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22 Feb 2023, 06:43
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If a data set A consists of all x such that \((x^2 - 5x + 5)^{(x^2 + 8x)} = 1\), then what is the median of A ?
A. \(-4\)
B. \(0.5\)
C. \(1\)
D. \(1.5\)
E. \(2.5\)
A. \(-4\)
B. \(0.5\)
C. \(1\)
D. \(1.5\)
E. \(2.5\)
22 Feb 2023, 06:43
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Official Solution:
If a data set A consists of all x such that \((x^2 - 5x + 5)^{(x^2 + 8x)} = 1\), then what is the median of A ?
A. \(-4\)
B. \(0.5\)
C. \(1\)
D. \(1.5\)
E. \(2.5\)
Three cases are possible for given equation to hold true:
CASE 1: the base is 1, because \(1^m=1\), for all \(m\)
\(x^2 - 5x + 5=1\);
\(x^2 - 5x + 4=0\);
\(x=1\) or \(x=4\).
CASE 2: the exponent is 0, because \(n^0=1\), for any nonzero \(n\).
\(x^2 + 8x=0\);
\(x(x + 8)=0\);
\(x=0\) or \(x=-8\).
CASE 3: the base is -1 and the exponent is even, because \((-1)^k=1\), for any even \(k\).
\(x^2 - 5x + 5=-1\);
\(x^2 - 5x + 6=0\);
\(x=2\) or \(x=3\).
We need to check for which of these values of \(x\), is the exponent even:
When \(x=2\), then the exponent, \(x^2 + 8x=20=even\). Valid.
When \(x=3\), then the exponent, \(x^2 + 8x=33=odd\). Not valid.
Therefore, data set A is {-8, 0, 1, 2, 4 } and its median is 1.
Answer: C
If a data set A consists of all x such that \((x^2 - 5x + 5)^{(x^2 + 8x)} = 1\), then what is the median of A ?
A. \(-4\)
B. \(0.5\)
C. \(1\)
D. \(1.5\)
E. \(2.5\)
Three cases are possible for given equation to hold true:
CASE 1: the base is 1, because \(1^m=1\), for all \(m\)
\(x^2 - 5x + 5=1\);
\(x^2 - 5x + 4=0\);
\(x=1\) or \(x=4\).
CASE 2: the exponent is 0, because \(n^0=1\), for any nonzero \(n\).
\(x^2 + 8x=0\);
\(x(x + 8)=0\);
\(x=0\) or \(x=-8\).
CASE 3: the base is -1 and the exponent is even, because \((-1)^k=1\), for any even \(k\).
\(x^2 - 5x + 5=-1\);
\(x^2 - 5x + 6=0\);
\(x=2\) or \(x=3\).
We need to check for which of these values of \(x\), is the exponent even:
When \(x=2\), then the exponent, \(x^2 + 8x=20=even\). Valid.
When \(x=3\), then the exponent, \(x^2 + 8x=33=odd\). Not valid.
Therefore, data set A is {-8, 0, 1, 2, 4 } and its median is 1.
Answer: C
