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Re: x(4 – x) [#permalink]
Niraj137 wrote:
We can easily say that factors of \(x(4-x)\) are \(x\) and \((4-x)\)

Quantity B is equal to 6 which is a positive number.

For \(x(4-x)\) to be positive, either both \(x\) and \((4-x)\) must be positive or both must be negative.

Case I - Both are negative

if \(x\) is negative, i.e, \(x<0\), that will basically force \((4-x)\) to be positive, so both cannot be negative.

If one of \(x\) or \((4-x)\) is negative, then the other is bound to be positive. But the product will be negative overall which will be less than 6 which is a positive number.

Case II - Both are positive

For \((4-x)\) to be positive, \(4 > x\) but \(x\) is also positive, which means \(0 < x < 4\)

Let us say there exist a number \((4 - e)\) which is closest number to \(4\). Meaning its the number just left to \(4\) on the number line, including decimal points.

If \(x = (4-e)\), then \(x(4-x) = (4-e)(4 - (4-e)) = (4-e)(e)\)

Now it does seem like we are back to square one, but we are not. Remember \((4-e)\) is closest to \(4\) including decimal points, which means \(0<e<1\), meaning \(e\) is basically of the form \(\frac{1}{a}\), where \(a>1\).

Since \((4-e)\) is already less than \(4\), multiplying it by \(e\) will result in a lesser number.

So basically \((4-e)e < 4\) which in turn is less than 6.

Hence, Answer is B


So this method (while incorrect, which I am noting so that anybody who reads this without previous context does not get confused and think it is correct) actually reminded me of a very interesting problem I was working on a few weeks ago that led me to a math stack exchange post about proving (a + b)^2 >= 4ab. I would link to it if I were allowed, but if you want to search for it, the solution I am referencing is by Benjamin Dickman and begins "Here is an alternative proof, which is circuitous but perhaps enjoyable as a curiosity". It uses a geometric proof by showing that a square gives the maximum area for the set of rectangles where the sum of the length and width is fixed.

The reason it relates to the above solution is that once you know that the maximum value of x must be in the region [0, 4] (this part of the above proof is correct), you can think of maximizing x(4-x) as being equivalent to maximizing the area of a rectangle where the sum of the length and width is 4. From basic geometry we know that must be a square, which is when the two sides are equal, i.e. x = 4-x --> x = 2. Then you can plug in and see 2 * (4 - 2) = 4.

I think this is a pretty interesting (albeit unnecessarily complicated) way to solve this problem, and it provides a geometric intuition for the solution.
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x(4 – x) [#permalink]
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Niraj137 wrote:
Scorplong wrote:

your answer only proves that f(x) < 4 for x in (0, 1) U (3, 4), it does not say anything about the value of x in [1, 3]. And unfortunately since the maximum of f(x) occurs at x=2, your answer does not prove a bound on the value of f.

Furthermore, even if the maximum of f occurred in the interval you discuss, you would still need to prove the maximum occurs in that interval. You can't just assume it does, otherwise you may assume incorrectly.


Well I could take that approach but is derivatives really included in GRE ? I can understand your reasoning completely and I do admit my mistake. But if not for derivatives how is one supposed to know what the maximum value for x(4-x) comes at ? If you know of any such method, I would like to learn it!


I don't want to spam messages, but my previous post actually provides a way to solve the maximum of x(4-x) without using any calculus and without completing the square. What's nice about this method is it is significantly faster to solve than either of the other two methods, although it is much more of a memorization trick.

The trick I suggested provides a way to solve the maximum of any quadratic represented in the form x(z-x) for any z > 0. Since maximizing the value of x(z-x) is equivalent to maximizing the area of a rectangle where the width and length sum to z, which is given by a square, we can solve for the maximum x through the following reasoning: first set the side lengths equal (x=z-x), then solve for x:
x = z-x
2x = z
x = z/2.

I hope this helps!
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Re: x(4 – x) [#permalink]
GreenlightTestPrep wrote:
Carcass wrote:




Quantity A
Quantity B
x(4 – x)
6


A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.


VERY tough question!! (165+)

I'm going to turn x(4 – x) into a perfect square
Before I do that, here are some other perfect squares:
x² + 6x + 9 = (x + 3)²
x² - 10x + 25 = (x - 5)²
x² - 4x + 4 = (x - 2)²
etc...

