Question

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:

(A) -1/2 and 1

(B) 1/2 and 3/4

(C) 1/2 and 1

(D) -1/2 and 3/2​

Answer 1

it is a good question

i think answer is a or d

Answer 2

Answer:

The answers is:

(A) -1/2 and 1.

Step-by-step explanation:

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:

$(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca\\\\As\ (a+b+c)^2>0\\\\Hence,\\\\1+2(ab+bc+ca)>0\\\\(ab+bc+ca)>-\dfrac{1}{2}$

Hence,

$\dfrac{-1}{2}<ab+bc+ca<1$