Free Practice Question

PS · Problem Solving

705-805

Roots

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

Answer Choices

Correct answer marked below

A

-1 < x < 0

B

x > 1

C

|x| < - 1

D

|x| > 1

x < -1

Correct

See the full step-by-step explanation

You can see the correct answer above. Sign in for free to unlock the complete worked solution.

Sign in Create free account

Track your performance and improve

Get detailed analytics, unlock full explanations, and move up difficulty tiers as you practice.

Start for free Sign in

Explanation

Preview

Official Explanation

Official

We have

\sqrt[3]{x} < \sqrt[5]{x}

.

Raising both sides to the 15th power (an odd power, which preserves the inequality direction) gives:

[\sqrt[3]{x}]^{15} < [\sqrt[5]{x}]^{15}

x^5 < x^3

x^5 - x^3 < 0

x^3(x^2 - 1) < 0

x^3(1 - x^2) < 0

We analyze the sign of

x^3(1 - x^2)

:

Case 1: If

x < 0

, then

x^3 < 0

. For the product to be negative, we need

1 - x^2 > 0

(so that the second factor is positive).

1 - x^2 > 0 \implies x^2 < 1 \implies -1 < x < 1

.

Combining with

x < 0

gives

-1 < x < 0

.

Case 2: If

x > 0

, then

x^3 > 0

. For the product to be negative, we need

1 - x^2 < 0

(so that the second factor is negative).

1 - x^2 < 0 \implies x^2 > 1 \implies x > 1

or

x < -1

.

Combining with

x > 0

gives

x > 1

.

Thus, the inequality

\sqrt[3]{x} < \sqrt[5]{x}

holds when

-1 < x < 0

or

x > 1

.

Among the answer choices, only option E (

x < -1

) is not in the solution set. Wait — careful: The question asks which inequality, if true, is sufficient alone to show the given inequality. That means we need a condition that guarantees

\sqrt[3]{x} < \sqrt[5]{x}

.

From our solution, the valid ranges are

-1 < x < 0

and

x > 1

.

Check each option:

(A)

-1 < x < 0

→ This is exactly one valid range. Sufficient.

(B)

x > 1

→ This is exactly the other valid range. Sufficient.

(C)

|x| < -1

→ Impossible, since absolute value cannot be less than a negative number. Not sufficient.

(D)

|x| > 1

→ This means

x > 1

or

x < -1

. The part

x < -1

does NOT satisfy the inequality (as seen in case 1: for

x < -1

,

1 - x^2 < 0

and

x^3 < 0

, product positive). So this option is not sufficient alone.

(E)

x < -1

→ As just noted, this does NOT satisfy the inequality. Not sufficient.

Thus, both (A) and (B) individually are sufficient. However, the problem likely expects only one correct answer. Let’s re‑examine the explanation provided: It concludes "Only option E is in the desired range." That seems contradictory to our analysis.…

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

Answer Choices

See the full step-by-step explanation

Track your performance and improve

Official Explanation

Read the full explanation

No comments yet

What it tests

Common trap

Unlock full explanations

Which of the following inequalities, if true, is sufficient alone to show that x3<x5\sqrt[3]{x} < \sqrt[5]{x}3x​<5x​ ?

Answer Choices

See the full step-by-step explanation

Track your performance and improve

Official Explanation

Read the full explanation

No comments yet

What it tests

Common trap

Unlock full explanations

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