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If F : R → (0, 2) Defined by `F (X) =(E^X - E^(X))/(E^X +E^(-x))+1`Is Invertible , Find F-1. - Mathematics

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Question

If f : R → (0, 2) defined by `f (x) =(e^x - e^(x))/(e^x +e^(-x))+1`is invertible , find f-1.

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Solution

 

Injectivity of f :
Let x and y be two elements of domain (R), such that

f (x) = f (y)

⇒ `(e^x - e^(-x))/(e^x -e^(-x)) +1 =(e^y - e^(-y))/(e^y -e^(-y)) + 1`

⇒`(e^x - e^(-x))/(e^x -e^(-x))= (e^y - e^(-y))/(e^y -e^(-y))`

⇒ `(e^-x(e^(2x) -1))/(e^-x(e^(2x)+1)) = (e^-y(e^(2y) -1))/(e^-y(e^(2y)+1)) `

⇒ `(e^(2x) -1)/(e^(2x) +1)  = (e^(2y) -1)/(e^(2y) +1)`

⇒ (e2x−1) (e2y+1) = (e2x+1) (e2y−1)

⇒ e2x+2y + e2x−e2y −1= e2x+2y − e2x + e2y − 1

⇒ 2 × e2x =2 × e2y

⇒ e2x = e2y

⇒ 2x = 2y

⇒ x = y

So, f is one-one.

Surjectivity of f:
Let y be in the co-domain (0,2) such that f(x) = y.

`(e^x - e^-x)/(e^x +e^-x) + 1 = y `

⇒ `(e^-x(e^(2x) -1))/(e^-x(e^(2x)+1))+1 = y`

⇒ `(e^-x(e^(2x) -1))/(e^-x(e^(2x)+1)) = y - 1`

⇒ `e^(2x) -1 = (y - 1) (e^(2y) + 1)`

⇒ `e^(2x) -1 = y xx e^(2x) +y - e^(2x) -1`

⇒ `e^(2x) = y xx e^(2x) + y -e^(2x)`

⇒ `e^(2x) (2- y) = y`

⇒ `e^(2x) = y/(2-y)`

⇒ `2x = log_e (y/(2-y))`

⇒ `x = 1/2  log_e (y/(2 -y)) in R` (domain)

So,  f is onto.

∴ f is a bijection and, hence, it is invertible.

Finding f  -1:

Let f−1 (x) = y           ...(1)

⇒ f (y) = x

⇒ `(e^y - e^-y)/(e^y + e^-y )+ 1 = x`

⇒ `(e^-y(e^(2y) -1))/(e^-y(e^(2y)+1)) + 1 = x`

⇒ `(e^-y(e^(2y) -1))/(e^-y(e^(2y)+1)) = x -1`

⇒ e2y −1 = ( x −1) ( e2y + 1 )

⇒ e2y − 1 = x × e2y + x − e2y − 1

⇒ e2y = x × e2y+ x − e2y

⇒ e2y ( 2 − x ) = x

⇒  `e^(2y)  = x/(2-x)`

⇒`2y = log_e  (x/(2-x))`

⇒`y =1/2 log_e  (x/(2-x)) in R`  (domain)

⇒`y =1/2 log_e  (x/(2-x)) = f^-1 (x)`  [from (1)]

` So,   f^-1  (x)  = 1/2  log_e  (x/(2-x))`

 

 

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Chapter 2: Functions - Exercise 2.4 [Page 69]

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RD Sharma Mathematics [English] Class 12
Chapter 2 Functions
Exercise 2.4 | Q 18 | Page 69

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