CBSE Class 12 Sample Paper for 2020 Boards

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

## A = R β {3} and B = R β {2}. Is the function f oneβone and onto? Is f invertible? If yes, then find its inverse

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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 oneβone 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 β6π₯_1+ 3π₯_2β9 = 2π₯_2 π₯_1 β6π₯_2+ 3π₯_1 β9 β6π₯_1+ 3π₯_2 = β6π₯_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¦β2))/((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^(β1) (π¦) = (3(π¦ + 1))/((π¦ β 2))