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Question

If a, b R distinct numbers satisfying |a1|+|b1|=|a|+|b|=|a+1|+|b+1|, then the minimum value of |ab| is



A

3

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B

0

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C

1

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D

2

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Solution

The correct option is A

2


|a1|+b1=ab=a+1+(b+1)

By analyzing (a - 1) & (b - 1) needs to be of different sign so as cancel off the 1 & be be equal |a| + |b|

11 aly, for (a + 1) & (b + 1) too,they need to be of different sign.

Therefore, a & b will be of different signs.

So, let a > 0 & b < 0.

|a1|+b1=ab=a+1+|b+1|

Equating 1st two. 

|a1|+b1=ab

For this to be true, a1,0a1

Equality last two

ab=a+1+|b+1|

For this to be true, b+10b1

(a,b)R(1,1)

|ab|mn=|1(1)|=2


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