UPSC CDS 1 2022 Math Paper

Given :
a2+b2+c2−ab−bc−ca
Where a,b and c are distinct real numbers (a≠b≠c)
Formula used:
a2+b2+c2−ab−bc−ca=‌

1

2

[(a−b)2+(b−c)2+(c−a)2]
Calculations :
It is given a,b and c are distinct real numbers
Let us take the value of a=1,b=2 and c=3(a<b<c)
We know that,
a2+b2+c2−ab−bc−ca=‌

1

2

[(a−b)2+(b−c)2+(c−a)2]
⇒‌

1

2

(1−2)2+(2−3)2+(3−1)2]
⇒(1∕2)(1+1+4)
⇒(6∕2)=3
⇒‌

1

2

[(a−b)2+(b−c)2+(c−a)2] is always positive
Therefore, (a2+b2+c2−ab−bc−ca) not depend on the statement - I and statement - II
∴ Neither Statement - I nor Statement - II is required to answer the question