positive values of $a$ for which the equation $\lfloor x+a \rfloor = \sin \left(\frac{\pi x}{2}\right)$ will have no solution The  positive values of $a$ for which the equation $\displaystyle \lfloor x+a \rfloor = \sin \left(\frac{\pi x}{2}\right)$ will have no solution is,
where  $\lfloor x \rfloor = $ floor function of $x$                    
Options
$(a)\;\;(0,1)$
$(b)\;\; (1,2)$
$(c)\;\; (0,2)$
$(d)$ None of these
$\bf{My\; Try::}$ Using the formula $x-1<\lfloor x \rfloor \leq x.,$ we get $\displaystyle (x+a)-1< \lfloor x+a \rfloor \leq (x+a)$
Now Given $\displaystyle \lfloor x+a \rfloor = \sin \left(\frac{\pi x}{2}\right).$ So $\displaystyle (x+a) - 1<\sin \left(\frac{\pi x}{2}\right)\leq (x+a)$
Now we will solve $\displaystyle (x+a)-1<\sin \left(\frac{\pi x}{2}\right)$ and $\displaystyle (x+a)\leq \sin \left(\frac{\pi x}{2}\right)$ for Common values of $x$
But I Did Understand How can I solve it
Help me
Thanks
 A: observe that $\sin(\frac{\pi x}{2})$ only takes integer values for integer $x$'s, hence we can restrict our attention to $x \in \mathbb{Z}$ and so the floor function becomes obsolete. Now, the situation is even better, since there aren't plenty of integer values $\sin$ can take - it's just $-1,0$ and $1$, therefore, since it's a test question the simplest way would be to check all the posibilities. Let me do point a) and I hope you'll finish the rest by yourself following what I do.
first let $a = 0$. we want an integer $x$ such that $x = \sin(\frac{\pi x}{2})$. all values we need to check are $1,0,-1$. $x=1$ works, so we're done, that can't be the answer. 
we could now cross answer a) as a wrong one, but nevertheless let me do the case of $a=1$ so that its more instructive
if $a = 1$ then we're looking for $x$ s.t. 
$$ x+ 1 = \sin(\frac{\pi x}{2})$$
because of what've said about the range of $\sin$ we only need to check $x = 0$, $x = -1$ and $x = -2$.
$x = 0$ won't work since it leads to 
$$1 = 0$$
$ x = -1$ doesn't work either cause we get
$$ 0 = -1$$
finally $x = -2$ gives
$$-1 = 0$$ which isn't true either, so for $a=1$ there are no solutions to your equation.
