COMEDK 2025 Afternoon Solved Paper

Concept:To evaluate the limit of a rational function with a square root, we factor and use rationalization to cancel the factor causing the zero denominator.Explanation:First, factor the denominator: $2x^2 + x - 3 = (x-1)(2x+3)$.Rewrite $\sqrt{x} - 1$ using its conjugate: $\sqrt{x} - 1 = \frac{x-1}{\sqrt{x}+1}$.Substitute these into the limit expression:$$L = \lim_{x\to1} \frac{\left(\frac{x-1}{\sqrt{x}+1}\right)(2x-3)}{(x-1)(2x+3)}.$$The factor $(x-1)$ cancels (since $x \neq 1$ during the limit process):$$L = \lim_{x\to1} \frac{2x-3}{(\sqrt{x}+1)(2x+3)}.$$Now substitute $x=1$ directly into the simplified expression:Numerator: $2(1)-3 = -1$.Denominator: $(\sqrt{1}+1)(2(1)+3) = (1+1)(5) = 2 \cdot 5 = 10$.Therefore, $L = \frac{-1}{10} = -\frac{1}{10}$.Answer:Option D: $-\frac{1}{10}$.