UPSC NDA I 2024 Mathematics Paper

To find the value of $m n$, we start with the general term in the expansion of $\left(m x + \frac{1}{x}\right)^n$, given by the binomial theorem:$T_{k+1} = \binom{n}{k} (m x)^{n-k} \left( \frac{1}{x} \right)^k$We need to find the 4 th term in this expansion, so let $k=3$. Substituting $k=3$ into the general term formula, we get:$T_4 = \binom{n}{3} (m x)^{n-3} \left( \frac{1}{x} \right)^3$Simplify the expression:$T_4 = \binom{n}{3} m^{n-3} x^{n-3} \cdot \frac{1}{x^3}$Combine the exponents of $x$ :$T_4 = \binom{n}{3} m^{n-3} x^{n-3-3} = \binom{n}{3} m^{n-3} x^{n-6}$Given that the 4 th term is $\frac{5}{2}$, this implies:$\binom{n}{3} m^{n-3} x^{n-6} = \frac{5}{2}$For the term to be a constant (i.e., not dependent on $x$ ), the exponent of $x$ must be zero:$n-6=0 \Rightarrow n=6$Substitute $n=6$ into the equation for the 4th term:$T_4 = \binom{6}{3} m^{6-3} x^0 = \frac{5}{2}$ This simplifies to:$\binom{6}{3} m^3 = \frac{5}{2}$Calculate $\binom{6}{3}$ :$\binom{6}{3} = \frac{6!}{3!3!} = 20$Substitute $\binom{6}{3}$ into the equation:$20 m^3 = \frac{5}{2}$Solve for $m^3$ :$m^3 = \frac{5}{2} \times \frac{1}{20} = \frac{5}{40} = \frac{1}{8}$Taking the cube root of both sides, we find:$m = \frac{1}{2}$Finally, calculate $m n$ :$m n = \left( \frac{1}{2} \right) \times 6 = 3$