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04 Feb 2019, 11:47
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
61% (02:12) correct
39%
(02:48)
wrong
based on 150
sessions
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If \(a\) and \(b\) are constants such that \(\,\,{{ax} \over {{x^2} - 1}} + {b \over {x - 1}} = {{2x - 1} \over {{x^2} - 1}}\,\,\) for every \(x > 1\), what is the value of \(a - b\) ?
(A) -2
(B) -1
(C) 0
(D) 2
(E) 4
(A) -2
(B) -1
(C) 0
(D) 2
(E) 4
KanishkM
Joined: 09 Mar 2018
Last visit: 18 Dec 2021
Posts: 765
Own Kudos:
Given Kudos: 123
Location: India
Schools: DeGroote'21 Desautels '21 NUS '21
04 Feb 2019, 12:02
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fskilnik
GMATH practice question (Quant Class 13)
If \(a\) and \(b\) are constants such that \(\,\,{{ax} \over {{x^2} - 1}} + {b \over {x - 1}} = {{2x - 1} \over {{x^2} - 1}}\,\,\) for every \(x > 1\), what is the value of \(a - b\) ?
(A) -2
(B) -1
(C) 0
(D) 2
(E) 4
Source: https://www.gmath.net
If \(a\) and \(b\) are constants such that \(\,\,{{ax} \over {{x^2} - 1}} + {b \over {x - 1}} = {{2x - 1} \over {{x^2} - 1}}\,\,\) for every \(x > 1\), what is the value of \(a - b\) ?
(A) -2
(B) -1
(C) 0
(D) 2
(E) 4
Source: https://www.gmath.net
ax + b x + b = 2x -1 , for x > 1
a - b = ?
(a+b)x + b = 2x - 1
b = -1, then a+b = 2
a = 3
Now a - b = 3 +1 = 4
E
04 Feb 2019, 14:33
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fskilnik
GMATH practice question (Quant Class 13)
If \(a\) and \(b\) are constants such that \(\,\,{{ax} \over {{x^2} - 1}} + {b \over {x - 1}} = {{2x - 1} \over {{x^2} - 1}}\,\,\) for every \(x > 1\), what is the value of \(a - b\) ?
(A) -2
(B) -1
(C) 0
(D) 2
(E) 4
Source: https://www.gmath.net
\(? = a - b\)If \(a\) and \(b\) are constants such that \(\,\,{{ax} \over {{x^2} - 1}} + {b \over {x - 1}} = {{2x - 1} \over {{x^2} - 1}}\,\,\) for every \(x > 1\), what is the value of \(a - b\) ?
(A) -2
(B) -1
(C) 0
(D) 2
(E) 4
Source: https://www.gmath.net
\({{ax} \over {{x^2} - 1}} + {b \over {x - 1}} = {{2x - 1} \over {{x^2} - 1}}\,\,\,\,\, \Rightarrow \,\,\,\,\,{{ax + b\left( {x + 1} \right)} \over {{x^2} - 1}} = {{2x - 1} \over {{x^2} - 1}}\,\,\,\,\, \Rightarrow \,\,\,\,\,x\left( {a + b - 2} \right) + b + 1 = 0\,\,,\,\,{\rm{for}}\,\,{\rm{all}}\,\,x > 1\,\,\,\,\left( * \right)\)
\(\left( * \right)\,\,\,\, \Rightarrow \,\,\,\,\left\{ \matrix{\\
\,b + 1 = 0\,\,\,\, \Rightarrow \,\,\,\,b = - 1 \hfill \cr \\
\,a + b - 2 = 0 \hfill \cr} \right.\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,a = 3\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,? = 4\)
The correct answer is therefore (E).
We follow the notations and rationale taught in the GMATH method.
Regards,
Fabio.
