UPSC NDA I 2024 Mathematics Paper

Concept:Simplify the given expression using algebraic identities and then find the coefficient of $x^{10}$ in the resulting polynomial.Explanation:Rewrite the second factor: $\left(2 - x^2 - \frac{1}{x^2}\right)^{-5}$.Observe that $2 - x^2 - \frac{1}{x^2} = -\left(x^2 + \frac{1}{x^2} - 2\right)$.But $x^2 + \frac{1}{x^2} - 2 = \left(x - \frac{1}{x}\right)^2$, so the expression becomes $-\left(x - \frac{1}{x}\right)^2$.Thus the second factor is $-\left(x - \frac{1}{x}\right)^2$ raised to the power $-5$, i.e., $-\left(x - \frac{1}{x}\right)^{-10}$.Rewrite $\left(x - \frac{1}{x}\right)^{-10} = \left(\frac{x^2 - 1}{x}\right)^{-10} = \left(\frac{x}{x^2 - 1}\right)^{10} = \frac{x^{10}}{(x^2 - 1)^{10}}$.Therefore the entire expression becomes $\left(1 - x^2\right)^{20} \times \left( - \frac{x^{10}}{(x^2 - 1)^{10}} \right)$.Note that $1 - x^2 = -(x^2 - 1)$, so $\left(1 - x^2\right)^{20} = (x^2 - 1)^{20}$ (since the exponent is even).Hence the product simplifies to $(x^2 - 1)^{20} \times \left( - \frac{x^{10}}{(x^2 - 1)^{10}} \right) = - x^{10} (x^2 - 1)^{10}$.The coefficient of $x^{10}$ in the product is the constant term (coefficient of $x^0$) in $-(x^2 - 1)^{10}$ multiplied by $x^{10}$.The constant term in $(x^2 - 1)^{10}$ is $(-1)^{10} = 1$ (since the expansion is $\sum_{k=0}^{10} \binom{10}{k} (x^2)^{10-k} (-1)^k$, the constant term occurs when $x^2$ term has exponent 0, i.e., $k=10$, giving $\binom{10}{10}(-1)^{10}=1$).Multiplying by the negative sign gives $-1$.Thus the coefficient of $x^{10}$ is $-1$.Answer:$\boxed{-1}$