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Re: x(4 – x) [#permalink]
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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]
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What panab01 pointed out is ONE of the most common mistakes made by the students.

Keep it in mind. Always.

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Re: x(4 – x) [#permalink]
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]
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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.

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Re: x(4 – x) [#permalink]
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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]
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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]
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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]
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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]
@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]
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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,
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Re: x(4 – x) [#permalink]
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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]
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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]
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]
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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]
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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]
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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]
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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