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

JEE MATHEMATICS MCQ: FUNCTIONS

1 Easy

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Analyzing +4 | -1

f\left( x \right) = x + \sqrt {{x^2}}

is a function from

R\;\;\mathrm{to}\;R

, then

f\left( x \right)

View answer and explanation

Correct option: D

None of these

Option D (None of these) is correct because the function simplifies to either 0 or 2x, depending on the value of x, and is therefore not injective, surjective, or bijective.

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