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Question

Let AZ and a function f:AB be defined as f(x)=|x|1|x|+1 2+|x|2|x|. Then

A
n(A)2
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B
n(A)=2
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C
n(B)=1
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D
n(B)1
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Solution

The correct option is D n(B)1
For domain,
|x|1|x|+10
|x|1 (1)
and |x|+2|x|20
|x|<2 (2)

From (1) and (2),
x(2,1][1,2)
A={1,1}
Now, for x=±1, we have f(±1)=3
So, codomain B can have at least one element.

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