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Question

If |log2x+1|+1(log2x)2=log2x+(log2x)2, then the true set of values of x is {λ}[μ,). Then


A
2λ3=0
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B
μ2=0
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C
2λμ=0
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D
4λμ=0
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Solution

The correct options are
B μ2=0
D 4λμ=0
|log2x+1|+1(log2x)2=log2x+(log2x)2 
|log2x+1|+(log2x)21=log2x+(log2x)2 
(log2x+1)((log2x)21)0
(log2x+1)2(log2x1)0
log2x=1 or log2x1
x{12}[2,)
λ=12,μ=2.

Mathematics

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