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Question

If a+b+c=2 , a2+b2+c2=1 and abc=3 then 1a+1b+1c is equal to

A
0
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B
12
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C
1
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D
none of these
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Solution

The correct option is A 12
Given that,

a+b+c=2,a2+b2+c2=1abc=3

Considering a+b+c=2 and squaring both sides
a2+b2+c2+2ab+2bc+2ca=4

2(ab+bc+ca)=41

2(a)(b)(c)(1a+1b+1c)=3

2×3×(1a+1b+1c)=3

1a+1b+1c=12

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