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25 May 2020, 02:42
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
32% (02:31) correct
68%
(02:22)
wrong
based on 146
sessions
History
Date
Time
Result
Not Attempted Yet
If \(a^2 + b^2 + c^2 = 1\), then what is the sum of the minimum possible value and the maximum possible value of \(ab + bc + ca\)?
A. -1/2
B. 0
C. 1/2
D. 1
E. 3/2
Are You Up For the Challenge: 700 Level Questions
25 May 2020, 09:42
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square of any number is non-negative
\((a-b)^2 = a^2+b^2-2ab ≥ 0\)......(1)
\((b-c)^2 = b^2+c^2-2bc ≥ 0\)........(2)
\((c-a)^2 = c^2+a^2-2ca ≥ 0\)........(3)
Add (1), (2) and (3)
\(2a^2+2b^2+2c^2 - 2ab-2bc -2ca ≥ 0\)
1 ≥ ab+bc+ca
\((a+b+c)^2 = a^2+b^2+c^2 + 2ab+2bc+2ca ≥ 0\)
\(1+ 2ab+2bc+2ca ≥ 0\)
\(ab+bc +ca ≥ -\frac{1}{2}\)
Sum of maximum and minimum value of ab+bc+ca \(= -\frac{1}{2} +1 = \frac{1}{2}\)
C
\((a-b)^2 = a^2+b^2-2ab ≥ 0\)......(1)
\((b-c)^2 = b^2+c^2-2bc ≥ 0\)........(2)
\((c-a)^2 = c^2+a^2-2ca ≥ 0\)........(3)
Add (1), (2) and (3)
\(2a^2+2b^2+2c^2 - 2ab-2bc -2ca ≥ 0\)
1 ≥ ab+bc+ca
\((a+b+c)^2 = a^2+b^2+c^2 + 2ab+2bc+2ca ≥ 0\)
\(1+ 2ab+2bc+2ca ≥ 0\)
\(ab+bc +ca ≥ -\frac{1}{2}\)
Sum of maximum and minimum value of ab+bc+ca \(= -\frac{1}{2} +1 = \frac{1}{2}\)
C
General Discussion
25 May 2020, 07:07
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Quote:
a^2 + b^2 + c^2 = 1
maximum:
1/2^2+1/2^2+√2/2^2=1
1/4+1/4+1/2=1
1/2*1/2+2(1/2*√2/2)=,
1/4+√2/2=(1+2√2)/4=0.95~
minimum:
1/2^2+1/2^2+(-√2/2)^2=1
1/4+1/4+1/2=1
1/2*1/2+2(1/2)(-√2/2)=,
1/4+(-√2/2)=(1-2√2)/4=-0.45~
-1/2*1/2+(-0.5)*(√2/2)+0.5(√2/2)=√2/2(0
1/4+(-√2/2)=(1-2√2)/4=-0.45~
sum max+min = 0.5
ans (C)
