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JEE MATHEMATICS MCQ: FUNCTIONS

1 Easy
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Application +4 | -1
For θ>π3\theta > \frac{\pi}{3}, the value of f(θ)=sec⁡2θ+cos⁡2θf(\theta)=\sec ^{2} \theta+\cos ^{2} \theta always lies in the interval
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Correct option: D
[2,∞)[2, \infty)
Option D ([2,∞)[2, \infty)) is correct because the minimum value of the function occurs when sec²θ = cos²θ, which happens when cos²θ = 1 or cos²θ = 0, but since θ > π/3, the minimum value is 2.

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