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x>x^2 and y=3-|x-2|
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13 May 2025, 00:25
13 May 2025, 00:25
Expert Reply
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Question Stats:
50% (01:29) correct
50% (01:27) wrong based on 12 sessions
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\(x>x^2\) and \(y=3-|x-2|\)
Quantity A |
Quantity B |
y |
2 |
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Re: x>x^2 and y=3-|x-2|
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13 May 2025, 20:05
13 May 2025, 20:05
1
for absolute value |x−2|,
x-2, if x>=2
2-x, if x<2
Now, from x>x^2
critical points of x=(0,1)
therefore, 0<x<1
Now,
y=3-2+x since x<2
the range of y would be 1<y<2
Note that, y must be less than 2.
Hence, B
x-2, if x>=2
2-x, if x<2
Now, from x>x^2
critical points of x=(0,1)
therefore, 0<x<1
Now,
y=3-2+x since x<2
the range of y would be 1<y<2
Note that, y must be less than 2.
Hence, B
Re: x>x^2 and y=3-|x-2|
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16 May 2025, 04:15
16 May 2025, 04:15
Expert Reply
We are given:
1. $\(x>x^2\)$ (which implies $\(0<x<1\)$ )
2. $\(y=3-|x-2|\)$
We need to compare:
- Quantity A: $y$
- Quantity B: 2
Step 1: Solve $\(x>x^2\)$
This inequality simplifies to:
$$
\(\begin{gathered}
x^2-x<0 \\
x(x-1)<0
\end{gathered}\)
$$
The solution is $\(0<x<1\)$.
Step 2: Analyze $\(y=3-|x-2|\)$
Given $\(0<x<1\)$, let's evaluate $\(|x-2|\)$ :
- Since $\(x<1, x-2<-1\)$, so $\(|x-2|=2-x\)$.
Thus:
$$
\(y=3-(2-x)=1+x\)
$$
Step 3: Determine the Range of $y$
Since $\(0<x<1\)$ :
$$
\(\begin{gathered}
1+0<y<1+1 \\
1<y<2
\end{gathered}\)
$$
Step 4: Compare $y$ and 2
From the range $\(1<y<2\)$ :
- $y$ is always less than 2 (since $y$ approaches but never reaches 2 ).
Verification with Examples
1. For $\(x=0.5\)$ :
$$
\(y=3-|0.5-2|=3-1.5=1.5\)
$$
$$
\(1.5<2 \rightarrow A<B .\)
$$
2. For $\(x \rightarrow 1^{-}\)$:
$$
\(y \rightarrow 3-|1-2|=2^{-}\)
$$
Conclusion
For all valid $\(x(0<x<1)\)$, Quantity A ( $\(y\)$ ) is less than Quantity B (2).
1. $\(x>x^2\)$ (which implies $\(0<x<1\)$ )
2. $\(y=3-|x-2|\)$
We need to compare:
- Quantity A: $y$
- Quantity B: 2
Step 1: Solve $\(x>x^2\)$
This inequality simplifies to:
$$
\(\begin{gathered}
x^2-x<0 \\
x(x-1)<0
\end{gathered}\)
$$
The solution is $\(0<x<1\)$.
Step 2: Analyze $\(y=3-|x-2|\)$
Given $\(0<x<1\)$, let's evaluate $\(|x-2|\)$ :
- Since $\(x<1, x-2<-1\)$, so $\(|x-2|=2-x\)$.
Thus:
$$
\(y=3-(2-x)=1+x\)
$$
Step 3: Determine the Range of $y$
Since $\(0<x<1\)$ :
$$
\(\begin{gathered}
1+0<y<1+1 \\
1<y<2
\end{gathered}\)
$$
Step 4: Compare $y$ and 2
From the range $\(1<y<2\)$ :
- $y$ is always less than 2 (since $y$ approaches but never reaches 2 ).
Verification with Examples
1. For $\(x=0.5\)$ :
$$
\(y=3-|0.5-2|=3-1.5=1.5\)
$$
$$
\(1.5<2 \rightarrow A<B .\)
$$
2. For $\(x \rightarrow 1^{-}\)$:
$$
\(y \rightarrow 3-|1-2|=2^{-}\)
$$
Conclusion
For all valid $\(x(0<x<1)\)$, Quantity A ( $\(y\)$ ) is less than Quantity B (2).
