Problem Solving

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

OA is D

this is plug-in and try q.
try x<1
x>1
x>2
out of all you should notice only x<1 is valid for I and II with x=0.5

Akansha wrote: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

OA is D

pemdas wrote:this is plug-in and try q.
try x<1
x>1
x>2
out of all you should notice only x<1 is valid for I and II with x=0.5

The question should read "1/x, 2x, and x^2", and in this case II isn't valid for x=0.5. II with 0.5 gives (1/4)<2<1.

Since we know x is positive, we're allowed to multiply the inequalities by x.
1) 2x<1/x = 2x^2<1 = x<sqr(1/2)
Since we also know that the square of a fraction less than 1 is less than that fraction, we know x^2 < 2x < 1/x when x<sqr(1/2)
2) x^2<1/x = x^3<1, so x<1
1/x<2x = 1<2x^2 = sqr(1/2)<x.
So x^2 <1/x < 2x when sqr(1/2)<x<1
3) x^2<1/x = x^3<1, so x<1
2x<x^2 = 0<x^2-2x. x^2-2x is negative between x=0 and x=2, so this conflicts with x^2<1/x.

So only I and II.

hi SoCan, it says try and plug-in and x<1 with x=0.5 as solution possible but not exact for II correctly revised

why don't you go ahead and divide by x "1/x, 2x, and x^2" to get 1/x^2, 2, and x
then it's obvious that 2x (now 2 after division by x) can't be the least value here (III) with x<1
and with x>1 1/x is always less (not greater) than x

Akansha wrote: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

OA is D

I received a PM asking me to comment.

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, we 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.