UPSC NDA I 2022 Mathematics Solved Paper

Concept:Solve a functional equation by creating a system of linear equations in $f(x)$ and $f\!\left(\frac{1}{x}\right)$.Eliminate $f\!\left(\frac{1}{x}\right)$ to isolate $f(x)$.Explanation:First, simplify the right-hand side using $(a+b)(a-b) = a^2 - b^2$.$(2x + \frac{1}{x})(2x - \frac{1}{x}) = 4x^2 - \frac{1}{x^2}$.Thus the given equation becomes:$4f(x) - f\!\left(\frac{1}{x}\right) = 4x^2 - \frac{1}{x^2}$   ...(1)Replace $x$ by $\frac{1}{x}$ in equation (1):$4f\!\left(\frac{1}{x}\right) - f(x) = \frac{4}{x^2} - x^2$   ...(2)Multiply equation (1) by 4:$16f(x) - 4f\!\left(\frac{1}{x}\right) = 16x^2 - \frac{4}{x^2}$   ...(3)Add equation (3) and equation (2):$[16f(x) - 4f(\frac{1}{x})] + [4f(\frac{1}{x}) - f(x)] = (16x^2 - \frac{4}{x^2}) + (\frac{4}{x^2} - x^2)$.The terms $-4f(\frac{1}{x})$ and $+4f(\frac{1}{x})$ cancel.Left side simplifies to $15f(x)$.Right side simplifies to $15x^2$ because $-\frac{4}{x^2} + \frac{4}{x^2} = 0$.So $15f(x) = 15x^2$.Divide both sides by 15: $f(x) = x^2$.Now substitute $x = 2$:$f(2) = 2^2 = 4$.Answer:Option D: $4$.