UPSC NDA I and II 2020 Math Solved Paper

Concept:A rational function is a function of the form $\frac{P(x)}{Q(x)}$ where $P(x)$ and $Q(x)$ are polynomials and $Q(x) \neq 0$.After integration, the result must not contain logarithmic or irrational terms to be a rational function.Explanation:We start with $f(x) = \frac{3x^{2}+8-4k}{x}$.Simplify by dividing each term in the numerator by $x$:$$f(x) = 3x + \frac{8-4k}{x}$$Now integrate with respect to $x$:$$\int f(x) \, dx = \int \left( 3x + \frac{8-4k}{x} \right) dx$$$$= \frac{3x^{2}}{2} + (8-4k) \ln |x| + C$$For the result to be a rational function, the logarithmic term $(8-4k) \ln |x|$ must disappear.This occurs when the coefficient is zero: $8-4k = 0$.Solve for $k$: $8 = 4k$ gives $k = 2$.When $k = 2$, the integral becomes $\frac{3x^{2}}{2} + C$, which is a polynomial and hence a rational function.Answer:The value of $k$ is $2$, which corresponds to option C.