Given: x(4 – x) = 4x - x²
= -x² + 4x
= -1(x² - 4x)

What do we need to add to x² - 4x to make it a perfect square?
We need to add 4 to it to get x² - 4x + 4, which is equal to (x - 2)²
Of course, we can't just randomly add 4 to the given expression, since that totally changes the expression.
Instead, we're going to add 0 to the given expression. This is fine since adding 0 does not change anything.
HOWEVER, we're going to add 0 in a very SPECIAL WAY. We're going to add + 4 - 4 to the expression.
This is fine, since adding + 4 - 4 to the expression is the same as adding 0 to the expression.

We get: x(4 – x) = 4x - x²
= -x² + 4x
= -1(x² - 4x)
= -1(x² - 4x + 4 - 4)
= -1(x² - 4x + 4) + 4 [to remove -4 from the brackets, I had to multiply it by -1, since we are multiplying everything in the brackets by -1]
= -1(x - 2)² + 4

So, we can now write the following:
Quantity A: -1(x - 2)² + 4
Quantity B: 6

At this point, we must recognize that 4 is the GREATEST possible value of -1(x - 2)² + 4
We know this, because (x - 2)² is always greater than or equal to 0
So, -1(x - 2)² is always less than or equal to 0
So, the greatest value of -1(x - 2)² is 0. This occurs when x = 2
If 0 is the greatest possible value of -1(x - 2)², then 4 is the greatest possible value of -1(x - 2)² + 4

So, we get:
Quantity A: some number less than or equal to 4
Quantity B: 6

Answer:
Show: ::
B


Phew!!!!

great explanation!!!
i was wondering if ans has to be option D then which will be possible case that will led to option D.
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Re: x(4 – x) [#permalink]
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void wrote:
great explanation!!!
i was wondering if ans has to be option D then which will be possible case that will led to option D.

If Quantity B were 4, then the correct answer would be D

Similarly, if Quantity B were 3, the correct answer would also be D

etc
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x(4 – x) [#permalink]
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Tip: in QC never fall a prey to D choice, but MANIPULATE as much as possible

:)
Quantity A
Quantity B
\(4x - x^2 - 4\)
\(6 - 4\)


Quantity A
Quantity B
\(-(x - 2)^2\)
\(2\)


It's clear that a negative of the square is negative (or zero), hence answer is B.


Carcass wrote:

This question is part of GREPrepClub - The Questions Vault Project





Quantity A
Quantity B
\(x(4 - x)\)
\(6\)


A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
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x(4 x) [#permalink]
1
In general, \(x(b-x)\) or \(-x(x-b)\) represents parabolas that are concave down (curving downward) between x-intercepts 0 and b. Their vertex/peak is always at \((b/2,b)\). Incidentally, \(x(x-b)\) are concave up (curving upward).

In the problem we obtain \(b=4\) from \(-x(x-4)\). Hence the peak of the curve is at 4 and it never reaches 6.

B is the answer.
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x(4 x) [#permalink]
1
You can do this by plugging in numbers as well. And we don't have to do much, just plugging in numbers from 1 to 3 to realize that the result will always be a positive number less than 6. After that, the results become negative, and therefore are less than 6.

When x=0, x(4-x) = 0 and is less than 6.

Now in the case of x being a fraction, we can be sure that even the largest fraction (whose decimal equivalent would be 0.9999999) can only reduce the value of 4 further. But such a fraction would actually be acting on (4-x) which will be close to 3 already, prior to further reduction. If x was a very tiny fraction, say 0.0000000001, it would reduce the value in the bracket much much more, definitely very less than a thousandth of a four.

When x is negative, irrespective of whether it is an integer or fraction, the result will always be less than 6.

A lot to process here, but you can conclude that B is always greater.
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x(4 x) [#permalink]
= -1(x² - 4x + 4) + 4 [to remove -4 from the brackets, I had to multiply it by -1, since we are multiplying everything in the brackets by -1]
= -1(x - 2)² + 4

Hey Brent I dont understand how you made -1(x^2 -4x +4 - 4) into -1(x^2 -4x +4) + 4.. please elaborate
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Re: x(4 x) [#permalink]
2
Ykayonu wrote:
= -1(x² - 4x + 4) + 4 [to remove -4 from the brackets, I had to multiply it by -1, since we are multiplying everything in the brackets by -1]
= -1(x - 2)² + 4

Hey Brent I dont understand how you made -1(x^2 -4x +4 - 4) into -1(x^2 -4x +4) + 4.. please elaborate



-1(x^2-4x+4-4)

-x^2+4x-4+4

(-x^2+4x-4)+4

-1(x^2-4x+4) + 4

Hope that clarifies
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Re: x(4 x) [#permalink]
thanks it does
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