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Question Number 112320 by mohammad17 last updated on 07/Sep/20

Commented by mohammad17 last updated on 07/Sep/20

$$helpmesirinQ1$$

Answered by john santu last updated on 07/Sep/20

$$\left(1\right)let{2}^{\cos t}=m\Rightarrow \frac{1}{m-1}>\frac{1}{1-\frac{1}{2}m}$$$$\frac{1}{m-1}>\frac{2}{2-m}\Rightarrow \frac{1}{m-1}-\frac{2}{2-m}>0$$$$\Rightarrow \frac{1}{m-1}+\frac{2}{m-2}>0$$$$\Rightarrow \frac{m-2+2m-2}{(m-1)(m-2)}>0$$$$\Rightarrow \frac{3m-4}{(m-1)(m-2)}>0$$$$\Rightarrow 1<m<\frac{4}{3}\cup m>2$$$$\Rightarrow 1<2^{\cos t}<\frac{4}{3}\cup 2^{\cos t}>2$$$$\Rightarrow 0<\cos t<2-\log {}_2\left(3\right)\cup \cos t>1$$$$for\cos t>1,t\in \mathbb{R},t=\varnothing$$$$\therefore 0<\cos t<0.415$$

Commented by mohammad17 last updated on 07/Sep/20

$$thankyousir$$

Answered by 1549442205PVT last updated on 08/Sep/20

$$\mathrm{Put}{2}^{\mathrm{cost}}=\mathrm{x}.\mathrm{Then}\frac{1}{\mathrm{x}-1}>\frac{1}{1-\frac{1}{2}\mathrm{x}}$$$$\iff \frac{1}{\mathrm{x}-1}>\frac{2}{2-\mathrm{x}}\iff \frac{1}{\mathrm{x}-1}-\frac{2}{2-\mathrm{x}}>0$$$$\frac{2-\mathrm{x}-2(\mathrm{x}-1)}{(\mathrm{x}-1)(2-\mathrm{x})}>0\iff \frac{4-3\mathrm{x}}{(\mathrm{x}-1)(2-\mathrm{x})}>0$$$$\iff \frac{3\mathrm{x}-4}{(\mathrm{x}-1)(\mathrm{x}-2)}>0\iff \frac{\mathrm{x}-4/3}{(\mathrm{x}-1)(\mathrm{x}-2)}>0$$$$\iff \mathrm{x}\in (1;\frac{4}{3})\cup (2;+\mathrm{\infty })$$$$\mathit{a})\mathit{x}\in (1,\frac{}{})\iff 1{<}^{\mathit{c}\mathit{o}\mathit{s}\mathit{t}}<\frac{}{}$$$$\iff 0<\mathbf{cost}<\frac{\mathbf{ln}\left(4/3\right)}{\mathbf{ln}2}\approx 0.4150=\mathbf{cos}65°{29}^{\prime }$$$$\iff \mathrm{t}\in (-90°+\mathrm{k}360°;-65°{28}^{\prime }+\mathrm{k}360)\cup (65°{28}^{\prime }+\mathrm{k}360°;90°+\mathrm{k}360)$$$$\mathbf{b})\mathbf{x}\in (2;+\mathrm{\infty })\iff 2^{\mathbf{cost}}>2\iff \mathbf{cost}>1$$$$\mathbf{inequality}\mathbf{has}\mathbf{no}\mathbf{roots}$$$$\mathbf{Thus},\mathbf{given}\mathbf{inequality}\mathbf{has}\mathbf{roots}\mathbf{are}:$$$$\mathbf{t}\in (-90°+\mathbf{k}360°;-65°{28}^{\prime }+\mathbf{k}360)\cup$$$$(65°{28}^{\prime }+\mathbf{k}360°;90°+\mathbf{k}360°)$$

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