If 0 < x < 1, which of the following inequalities must be true ?

If 0 < x < 1, which of the following inequalities must be true ?

I. \(x^5 < x^3\)
II. \(x^4 + x^5 < x^3 + x^2\)
III. \(x^4 - x^5 < x^2 - x^3\)

A. None
B. I only
C. II only
D. I and II only
E. I, II and III

If 0 < x < 1, then x > x^2 > x^3 > x^4 > x^5 ...

I. \(x^5 < x^3\). True.

II. \(x^4 + x^5 < x^3 + x^2\). Each term on the left hand side is less than each term on the right hand side, thus LHS < RHS.

III. \(x^4 - x^5 < x^2 - x^3\) --> \(x^4(1 - x) < x^2(1 - x)\). Since 0 < x < 1, then 1 - x > 0, so we can reduce by it: \(x^4 < x^2\). True.

Answer: E.

Hope it's clear.
_________________

Hi sinhap07,

Please find the algebraic approach for the question

I. \(x^5 <x^3\)

\(x^3(x^2 - 1) < 0\) i.e. \(x^3 (x -1) (x +1) < 0.\)

Using the wavy line method we know that for \(0 < x < 1\) the inequality holds true.



II.\(x^4 + x^5 < x^3 + x^2\)

\(x^5 - x^3 + x^4 - x^2 < 0\) i.e. \(x^2(x -1) (x + 1)^2 < 0\)

Using wavy line method we know that for \(0 < x < 1\) the inequality holds true



III.\(x^4 - x^5 < x^2 - x^3\)

\(x^4 - x^2 - x^5 + x^3 < 0\) i.e. \(x^2 (x + 1)(x - 1) (1 -x) < 0\)

\(x^2 (x -1)^2(x + 1) > 0\)

Using wavy line method we know that for \(0 < x < 1\) the inequality holds true



Thus all the three inequalities are true for the range \(0 < x < 1\)

Hope it's clear

Regards
Harsh
_________________

The answer is E

I: x^5-x^3 = x^3*(x-1)*(x+1) < 0 since x > 0 and x < 1
II: x^4+x^5-x^3-x^2 = x^2*(x+1)*(x-1)*(x+1) < 0 for same reasons
III: x^4-x^5-x^2+x^3 = x^2*(1-x)*(x-1)*(x+1) = -x^2*(x-1)^2*(x+1) < 0

My two cents:

I beeter if you divided by x^2: x^2+x^3<x+1 -> always
II better if you divided by x^2: x^2-x^3<1-x^2 always
III dividing by x^2: x^2-x^3<1-x^2
if x=0.1 thus 0.01-0.001<1-0.01 always
if x=0.9 thus 0.81-0.729<1-0.9 // 0.081<1 always

E

OA?

Cheers

I also believe the answer is E since even the third equation,
x^4 - x^5 < x^2 - x^3
= x^4 + x^3 < x^2 + x^5
=x^2 + x < 1+x^3
which is true since the value on the right hand side will be greater than 1.

thanks
SM

i will also go with E

raising a decimal number to any power reduces the value of the number.
so .1^2=.01
and .1^3=.001

can be solved by picking numbers

Bunuel

Can we do this question algebraically without substituting fractions? By trying to prove that each of the statements given indeed have x values within the range of 0 and 1 on the number line?

Bunuel

Can we do this question algebraically without substituting fractions? By trying to prove that each of the statements given indeed have x values within the range of 0 and 1 on the number line?[/quote]

Hi sinhap07,

Please find the algebraic approach for the question

I. \(x^5 <x^3\)

\(x^3(x^2 - 1) < 0\) i.e. \(x^3 (x -1) (x +1) < 0.\)

Using the wavy line method we know that for \(0 < x < 1\) the inequality holds true.



II.\(x^4 + x^5 < x^3 + x^2\)

\(x^5 - x^3 + x^4 - x^2 < 0\) i.e. \(x^2(x -1) (x + 1)^2 < 0\)

Using wavy line method we know that for \(0 < x < 1\) the inequality holds true



III.\(x^4 - x^5 < x^2 - x^3\)

\(x^4 - x^2 - x^5 + x^3 < 0\) i.e. \(x^2 (x + 1)(x - 1) (1 -x) < 0\)

\(x^2 (x -1)^2(x + 1) > 0\)

Using wavy line method we know that for \(0 < x < 1\) the inequality holds true



Thus all the three inequalities are true for the range \(0 < x < 1\)

Hope it's clear

Regards
Harsh[/quote]


Thanks Harsh for your reply. Can you please throw some more light on the wave method? Of what I knw, signs usually alternate but what I see in your workings, signs don't always alternate. Can you explain it?
Second, why are we ignoring the negative portions say in statement 1 that makes it less than -1. It shows x values dont only pertain to 0 and 1 region. So how can stmt 1 be sufficient?

