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Question

If y=4sin2θcos2θ, then y lies in the interval .........

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Solution

Ans. [1,5]. We have
y=4sin2θcos2θ
=4sin2θ(12sin2θ)=6sin2θ1
y+16=sin2θ
But 0sin2θ1. Hence 0y+161,
or 0y+16 or 1y5y[1,5]

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