y} & \boldsymbol{z} \\
\mathbf{1} & \mathbf{5}
\end{array}\right]\)

(ii) \(\left[\begin{array}{cc}
x+y & 2 \\
5+z & x y
\end{array}\right]=\left[\begin{array}{ll}
6 & 2 \\
5 & 8
\end{array}\right]\)

(iii) \(\left[\begin{array}{c}
x+y+z \\
x+z \\
y+z
\end{array}\right]=\left[\begin{array}{l}
9 \\
5 \\
7
\end{array}\right]\)
Solution.
(i) \(\left[\begin{array}{ll}
\mathbf{4} & \mathbf{3} \\
\boldsymbol{x} & 5
\end{array}\right]=\left[\begin{array}{ll}
\boldsymbol{y} & \boldsymbol{z} \\
\mathbf{1} & \mathbf{5}
\end{array}\right]\)
As the given matrices are equal, their corresponding elements are also equal.
Comparing the corresponding eleme nts, we get x = 1, y = 4, and z = 3

PSEB 12th Class Maths Solutions Chapter 3 Matrices Ex 3.1

(ii) \(\left[\begin{array}{cc}
x+y & 2 \\
5+z & x y
\end{array}\right]=\left[\begin{array}{ll}
6 & 2 \\
5 & 8
\end{array}\right]\)
As the given matrices are equal, their corresponding elements are also equal.
Comparing the corresponding elements, we get x + y = 6, xy = 8, 5 + z = 5
Now, 5 + z = 5
⇒ z = 0
We know that,
(x – y)2 = (x + y)2 – 4xy
⇒ (x – y)2 = 36 – 32 = 4
⇒ x – y = ±2
Now, when x – y – 2 and x + y = 6, we get x = 4 and y = 2
When x – y = – 2 and x + y = 6, we get x = 2 and y = 4
∴ x = 4, y = 2, and z = 0 or x = 2, y = 4, and z = 0.

(iii) \(\left[\begin{array}{c}
x+y+z \\
x+z \\
y+z
\end{array}\right]=\left[\begin{array}{l}
9 \\
5 \\
7
\end{array}\right]\)
As the two matrices are equal, their corresponding elements are also equal.
Comparing the corresponding elements, we get
x + y + z = 9 ………..(i)
x + z = 5 ………….(ii)
y + z = 7
From Eqs. (i) and (ii), we have
y = 9 – 5
⇒ y = 4
Then, from Eq. (iii), we have:
4 + z = 7
⇒ z = 3
Now, x + z = 5
⇒ x = 5 – 3 = 2
∴ x – 2, y = 4 and z = 3.

PSEB 12th Class Maths Solutions Chapter 3 Matrices Ex 3.1

Question 7.
Find the value of a, b, c and d from the following equation:
\(\left[\begin{array}{cc}
a-b & 2 a+c \\
2 a-b & 3 c+d
\end{array}\right]=\left[\begin{array}{cc}
-1 & 5 \\
0 & 13
\end{array}\right]\)
Solution.
We have,
\(\left[\begin{array}{cc}
a-b & 2 a+c \\
2 a-b & 3 c+d
\end{array}\right]=\left[\begin{array}{cc}
-1 & 5 \\
0 & 13
\end{array}\right]\)
As the two matrices are equal, their corresponding elements are also equal.
Comparing the corresponding elements, we get
a – b = – 1 ……………(i)
2a – b = 0 …………….(ii)
2a + c = 5 …………..(iii)
3c + d = 13 ………..(iv)
From Eq. (ii), we have
b = 2a
Then, from Eq. (i), we have
a – 2a = – 1
⇒ a -1
⇒ b = 2
Now, from Eq. (iii), we have
2 x 1 + c = 5
⇒ c = 5 – 2 = 3
From Eq. (iv) we have
3 × 3 + d = 13
⇒ 9 + d = 13
⇒ d = 13 – 9 = 4
Hence, a = 1, b = 2, c = 3 and d = 4.

PSEB 12th Class Maths Solutions Chapter 3 Matrices Ex 3.1

Question 8.
A = [aij]m × n is a square matrix, if
(A) m < n (B) m > n
(C) m = n
(D) None of these
Solution.
It is known that a given matrix is said to be a square matrix, if the number of rows is equal to the number of columns.
Therefore, A = [aij]m × n is a square matrix, if m – n.
Hence, the correct answer is (C).

Question 9.
Which of the given values of x and y make the following pair of matrices equal?
\(\left[\begin{array}{cc}
3 x+7 & 5 \\
y+1 & 2-3 x
\end{array}\right]=\left[\begin{array}{cc}
0 & y-2 \\
8 & 4
\end{array}\right]\)
(A) x = \(\frac{-1}{3}\), y = 7

(B) Not possible to find

(C) y = 7, x = \(\frac{-2}{3}\)

(D) x = \(\frac{-1}{3}\), y = \(\frac{-2}{3}\)
Solution.
It is given that \(\left[\begin{array}{cc}
3 x+7 & 5 \\
y+1 & 2-3 x
\end{array}\right]=\left[\begin{array}{cc}
0 & y-2 \\
8 & 4
\end{array}\right]\)
On equating the corresponding elements, we get
3x + y = 0
⇒ x = – 3
5 = y – 2
⇒ y = 7
y + 1 = 8
⇒ y = 7
2 – 3x = 4
⇒ x = \(\frac{-2}{3}\)

We find that on comparing the corresponding elements of the two matrices, we get two different values of x, which is not possible.
Hence, it is not possible to find the values of x and y for which the given matrices are equal.
Hence, the correct answer is (B).

PSEB 12th Class Maths Solutions Chapter 3 Matrices Ex 3.1

Q. 10.
The number of all possible matrices of order 3 x 3 with each entry 0 or 1 is
(A) 27
(B) 18
(C) 81
(D) 512
Solution.
The given matrix of the order 3 × 3 has 9 elements and each of these elements can be either 0 or 1.
Now, each of the 9 elements can be filled in two possible ways.
Therefore, by the multiplication principle, the required number of possible matrices is 29 = 512.
Hence, the correct answer is (D).