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Question Number 18402 by Tinkutara last updated on 20/Jul/17

If P_n  = cos^n  θ + sin^n  θ, θ ∈ [0, (π/2)], n ∈  (−∞, 2), then minimum of P_n  will be  (1) 1  (2) (1/2)  (3) (√2)  (4) (1/(√2))

$$\mathrm{If}\:{P}_{{n}} \:=\:\mathrm{cos}^{{n}} \:\theta\:+\:\mathrm{sin}^{{n}} \:\theta,\:\theta\:\in\:\left[\mathrm{0},\:\frac{\pi}{\mathrm{2}}\right],\:{n}\:\in \\ $$$$\left(−\infty,\:\mathrm{2}\right),\:\mathrm{then}\:\mathrm{minimum}\:\mathrm{of}\:{P}_{{n}} \:\mathrm{will}\:\mathrm{be} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{1} \\ $$$$\left(\mathrm{2}\right)\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\left(\mathrm{3}\right)\:\sqrt{\mathrm{2}} \\ $$$$\left(\mathrm{4}\right)\:\frac{\mathrm{1}}{\sqrt{\mathrm{2}}} \\ $$

Answered by Tinkutara last updated on 28/Jul/17

(sin θ)^(−n)  ≥ ... ≥ (sin θ)^(−1)  ≥ sin θ ≥ sin^2  θ  (cos θ)^(−n)  ≥ ... ≥ (cos θ)^(−1)  ≥ cos θ ≥ cos^2  θ  (sin θ)^(−n)  + (cos θ)^(−n)  > sin^2  θ + cos^2  θ = 1  ∴ Minimum of P_n  = 1

$$\left(\mathrm{sin}\:\theta\right)^{−{n}} \:\geqslant\:...\:\geqslant\:\left(\mathrm{sin}\:\theta\right)^{−\mathrm{1}} \:\geqslant\:\mathrm{sin}\:\theta\:\geqslant\:\mathrm{sin}^{\mathrm{2}} \:\theta \\ $$$$\left(\mathrm{cos}\:\theta\right)^{−{n}} \:\geqslant\:...\:\geqslant\:\left(\mathrm{cos}\:\theta\right)^{−\mathrm{1}} \:\geqslant\:\mathrm{cos}\:\theta\:\geqslant\:\mathrm{cos}^{\mathrm{2}} \:\theta \\ $$$$\left(\mathrm{sin}\:\theta\right)^{−{n}} \:+\:\left(\mathrm{cos}\:\theta\right)^{−{n}} \:>\:\mathrm{sin}^{\mathrm{2}} \:\theta\:+\:\mathrm{cos}^{\mathrm{2}} \:\theta\:=\:\mathrm{1} \\ $$$$\therefore\:\mathrm{Minimum}\:\mathrm{of}\:{P}_{{n}} \:=\:\mathrm{1} \\ $$

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