Class 9 Maths Polynomials Exercise 2.2 Solutions in Assamese Medium

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Exercise 2.2 in Assamese Medium

 

 

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Exercise 2.2 in Assamese Medium

 





 

Assamese Medium NCERT Solutions for  Class 9 Maths Chapter 2 Polynomials Topics and Subtopics:


  • Polynomials
  • Introduction
  • Polynomials In One Variable
  • Zeroes Of A Polynomial
  • Remainder Theorem
  • Factorization Of Polynomials
  • Algebraic Identities
  • Summary

 

Polynomials Chapter  Class 9 has total 5 Exercise. Just click on the exercise wise links given below to practice the Maths solutions for the respective exercise


বহুপদSolutions Link
অনুশীলনী 2.1 (Exercise 2.1)সমাধান
অনুশীলনী 2.2 (Exercise 2.2)সমাধান


NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Exercise 2.3 in Assamese Medium


  • Chapter 2 Polynomials Solutions Exercise 2.1
  • Chapter 2 Polynomials Solutions Exercise 2.2
  • Chapter 2 Polynomials Solutions Exercise 2.3
  • Chapter 2 Polynomials Solutions Exercise 2.4
  • Chapter 2 Polynomials Solutions Exercise 2.5
  • Extra Questions for Polynomials in Assamese Medium



Class 9 Maths Polynomials Exercise 2.2 Solutions in Assamese Medium

Class 9 Maths Exercise 2.2 Question 1 Solution Assamese medium

       1.     5x – 4x2 + 3 বহুপদৰ মান নিৰ্ণয় কৰা যেতিয়া –

   

সমাধান:

  

(ii)                            x = - 1

 ∴ x ৰ মান 5x – 4x2 + 3 ত বহুৱাই পাওঁ,

                 5 × (-1) – 4 × (-1)2 + 3

           = - 5 - 4 × 1 + 3

           = - 5 – 4 + 3

           = - 9 + 3

           = - 6


(ii)                          x = 2

∴ x ৰ মান 5x – 4x2 + 3 ত বহুৱাই পাওঁ,

               5 × 2 – 4 × 22 + 3

          = 10 - 4 × 4 + 3

          = 10 – 16 + 3

          = 13 – 16

          = - 3


Class 9 Maths Exercise 2.2 Question 2 Solution Assamese medium

      2.     তলৰ বহুপদবোৰৰ প্ৰত্যেকৰ বাবে P(0), P(1) আৰু P (2) নিৰ্ণয় কৰা –

(i)                         P(y) = y2 – y + 1

সমাধান:

                      P(y) = y2 – y + 1

            ∴ P(1) = 12 – 1 + 1        (y = 1 ধৰি)

                       = 1 – 1 + 1

                       = 1

          ∴ P(0) = 02 – 0 + 1           (y = 0 ধৰি)

                      = 0 – 0 + 1

                      = 1

         ∴P(2)   = 22 – 2 + 1          (y = 2 ধৰি)

                        = 4 – 2 + 1

                        = 5 – 2

                        = 3 




          ∴ P(0) = 2 + 0+ 2 × 02 - 03                    (t = 0 ধৰি)

                      = 2 + 0 + 2 × 0 – 0

                      = 2 + 0 + 0 - 0

                      = 2

          ∴ P(1) = 2 + 1 + 2 × 12 - 13           (t = 1 ধৰি)

                      = 2 + 1 + 2 × 1-1

                      = 2 + 1 + 2 – 1

                      = 4

          ∴ P(2) =  2 + 2 + 2 × 22 – 23             (t = 2 ধৰি)

                      = 2 + 2 + 2 × 4 – 8

                      = 2 + 2 + 8 – 8

                      = 4


 







Class 9 Maths Exercise 2.2 Question 3 Solution Assamese medium

      3.      à¦•াষৰ উল্লখিত মানবোৰ বহুপদটোৰ শূন্য হয়নে নহয় সত্যপনা কৰা ।