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

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

Hey,

I don't think there was any need to find the value of x in this case. As correctly pointed by you the value of the expression will be maximum when the denominator is minimum and to do that we need to need ensure that \(2x^2 - 8 = 0\)

Thus the maximum value of the expression is :

\(22/(4x^4-32x^2+75)\)
\(={22}{(2x^2 - 8)^2 + 11}\)
\(=22/11\)
\(= 2\)

I think it just a coincidence that the value of x and the value of the expression is 2. And maybe that is why you did not notice that my mistake you found the value of x instead of the value of the expression. Just thought to let you know.


Thanks,
Saquib
Quant Expert
e-GMAT

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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

In order to maximize the expression the denominator should be minimum.

For a quadratic equation to be minimum the d/dx[(4x^4-32x^2+75)]=16x^3 - 64x=0
Which gives x=+2,-2
Substitute this in the original expression and we get answer as (c).

Can someone please break down how

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

Thanks!

you need to think about \((a-b)^2 = a^2 - 2ab + b^2\)
\(4x^4−32x^2+75\)

\(4x^4\) can be weritten as \(( 2 x^2 ) ^ 2\) = \(a^2\)

\(- 32x^2\) = 2 * \(2x^2\) * (-8) = 2ab (you need 32/4 = 8)

\(b^2 = -8^2 = 64\) (you can separate 75 into 64 + 11)

Therefore \((2x^2 - 8)^ + 11\)

Thanks pandeyashwin, very simple/clear

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