## Let f: A β B be a function defined as f(x) = (2x + 3)/(x - 3) , where

## A = R β {3} and B = R β {2}. Is the function f oneβone and onto? Is f invertible? If yes, then find its inverse

CBSE Class 12 Sample Paper for 2020 Boards

Paper Summary

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Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10 Important

Question 11 Important

Question 12

Question 13

Question 14 (OR 1st Question)

Question 14 (OR 2nd Question) Important

Question 15 (OR 1st Question)

Question 15 (OR 2nd Question) Important

Question 16

Question 17 Important

Question 18 (OR 1st Question)

Question 18 (OR 2nd Question)

Question 19

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Question 21 (OR 1st Question) Important

Question 21 (OR 2nd Question) Important

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Question 24 (OR 1st Question)

Question 24 (OR 2nd Question)

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Question 26 Important

Question 27 You are here

Question 28 (OR 1st Question) Important

Question 28 (OR 2nd Question)

Question 29

Question 30 Important

Question 31 (OR 1st Question) Important

Question 31 (OR 2nd Question)

Question 32 Important

Question 33 (OR 1st Question) Important

Question 33 (OR 2nd Question) Important

Question 34

Question 35 (OR 1st Question) Important

Question 35 (OR 2nd Question) Important

Question 36 Important

Class 12

Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

Last updated at Oct. 18, 2019 by Teachoo

Question 27 Let π: A β B be a function defined as π(π₯)=(2π₯ + 3)/(π₯ β 3) , where A = R β {3} and B = R β {2}. Is the function f oneβone and onto? Is f invertible? If yes, then find its inverse π(π₯)=(2π₯ + 3)/(π₯ β 3) Checking one-one Let π₯_1 , π₯_2 β A π(π₯_1 )=(2π₯_1+ 3)/(π₯_1β 3) π(π₯_2 )=(2π₯_2+ 3)/(π₯_2 β 3) One-one Steps 1. Calculate f(x1) 2. Calculate f(x2) 3. Putting f(x1) = f(x2) we have to prove x1 = x2 Putting π(π₯_1 ) = π(π₯_2 ) (2π₯_1+ 3)/(π₯_1β 3) = (2π₯_2+ 3)/(π₯_2 β 3) (2π₯_1+ 3) (π₯_2 β 3) = (2π₯_2+ 3) (π₯_1 β 3) 2π₯_1 "(" π₯_2 β 3")"+ 3 (π₯_2 β 3) = 2π₯_2 "(" π₯_1 β 3")"+ 3(π₯_1 β 3) 2π₯_1 π₯_2 β6π₯_1+ 3π₯_2β9 = 2π₯_2 π₯_1 β6π₯_2+ 3π₯_1 β9 β6π₯_1+ 3π₯_2 = β6π₯_2+ 3π₯_1 3π₯_2+6π₯_2 = 3π₯_1+6π₯_1 9π₯_2= 9π₯_1 π₯_2= π₯_1 If π(π₯_1 ) = π(π₯_2 ) , then π₯_1= π₯_2 β΄ f is one-one Checking onto π(π₯)=(2π₯ + 3)/(π₯ β 3) Let y β B where y = (2π₯ + 3)/(π₯ β 3) y(x β 3) = 2x + 3 xy β 3y = 2x + 3 xy β 2x = 3y + 3 x(y β 2) = 3(y + 1) x = (3(π¦ + 1))/((π¦ β 2)) For y = 2 , x is not defined But it is given that y β R β {2} Hence , x = (3(π¦ + 1))/((π¦ β 2)) β R β {3} Now, Checking for y = f(x) Putting value of x in f(x) f(x) = f((3(π¦ + 1))/((π¦ β 2))) = (2(3(π¦ + 1)/((π¦ β 2) ))+ 3)/((3(π¦ + 1)/((π¦ β 2) ))β 3) = (((6(π¦ + 1)+3(π¦β2))/((π¦ β 2) )))/((3(π¦ + 1)/((π¦ β 2) ))β 3) Also, y = f(x) Putting y in f(x) Hence, f is onto = (2(3(π¦ + 1)/((π¦ β 2) ))+ 3)/((3(π¦ + 1)/((π¦ β 2) ))β 3) = (((6(π¦ + 1) + 3(π¦ β 2))/((π¦ β 2) )))/(((3(π¦ + 1) β 3(π¦ β 2))/((π¦ β 2) )) ) = (6(π¦ + 1) + 3(π¦ β 2))/(3(π¦ + 1) β 3(π¦ β 2)) = (6π¦ + 6 + 3π¦ β 6)/(3π¦ + 3 β 3π¦ + 6) = 9π¦/9 = y Thus, for every y β B, there exists x β A such that f(x) = y Hence, f is onto Since f(x) is one-one and onto, So, f(x) is invertible And Inverse of x = π^(β1) (π¦) = (3(π¦ + 1))/((π¦ β 2))