UPSC CDS 2 2021 Math Paper

Concept:Subtract the two equations to eliminate $A$ and solve for $B$.Explanation:We are given:$A + B = \frac{x^2 - 8}{x + 2}$ and $A - B = \frac{-x^2 + 2x + 4}{x + 2}$.Subtract the second equation from the first:$(A + B) - (A - B) = \frac{x^2 - 8}{x + 2} - \frac{-x^2 + 2x + 4}{x + 2}$.This simplifies to $2B = \frac{x^2 - 8 + x^2 - 2x - 4}{x + 2}$.So $2B = \frac{2x^2 - 2x - 12}{x + 2} = \frac{2(x^2 - x - 6)}{x + 2}$.Divide both sides by 2: $B = \frac{x^2 - x - 6}{x + 2}$.Factor the numerator: $x^2 - x - 6 = (x - 3)(x + 2)$.Thus $B = \frac{(x - 3)(x + 2)}{x + 2} = x - 3$, provided $x \neq -2$.Now check which option equals $x - 3$:• Option A: $\frac{x^2 - 4}{x^2 + 4x + 4} \neq x - 3$.• Option B: $\frac{x^2 - 4}{x^2 - 4x + 4} \neq x - 3$.• Option C: $\frac{2x^2 - 7x + 3}{2x - 1}$. Factor numerator: $2x^2 - 7x + 3 = (2x - 1)(x - 3)$. So $\frac{(2x - 1)(x - 3)}{2x - 1} = x - 3$ (if $x \neq \frac12$). This matches $B$.• Option D: $\frac{2x^2 + 7x - 3}{2x - 1}$ does not simplify to $x - 3$.Answer:Option C.