Hi sinhap07,

You are right when you say signs usually alternate around zero points of the inequality but they only alternate when the expression is sign dependent i.e. for odd powers. For example sign of \(x^3\) is dependent on the sign of \(x\) but sign of \(x^2\) is independent of sign of \(x\). Hence for expressions, signs of which are independent of the signs of their base variable, the wave would not alternate but rather bounce back to the same region it is currently in.

I would also suggest you to go through this post on Wavy Line method which will help you understand it better.

For your second query, in st-I the range of \(x\) for which the inequality holds is \(0 < x < 1\) or \(x < -1\) as seen from the wavy line diagram. However as the question asks us if the inequality is true for the range \(0 < x < 1\), we are not concerned about the other possible values of x for which the inequality is true. Once the inequality satisfies the range \(0 < x < 1\), we have our answer.

Hope it's clear

Regards
Harsh
_________________

Ok Harsh, but by the same logic haven't been able to crack the question below? Can you help?

If –1 < x < 1 and x ≠ 0, which of the following inequalities must be true?
I. x3 < x
II. x2 < |x|
III. x4 – x5 > x3 – x2
(A) I only
(B) II only
(C) III only
(D) II and III only
(E) I, II, and III

Hi sinhap07,

Please refer to this post where Japinder has presented the algebraic approach using the wavy line method.

Request you to go through it and let us know if you have trouble at any point of the solution

Regards
Harsh
_________________

Ok Harsh, but by the same logic haven't been able to crack the question below? Can you help?

If –1 < x < 1 and x ≠ 0, which of the following inequalities must be true?
I. x3 < x
II. x2 < |x|
III. x4 – x5 > x3 – x2
(A) I only
(B) II only
(C) III only
(D) II and III only
(E) I, II, and III[/quote]

Hi sinhap07,

Please refer to this post where Japinder has presented the algebraic approach using the wavy line method.

Request you to go through it and let us know if you have trouble at any point of the solution

Regards
Harsh[/quote]

Hey Harsh

Saw Japinder's solution and posted some concerns. Haven't managed a reply yet. Pls find below my comments.

Japinder two concerns:
1. why have we not flipped the inequality sign for stmt 2 when changing the sign to negative?
2. for stmt 3, we have x values less than 1 till negative infinity. how can this be sufficient as the range we want it to be is from -1 to +1

Hi sinhap07

Please refer to the post here where Japinder has addressed your concerns.

Also restraint from posting doubts of a question into another question's thread for avoiding confusion to others.

Hope this helps

Regards
Harsh
_________________

I've started with I: just pick x=1/2 --> 1/32<1/8. Now look at the answer choices -which would be easier to test. As we already know "I" is correct, now, if we test III and it's correct, we don't need to test II anymore, because the only answer choice that has I and III is E.
If you manipulate III as stated in many solutions here, one can see that it must be true --> Answer E (we could save some time by picking III after I)

There is a clear indication that x is a positive decimal (fraction) Let us say 0.5
When a positive decimal is squared it becomes smaller than the original value 0.5>0.25
Every subsequent increase in exponent makes a decimal smaller than the earlier one 0.5>0.25>0.125

I. \(x^5 < x^3\) TRUE

II. \(x^4 + x^5 < x^3 + x^2\)
\(x^4(x+1) < x^2(x+1)\) cancel (x+1 from both side)
\(x^4< x^2\) TRUE

III. \(x^4 - x^5 < x^2 - x^3\)
\(x^4(1-x) < x^2(1-x)\) (cancel (1-x) from both side
\(x^4 < x^2\) TRUE

ANSWER IS E (I,II, and II)

first teo are obvious....
for 3rd....it is logical that...difference between X^4-X5 will be less than X^2-X^3.

When you get handed a gift . . .
I looked at the answer choices and decided to test option III, which is included only in Answer E. If option III must be true, the answer is E. (And if III is not part of the answer, given the answer choices, I would be in no worse shape than if I had started with option I.)

Let x = \(\frac{1}{2}\)

III. \(x^4 - x^5 < x^2 - x^3\)

\(\frac{1}{16} - \\
\frac{1}{32} < \frac{1}{4} - \frac{1}{8}\) ??

LHS: \(\frac{2}{32} - \frac{1}{32} = \frac{1}{32}\)

RHS: \(\frac{2}{8} - \frac{1}{8} = \frac{1}{8}\)

\(\frac{1}{32} < \frac{1}{8}\) TRUE

ANSWER E
_________________

0 < x < 1 => Improper Fractions. Property: x > x^2 > x^3 ...

1) x^5 < x^3 => MBT

2) x^4 (1+x) < x^2 (1+x)
Since 1+x > 0, we can cancel it off from both sides
x^4 < x^2 => MBT

3) x^4 (1-x) < x^2 (1-x)
Since 1-x > 0, we can cancel it off from both sides
x^4 < x^2 => MBT

ANSWER: E

Hey Guys

I have a quick Qs for option -
III. x4−x5<x2−x3

The Qs fails when I take x2 to LHS and x5 to RHS-

x4-x2 <x5-x3
x2(x2 -1) < x3(x2 -1)

So here x2 < x3 : wiz incorrect

Could anyone please explain this ?

Thanks in Advance!