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Question

The value of limxπ/2tan2x(2sin2x+3sinx+4sin2x+6sinx+2) is equal to:

A
110
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B
111
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C
112
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D
18
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Solution

The correct option is C 112
limxπ/2tan2x(2sin2x+3sinx+4sin2x+6sinx+2)
Rationalizing,
=limxπ/2tan2xsin2x3sinx+22sin2x+3sinx+4+sin2x+6sinx+2
=limxπ/2tan2x(2sinx)(1sinx)2sin2x+3sinx+4+sin2x+6sinx+2
=limxπ/2tan2x(2sinx)cos2x(1+sinx)[2sin2x+3sinx+4+sin2x+6sinx+2]
=limxπ/2sin2x(2sinx)(1+sinx)(2sin2x+3sinx+4+sin2x+6sinx+2)=112

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