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abhiyoo
The value of \(\frac{1}{x} - \frac{4}{\sqrt{x}} + 4\) is the least when x =

A. 1/(6^1/2)
B. 1/(4^1/2)
C. 1/(2^1/2)
D. 1/2
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.

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

But k=sqrt(x), so x should be 1/4 and that option is not in the answers...
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abhiyoo
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

The use of 4, which is 2^2 and x, which is \(\sqrt{x}^2\) should tell you that we are looking at the formula (a-b)^2

\(\frac{1}{x} - \frac{4}{\sqrt{x}} + 4\)

\(\sqrt{\frac{1}{x}}^2 -2*2 \frac{1}{\sqrt{x}} + 2^2\)

\(( \frac{1}{\sqrt{x}} -2)^2\)

The least value of a square is 0, and that will happen when \(( \frac{1}{\sqrt{x}} =2)\)

Squaring both sides,

\(\frac{1}{x}=4\)

\(x=\frac{1}{4}\)

B
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