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Question

The set of values of k for which x2kx+sin1(sin4)>0 for all real x is

A
ϕ
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B
(2,2)
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C
R
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D
none of these
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Solution

The correct option is C ϕ
Given, x2kx+sin1(sin4)>0
Now,
sin1(sin4)=sin1(sin(π+(4π)))=π4
So, the given inequality is
x2kx+π4>0
D<0
k24(π4)<0
k2+164π<0
which is false all the time.

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