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Q.

Range of f(θ)=cos2⁡θcos2⁡θ+1+2sin2⁡θ is

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a

[3/4,1]

b

[3/16,1]

c

[3/4,7/4]

d

[7/4,2]

answer is D.

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Detailed Solution

f(θ)=cos2⁡θcos2⁡θ+1+2sin2⁡θ=cos4⁡θ+cos2⁡θ+sin2⁡θ+sin2⁡θ=cos4⁡θ+1+1−cos2⁡θ=cos2⁡θ−122+2−14=cos2⁡θ−122+74fmin. =74fmax.=14+74=2
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Range of f(θ)=cos2⁡θcos2⁡θ+1+2sin2⁡θ is