SSC Selection Post 2024 PYQ — Question 54

SSC Selection Post - Higher Secondary Level - 20 Jun 2024 - Shift 2 • Quantitative Aptitude • Q54

if x2∔1x2=119(x>0)^{x^2}\dotplus \frac{1}{x^2}= 119\left ( x> 0 \right ) , then the value of (x3∔1x3)\left ( x^3\dotplus \frac{1}{x^3} \right )is:

Show answer & explanation

Correct answer: B

Explanation

The correct answer is Option 2 i.e. 1298.

x2 + 1/x2 = 119

⇒ x + 1/x = √(119 + 2) = √121 = 11

So, x3 + 1/x3 = 113 - 3(11) = 1331 - 33 = 1298