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22 May 2020, 07:53
Kudos
Bookmarks
E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
58% (01:12) correct
42%
(01:31)
wrong
based on 1494
sessions
History
Date
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Not Attempted Yet
\(\frac{x + 1}{x^2 - 1} - \frac{x - 2}{x^2 - 4x + 4}\)
The values of x for which the expression above is NOT defined are
A. 1 and 2 only
B. –1 and 2 only
C. –1 and –2 only
D. –1, 1, and –2
E. –1, 1, and 2
PS21172
The values of x for which the expression above is NOT defined are
A. 1 and 2 only
B. –1 and 2 only
C. –1 and –2 only
D. –1, 1, and –2
E. –1, 1, and 2
PS21172
15 Jun 2020, 05:40
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Kritisood
GMATinsight
\(\frac{x + 1}{x^2 - 1} - \frac{x - 2}{x^2 - 4x + 4}\)
if i solve this further i get
\(\frac{x+1}{(x+1)(x-1)} - \frac{(x-2)}{(x-2)^2}\)
cancelling like terms
\(\frac{1}{x-1} - \frac{1}{x-2}\)
now the only values that give denominator 0 are 1 and 2
i know im missing -1 here, but could you help with where I went wrong? which term did I cancel that I wasn't supposed to cancel and why?
\(\frac{x + 1}{x^2 - 1} - \frac{x - 2}{x^2 - 4x + 4}\)
if i solve this further i get
\(\frac{x+1}{(x+1)(x-1)} - \frac{(x-2)}{(x-2)^2}\)
cancelling like terms
\(\frac{1}{x-1} - \frac{1}{x-2}\)
now the only values that give denominator 0 are 1 and 2
i know im missing -1 here, but could you help with where I went wrong? which term did I cancel that I wasn't supposed to cancel and why?
Kritisood
Canceling Like terms is acceptable ONLY
- If the term to be canceled is NON-ZERO
You are canceling the term (x+1) without knowing whether it's NON-ZERO - \(This is the mistake\)
Since that term also may be 0 bring the situation of \(\frac{0}{0}\) therefore it will be WRONG to cancel it.
SImilar example
\(x^2-2x = 0\) does NOT mean \(
\(x^2-2x = 0\) means \(x*(x-2) = 0\) i.e. x = 0 or 2
I hope this help!
22 May 2020, 08:11
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Bunuel
\(\frac{x + 1}{x^2 - 1} - \frac{x - 2}{x^2 - 4x + 4}\)
The values of x for which the expression above is NOT defined are
A. 1 and 2 only
B. –1 and 2 only
C. –1 and –2 only
D. –1, 1, and –2
E. –1, 1, and 2
PS21172
The values of x for which the expression above is NOT defined are
A. 1 and 2 only
B. –1 and 2 only
C. –1 and –2 only
D. –1, 1, and –2
E. –1, 1, and 2
PS21172
Key concept: A rational expression (aka fraction) is undefined when the denominator equal zero
So let's check each denominator individually and see what values of x make that denominator equal to zero.
If \(x^2 - 1 = 0\), then \(x^2 = 1\), which means \(x=1\) and \(x=-1\)
In other words, the expression is not defined when \(x=1\) and when \(x=-1\)
If \(x^2 - 4x + 4 = 0\), then we can factor the left side to get: \((x-2)(x-2)= 0\)
This means \(x = 2\)
So, the expression is not defined when \(x=2\)
So the expression is undefined when \(x=1\), \(x=-1\) and \(x=2\)
Answer: E
Cheers,
Brent
