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24 May 2021, 10:39
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
55% (01:56) correct
45%
(01:58)
wrong
based on 609
sessions
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Not Attempted Yet
The value of \(\frac{1}{x} - \frac{4}{\sqrt{x}} + 4\) is the least when x =
A. 1/6
B. 1/4
C. 1/2
D. 2/3
E. 1
A. 1/6
B. 1/4
C. 1/2
D. 2/3
E. 1
The E-gmat has mentioned the answer B as that leads to 0 as the value.
But using simple plug in method the answer for option A is -0.18
Can someone help me explain the issue here. and irrespective of the approach of algebric expression or plug in method the answer should be the same.
I have attached the photo of the question to clear the questions asked.
But using simple plug in method the answer for option A is -0.18
Can someone help me explain the issue here. and irrespective of the approach of algebric expression or plug in method the answer should be the same.
I have attached the photo of the question to clear the questions asked.
25 May 2021, 08:06
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Asked: The value of \(\frac{1}{x} - \frac{4}{\sqrt{x}} + 4\) is the least when x =
Let \(t = \frac{1}{\sqrt{x}}\)
\(E(x) = \frac{1}{x} - \frac{4}{\sqrt{x}}+ 4 = t^2 - 4t + 4 = (t-2)^2 \)
E(x) is least when t = 2
\(\frac{1}{\sqrt{x}} = 2\)
\(\sqrt{x} = \frac{1}{2}\)
\(x = \frac{1}{4}\)
IMO B
Let \(t = \frac{1}{\sqrt{x}}\)
\(E(x) = \frac{1}{x} - \frac{4}{\sqrt{x}}+ 4 = t^2 - 4t + 4 = (t-2)^2 \)
E(x) is least when t = 2
\(\frac{1}{\sqrt{x}} = 2\)
\(\sqrt{x} = \frac{1}{2}\)
\(x = \frac{1}{4}\)
IMO B
24 May 2021, 12:09
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abhiyoo
Let us get rid of the root term by substituting x = k^2, hence root(x) = k
Thus, the expression that we need to minimize, becomes:
\(\frac{1}{k^2} - \frac{4}{k} + 4 \)
\(= (\frac{1}{k^2}) * [1 - 4k + 4k^2] \)
\(= (\frac{1}{k^2}) * (2k - 1)^2\)
\(= [\frac{(2k - 1)}{k}]^2\)
\(= (2 - \frac{1}{k})^2\)
This is a square term. Hence, the minimum value of this cannot be negative; the best it can become is 0, when:
2 - 1/k = 0
=> 1/k = 2
=> k = 1/2, which is the same as Option B
