जर x 2 - 7x + 1 = 0, आणि 0 < x < 1 असेल, तर x 2 - \(\frac{1}{x^2}\) चे मूल्य किती आहे?

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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. \(२८ \sqrt{5}\)
  4. \(-28 \sqrt{5}\)

Answer (Detailed Solution Below)

Option 2 : \(-21 \sqrt{5}\)
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Detailed Solution

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वापरलेले सूत्र:

x + (1/x) = a

नंतर x - (1/x) = √(a 2 - 4)

गणना:

⇒ x 2 - 7x + 1 = 0

x ने भागल्यावर आपल्याला मिळते:

⇒ x - 7 + (1/x) = 0

⇒ x + (1/x) = 7

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

[येथे 0 < x < 1 त्यामुळे x - (1/x) < 0]

⇒ x 2 - (1/x 2 ) = [x - (1/x)] [x + (1/x)]

⇒ 7 x (-3√5)

⇒ -21√5

∴ बरोबर उत्तर आहे - 21√5.

चूक गुण

येथे 0 < x < 1

⇒ 1/x > 1

x - 1/x < 0

अशा प्रकारे, x 2 - (1/x 2 ) = [x - (1/x)] [x + (1/x)] < 0

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