x2 - 7x + 1 = 0, మరియు 0 < x < 1 అయితే, x2 - \(\frac{1}{x^2}\) విలువ ఎంత?

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SSC CGL 2023 Tier-I Official Paper (Held On: 19 Jul 2023 Shift 3)
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  1. \(21 \sqrt{5}\)
  2. \(-21 \sqrt{5}\)
  3. \(28 \sqrt{5}\)
  4. \(-28 \sqrt{5}\)

Answer (Detailed Solution Below)

Option 2 : \(-21 \sqrt{5}\)
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PYST 1: SSC CGL - English (Held On : 11 April 2022 Shift 1)
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Detailed Solution

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ఉపయోగించిన సూత్రం:

x + (1/x) = a

తర్వాత x - (1/x) = √(a2 - 4)

గణన :

⇒ x2 - 7x + 1 = 0

x ద్వారా భాగిస్తే మనకు లభిస్తుంది:

⇒ x - 7 + (1/x) = 0

⇒ x + (1/x) = 7

ఇప్పుడు, (x - 1/x) = -√(49 - 4) = - √45 = - 3√5

[ఇక్కడ 0 < x < 1 కాబట్టి x - (1/x) < 0]

⇒ x2 - (1/x2 ) = [x - (1/x)] [x + (1/x)]

⇒ 7 x (-3√5)

⇒ -21√5

∴ సరైన సమాధానం - 21√5.

 Mistake Points

ఇక్కడ 0 < x < 1

⇒ 1/x > 1

x - 1/x < 0

అందువలన, x2 - (1/x2 ) = [x - (1/x)] [x + (1/x)] < 0

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