Finding Inverse

Chapter 1 Class 12 Relation and Functions
Concept wise

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Ex 1.3, 6 Show that f: [β1, 1] β R, given by f(x) = π₯/(π₯ + 2) is one-one. Find the inverse of the function f: [β1, 1] β Range f. (Hint: For y β Range f, y = f(x) = π₯/(π₯ + 2) , for some x in [β1, 1], i.e., x = 2π¦/(1 β π¦) ) f(x) = x/(x+2) Check one-one f(x1) = π₯1/(π₯1 + 2) f(x2) = π₯2/(π₯2 + 2) Rough One-one Steps: 1. Calculate f(x1) 2. Calculate f(x2) 3. Putting f(x1) = f(x2) we have to prove x1 = x2 Putting f(x1) = f(x2) π₯1/(π₯1 + 2) = π₯2/(π₯2 + 2) x1(x2 + 2) = x2(x1 + 2) x1x2 + 2x1 = x2x1 + 2x2 x1x2 β x2x1 + 2x1 = 2x2 0 + 2x1 = 2x2 2x1 = 2x2 x1 = x2 Hence, if f(x1) = f(x2) , then x1 = x2 β΄ f is one-one Checking onto f(x) = π₯/(π₯ + 2) Putting f(x) = y y = π₯/(π₯ + 2) y(x + 2) = x yx + 2y = x yx β x = β2y x(y β 1) = β2y x = (β2π¦ )/(π¦ β1) x = (β2π¦ )/(β1(βπ¦ + 1) ) x = (2π¦ )/((1 β π¦) ) Now, Checking for y = f(x) Putting value of x in f(x) f(x) = f((2π¦ )/((1 β π¦) )) = ((2π¦ )/((1 β π¦) ))/((2π¦ )/((1 β π¦) ) + 2) = ((2π¦ )/((1 β π¦) ))/((2π¦ + 2(1 β π¦) )/((1 β π¦) )) = 2π¦/(2π¦ + 2 β 2π¦) = y Thus, for every y β Range f, there exists x β [β1, 1] 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¦) = (2π¦ )/((1 β π¦) ) , y β  1 Note: Here, y β Range f is important Inverse is not defined for y β R Because denominator in (2π¦ )/((1 β π¦) ) will be 0 if y = 1