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Question

If 22πsin1x2(a+2)2πsin1x+8a<0 for atleast one real x, then

A
18a<2
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B
a<2
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C
aR{2}
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D
a[0,18)(2,)
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Solution

The correct option is D a[0,18)(2,)
22π/sin1x2(a+2)2π/sin1x+8a<0

(2π/sin1x4)(2π/sin1x2a)<0

Now, 2π/sin1x(0,14][4,)

for 2π/sin1x(0,14].We have

(2π/sin1x4)<0
22π/sin1x2a>0

or 0a<18
Similarly, for 2π/sin1x[4,),a>2
So a[0,18)(2,)

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