BrentGMATPrepNow wrote:
parkhydel wrote:
If \(y = \frac{|3x - 5|}{-x^2 - 3}\) for what value of x will the value of y be greatest?
A. -5
B. -3/5
C. 0
D. 3/5
E. 5/3
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 NEGATIVEIn 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 denominator, |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
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