Correct Answer - Option 1 :
\(- \dfrac{1}{2} \ and \ 1\)
Given :
The sum of Squares of 3 distinct numbers = 1
The sum of their product taken two at a time , lies in between
Calculation :
Let a, b and c be three numbers such that:
a2+b2+c2 = 1_____(1)
Also,
we are asked to find the range in which (ab +bc + ca) lies.
Clearly from equation (1) we could observe that:
a ,b, c lie between -1 and 1
hence ,a.b
Also, when a=1 and b=c=0 (Then equation (1) is satisfied)
then, ab + bc + ca = 0
Hence, option: (B) and (C) are discarded.
Also, we know that:
⇒ (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc
As (a + b + c)2 > 0
Hence,
⇒1 + 2(ab + bc + ca) > 0
⇒(ab + bc + ca) > -1/2
Hence,
⇒ -1/2 <ab + bc + ca <1
∴ The required answer is -1/2 and 1