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Question

4. The function f : f:R{xR:1<x<1} defined by f(x)= x1+|x| is

A
one-one and into
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B
one-one and onto
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C
Many-one and into
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D
Many-one and onto
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Solution

The correct option is B one-one and onto
f:R{xR:1<x<1}definedbyf(x)=x1+|x|forx>0|x|=xforx<0|x|=xasx,f(x)=f(x)=x1+|x|=11x+1=1x1,f(x)=x1|x|=11x1=1foreverydifferentx,f(x)hasdifferentvalue.f(x1)=f(x2)x11+x1=x21+x2x1+x1x2=x2+x2x1x1=x2
so function is one-one and range is equal to co-domain so function is onto.

943393_1017374_ans_bf5685997dbd41f2b4e252ee1f99266b.png

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