UPSC CDS 2 2021 Math Paper

Concept:Simplify each inequality algebraically to find the condition for truth, then check the given restrictions.Explanation:For statement 1: $\frac{a^{2}-b^{2}}{a^{2}+b^{2}} > \frac{a-b}{a+b}$ with $a>b>0$.Factor $a^{2}-b^{2}=(a-b)(a+b)$ and cross-multiply (all terms positive):$(a-b)(a+b)(a+b) > (a-b)(a^{2}+b^{2})$.Cancel $(a-b)>0$: $(a+b)^{2} > a^{2}+b^{2}$.This gives $a^{2}+2ab+b^{2} > a^{2}+b^{2}$, i.e. $2ab > 0$, which is always true for $a>b>0$.Hence statement 1 is correct.For statement 2: $\frac{a^{3}+b^{3}}{a^{2}+b^{2}} > \frac{a^{2}+b^{2}}{a+b}$ with $a>b>0$.Cross-multiply: $(a^{3}+b^{3})(a+b) > (a^{2}+b^{2})^{2}$.Expand LHS: $a^{4}+a^{3}b+ab^{3}+b^{4} = a^{4}+b^{4}+ab(a^{2}+b^{2})$.RHS: $a^{4}+2a^{2}b^{2}+b^{4}$.Inequality becomes $ab(a^{2}+b^{2}) > 2a^{2}b^{2}$.Divide by $ab>0$: $a^{2}+b^{2} > 2ab$, i.e. $(a-b)^{2} > 0$.This holds for any $a \neq b$, so it is true for both $a>b>0$ and $b>a>0$.The statement claims it is true "only when $a>b>0$". That is false because it also holds for $b>a>0$.Therefore statement 2 is incorrect.Answer:Only statement 1 is correct. Option A.