f(x) = x + sqrt{x^2} is a function from R to | vedclass

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(D) Given the function $f(x) = x + \sqrt{x^2} = x + |x|$.We can express this function as:$f(x) = \begin{cases} x + x = 2x, & 	ext{if } x \ge 0 \ x -

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None of these

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(D) Given the function $f(x) = x + \sqrt{x^2} = x + |x|$.We can express this function as:$f(x) = \begin{cases} x + x = 2x, & 	ext{if } x \ge 0 \ x - x = 0, & 	ext{if } x $1$. Check for Injectivity (One-one):For $x $2$. Check for Surjectivity (Onto):The range of $f(x)$ is $[0, \infty)$. Since the codomain is $R$ (all real numbers) and the range is not equal to the codomain,the function is not surjective.Since the function is neither injective nor surjective,it is not bijective.Therefore,the correct option is $(d)$.

$f(x) = x + \sqrt{x^2}$ is a function from $R \to R$,then $f(x)$ is

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