UPSC NDA I 2017 Mathematics Solved Paper

Concept:The problem involves function substitution and algebraic simplification to find the ratio of two function values.Explanation:Start with $f(x) = \frac{x}{x-1}$.Compute $f(a)$ by substituting $x=a$: $f(a) = \frac{a}{a-1}$.Compute $f(a+1)$: $f(a+1) = \frac{a+1}{(a+1)-1} = \frac{a+1}{a}$.Now evaluate the ratio $\frac{f(a)}{f(a+1)} = \frac{\frac{a}{a-1}}{\frac{a+1}{a}}$.Simplify by multiplying numerator and denominator: $\frac{a}{a-1} \times \frac{a}{a+1} = \frac{a^2}{(a-1)(a+1)}$.Recognize that $(a-1)(a+1) = a^2 - 1$, so the ratio becomes $\frac{a^2}{a^2 - 1}$.Observe that $f(a^2) = \frac{a^2}{a^2 - 1}$.Thus, $\frac{f(a)}{f(a+1)} = f(a^2)$.Compare with the given options to identify the correct match.Answer:The expression simplifies to $f(a^{2})$, which corresponds to option B.