UPSC CDS 2 2020 Math Paper

Concept:The problem involves simplifying a complex algebraic fraction expression using factorization and finding a common denominator.Explanation:Step 1: Start with the given expression: $\frac{a^{2}+ac}{a^{2}c - c^{3}} - \frac{a^{2} - c^{2}}{a^{2}c + 2ac^{2} + c^{3}} - \frac{2c}{a^{2} - c^{2}} + \frac{3}{a+c}$ Step 2: Factor each denominator where possible. First term denominator: $a^{2}c - c^{3} = c(a^{2} - c^{2}) = c(a-c)(a+c)$ First term numerator: $a^{2}+ac = a(a+c)$ So first term becomes: $\frac{a(a+c)}{c(a-c)(a+c)} = \frac{a}{c(a-c)}$ Step 3: Second term denominator: $a^{2}c + 2ac^{2} + c^{3} = c(a^{2} + 2ac + c^{2}) = c(a+c)^{2}$ Second term numerator: $a^{2} - c^{2} = (a-c)(a+c)$ Thus second term becomes: $\frac{(a-c)(a+c)}{c(a+c)^{2}} = \frac{a-c}{c(a+c)}$ Step 4: Third term denominator: $a^{2} - c^{2} = (a-c)(a+c)$ So third term is: $\frac{2c}{(a-c)(a+c)}$ Now the expression is: $\frac{a}{c(a-c)} - \frac{a-c}{c(a+c)} - \frac{2c}{(a-c)(a+c)} + \frac{3}{a+c}$ Step 5: Combine all terms over a common denominator $c(a-c)(a+c)$: $\frac{a(a+c) - (a-c)^{2} - 2c^{2} + 3c(a-c)}{c(a-c)(a+c)}$ Step 6: Expand the numerator: $a(a+c) = a^{2} + ac$ $(a-c)^{2} = a^{2} - 2ac + c^{2}$ $3c(a-c) = 3ca - 3c^{2}$ So numerator becomes: $a^{2} + ac - (a^{2} - 2ac + c^{2}) - 2c^{2} + 3ca - 3c^{2}$ = $a^{2} + ac - a^{2} + 2ac - c^{2} - 2c^{2} + 3ac - 3c^{2}$ = $(ac + 2ac + 3ac) + (-c^{2} - 2c^{2} - 3c^{2})$ = $6ac - 6c^{2}$ Step 7: Factor $6c$ out of the numerator: $6c(a - c)$ The denominator is $c(a-c)(a+c)$. Cancel $c$ and $(a-c)$: $\frac{6c(a-c)}{c(a-c)(a+c)} = \frac{6}{a+c}$ Thus the simplified expression is $\frac{6}{a+c}$, which matches option D.Answer:Option D: $\frac{6}{a+c}$