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# What is the minimum value of the expression

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e-GMAT Representative
Joined: 04 Jan 2015
Posts: 3019
What is the minimum value of the expression  [#permalink]

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31 Jul 2019, 05:17
00:00

Difficulty:

85% (hard)

Question Stats:

38% (01:40) correct 63% (02:33) wrong based on 40 sessions

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Algerba Question Series- Question 3

What is the minimum value of the expression $$\frac{1}{x^2}$$– $$\frac{1}{2x}$$ + $$\frac{5}{4}$$?

A. $$\frac{-5}{4}$$
B. $$\frac{-19}{16}$$
C. 0
D. $$\frac{19}{16}$$
E. $$\frac{5}{4}$$

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Joined: 03 Jun 2019
Posts: 954
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What is the minimum value of the expression  [#permalink]

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31 Jul 2019, 08:41
1
EgmatQuantExpert wrote:

Algerba Question Series- Question 3

What is the minimum value of the expression $$\frac{1}{x^2}$$– $$\frac{1}{2x}$$ + $$\frac{5}{4}$$?

A. $$\frac{-5}{4}$$
B. $$\frac{-19}{16}$$
C. 0
D. $$\frac{19}{16}$$
E. $$\frac{5}{4}$$

What is the minimum value of the expression $$\frac{1}{x^2}– \frac{1}{2x} + \frac{5}{4}$$
$$t^2– \frac{t}{2} + \frac{5}{4}$$ where $$t=\frac{1}{x}$$
$$t^2 - \frac{2t}{4} + \frac{1}{16} +\frac{5}{4}-\frac{1}{16}$$
$$=(t-\frac{1}{4})^2 + \frac{19}{16}$$
=\frac{19}{16} since minimum value of expression is when $$(t-\frac{1}{4})^2 =0$$

IMO D
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Re: What is the minimum value of the expression  [#permalink]

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05 Aug 2019, 22:01

Solution

Given:
• An expression $$\frac{1}{x^2} – \frac{1}{2x} + \frac{5}{4}$$

To find:
• The minimum value of the expression $$\frac{1}{x^2} – \frac{1}{2x} + \frac{5}{4}$$

Approach and Working Out:
• $$\frac{1}{x^2} – \frac{1}{2x} + \frac{5}{4} = (\frac{1}{x})^2 – 2 * \frac{1}{x} * \frac{1}{4} + (\frac{1}{4})^2 + \frac{19}{16} = (\frac{1}{x} – \frac{1}{4})^2 + \frac{19}{16}$$

Therefore, the minimum value of the expression $$\frac{1}{x^2} – \frac{1}{2x} + \frac{5}{4} = 0 + \frac{19}{16} = \frac{19}{16}$$

Hence, the correct answer is Option D.

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Re: What is the minimum value of the expression   [#permalink] 05 Aug 2019, 22:01
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