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x(4 – x)

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x(4 – x) [#permalink] New post   15 Jun 2017, 08:18
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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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B
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Re: x(4 – x) [#permalink] New post   19 Jun 2017, 10:10
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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!!!!
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wongpcla
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Re: x(4 – x) [#permalink] New post   19 Dec 2017, 22:48
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I just simply plug number in,
when X is negative, X=-10
-10(4-(-10))=negative
when x is positive, x=10
10(4-10)= negative

B is always greater
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Re: x(4 – x) [#permalink] New post   19 Dec 2017, 23:23
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wongpcla wrote:
I just simply plug number in,
when X is negative, X=-10
-10(4-(-10))=negative
when x is positive, x=10
10(4-10)= negative

B is always greater


This time you got Lucky,
However in this type of Ques you have to consider every possible aspects. Since no information is given for x, so it can be negative, positive , fraction or integer
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Re: x(4 – x) [#permalink] New post   20 Dec 2017, 14:04
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What panab01 pointed out is ONE of the most common mistakes made by the students.

Keep it in mind. Always.

Regards
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Re: x(4 – x) [#permalink] New post   20 Dec 2017, 17:58
If I plug four different number
(negative, positive , fraction or integer)
Can I avoid all the calculation?
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Re: x(4 – x) [#permalink] New post   21 Dec 2017, 16:37
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wongpcla wrote:
If I plug four different number
(negative, positive , fraction or integer)
Can I avoid all the calculation?



I think no. At least to a minimum extent.

Regards
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Re: x(4 – x) [#permalink] New post   01 Mar 2018, 05:18
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In such questions, always play with the extremes. i.e
Plug in 0, -ve (betwen 0 and -1), -ve( less than -1), +ve(between 0 and 1) and +ve (greater than 1)

There's how you will damn sure about your answer.

So in this question above, we need to solve in like an inequality expression as below:

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

We can add x^2 on both Quantities, the inequality will not effect, i.e

Quantity A: 4x
Quantity B: 6 + x^2

Similarly, we can add 6 on both Quantities, the inequality will not effect, i.e

Quantity A: 4x - 6
Quantity B: x^2

Now, if you put x = 0, 0.1, 10, -0.1 and -10

Actually, don't put -ve values, because in all negative values, Quantity B will be greater being positive.

Also, don't put 0, because Quantity A will be negative, while Quantity B remain 0.

Finally, only you need to put any +ve value, Quantity B will always greater.

So, Choice B is correct.

It was a good question.
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Re: x(4 – x) [#permalink] New post   28 Mar 2018, 11:06
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Answer: B
There is no limit for x.
If x is negative:
x(4-x) = 4x - x^2 is always negative because x^2 is always positive and thus -x^2 is negative, also 4x is negative.

If x is positive:
x(4-x) = 4x - x^2
For x = 4 the equation is 0. For 1 <= x <=3 it is positive and for values more than 4 for x, it is negative. Because x^2 is bigger for 4x.
x =1 4x-x^2 = 2
x =2 4x-x^2 = 4
x =3 4x-x^2 = 3
So either the equation is negative or is less than 6 ( 2,3,4). B is bigger than A.
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Re: x(4 – x) [#permalink] New post   19 Apr 2018, 05:18
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We can find the maximum value of the expression x(4-x) by using the identity: Arithmetic Mean > Geometric Mean:
So,
AM > GM
(x + (4-x))/2 > \sqrt{x(4-x)}
2 > \sqrt{x(4-x)}
Squaring both sides
4 > x(4-x)
So, Quantity A would be less than 4. Hence Quantity B is larger


Another way of doing it is using a little bit of Calculus
The expression x(4-x) would be maximum/ minimum when we put first derivative of x(4-x) = 0
First Derivative: 4 - 2x = 0 --> x =2

So to see if the expression would have a maximum or minimum value at x =2, we see the second derivative of x(4-x). If the second derivative is negative, the expression x(4-x) would have the maximum value at 2 and if the second derivative is positive, the expression would have the minimum value at 2.
Second Derivative: -2

Hence the expression x(4-x) has the maximum value at 2 and that is 2(4-2) = 4
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Re: x(4 – x) [#permalink] New post   03 Feb 2020, 01:28
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We can compare the two quantities with a placeholder:

4x -x^2 ? 6 | -6
-x^2 +4x -6 ? 0 | take quadratic formula

--> see that -b + 4ac = -8 --> therefore no solution for x, either all outcomes are positive or negative

--> x= 1 f(1)= - 3
--> x= -1 f(-1)= -11

--> therefore the quadratic formula -x^2 +4x -6 is always negative, hence 6 is greater --> B is the correct answer
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Re: x(4 – x) [#permalink] New post   20 Feb 2020, 17:23
@GreenlightTestPrep
@Carcass

Why isn´t correct just to plug 0 in the first term? x(4-0) .... 0(4-0) = 0 and B would be greater?

