If sinxα$$+cosx$$α≥$$1$$, $$0$$<α<π$$2$$, then

[$$SolutionStep1$$]:sinxα$$+cosx$$α≥$$1$$, $$0$$ $$1$$ if x<$$2$$ Thus x≤$$2$$ i.e. x∈$$(-$$∞,$$2$$]

If sinxα$$+cosx$$α≥$$1$$, $$0$$<α<π$$2$$, then

Correct option is 2x∈−∞,2

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$If \sin^{x} α + \cos^{x} α \geq 1, 0 < α < \frac{π}{2}, then$

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x∈[2,∞)

x∈−∞,2

x∈[−1,1]

x∈ℝ

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detailed solution

Correct option is B

[SolutionStep1]:sinxα+cosxα≥1, 0<α<π2 Since equality holds for x=2 If x<2, both cosα and sinα increase (being + ve fraction). cosxα+sinxα>1 if x<2 Thus x≤2 i.e. x∈(-∞,2]

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If sinxα+cosxα≥1, 0<α<π2, then

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$If \sin^{x} α + \cos^{x} α \geq 1, 0 < α < \frac{π}{2}, then$