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Updated on: 16 Jun 2017, 15:11
Originally posted by Lolaergasheva on 19 Feb 2011, 09:34.
Last edited by Bunuel on 16 Jun 2017, 15:11, edited 2 times in total.
Last edited by Bunuel on 16 Jun 2017, 15:11, edited 2 times in total.
Renamed the topic and edited the question.
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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)!
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83% (01:21) correct
17%
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For all numbers x such that x ≠ 1, if g(x) is defined by \(g(x)= \frac{x^2 + 2}{x - 1}\), then \((\frac{1}{g(2)})*(\frac{1}{g(x)})=\)
A. \(\frac{6(x-1)}{x^2+2}\)
B. \(\frac{6(x^2+2)}{x-1}\)
C. \(\frac{x^2+2}{2(x-1)}\)
D. \(\frac{x-1}{6(x^2+2)}\)
E. \(\frac{x^2+2}{6(x-1)}\)
A. \(\frac{6(x-1)}{x^2+2}\)
B. \(\frac{6(x^2+2)}{x-1}\)
C. \(\frac{x^2+2}{2(x-1)}\)
D. \(\frac{x-1}{6(x^2+2)}\)
E. \(\frac{x^2+2}{6(x-1)}\)
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19 Feb 2011, 09:49
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For all numbers x such that x ≠ 1, if g(x) is defined by \(g(x)= \frac{x^2 + 2}{x - 1}\), then \((\frac{1}{g(2)})*(\frac{1}{g(x)})=\)
A. \(\frac{6(x-1)}{x^2+2}\)
B. \(\frac{6(x^2+2)}{x-1}\)
C. \(\frac{x^2+2}{2(x-1)}\)
D. \(\frac{x-1}{6(x^2+2)}\)
E. \(\frac{x^2+2}{6(x-1)}\)
Given: \(g(x)=\frac{x^2+2}{x-1}\) (so \(\frac{1}{g(x)}=\frac{x-1}{x^2+2}\)). Question: \(\frac{1}{g(2)}*\frac{1}{g(x)}=\)
\(g(2)=\frac{2^2+2}{2-1}=6\) --> \(\frac{1}{g(2)}*\frac{1}{g(x)}=\frac{1}{6}*\frac{x-1}{x^2+2}=\frac{x-1}{6(x^2+2)}\).
Answer: D.
equal_115.PNG [ 57.82 KiB | Viewed 22311 times ]
A. \(\frac{6(x-1)}{x^2+2}\)
B. \(\frac{6(x^2+2)}{x-1}\)
C. \(\frac{x^2+2}{2(x-1)}\)
D. \(\frac{x-1}{6(x^2+2)}\)
E. \(\frac{x^2+2}{6(x-1)}\)
Given: \(g(x)=\frac{x^2+2}{x-1}\) (so \(\frac{1}{g(x)}=\frac{x-1}{x^2+2}\)). Question: \(\frac{1}{g(2)}*\frac{1}{g(x)}=\)
\(g(2)=\frac{2^2+2}{2-1}=6\) --> \(\frac{1}{g(2)}*\frac{1}{g(x)}=\frac{1}{6}*\frac{x-1}{x^2+2}=\frac{x-1}{6(x^2+2)}\).
Answer: D.
Attachment:
equal_115.PNG [ 57.82 KiB | Viewed 22311 times ]
19 Feb 2011, 09:59
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Thank you for fast reply, but why E is not an answer. I mean why you reversed, took reciprocal. sorry for my English.
