Square of an increasing function over an interval. Let $f:I\rightarrow R$ be an increasing function where I is an interval in R.Then
(a) $f^2$ is always increasing
(b) $f^2$ is always decreasing
(c)$f^2$ is constant$\Rightarrow  f$ is constant
(d)) $f^2$ may be neither increasing nor decreasing.
My try
If we take $I:[-1,1]$ and $f(x)=x$ then option (a) and (b) will be wrong. 
For option (c). Let$ f^2(x)=k$ then $f(x)= \sqrt k  \  \ or -  \sqrt k$. So (c) may be write. 
But the correct answer is option (a)
Can anyone explain.
This question was asked in entrance exam.Please see this image
 A: I am converting my comments into an answer and comments will be deleted shortly. 

For the given question the correct answer is (d) (you have already rejected (a), (b) via a correct argument and one of the comments to the question rejects (c)). So either there is a typo in question or in official key. Such printing mistakes are reasonable and in any case not a punishable offence :) :)
If on the other hand the question is about a function $f:I\to\mathbb{R} ^{+} $ which is increasing then it should be obvious that the correct answer is (a). This is because if $f$ is increasing then for any $x, y\in I$ with $x<y$ we have $f(x) \leq f(y) $ and since $f(x), f(y) $ are positive it follows that $f^{2}(x)\leq f^{2}(y)$ and thus $f^2$ is increasing. One can easily prove that the option (c) also holds.
Moreover one should notice that the ideas of increasing / decreasing functions does not necessarily require the functions concerned to be differentiable. And therefore it is not necessary to invoke any arguments based on derivatives to solve this problem. 
