Chapter 1 Class 12 Relation and Functions

Class 12
Important Questions for exams Class 12

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Misc 4 Show that function f: R β {x β R: β1 < x < 1} defined by f(x) = x/(1 + |π₯| ) , x β R is one-one and onto function. f: R β {x β R: β1 < x < 1} f(x) = x/(1 + |π₯| ) We know that |π₯| = {β( π₯ , π₯β₯0 @βπ₯ , π₯<0)β€ So, π(π₯)={β(π₯/(1 + π₯), π₯β₯0@&π₯/(1 β π₯), π₯<0)β€ For x β₯ 0 f(x1) = π₯_1/(1 + π₯_1 ) f(x2) = π₯_2/(1 + π₯_2 ) Putting f(x1) = f(x2) π₯_1/(1 + π₯_1 ) = π₯_2/(1 + π₯_2 ) π₯_1 (1 + π₯_2)= π₯_2 (1 + π₯_1) π₯_1+π₯_1 π₯_2= π₯_2 +π₯_2 π₯_1 π₯_1= π₯_2 For x < 0 f(x1) = π₯_1/(1 β π₯_1 ) f(x2) = π₯_2/(1 β π₯_2 ) Putting f(x1) = f(x2) π₯_1/(1 β π₯_1 ) = π₯_2/(1 β π₯_2 ) π₯_1 (1 β π₯_2)= π₯_2 (1 β π₯_1) π₯_1βπ₯_1 π₯_2= π₯_2 βπ₯_2 π₯_1 π₯_1= π₯_2 Hence, if f(x1) = f(x2) , then x1 = x2 β΄ f is one-one Checking onto π(π₯)={β(π₯/(1 + π₯), π₯β₯0@&π₯/(1 β π₯), π₯<0)β€ For x β₯ 0 f(x) = π₯/(1 + π₯) Let f(x) = y, "y = " π₯/(1 + π₯) y(1 + x) = x y + xy = x y = x β xy x β xy = y x(1 β y) = y x = π¦/(1 β π¦) For x < 0 f(x) = π₯/(1 β π₯) Let f(x) = y "y = " π₯/(1 β π₯) y(1 β x) = x y β xy = x y = x + xy x + xy = y x(1 + y) = y x = π¦/(1 + π¦) Thus, x = π¦/(1 β π¦) , for x β₯ 0 & x = π¦/(1 + π¦) , for x < 0 Here, y β {x β R: β1 < x < 1} i.e. Value of y is from β1 to 1 , β 1 < y < 1 If y = 1, x = π¦/(1 β π¦) will be not defined, If y = β1, x = π¦/(1 + π¦) will be not defined, But here β 1 < y < 1 So, x is defined for all values of y. & x β R β΄ f is onto Hence, f is one-one and onto.