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Question
Let A = {x &epsis; 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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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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