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Question

$f left parenthesis x right parenthesis equals x plus square root of x squared end root$ is a function from R to R, then $f left parenthesis x right parenthesis$ is

Injective
Surjective
Bijective
None of these

The correct answer is: None of these

Given, $f left parenthesis x right parenthesis equals x plus square root of x squared end root$
Since, this function is not defined
Given, $f left parenthesis x right parenthesis equals x cubed minus 1$
Let $x subscript 1 comma x subscript 2 element of R$
Now, $f open parentheses x subscript 1 close parentheses equals f open parentheses x subscript 2 close parentheses$
$table attributes columnalign right left right left right left right left right left right left columnspacing 0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em end attributes row cell not stretchy rightwards double arrow x subscript 1 superscript 3 minus 1 equals x subscript 2 superscript 3 minus 1 end cell row cell not stretchy rightwards double arrow x subscript 1 superscript 3 equals x subscript 2 superscript 3 end cell row cell not stretchy rightwards double arrow x subscript 1 equals x subscript 2 end cell end table$
$therefore f left parenthesis x right parenthesis$ is one-one. Also, it is onto as range of
Hence, it is a bijection.

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