SSC CGL 2023 PYQ — Question 72

SSC CGL - 17 July 2023 - Shift 1 • Quantitative Aptitude • Q72

If x2+1x2x^2 + \frac{1}{x^2}= 7, and 0 < x < 1, then find the value of x2−1x2x^2 - \frac{1}{x^2}.

Show answer & explanation

Correct answer: D

Explanation

The correct answer is option 4 i.e. -3√5.

  x2+1x2x^2 + \frac{1}{x^2}= 7

Add 2 on both sides

⇒ x2+1x2x^2 + \frac{1}{x^2}+ 2 = 7 + 2

 ⇒ (x + 1/x)2 = 9

 ⇒ (x + 1/x) = 3............1

  x2+1x2x^2 + \frac{1}{x^2}= 7

Subtract 2 on both sides

 ⇒  x2+1x2x^2 + \frac{1}{x^2}- 2 = 7 - 2

 ⇒ (x - 1/x)2 = 5

 ⇒ (x - 1/x) = √5 (We have given 0 < x < 1 So, 1/x > x)

 ⇒ (x - 1/x) = -√5............2

Multiplying equation 1 and 2

(x + 1/x)(x - 1/x) = -3√5

 ⇒ x2−1x2x^2 - \frac{1}{x^2} = -3√5