If \((x^2+\frac{1}{x^2})=7\), and 0 < x < 1, find the value of \(x^2-\frac{1}{x^2} \).

Question

Question

This question was previously asked in

SSC CGL 2023 Tier-I Official Paper (Held On: 17 Jul 2023 Shift 1)

Answer 1

Given:

x2 + (1/x2) = 7

Formula used:

x² + (1/x²) = P

then x + (1/x) = √(P + 2)

and x - (1/x) = √(P - 2)

⇒ x2 - (1/x2) = {x + (1/x)} × {x - (1/x)}

Calculation:

x2 + (1/x2) = 7

⇒ x + (1/x) = √(7 + 2) = √9

⇒ x + (1/x) = 3

⇒ x - (1/x) = -√(7 - 2)

⇒ x - (1/x) = - √5 {0 < x < 1}

x2 - (1/x2) = {x + (1/x)} × {x - (1/x)}

⇒ 3 × (- √5)

∴ The correct answer is - 3√5.

Mistake Points

Please note that

0 < x < 1

so

1/x > 1

so

x + 1/x > 1

and

x - 1/x < 0 (because 0 < x < 1 and 1/x > 1 so x - 1/x < 0)

so

(x - 1/x)(x + 1/x) < 0

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