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Question

If 2x2+2x+α<cosec1(cosec 6)+tan1(tan(5)) xR, then possible value(s) of α can be (where αZ)

A
3
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B
0
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C
2
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D
1
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Solution

The correct option is C 2
Given:
2x2+2x+α<cosec1(cosec 6)+tan1(tan(5))
we know that
cosec1(cosec x)=x2π ; 3π2<x<2π
tan1(tanx)=x+2π ; 2π<x<3π2
Then inequality becomes
2x2+2x+α<62π5+2π
2x2+2x+α1<0
ax2+bx+c<0 and a<0 D<0
4+4×2(α1)<0
α<12

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