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Question Number 106570 by bobhans last updated on 06/Aug/20

$$\left(1\right)\underset{0}{\stackrel{\mathrm{a}}{\int }}\frac{\sqrt{\mathrm{a}-\mathrm{x}}}{\sqrt{\mathrm{a}-\mathrm{x}}+\sqrt{\mathrm{x}}}\mathrm{dx}=?$$$$\left(\mathrm{a}\right)0\left(\mathrm{b}\right)\frac{\mathrm{a}}{2}\left(\mathrm{c}\right)\mathrm{a}\left(\mathrm{d}\right)2\mathrm{a}\left(\mathrm{e}\right)\frac{5}{2}\mathrm{a}$$$$\left(2\right)\underset{0}{\stackrel{\pi /4}{\int }}\frac{1-\tan \mathrm{x}}{1+\tan \mathrm{x}}\mathrm{dx}=?$$$$\left(\mathrm{a}\right)0\left(\mathrm{b}\right)\ln 2\left(\mathrm{c}\right)-\ln 2\left(\mathrm{d}\right)\pi \ln 2\left(\mathrm{e}\right)\frac{1}{2}\ln 2$$$$\left(3\right){(\sqrt{3}+2)}^{\mathrm{x}}>7-4\sqrt{3},\mathrm{find}\mathrm{the}\mathrm{solution}\mathrm{set}$$

Answered by bemath last updated on 06/Aug/20

$$\left(2\right)\frac{1-\tan \mathrm{x}}{1+\tan \mathrm{x}}=\frac{\cos \mathrm{x}-\sin \mathrm{x}}{\cos \mathrm{x}+\sin \mathrm{x}}$$$$=\frac{1-\sin 2\mathrm{x}}{\cos 2\mathrm{x}}=\sec 2\mathrm{x}-\tan 2\mathrm{x}$$$$\underset{0}{\stackrel{\pi /4}{\int }}(\sec 2\mathrm{x}-\tan 2\mathrm{x})\mathrm{dx}=$$$${[\frac{1}{2}\ln \mid \sec 2\mathrm{x}-\tan 2\mathrm{x}\mid -\frac{1}{2}\ln \mid \cos 2\mathrm{x}\mid ]}_0^{\frac{\pi }{4}}$$$$=\ln 2$$

Answered by john santu last updated on 06/Aug/20

$$\left(3\right){(2+\sqrt{3})}^{\mathrm{x}}>7-2\sqrt{12}$$$${(2+\sqrt{3})}^{\mathrm{x}}>{(2-\sqrt{3})}^2$$$${(2+\sqrt{3})}^{\mathrm{x}}>{\left(\frac{1}{2+\sqrt{3}}\right)}^2$$$${(2+\sqrt{3})}^{\mathrm{x}}>{(2+\sqrt{3})}^{-2}$$$$\mathrm{x}>-2$$

Answered by Dwaipayan Shikari last updated on 06/Aug/20

$${\int }_0^a\frac{\sqrt{a-x}}{\sqrt{a-x}+\sqrt{x}}dx={\int }_0^a\frac{\sqrt{x}}{\sqrt{x}+\sqrt{a-x}}dx=I$$$$2I={\int }_0^a\frac{\sqrt{a-x}+\sqrt{x}}{\sqrt{x}+\sqrt{a-x}}dx={\int }_0^{a}1dx=a$$$$I=\frac{a}{2}$$

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