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Question

Column IColumn IIa. For θR, if max{5sinθ+3sin(θα)}=7, then the set of possible values of α isp. 2nπ+3π4,nZ b. xnπ2 and (cosx)sin2x3sinx+2=1q. 2nπ±π3,nZ c. sinx+21/4cosx=0r. 2nπ+cos1(13),nZd. log5tanx=(log54)(log4(3sinx))s. no solution

Which of the following is the correct combination ?

A
ap,br,cq,ds
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B
ar,bq,cp,ds
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C
aq,bs,cp,dr
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D
as,br,cp,dq
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Solution

The correct option is C aq,bs,cp,dr
a. 5sinθ+3sin(θα)
=5sinθ+3(sinθcosαcosθsinα)
=(5+3cosα)sinθ3sinαcosθ

maxθ R{5sinθ+3sin(θα)}
=(5+3cosα)2+9sin2α
=34+30cosα

Therefore, 34+30cosα=72=49
cosα=12
α=2nπ±π3,nZ

b. (cosx)sin2x3sinx+2=1
sin2x3sinx+2=0
(sinx1)(sinx2)=0
sinx=1
But this does not satisfy the equation as 00=1 is absurd.

c. sinx+21/4cosx=0
sinx>0,
cosx<0
Also, sinx>0x lies in the 2nd quadrant.

sinx+21/4cosx=0
21/4cosx=sinx
2cos2x=sinx
2sin2x+sinx2=0
(2sinx1)(sinx+2)=0
sinx=12
x=2nπ+3π4,nZ

d. log5tanx=(log54)(log4(3sinx))
log5tanx=log5(3sinx)
For log to be defined,
tanx>0,sinx>0
x lies in the 1st quadrant.

log5tanx=log5(3sinx)
tanx=3sinx
cosx=13
x=2nπ+cos113,nZ

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