Which of the following inequalities, if true, is sufficient alone to

Question

Which of the following inequalities, if true, is sufficient alone to show that \sqrt {x} < \sqrt {x} ? (A) -1 < x < 0 (B) x > 1 (C) |x| < - 1 (D) |x| > 1 (E) x < -1

nick1816 · Answer

\sqrt {x} < \sqrt {x} Raising to the odd power won't change the sign of inequality. {x}]^{15} < {x}]^{15} x^5 < x^3 x^3(1-x^2) <0 case 1 if x<0, then 1 - x^2>0 x<-1 case 2 if x >0, then 1-x^2<0 0

ScottTargetTestPrep · Answer

Raising each term to the 15th power, we have:

x^5 < x^3

Thus, x must be either less than -1 or between 0 and 1. Of the given answer choices, only x < -1 is sufficient to show that x^(1/3) < x^(1/5).

Answer: E

sayan640 · Answer

X^5 - x^3 < 0 
x^3(x+1) (x-1) <0

Attached is the wave line diagram ; only valid range is x < -1 and x is between 0 and 1 because in these two ranges only the inequality becomes valid i.e negative.  egmat

E works and is the answer.

MartyMurray · Answer

Which of the following inequalities, if true, is sufficient alone to show that \(\sqrt[3]{x} < \sqrt[5]{x}\)?

We could just check the answer choices.

(A) \(-1 < x < 0\)

When \(x\) is a fraction less than \(1\), \(\sqrt[5]{x}\) is further from \(0\) than \(\sqrt[3]{x}\).

So, if \(-1 < x < 0\), then \(\sqrt[3]{x} > \sqrt[5]{x}\).

Eliminate.

(B) \(x > 1\)

Clearly incorrect since, if \(x > 1\), then \(\sqrt[3]{x} > \sqrt[5]{x}\).

Eliminate.

(C) \(|x| < - 1\)

Impossible.

Eliminate.

(D) \(|x| > 1\)

As we saw for (A), if \(x\) is between \(-1\) and \(0\), \(\sqrt[3]{x} > \sqrt[5]{x}\).

Eliminate.

(E) \(x < -1\)

For all values of \(x\) such that \(x < -1\), \(\sqrt[5]{x}\) is closer to \(0\), in other words, less negative, than \(\sqrt[3]{x}\).

So, if \(x < -1\), then \(\sqrt[3]{x} < \sqrt[5]{x}\).

Keep.

Correct answer: E

KarishmaB · Answer

When we raise an inequality to an odd power, the inequality sign does not change. It stays the same. Think about it. If both are positiive, both remain positive and the smaller magnitude remains smaller. If both are negative, both remain negative and the smaller magnitude remains smaller. So what was less negative stays less negative. If one is positive and the other negative then their signs remain as such. So negative one remains smaller. \sqrt {x} < \sqrt {x} When we raise it to the power 15, we get x^5 < x^3 Now you can use inequalities or focus on the number line. Try cases from x > 1 and 0 < x < 1 If x = 2, this obviously does hold true. But if x = 1/2, then it does. Since it works in the range 0 < x < 1, it will be valid in the range x < -1 too since we are dealing with odd powers. Hence 0 < x < 1 or x < -1 (A) -1 < x < 0 Not in our range (B) x > 1 Not in our range (C) |x| < - 1 Absolute value cannot be negative. Ignore. (D) |x| > 1 x > 1 is not in our range so not acceptable (E) x < -1 This is within our range so if it is true, x^5 is less than x^3. Answer (E) Method 2: Alternatively, if you do not want to raise both sides to the power 15, simply use test cases of 2^15 and 1/2^15

Which of the following inequalities, if true, is sufficient alone to

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Which of the following inequalities, if true, is sufficient alone to

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