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16 Sep 2014, 00:22
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
58% (01:08) correct
42%
(01:16)
wrong
based on 836
sessions
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Date
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Not Attempted Yet
If \(j \ne 0\), what is the value of \(j\) ?
(1) \(|j| = j^{-1}\)
(2) \(j^j = 1\)
(1) \(|j| = j^{-1}\)
(2) \(j^j = 1\)
16 Sep 2014, 00:22
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Official Solution:
If \(j \ne 0\), what is the value of \(j\) ?
(1) \(|j| = j^{-1}\):
\(|j| = \frac{1}{j}\);
\(|j|*j=1\).
Notice that \(j\) cannot be negative since in this case we would have \(|j|*j=\text{positive}*\text{negative}=\text{negative} \ne 1\). Therefore, \(j\) is positive. \(j > 0\) implies that \(|j| = j\), hence we get:
\(j^2=1\);
Since we know that \(j\) is positive, then \(j=1\).
Sufficient.
(2) \(j^j = 1\). Again only one solution: \(j=1\). Sufficient.
Answer: D
If \(j \ne 0\), what is the value of \(j\) ?
(1) \(|j| = j^{-1}\):
\(|j| = \frac{1}{j}\);
\(|j|*j=1\).
Notice that \(j\) cannot be negative since in this case we would have \(|j|*j=\text{positive}*\text{negative}=\text{negative} \ne 1\). Therefore, \(j\) is positive. \(j > 0\) implies that \(|j| = j\), hence we get:
\(j^2=1\);
Since we know that \(j\) is positive, then \(j=1\).
Sufficient.
(2) \(j^j = 1\). Again only one solution: \(j=1\). Sufficient.
Answer: D
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19 Sep 2014, 01:57
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As per ii) j^j = 1.
So 1 can also be equal to j^0.
Therefore, j^j = j^0. so same base power is equated which brings us to: j=0.
But 0^0 = 0 so that is not our solution.
hence j=1, as any number raised to same power if = 1 then that number is 1.
Am i right in assuming the following rule (highlighted one) based on the above calculations ?
Kindly shed some light on it.
So 1 can also be equal to j^0.
Therefore, j^j = j^0. so same base power is equated which brings us to: j=0.
But 0^0 = 0 so that is not our solution.
hence j=1, as any number raised to same power if = 1 then that number is 1.
Am i right in assuming the following rule (highlighted one) based on the above calculations ?
Kindly shed some light on it.
