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Question

Let A=x : -1x1 and f : AA such that fx=x|x|, then f is
(a) a bijection
(b) injective but not surjective
(c) surjective but not injective
(d) neither injective nor surjective

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Solution

Injectivity:
Let x and y be any two elements in the domain A.

Case-1: Let x and y be two positive numbers, such that
fx=fyxx=yyxx=yyx2=y2x=y

Case-2: Let x and y be two negative numbers, such that
fx=fyxx=yyx-x=y-y-x2=-y2x2=y2x=y

Case-3: Let x be positive and y be negative.
Then, xyfx=xx is positive and fy=yy is negativefxfySo, xyfxfy
So, f is one-one.

Surjectivity:
Let y be an element in the co-domain, such that y = f (x)
Case-1: Let y>0. Then, 0<y≤1y=fx=xx>0x>0x=xfx=yxx=yxx=yx2=yx=y A We do not get ±, as x>0Case-2: Let y<0. Then, -1≤y<0y=fx=xx<0x<0x=-xfx=yxx=yx-x=y-x2=yx2=-yx=--y A We do not get ±, as x>0
f is onto
f is a bijection.
So, the answer is (a).

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