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Question
Write whether f : R → R, given by `f(x) = x + sqrtx^2` is one-one, many-one, onto or into.
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Solution
\[f\left( x \right) = x + \sqrt{x^2} = x \pm x = 0 \text{ or 2x}\]
\[\text{So, each elementxin the domain may contain 2 images}.\]
\[\text{For example},\]
\[f\left( 0 \right) = 0 + \sqrt{0^2} = 0\]
\[f\left( - 1 \right) = - 1 + \sqrt{\left( - 1 \right)^2} = - 1 + \sqrt{1} = - 1 + 1 = 0\]
\[\text{Here, the image of 0 and -1 is } 0.\]
Hence, f is may-one.
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