If $f(x)=2 [x]+\cos x$ Then $f:R \to R$ is:

$(A)$ One-One and onto

$(B)$ One-One and into

$(C)$ Many-One and into

$(D)$ Many-One and onto

$[ .]$ represent floor function (also known as greatest integer function )

Clearly $f(x)$ is into as $2[x]$ is an even integer and $f(x)$ will not be able to achieve every real number.

Answer says option$(C)$ is correct but I cannot see $f(x)$ to be many-one as it does not look like that value of $f(x)$ is same for any two values of $x$

e.g. $f(x)= [x]+\cos x$, then $f(0)=f(\frac{\pi}{2})=1$ making the function many-one but can't see it happening for $f(x)= 2[x]+\cos x$

Could someone help me with this?

  • $\begingroup$ What does 'into' mean? $\endgroup$ – copper.hat May 27 '16 at 18:46
  • $\begingroup$ @copper.hat Into means that range is a proper subset of co-domain. $\endgroup$ – Akira May 27 '16 at 18:49

You are right with respect to surjectiveness (it is not onto).


For injectiveness (one to one), look in a neighbourhood around $x = 3\pi$ for example.


Note that $\pi \in (3,4)$ hence $f$ has a strict local minimum on $[3,4]$ at $\pi$. It follows that $f$ is not injective.

Note that if $x \ge 0$ then $f(x) >0$ and if $x< 0$ then $f(x) <0$. Hence $0$ is not in the range and so it is not surjective.


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