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Q.

If  f(θ)=|1+sin2θcos2θ4sin2θsin2θ1+cos2θ4sin2θsin2θcos2θ1+4sin2θ| then which of the following  is/are CORRECT?

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a

If  f(θ)=0  then values of  θ  in  [0,π] are 7π12,11π12

b

f(θ)=k  has atleast one solution if  k[2,6]

c

If  f(θ)=2 then number of values of  θ in  [0,2π] is ‘5’

d

maximum value of  f(θ) is ‘6’ 

answer is A, B, C, D.

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Detailed Solution

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(a) : Applying  R3R2R1  and  R3R3R1
 |1+sin2θcos2θ4sin2θ110101| (1+sin2θ)cos2θ(1)+4sin2θ(1) 1+sin2θ+cos2θ+4sin2θ 2+4sin2θ

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