What is wrong with doing that?

Regards!
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Re: x(4 – x) [#permalink] New post   20 Feb 2020, 17:45
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FCOCGALVAN wrote:
@GreenlightTestPrep
@Carcass

Why isn´t correct just to plug 0 in the first term? x(4-0) .... 0(4-0) = 0 and B would be greater?

What is wrong with doing that?

Regards!


Once you test one x-value, you're not done.

Yes, when x = 0, Quantity B is greater than Quantity A, but that doesn't necessarily mean the correct answer is B.
It could still be the case that the correct answer is D.

That's the problem with testing values. You can never be 100% certain have the correct answer until you get TWO CONFLICTING outcomes (e.g., in one case, quantity B is greater, and in another case, quantity A is greater).

Consider this rudimentary example:
Quantity A: x
Quantity B: 5

If we test x = 1, we find that Quantity B is greater than Quantity A, but this doesn't mean that we can now conclude that the correct answer is B.

The same applies to your solution.

Cheers,
Brent
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Re: x(4 – x) [#permalink] New post   20 Feb 2020, 17:48
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Because zero could bring you in unchartered territory.

We do know that zero is only even but neither positive nor negative.

When you test number, it is a rule of thumb to test the following

-1/2,-1,0,1/1/2

of course, in this case, zero has no sense because of a question that you could solve in one second using zero ......mmmmmhhh something is wrong. Too easy.

I hope what I said to make sense.

Regards
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Re: x(4 – x) [#permalink] New post   15 Aug 2020, 18:41
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..Quantity A...Quantity B
......x(4-x)...............6
......4x-x^2..............6
........0................x^2-4x+6
........0................x^2-4x+4+2.........(we complete the square to get vertex form of parabola)
........0................(x-2)^2+2..........(this is an upward facing parabola with vertex at (2,2))


The lowest y value of the parabola is 2. Therefore it will always be greater than 0.

Final Answer: B
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Re: x(4 – x) [#permalink] New post   12 Sep 2020, 08:24
My question is that
how to prepare for gre quants i am done with ets official and 5lb ,what is the good source for gre quants preparation and that to hard and tricky questions?
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Re: x(4 – x) [#permalink] New post   12 Sep 2020, 09:22
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Shubham51 wrote:
My question is that
how to prepare for gre quants i am done with ets official and 5lb ,what is the good source for gre quants preparation and that to hard and tricky questions?


We do have literally thousands of questions

here https://gre.myprepclub.com/forum/gre-quant ... -3523.html

here https://gre.myprepclub.com/forum/gre-quant ... 18001.html

and here https://gre.myprepclub.com/forum/free-gre- ... 18796.html

Hope this helps

Also use our tags to search for hard questions https://gre.myprepclub.com/forum/search.php

at the bottom
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x(4 – x) [#permalink] New post   Updated on: 16 Nov 2021, 16:05
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I would like to contribute a much simpler way to get the solution to this problem.

f(x) = x(4-x) = 4x - x^2

We can find an inflection point when the derivative is 0:
f'(x) = 4 - 2x = 0
x = 2

Then, taking the second derivative tells us if this is a min/max:
f''(x) = -2x
f''(2) = -4 < 0

So this is a maximum. Then we plug in f(2) = 2 * (4 - 2) = 4 < 6.

This is much more straightforward (at least for me) than previously given solutions.

Edit: I noticed that novice07 gave this solution before I did

Originally posted by Scorplong on 14 Nov 2021, 14:49.
Last edited by Scorplong on 16 Nov 2021, 16:05, edited 1 time in total.
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x(4 – x) [#permalink] New post   Updated on: 16 Nov 2021, 21:20
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The solution I posted was wrong. So I am not able to delete it somehow. So ignore please do ignore this post. Apologies from my side.

Originally posted by Niraj137 on 16 Nov 2021, 08:03.
Last edited by Niraj137 on 16 Nov 2021, 21:20, edited 1 time in total.
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Re: x(4 – x) [#permalink] New post   16 Nov 2021, 09:41
Niraj137

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.
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