\(\frac{22}{(4x^4-32x^2+75)}\)

This expression will take its maximum value when \(4x^4-32x^2+75\) will take its minimum value.

\(4x^4-32x^2+75 = (2x^2)^2 + 2*2x^2*8 + 64 + 11 = (2x^2 - 8)^2 + 11\)

If we get negative or positive value of \(2x^2 - 8\) it will be squared and become positive, hence increasing our denominator and decreasing value of our fraction. The only possible minimum can be achieved when \(2x^2 - 8 = 0\)

\(2x^2 - 8 =0\)

\(x^2 = 4\)

\(x= +/- 2\)

Answer C

Hi Saquib
Thanks for your kind advice. It was way beyond midnight and I did not put the solution to its logical end. What I’ve found in fact is the value of x that will maximize our fraction. In general the proper way to solve questions like that, as it’s done in analysis, is to find first and second derivatives of denominator and extreme values of our variable, put them into fraction to find its max or min value. But, calculus is beyond gmat, that’s why we have this case where we can simplify our expression to “perfect square plus 11” and, as you’ve correctly mentioned, it’s quite enough to take our perfect square equal to zero. I just had to put that x into our fraction, but that was unnecessary, 22/11 will be our answer. In general, this won’t do and we’ll need to go the long way.
Thanks again!
Merry holidays
Vitaliy

Can someone please break down how

\(4x^4\)−\(32x^2\)+75 becomes (\(2x^2\)−8)\(^{2}\)+11 ?

Thanks!

Thanks pandeyashwin, very simple/clear