Let S=\{\frac{a^2+b^2+c^2}{ab+bc+ca}:a,b,c\inR,ab +bc +ca\neq0\}, where R is the set of real numbers. Then S equals

Correct Answer is : (–∞,–2]∪[1,∞)

(–\infty, – 1] \cup [1, \infty)

(–\infty, 0] \cup [0, \infty)

(–\infty, – 1] \cup [2, \infty)

(–\infty, – 2] \cup [1, \infty)

KVPY SX SB Exam 19-Nov-2017 Question Paper

Question : 3 of 120

Marks: +1, -0

Let

S=\{{a^2+b^2+c^2}/{ab+bc+ca}:a,b,c∈R,ab

+bc +ca≠0\}

, where R is the set of real numbers. Then S equals

Solution:

Case-I

(a-b)^ 2 + (b -c)^2 + (c -a)^2 ≥ 0

a^ 2 + b^2 + c^2 - ab - bc - ca≥ 0

a ^ 2 + b^2 + c^2 ≥ ab + bc +ca

If ab + bc + ca > 0
Then

{ a ^ 2 + b^2 + c^2}/{ab + bc +ca}≥1

Case-II
Let ab + bc + ca < 0

(a+b+c)^2/{{ab + bc +ca}}≤0

{a^ 2 + b^2 + c^2+2(ab + bc +ca)}/ {ab + bc +ca}≤0

{ a ^ 2 + b^2 + c^2}/{ab + bc +ca}+2≤0

{ a ^ 2 + b^2 + c^2}/{ab + bc +ca}≤-2

So, Range is

(– ∞, –2] ∪ [1, ∞)

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