# If, the sum of Squares of 3 distinct numbers is 1, then the sum of their product taken two at a time, lies in between:

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If, the sum of Squares of 3 distinct numbers is 1, then the sum of their product taken two at a time, lies in between:
1. $- \dfrac{1}{2} \ and \ 1$
2. $\dfrac{1}{2} \ and \ \dfrac{3}{4}$
3. $\dfrac{1}{2} \ and \ 1$
4. $- \dfrac{1}{2} \ and \ \dfrac{3}{2}$

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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+c= 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 = a+ b2 + c+ 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