COMEDK 2025 Morning Solved Paper

Concept:Use rationalization and equate real and imaginary parts of a complex equation.Explanation:Given: $\frac{x-1}{3+i} + \frac{y-1}{3-i} = i$Rationalize each denominator:$\frac{x-1}{3+i} = \frac{(x-1)(3-i)}{(3+i)(3-i)} = \frac{(x-1)(3-i)}{10}$$\frac{y-1}{3-i} = \frac{(y-1)(3+i)}{(3-i)(3+i)} = \frac{(y-1)(3+i)}{10}$Add both fractions: $\frac{(x-1)(3-i)+(y-1)(3+i)}{10} = i$Multiply both sides by 10: $(x-1)(3-i)+(y-1)(3+i) = 10i$Expand each term:$(x-1)(3-i) = 3x-3 -i x + i = (3x-3) + i(1-x)$$(y-1)(3+i) = 3y-3 + i y - i = (3y-3) + i(y-1)$Sum real parts: $(3x-3)+(3y-3) = 3x+3y-6$Sum imaginary parts: $i(1-x + y-1) = i(y-x)$So equation becomes: $(3x+3y-6) + i(y-x) = 10i$Equate real and imaginary components:Real: $3x+3y-6 = 0 \implies x+y = 2$Imaginary: $y-x = 10$Solve the system: adding gives $2y = 12 \implies y=6$, then $x = 2-6 = -4$Thus $(y,x) = (6,-4)$.Answer:$(y,x) = (6,-4)$, which corresponds to Option C.