Key idea: \(x^2\) is always greater than or equal to 0
This means \(-x^2\) is always less than or equal to 0
And this means \(-x^2 - 3\) is always NEGATIVE
In other words, the denominator of \(y\) is most definitely NEGATIVE

From here, there are two possible cases:
case i: If the numerator of \(y\) is NEGATIVE, then y is POSITIVE
case ii: If the numerator of \(y\) is POSITIVE, then y is NEGATIVE

Since we're trying to MAXIMIZE the value of y, we need the denominator to be negative (which would make y POSITIVE)
However, the numerator, |3x - 5|, will always be greater than or equal to 0.
Since we can't make the numerator NEGATIVE, the only way to maximize the value of y is make the numerator ZERO
In other words: \(|3x - 5|=0\)
Which means \(3x -5=0\)
So \(3x =5\)
Ergo \(x = \frac{5}{3} \)

Answer: E

Cheers,
Brent
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|3x-5| is always non-negative and -(x^2+3) is alway negative

\( \frac{|3x - 5|}{(-x^2 - 3)} ≤ 0\)

Hence maximum value of \(\frac{|3x - 5|}{-x^2 - 3}\) is 0.

\(|3x-5| = 0\)

\(x= \frac{5}{3}\)

We have \(y = \frac{|3x - 5|}{-x^2 - 3}\)

The denominator, \(-x^2 - 3\) MUST be Negative

The Numerator, \(|3x - 5|\) can NOT be Negative

i.e. \(y = \frac{|3x - 5|}{-x^2 - 3}\) MUST be Non-Positive

i.e. 3x-5 = 0
i.e. y is maximum at \(x = 5/3\) and \(y_{max}=0\)

Answer: Option E
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Confused of signs ? just substitute X= options

arrive at Solution which gives greatest takes 1 to 2 mins. But helps if u are confused

E.5/3 gives greatest value when substituted

HOPE this helps.
THANKS
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Segment the numerator and denominator to look for extreme combination: |3x-5|>=0 and \(-x^2-3<=-3\), and hence y<=0.

Thus the greatest value of y is 0 --> 3x-5 = 0 --> x = \(5/3\)

\(y = \frac{+ numerator }{ - denominator}\)
Hence y will always be negative except when the numerator is zero which is the largest value y can take.

\(\implies\) |3x-5| = 0
\(\implies x = \frac{5}{3}\)

Answer E.
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Hi BrentGMATPrepNow Brent, the explanation is nice i just have a few questions:

When i saw |3x - 5| i thought i should consider two case

3x-5 or -3x+5 so I kinda confused...

3x=5 or -3x=-5

x = 5/3 or x = 5/3 (negative/negative =positive) both options give the same results thats strange to me Why ?

General question: if the absolute value by definition is distance from 0 and the distance is always positive or zero. why do we consider two cases positive and negative

p.s. i think you meant \(-x^2 - 3\) in the highlighted part

You have taken the equation |3x - 5| = 0 and concluded that EITHER 3x - 5 = 0 OR -(3x - 5) = 0
Noticed that both of these equations are the same.
If we take -(3x - 5) = 0 and multiply both sides by -1, we got the equivalent equation 3x - 5 = -0, which is identical to the equation 3x - 5 = 0

Thanks for the heads-up about \(-x^2 - 3\)
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BrentGMATPrepNow this time i picked A because i thought since Absolute value is always positive i can pick -5 so and in denominator since it it is --
-x^2 after plugging in -5, the sign will change into positive - *-5^2 = 25

So
\(y = \frac{|3 (-5) - 5|}{- -5^2 - 3}\)

= 10/22 greater than 0

why its wrong with it?

Think of -x² as -(x²)
So, -x² - 3 = -(x²) - 3
So, when we plug in x = -5, we get: -(x²) - 3 = -((-5)²) - 3 = -(25) - 3 = -28

Does that help?
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IMO - E
I did it by subsituting the given value of X in option and found the largest value for Y is obtained by option E (5/3)
also realized, is denominator is negative and numerator is positive,
to obain highest value Y when the numerator is zero

Denominator is always negative -x^2 - 3

Max value can be 0

x = 5/3

E

Hi Brent,

Thank you for the great explanation. I understood the concept based on your walk-through.

Although, I did catch one issue with your one statement and please correct me if I am wrong.



Shouldn't the statement above read "Numerator" instead of "Denominator"?

Regards,
Sud

Hi Brent,

I have a question regarding the modulus of the numerator.....since this is an absolute value function, the sign of the numerator will depend on the range of X. In this case, if X<5/3 the function should be written as -(3X-5), which makes the numerator negative for all range of values of X<5/3. So there can be values greater than 0 now.