Question 28

If $$x = 2026$$ and $$y = \frac{1}{1013}$$, then the value of $$\left( \frac{x^2 + 1}{y}\right)\left(\frac{y^2 + 1}{x}\right) + \left( \frac{x^2 - 1}{y}\right)\left( \frac{y^2 - 1}{x}\right)$$ is

Correct Answer: 5

Solution

Using $$(A + B)(C + D) + (A - B)(C - D) = 2(AC + BD)$$ with $$A = x^2$$, $$B = \frac{1}{y}$$, $$C = y^2$$, $$D = \frac{1}{x}$$, the expression equals $$2\left(x^2y^2 + \frac{1}{xy}\right)$$. Here $$xy = 2026 \times \frac{1}{1013} = 2$$, so $$x^2y^2 = 4$$ and $$\frac{1}{xy} = \frac{1}{2}$$. The value is $$2\left(4 + \frac{1}{2}\right) = 9$$.

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