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Question

If a, b, c all are non-zero and unequal and ∣ ∣1+a1111+b1111+c∣ ∣=0, then 1+1a+1b+1c is equal to

A
abc
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B
1abc
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C
1a+1b+1c
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D
0
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Solution

The correct option is D 0
abc0

∣ ∣1+a1111+b1111+c∣ ∣=0

abc∣ ∣ ∣ ∣ ∣ ∣1a+11a1a1b1b+11b1c1c1c+1∣ ∣ ∣ ∣ ∣ ∣=0
R1R1+R2+R3

(abc)(1+1a+1b+1c)∣ ∣ ∣ ∣ ∣1111b1b11b1c1c1c+1∣ ∣ ∣ ∣ ∣=0
C1C1C2;C2C2C3

(abc)(1+1a+1b+1c)∣ ∣ ∣ ∣ ∣001111b011c+1∣ ∣ ∣ ∣ ∣=0
(abc)(1+1a+1b+1c)=0
abc0 1+1a+1b+1c=0
Ans: D

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