Product of cdf and pdf of normal distribution. 
A continuos random variable $X$ has the density
  $$
f(x) = 2\phi(x)\Phi(x), ~x\in\mathbb{R}
$$
  then
(A) $E(X) > 0$
(B) $E(X) < 0$
(C) $P(X\leq 0) > 0.5$
(D) $P(X\ge0) < 0.25$
\begin{eqnarray}
\Phi(x) &=& \text{Cumulative distribution function of } N(0,1)\\
\phi(x) &=& \text{Density function of } N(0, 1)
\end{eqnarray}

I don't have a slightest clue where to start with. Can someone give me a little push. I saw some answers on same question like this but I didn't understand how should I integrate it when calculating expectation. 

This question requires no calculations.
You should not integrate anything to answer it.
The Key
If
$\phi(x)$ is the density function of a distribution $D$,
and
$\Phi(x)$ is the cumulative distribution function of $D$,
then
the density $f(x) = 2\phi(x)\Phi(x)$
corresponds to the distribution of a random variable $X$ defined 
as the greater of two random variables (say $D_1,D_2$) picked according to $D$.
Why you would see this
To see this from the formula for $f$, see that $\phi(x)$ corresponds to the chance that $D_1$ has a certain value, and $\Phi(x)$ is the chance that another sample $D_2$ has a lesser value.  Finally, the factor of 2 is for the alternative case that actually $D_2$ had this value and was greater, so together, $f(x)$ is the chance that $x$ is the greater of two randomly picked values.
Once you see this, answering the question is easy and requires no calculations.