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What is the minimum value of the expression

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What is the minimum value of the expression  [#permalink]

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New post 31 Jul 2019, 05:17
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D
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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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What is the minimum value of the expression  [#permalink]

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New post 31 Jul 2019, 08:41
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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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New post 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.

Answer: D

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