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Let a = {X andEpsis; R | −1 ≤ X ≤ 1} and Let F : a → A, G : a → a Be Two Functions Defined by F(X) = X2 and G(X) = Sin (π X/2). Show that G−1 Exists but F−1 Does Not Exist. Also, Find G−1.

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Question

Let A = {x &epsis; R | −1 ≤ x ≤ 1} and let f : A → Ag : A → A be two functions defined by f(x) = x2 and g(x) = sin (π x/2). Show that g−1 exists but f−1 does not exist. Also, find g−1.

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Solution

f is not one-one because

f (−1) = (−1)2 = 1

and f (1) = 12 = 1

⇒ -1 and 1 have the same image under f.

⇒ f is not a bijection.
So, f -1 does not exist.

Injectivity of g:

Let x and y be any two elements in the domain (A), such that

g (x) = g (y)

⇒ `sin   ((πx)/2) = sin ((πy)/2) `

⇒ `((πx)/2) = ((πy)/2)`

⇒ x = y

So, g is one-one.

Surjectivity of g :

Range of g = ` [ sin  ((π(-1))/2) , sin  ((π(1))/2) ]`

` =  [ sin  ((-π)/2) , sin  (π/2) ]` = [−1, 1] = A(co-domain of g)

⇒ g is onto.
⇒ g is a bijection.
So, g-1 exists.

Also,

let g−1 (x) = y                      ...(1)

⇒ g (y) = x

⇒ `sin  ((xy)/2) = x`

⇒ `y = 2/π  sin^-1  x `

⇒ `g^-1 (x) = 2/π  sin^-1 x`         [from (1)]

 

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Chapter 2: Functions - Exercise 2.4 [Page 69]

APPEARS IN

R.D. Sharma Mathematics Volume 1 and 2 [English] Class 12
Chapter 2 Functions
Exercise 2.4 | Q 20 | Page 69

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