UPSC CDS 1 2022 Math Paper

Concept:The expression $a^2 + b^2 + c^2 - ab - bc - ca$ simplifies to $\frac{1}{2}[(a-b)^2 + (b-c)^2 + (c-a)^2]$, which is always positive for distinct real numbers $a, b, c$.Explanation:Step 1: Use the standard identity: $a^2 + b^2 + c^2 - ab - bc - ca = \frac{1}{2}[(a-b)^2 + (b-c)^2 + (c-a)^2]$.Step 2: Since $a, b, c$ are distinct real numbers, each squared difference $(a-b)^2$, $(b-c)^2$, $(c-a)^2$ is strictly greater than zero.Step 3: The sum of these three positive terms is positive, and multiplying by $\frac{1}{2}$ still gives a positive number.Step 4: No condition on the order or sum of $a, b, c$ (like $a > b > c$ or $a+b+c=0$) is needed to guarantee the positivity.Step 5: Therefore, neither Statement I nor Statement II is necessary to answer the question.Answer:D. Neither Statement – I nor Statement – II is required to answer the question.