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Question

If (sin1x+sin1w)(sin1y+sin1z)=π2, then D=xN1yN2zN3wN4(N1,N2,N3,N4ϵN)

A
has a maximum value of 2
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B
has a minimum value of 0
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C
16 different D are possible
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D
has a minimum value of -2
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Solution

The correct options are
A has a maximum value of 2
C 16 different D are possible
D has a minimum value of -2
(sin1x+sin1w)(sin1y+sin1z)=π2
sin1x+sin1w=sin1y+sin1z=π
or sin1x+sin1w=sin1y+sin1z=π
x=y=w=z=1 or x=y=w=z=1


Hence, the maximum value of 1N11N21N31N4=1111=2 and minimum value 1111=2

Also, there are 16 different determinants as each place value is either 1 or -1.

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