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06 Sep 2022, 06:44
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A
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Difficulty:
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Question Stats:
35% (02:37) correct
65%
(02:45)
wrong
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If n is an integer such that \(\frac{1}{9} < \frac{1}{(n^2-1)} < \frac{1}{2}\), what is the value of n?
(1) 1/3 > 1/(1 - n) > 1/7
(2) n is not an even integer.
(1) 1/3 > 1/(1 - n) > 1/7
(2) n is not an even integer.
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06 Sep 2022, 09:48
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Asked: If n is an integer such that \(\frac{1}{9} < \frac{1}{(n^2-1)} < \frac{1}{2}\), what is the value of n?
1/(n^2-1) > 1/9
1/(n^2 - 1) - 1/9 > 0
{9 - (n^2 - 1)}/9(n^2 - 1) > 0
(10 - n^2)/(n^2 - 1) > 0
\(\frac{(n + \sqrt{10})(n-\sqrt{10})}{(n-1)(n+1)} < 0\)
\(-4 < -\sqrt{10} < n < -1\) or
\(1 < n < \sqrt{10} < 4\)
n = {-3,-2,2,3}
1/(n^2 - 1) < 1/2
1/(n^2 - 1) - 1/2 < 0
{2 - (n^2 - 1)}/2(n^2 - 1) < 0
(3 - n ^2)/(n^2 - 1) < 0
(n^2 - 3)/(n^2 - 1) > 0
\(n < - \sqrt{3}\) or
-1 < n < 1 or
\(n > \sqrt{3}\)
n = {-3-2,2,3}
(1) 1/3 > 1/(1 - n) > 1/7
-7 < n - 1 < -3
-6 < n < -2
n = -3
SUFFICIENT
(2) n is not an even integer.
n = -3 or 3
NOT SUFFICIENT
IMO A
1/(n^2-1) > 1/9
1/(n^2 - 1) - 1/9 > 0
{9 - (n^2 - 1)}/9(n^2 - 1) > 0
(10 - n^2)/(n^2 - 1) > 0
\(\frac{(n + \sqrt{10})(n-\sqrt{10})}{(n-1)(n+1)} < 0\)
\(-4 < -\sqrt{10} < n < -1\) or
\(1 < n < \sqrt{10} < 4\)
n = {-3,-2,2,3}
1/(n^2 - 1) < 1/2
1/(n^2 - 1) - 1/2 < 0
{2 - (n^2 - 1)}/2(n^2 - 1) < 0
(3 - n ^2)/(n^2 - 1) < 0
(n^2 - 3)/(n^2 - 1) > 0
\(n < - \sqrt{3}\) or
-1 < n < 1 or
\(n > \sqrt{3}\)
n = {-3-2,2,3}
(1) 1/3 > 1/(1 - n) > 1/7
-7 < n - 1 < -3
-6 < n < -2
n = -3
SUFFICIENT
(2) n is not an even integer.
n = -3 or 3
NOT SUFFICIENT
IMO A
07 Sep 2022, 02:45
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Please tell me if I'm wrong.
n is an integer
1/9<1/(n^2-1)<1/2 => 9>(n^2-1)>2
(a) 9>(n^2-1) => 10>n^2
(b) (n^2-1)>2 => n^2>3
=> 10>n^2>3
=> n^2 can only be 4 or 9 => n=-2, 2, -3, 3
Let's go to the options:
(1)1/3 > 1/(1 - n) > 1/7 => 3<1-n<7 => -2<n<-6
=> n=-3 => sufficient
(2) => not sufficient because we already solved that n can be integer from the beginning
n is an integer
1/9<1/(n^2-1)<1/2 => 9>(n^2-1)>2
(a) 9>(n^2-1) => 10>n^2
(b) (n^2-1)>2 => n^2>3
=> 10>n^2>3
=> n^2 can only be 4 or 9 => n=-2, 2, -3, 3
Let's go to the options:
(1)1/3 > 1/(1 - n) > 1/7 => 3<1-n<7 => -2<n<-6
=> n=-3 => sufficient
(2) => not sufficient because we already solved that n can be integer from the beginning
