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-1<x<1 and x is different from zero
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05 May 2025, 14:21
05 May 2025, 14:21
Expert Reply
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Question Stats:
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100% (01:05) wrong based on 1 sessions
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\(-1<x<1\) and \(x \neq 0\)
Quantity A |
Quantity B |
\(x^3+x^5\) |
\( x^2+x^4 \) |
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Re: -1<x<1 and x is different from zero
[#permalink]
09 May 2025, 04:30
09 May 2025, 04:30
Expert Reply
Step 1: Simplify the Comparison
1. Factor Both Quantities:
Quantity A: $\(x^3\left(1+x^2\right)\)$
Quantity B: $\(x^2\left(1+x^2\right)\)$
2. Divide Both Sides by \($x^2\left(1+x^2\right)\)$ :
- Since $\(x^2>0\)$ and $\(1+x^2>0\)$ for all $\(x \neq 0\)$, the inequality direction remains unchanged.
- Simplified comparison: $x$ vs. 1 .
Step 2: Analyze for $\(-1<x<1\)$
1. Case 1: $\(0<x<1\)$
- $x<1$, so $\(x^3+x^5<x^2+x^4\)$.
- Example: $\(x=0.5\)$
- Quantity A: $\(0.125+0.03125=0.15625\)$
- Quantity B: $\(0.25+0.0625=0.3125 \rightarrow \mathbf{A}<\mathbf{B}\)$.
2. Case 2: $\(-1<x<0\)$
- $x<0$, making $\(x^3+x^5\)$ negative, while $\(x^2+x^4\)$ is positive.
- Example: $\(x=-0.5\)$
- Quantity A: $\(-0.125-0.03125=-0.15625\)$
- Quantity B: $\(0.25+0.0625=0.3125 \rightarrow \mathbf{A}<\mathbf{B}\)$.
Conclusion
For all $\(-1<x<1\)$ (excluding $x=0$ ), Quantity $\(\mathbf{B}\)$ is greater than Quantity A.
1. Factor Both Quantities:
Quantity A: $\(x^3\left(1+x^2\right)\)$
Quantity B: $\(x^2\left(1+x^2\right)\)$
2. Divide Both Sides by \($x^2\left(1+x^2\right)\)$ :
- Since $\(x^2>0\)$ and $\(1+x^2>0\)$ for all $\(x \neq 0\)$, the inequality direction remains unchanged.
- Simplified comparison: $x$ vs. 1 .
Step 2: Analyze for $\(-1<x<1\)$
1. Case 1: $\(0<x<1\)$
- $x<1$, so $\(x^3+x^5<x^2+x^4\)$.
- Example: $\(x=0.5\)$
- Quantity A: $\(0.125+0.03125=0.15625\)$
- Quantity B: $\(0.25+0.0625=0.3125 \rightarrow \mathbf{A}<\mathbf{B}\)$.
2. Case 2: $\(-1<x<0\)$
- $x<0$, making $\(x^3+x^5\)$ negative, while $\(x^2+x^4\)$ is positive.
- Example: $\(x=-0.5\)$
- Quantity A: $\(-0.125-0.03125=-0.15625\)$
- Quantity B: $\(0.25+0.0625=0.3125 \rightarrow \mathbf{A}<\mathbf{B}\)$.
Conclusion
For all $\(-1<x<1\)$ (excluding $x=0$ ), Quantity $\(\mathbf{B}\)$ is greater than Quantity A.
gmatclubot
Re: -1<x<1 and x is different from zero [#permalink]
09 May 2025, 04:30
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