Need explanation

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hi guyzzz... i need explanation in this sum;... please help


"If n is positive, which of the following could be the correct ordering of
1/x, 2x, and x2?
x2 < 2x < 1/x
x2 < 1/x < 2x
2x < x2 < 1/x
a. None
b. I
c. III
d. I and II
e. I, II, and III

[MBA.Aspirant]

hi guyzzz... i need explanation in this sum;... please help


"If n is positive, which of the following could be the correct ordering of
1/x, 2x, and x2?
x2 < 2x < 1/x
x2 < 1/x < 2x
2x < x2 < 1/x
a. None
b. I
c. III
d. I and II
e. I, II, and III

n is positive so it could be an integer, fraction or decimal

I) x2 < 2x < 1/x

if x = 1/2

then 1/2^2 < 2(1/2) < 1/1/2

1/4 < 1 < 2

so I is correct

II) x2 < 1/x < 2x

if x =2

4 < 1/2 < 4 no

if x= 1/2

1/4 < 1/1/2 < 2(1/2)
1/4 < 2 < 1 no

II is incorrect


III) 2x < x2 < 1/x

If x= 2

2*2 < 2^2 < 1/2 no

if x =1/2

2 (1/2)< 1/2^2 < 1/1/2 no

III is also incorrect

The correct one is only I, so choose B

[Roy@MasterGmat]

This question can and should be solved by plugging in values. However, note that we were asked "which of the following could be the correct ordering of", meaning a statement is to be considered correct even if only one plug in satisfies it.

Statement 1 is satisfied by x=1/2:

(1/2)^2 < 2*(1/2) < 1/(1/2)

1/4 < 1 < 2

Statement 2 is a bit more tricky, but it's satisfied by x=3/4:

(3/4)^2 < 1/(3/4) < 2*(3/4)

9/16 < 4/3 < 6/4

Statement 3 is never true. There are two ways to reach this conclusion: The first is simply trying a few plug ins - it never works. The second option is understanding the logic: if 1/x is greater then 2x, x must be smaller then one. That means that x^2 can't be greater then 2x, since x^2 is smaller then x, whereas 2x is greater then x (try this yourself- any number smaller then one becomes smaller when raised to the power of 2, and becomes larger when multiplied by 2).

The correct answer is d.

There is also a (much) more algebraic solution to this problem. I'd be glad to post it if you're interested.

[GMATGuruNY]

If x is positive, which of the following could be the correct ordering of 1/x, 2x, and x²?

I. x² < 2x < 1/x
II. x² < 1/x < 2x
III. 2x < x² < 1/x

a. None
b. I
c. III
d. I and II
e. I, II, and III

Determine the critical points by setting the expressions equal to each other:

1/x = 2x
2x² = 1
x² = 1/2
x = √(1/2) = 1/√2 ≈ 1/1.4 ≈ 10/14 ≈ 5/7.

1/x = x²
x^3 = 1
x = 1.

2x = x²
x=2
(We can divide by x because x>0.)

The critical points are x=5/7, x=1, x=2.
These critical points indicate where two of the expressions are equal.
Thus, to the right and left of each critical point, the value of one expression must be greater than the value of another.

To determine which answer choices are possible, plug in one value to the left and one value to the right of each critical point.

x < 5/7:
If x=1/2, then:
1/x = 2.
x² = 1/4.
2x = 1.
Since x² < 2x < 1/x, we know that I could be true.
Eliminate A and C.

5/7 < x < 1:
If x = 3/4, then:
1/x = 4/3.
x² = 9/16.
2x = 3/2.
Since x² < 1/x < 2x, we know that II could be true.
Eliminate B.

In III, the largest value listed is 1/x.
For 1/x to be the largest value, x would have to be a fraction.
Having tried a fraction on each side of the critical point of 5/7, we know that there is no way that III could be true.

The correct answer is D.

[n@resh]

If x is positive, which of the following could be the correct ordering of 1/x, 2x, and x²?

I. x² < 2x < 1/x
II. x² < 1/x < 2x
III. 2x < x² < 1/x

a. None
b. I
c. III
d. I and II
e. I, II, and III

Determine the critical points by setting the expressions equal to each other:

1/x = 2x
2x² = 1
x² = 1/2
x = √(1/2) = 1/√2 ≈ 1/1.4 ≈ 10/14 ≈ 5/7.

1/x = x²
x^3 = 1
x = 1.

2x = x²
x=2
(We can divide by x because x>0.)

The critical points are x=5/7, x=1, x=2.
These critical points indicate where two of the expressions are equal.
Thus, to the right and left of each critical point, the value of one expression must be greater than the value of another.

To determine which answer choices are possible, plug in one value to the left and one value to the right of each critical point.

x < 5/7:
If x=1/2, then:
1/x = 2.
x² = 1/4.
2x = 1.
Since x² < 2x < 1/x, we know that I could be true.
Eliminate A and C.

5/7 < x < 1:
If x = 3/4, then:
1/x = 4/3.
x² = 9/16.
2x = 3/2.
Since x² < 1/x < 2x, we know that II could be true.
Eliminate B.

In III, the largest value listed is 1/x.
For 1/x to be the largest value, x would have to be a fraction.
Having tried a fraction on each side of the critical point of 5/7, we know that there is no way that III could be true.

The correct answer is D.

If i consider the other critical point, i.e. 1 ( other than 5/7 )

Then we check for 1 < x < 1, i.e. x = 1/2 and x = 3/2 i.e. one value to the left and one value to the right of each critical point.

if x = 1/2, 1/4 < 1 < 2 ..Hence I is right
if x = 3/2, 9/4 3 2/3 => 1/x < x^2 < 2x ... here II won't satisfy, right?!

please let me know where i was wrong?

[kumadil2011]

Hi mitch..

Could you please give us a tip, on which problems types we need to set up critical points..
Because i start do so in all the inequality problems but later i realize the i should not have used that method..

appreciate your help