यदि x- 7x + 1 = 0, और 0 < x < 1,  x\(\frac{1}{x^2}\) का मान क्या है?

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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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Detailed Solution

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प्रयुक्त सूत्र:

x + (1/x) = a

तब x - (1/x) = √(a2 - 4)

गणना: 

⇒ x2 - 7x + 1 = 0

x से विभाजित करने पर हमें प्राप्त होता है:

⇒ x - 7 + (1/x) = 0

⇒ x + (1/x) = 7

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

[यहाँ 0 < x < 1 अत: x - 1/x < 0] 

⇒ x2 - (1/x2) = (x - 1/x)(x + 1/x)

⇒ 7 × (-3√5)

⇒ -21√5

∴ सही उत्तर - 21√5 है।

Mistake Points

यहाँ 0 < x < 1

⇒ 1/x > 1

⇒ x - 1/x < 0

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

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