UPSC CDS II 2025 Mathematics Paper

Concept:When dividing a polynomial $P(x)$ by another polynomial $D(x)$, the remainder $R(x)$ must have degree less than that of $D(x)$. Here $D(x)=x^{2}+1$ (degree 2), so remainder is of the form $ax+b$.Explanation:Let $x^{6} = (x^{2}+1)Q(x) + (ax+b)$. Set $x^{2} = -1$ (since $x^{2}+1=0$). Then $x^{6} = (x^{2})^{3} = (-1)^{3} = -1$.Substitute into the equation: $-1 = (0)Q(x) + a x + b$ for any $x$ satisfying $x^{2}=-1$.Taking $x=i$ (the imaginary unit), we get $-1 = a i + b$. Comparing real and imaginary parts: $b = -1$, $a = 0$.Thus remainder is $0 \cdot x + (-1) = -1$.Answer:$-1$   (Option A)