Evaluating different inputs into a given system We need to find which signal wont be the same after going through this sytem.

My approach is this one:

I took the filters frequency and did 2pi * 1/4 and gave me this "limit" 1.5707

(2pi because the periodicity and 1/4 because the filter "bandwith"?)

Then i proceed with the inputs evaluating them this way:

a) 4 * pi/7 = 1.7951 
b) 4 * pi/9 = 1.3962
c) 4 * pi/8 = 1.5707
d) idk

The result "a", which is out of the "limit" is the one that does not follow the rule y[n]=x[n] and thus, the correct answer. (It is the actual correct answer)

Did i have luck there with the math or the approach is correct?
The output will be equal to the input $(y[n]=x[n])$ if no aliasing occurs as a result of downsampling. Which means $X(e^{j\omega})=0$ for $\frac{\pi}{4} \leq |\omega| \leq \pi$.   
a) $X(e^{j\omega})$ has impulses at $\omega = \pm \frac{2\pi}{7}$ $=&gt;$ aliasing exists $=&gt;$ $y[n]\neq x[n]$ 
b) $X(e^{j\omega})$ has impulses at $\omega = \pm \frac{2\pi}{9}$ $=&gt;$ no aliasing $=&gt;$ $y[n]= x[n]$ 
c) $X(e^{j\omega})$ has impulses at $\omega = \pm \frac{\pi}{8}$ $=&gt;$ no aliasing $=&gt;$ $y[n]= x[n]$ 
d) You need to sketch $X(e^{j\omega})$ to see if aliasing exists  hint: the impulse response of your filter is $h[n]=\frac{sin(\pi n/4)}{\pi n}$