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which of the following has the least value?
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16 May 2021, 02:31
16 May 2021, 02:31
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If \(n \leq -5\) , which of the following has the least value?
A. \(\frac{1}{(n+3)^2
}\)
B. \(- \frac{1}{(n+3)^2
}\)
C. \(\frac{1}{n+3}
\)
D. \(\frac{1}{n+4}\)
E. \(\frac{1}{n-4}\)
A. \(\frac{1}{(n+3)^2
}\)
B. \(- \frac{1}{(n+3)^2
}\)
C. \(\frac{1}{n+3}
\)
D. \(\frac{1}{n+4}\)
E. \(\frac{1}{n-4}\)
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which of the following has the least value?
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16 May 2021, 05:46
16 May 2021, 05:46
1
The easiest way to solve this problem is to substitute a value of n which satisfies \(n \leq -5\)
Lets take n = -5 and put in all the option choices
A. \(\frac{1}{(n+3)^2}\)
n = -5
=> \(\frac{1}{(n+3)^2}\) = \(\frac{1}{(-5+3)^2}\) = \(\frac{1}{(-2)^2}\) = \(\frac{1}{4}\) = 0.25
B. \(- \frac{1}{(n+3)^2}\)
n = -5
=> \(- \frac{1}{(n+3)^2}\) => this value is same as option A but with negative sign. So value will be -0.25
C. \(\frac{1}{n+3}\)
n=-5
=> \(\frac{1}{n+3}\) = \(\frac{1}{-5+3}\) = \(\frac{1}{-2}\) = -\(\frac{1}{2}\) = -0.5
D. \(\frac{1}{n+4}\)
n=-5
=> \(\frac{1}{n+4}\) = \(\frac{1}{-5+4}\) = \(\frac{1}{-1}\) = -1
E. \(\frac{1}{n-4}\)
n=-5
=> \(\frac{1}{n-4}\) = \(\frac{1}{-5-4}\) = \(\frac{1}{-9}\) = -\(\frac{1}{9}\) = -0.11
Clearly, option D is the least value.
Hope it helps!
Lets take n = -5 and put in all the option choices
A. \(\frac{1}{(n+3)^2}\)
n = -5
=> \(\frac{1}{(n+3)^2}\) = \(\frac{1}{(-5+3)^2}\) = \(\frac{1}{(-2)^2}\) = \(\frac{1}{4}\) = 0.25
B. \(- \frac{1}{(n+3)^2}\)
n = -5
=> \(- \frac{1}{(n+3)^2}\) => this value is same as option A but with negative sign. So value will be -0.25
C. \(\frac{1}{n+3}\)
n=-5
=> \(\frac{1}{n+3}\) = \(\frac{1}{-5+3}\) = \(\frac{1}{-2}\) = -\(\frac{1}{2}\) = -0.5
D. \(\frac{1}{n+4}\)
n=-5
=> \(\frac{1}{n+4}\) = \(\frac{1}{-5+4}\) = \(\frac{1}{-1}\) = -1
E. \(\frac{1}{n-4}\)
n=-5
=> \(\frac{1}{n-4}\) = \(\frac{1}{-5-4}\) = \(\frac{1}{-9}\) = -\(\frac{1}{9}\) = -0.11
Clearly, option D is the least value.
Hope it helps!
Re: which of the following has the least value?
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16 May 2021, 09:56
16 May 2021, 09:56
Expert Reply
BrushMyQuant wrote:
The easiest way to solve this problem is to substitute a value of n which satisfies \(n \leq -5\)
Lets take n = -5 and put in all the option choices
A. \(\frac{1}{(n+3)^2}\)
n = -5
=> \(\frac{1}{(n+3)^2}\) = \(\frac{1}{(-5+3)^2}\) = \(\frac{1}{(-2)^2}\) = \(\frac{1}{4}\) = 0.25
B. \(- \frac{1}{(n+3)^2}\)
n = -5
=> \(- \frac{1}{(n+3)^2}\) => this value is same as option A but with negative sign. So value will be -0.25
C. \(\frac{1}{n+3}\)
n=-5
=> \(\frac{1}{n+3}\) = \(\frac{1}{-5+3}\) = \(\frac{1}{-2}\) = -\(\frac{1}{2}\) = -0.5
D. \(\frac{1}{n+4}\)
n=-5
=> \(\frac{1}{n+4}\) = \(\frac{1}{-5+4}\) = \(\frac{1}{-1}\) = -1
E. \(\frac{1}{n-4}\)
n=-5
=> \(\frac{1}{n-4}\) = \(\frac{1}{-5-4}\) = \(\frac{1}{-9}\) = -\(\frac{1}{9}\) = -0.11
Clearly, option D is the least value.
Hope it helps!
Lets take n = -5 and put in all the option choices
A. \(\frac{1}{(n+3)^2}\)
n = -5
=> \(\frac{1}{(n+3)^2}\) = \(\frac{1}{(-5+3)^2}\) = \(\frac{1}{(-2)^2}\) = \(\frac{1}{4}\) = 0.25
B. \(- \frac{1}{(n+3)^2}\)
n = -5
=> \(- \frac{1}{(n+3)^2}\) => this value is same as option A but with negative sign. So value will be -0.25
C. \(\frac{1}{n+3}\)
n=-5
=> \(\frac{1}{n+3}\) = \(\frac{1}{-5+3}\) = \(\frac{1}{-2}\) = -\(\frac{1}{2}\) = -0.5
D. \(\frac{1}{n+4}\)
n=-5
=> \(\frac{1}{n+4}\) = \(\frac{1}{-5+4}\) = \(\frac{1}{-1}\) = -1
E. \(\frac{1}{n-4}\)
n=-5
=> \(\frac{1}{n-4}\) = \(\frac{1}{-5-4}\) = \(\frac{1}{-9}\) = -\(\frac{1}{9}\) = -0.11
Clearly, option D is the least value.
Hope it helps!
Thank you Sir
Amazing explanation
And Yes the answer is D
