​Match the LIST–I with LIST–II:​

​LIST–I (Types of knowledge)​	​LIST–II (Assessment Literacy)​
A. $$\text{If, } x + \frac{1}{x} = 2, \quad \text{then } \left( x - \frac{1}{x} \right) = ?$$​​	I. 11
B. $$\text{If, } x - y = 1 \text{ and } x^2 + y^2 = 41, \quad \text{then } x + y = ?$$​​	II. 2
C. $$\text{If, } \left(1 - \frac{1}{x}\right) = 2, \quad \text{then } 1 + \frac{1}{x^2} = ?$$​​	III. 0
D. $$\text{If, } \left(x - \frac{1}{x}\right) = 3, \quad \text{then } x^2 + \frac{1}{x^2} = ?$$​​	IV. $$\pm 9$$​​

​Choose the correct answer from the options given below:​

Correct option is D

Solution:

$$\textbf{Problem A:} \\\ \\\text{Given: } x + \frac{1}{x} = 2 \\\ \\\text{Find: } \left( x - \frac{1}{x} \right) = ? \\\ \\\text{Solution:} \\\ \\(x + \frac{1}{x})^2 = 2^2 \\\ \\x^2 + 2 \cdot x \cdot \frac{1}{x} + \frac{1}{x^2} = 4 \\\ \\x^2 + 2 + \frac{1}{x^2} = 4 \\\ \\x^2 + \frac{1}{x^2} = 2 \\\ \\(x - \frac{1}{x})^2 = x^2 - 2 + \frac{1}{x^2} = (x^2 + \frac{1}{x^2}) - 2 = 2 - 2 = 0 \\\ \\x - \frac{1}{x} = 0 \\\ \\\text{Answer: } 0 \\\ \\\boxed{\text{Match: A} \to \text{III}}\\\ \\\textbf{Problem B:} \\\ \\\text{Given: } x - y = 1, \quad x^2 + y^2 = 41 \\\ \\\text{Find: } x + y = ? \\\text{Solution:} \\\ \\(x - y)^2 = 1^2 \\\ \\x^2 - 2xy + y^2 = 1 \\\ \\(x^2 + y^2) - (x^2 - 2xy + y^2) = 41 - 1 \\\ \\2xy = 40 \Rightarrow xy = 20 \\\ \\(x + y)^2 = x^2 + 2xy + y^2 = (x^2 + y^2) + 2xy = 41 + 40 = 81 \\\ \\x + y = \pm 9 \\\ \\\text{Answer: } \pm 9 \\\ \\\boxed{\text{Match: B} \to \text{IV}}$$

​Problem C:

Given:
1 − (1/x) = 2

Find:
1 + (1/x²) = ?

Solution:

1 − (1/x) = 2

−(1/x) = 1 → (1/x) = −1 → x = −1

(1/x²) = (−1)² = 1

1 + (1/x²) = 1 + 1 = 2

Answer: 2

Match: C → II

Problem D:

Given:
x − (1/x) = 3

Find:
x² + (1/x²) = ?

Solution:
(x − (1/x))² = 3²

x² − 2 + (1/x²) = 9

​$$\frac{1}{x^2} = 9 + 2 = 11 \\\text{Answer: } 11 \\\ \\\boxed{\text{Match: D} \to \text{I}}\\$$​​

​Final Matching:
D) A–III, B–IV, C–II, D–I