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26 Jul 2013, 15:58
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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:
15%
(low)
Question Stats:
75% (01:04) correct
25%
(01:09)
wrong
based on 493
sessions
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Which of the following describes all values of n for which \(n^2-1\geq{0}\)
(A) \(n\geq{1}\)
(B) \(n\leq{1}\)
(C) \(0\leq{n}\leq{1}\)
(D) \(n\leq{-1}\) or \(n\geq{1}\)
(E) \(-1\leq{n}\leq{1}\)]
Disclaimer: I have used the Search Box Before Posting. I used the first sentence of the question or a string of words exactly as they show up in the question below for my search. I did not receive an exact match for my question.
Source: Veritas Prep; Book 04
Chapter: Homework
Topic: Algebra
Question: 77
Question: Page 210
Edition: Third
(A) \(n\geq{1}\)
(B) \(n\leq{1}\)
(C) \(0\leq{n}\leq{1}\)
(D) \(n\leq{-1}\) or \(n\geq{1}\)
(E) \(-1\leq{n}\leq{1}\)]
Disclaimer: I have used the Search Box Before Posting. I used the first sentence of the question or a string of words exactly as they show up in the question below for my search. I did not receive an exact match for my question.
Source: Veritas Prep; Book 04
Chapter: Homework
Topic: Algebra
Question: 77
Question: Page 210
Edition: Third
Answer:
Equation: \(n^2-1\geq{0}\)
\((n+1).(n-1)\geq{0}\)
The above inequality can be broken down into the following two inequalities.
\((n+1) \geq{0}\) and \((n-1)\geq{0}\)
\((n+1) \geq{0}\)
\(n\geq{-1}\)
\((n-1) \geq{0}\)
\(n\geq{1}\)
The Official Answer is D. Why am i not getting the \(n\leq{-1}\) ? What am i doing wrong above in my calculation ?
Equation: \(n^2-1\geq{0}\)
\((n+1).(n-1)\geq{0}\)
The above inequality can be broken down into the following two inequalities.
\((n+1) \geq{0}\) and \((n-1)\geq{0}\)
\((n+1) \geq{0}\)
\(n\geq{-1}\)
\((n-1) \geq{0}\)
\(n\geq{1}\)
The Official Answer is D. Why am i not getting the \(n\leq{-1}\) ? What am i doing wrong above in my calculation ?
26 Jul 2013, 16:09
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hb
You are missing the case when both multiples are negative:
\((n+1) \leq{0}\) --> \(n\leq{-1}\);
\((n-1) \leq{0}\) --> \(n\leq{1}\).
Common range \(n\leq{-1}\).
This can be solved in another way:
\(n^2-1\geq{0}\) --> \(n^2\geq{1}\) --> \(n\leq{-1}\) or \(n\geq{1}\).
Or:
\(n^2-1\geq{0}\) --> \(n^2\geq{1}\) --> \(|n|\geq{1}\) --> \(n\leq{-1}\) or \(n\geq{1}\).
Answer: D.
Solving inequalities:
x2-4x-94661.html#p731476
inequalities-trick-91482.html
data-suff-inequalities-109078.html
range-for-variable-x-in-a-given-inequality-109468.html
everything-is-less-than-zero-108884.html
graphic-approach-to-problems-with-inequalities-68037.html
inequations-inequalities-part-154664.html
inequations-inequalities-part-154738.html
Hope it helps.
General Discussion
29 Oct 2015, 02:55
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I tried to solve it using this method.But not able to figure out where i went wrong.any help would be much appreciated.
n^2– 1 ≥ 0
(n+1)(n-1)≥0
n+1≥0 therefore n≥-1 -->1
n-1≥0 therefore n≥1 -->2
Combining 1 & 2, n≥1 is my solution but its wrong as per the official answer.
n^2– 1 ≥ 0
(n+1)(n-1)≥0
n+1≥0 therefore n≥-1 -->1
n-1≥0 therefore n≥1 -->2
Combining 1 & 2, n≥1 is my solution but its wrong as per the official answer.
