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Q.

If fα=limx→2  sinxα+cosxα1x-2    for  α∈0,π2, then

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a

f(0) = 1

b

fπ2=1

c

fα=cosαcos2α.sinαsin2a   if  α∈0,π2

d

fα=sinαsin2αcosαcos2α   if   α∈0,π2

answer is A.

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Detailed Solution

fα=limx→2  sinxα+cosxα1x-2                 1∞  form=elimx→2sinxα+cosxα-1x-2,    α∈0,π21,                     α=0,π2Now,   elimx→2  sinxα+cosca-1x-2=elimx→2. sinxα+cosxα-sin2α-cos2αx-2                                         = elimx→2 sinxα(sinx-2α-1) +  cos2α(cosx-2α-1)x-2                                         = e sin2αlogesinα+cos2αlogecosα                                         = elogesinαsin2α+logecosαcos2α                                         = elogesinαsin2αcosαcos2α                                         = sinαsin2αcosαcos2α∴   fx=cos αcos2α.sinαsin2α,   a∈0,π21,                              α=0,π2
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