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Question

The function : R[12,12] defined as f(x)=x1+x2 is

A
invertible
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B
injective but not surjective
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C
surjective but not injective
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D
neither injective nor surjective
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Solution

The correct option is C surjective but not injective
We have, f(x)=x1+x2
f(1x)=1x1+1x2=x1+x2=f(x) f(12)=f(2) or f(13)=f(3) and so on
So, f(x) is many – one function.
Again, let y=f(x)y=x1+x2
y+x2y=xyx2x+y=0As xϵR (1)24(y)(y)0 14y20 yϵ[12,12]
Range = Codomain =[12, 12]

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