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Question Number 106570 by bobhans last updated on 06/Aug/20
(1)∫_0 ^a ((√(a−x))/( (√(a−x))+(√x))) dx =? (a) 0 (b) (a/2) (c) a (d) 2a (e) (5/2)a (2) ∫_0 ^(π/4) ((1−tan x)/(1+tan x)) dx =? (a) 0 (b) ln 2 (c) −ln 2 (d) πln 2 (e)(1/2)ln 2 (3) ((√3)+2)^x > 7−4(√3) , find the solution set
$$\left(\mathrm{1}\right)\underset{\mathrm{0}} {\overset{\mathrm{a}} {\int}}\:\frac{\sqrt{\mathrm{a}−\mathrm{x}}}{\:\sqrt{\mathrm{a}−\mathrm{x}}+\sqrt{\mathrm{x}}}\:\mathrm{dx}\:=? \\ $$$$\left(\mathrm{a}\right)\:\mathrm{0}\:\:\:\:\:\:\left(\mathrm{b}\right)\:\frac{\mathrm{a}}{\mathrm{2}}\:\:\:\:\:\:\left(\mathrm{c}\right)\:\mathrm{a}\:\:\:\:\:\:\:\:\left(\mathrm{d}\right)\:\mathrm{2a}\:\:\:\:\:\:\left(\mathrm{e}\right)\:\frac{\mathrm{5}}{\mathrm{2}}\mathrm{a} \\ $$$$\left(\mathrm{2}\right)\:\underset{\mathrm{0}} {\overset{\pi/\mathrm{4}} {\int}}\frac{\mathrm{1}−\mathrm{tan}\:\mathrm{x}}{\mathrm{1}+\mathrm{tan}\:\mathrm{x}}\:\mathrm{dx}\:=? \\ $$$$\left(\mathrm{a}\right)\:\mathrm{0}\:\:\:\:\left(\mathrm{b}\right)\:\mathrm{ln}\:\mathrm{2}\:\:\:\:\:\left(\mathrm{c}\right)\:−\mathrm{ln}\:\mathrm{2}\:\:\:\:\:\left(\mathrm{d}\right)\:\pi\mathrm{ln}\:\mathrm{2}\:\:\:\left(\mathrm{e}\right)\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\mathrm{2} \\ $$$$\left(\mathrm{3}\right)\:\left(\sqrt{\mathrm{3}}+\mathrm{2}\right)^{\mathrm{x}} \:>\:\mathrm{7}−\mathrm{4}\sqrt{\mathrm{3}}\:,\:\mathrm{find}\:\mathrm{the}\:\mathrm{solution}\:\mathrm{set} \\ $$
Answered by bemath last updated on 06/Aug/20
(2) ((1−tan x)/(1+tan x)) = ((cos x−sin x)/(cos x+sin x)) = ((1−sin 2x)/(cos 2x)) = sec 2x−tan 2x ∫_0 ^(π/4) (sec 2x−tan 2x)dx = [(1/2)ln ∣sec 2x−tan 2x∣−(1/2)ln ∣cos 2x∣ ]_( 0) ^(π/4) = ln 2
$$\left(\mathrm{2}\right)\:\frac{\mathrm{1}−\mathrm{tan}\:\mathrm{x}}{\mathrm{1}+\mathrm{tan}\:\mathrm{x}}\:=\:\frac{\mathrm{cos}\:\mathrm{x}−\mathrm{sin}\:\mathrm{x}}{\mathrm{cos}\:\mathrm{x}+\mathrm{sin}\:\mathrm{x}} \\ $$$$\:\:\:=\:\frac{\mathrm{1}−\mathrm{sin}\:\mathrm{2x}}{\mathrm{cos}\:\mathrm{2x}}\:=\:\mathrm{sec}\:\mathrm{2x}−\mathrm{tan}\:\mathrm{2x} \\ $$$$\underset{\mathrm{0}} {\overset{\pi/\mathrm{4}} {\int}}\left(\mathrm{sec}\:\mathrm{2x}−\mathrm{tan}\:\mathrm{2x}\right)\mathrm{dx}\:= \\ $$$$\left[\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\mid\mathrm{sec}\:\mathrm{2x}−\mathrm{tan}\:\mathrm{2x}\mid−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\mid\mathrm{cos}\:\mathrm{2x}\mid\:\right]_{\:\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \\ $$$$=\:\mathrm{ln}\:\mathrm{2} \\ $$
Answered by john santu last updated on 06/Aug/20
(3) (2+(√3))^x > 7−2(√(12)) (2+(√3))^x > (2−(√3))^2 (2+(√3))^x > ((1/(2+(√3))))^2 (2+(√3))^x > (2+(√3))^(−2) x > −2
$$\left(\mathrm{3}\right)\:\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)^{\mathrm{x}} \:>\:\mathrm{7}−\mathrm{2}\sqrt{\mathrm{12}} \\ $$$$\:\:\:\:\:\:\:\:\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)^{\mathrm{x}} \:>\:\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)^{\mathrm{x}} \:>\:\left(\frac{\mathrm{1}}{\mathrm{2}+\sqrt{\mathrm{3}}}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)^{\mathrm{x}} \:>\:\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)^{−\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{x}\:>\:−\mathrm{2}\: \\ $$
Answered by Dwaipayan Shikari last updated on 06/Aug/20
∫_0 ^a ((√(a−x))/( (√(a−x))+(√x)))dx=∫_0 ^a ((√x)/( (√x)+(√(a−x))))dx=I 2I=∫_0 ^a (((√(a−x))+(√x))/( (√x)+(√(a−x))))dx=∫_0 ^a 1dx=a I=(a/2)
$$\int_{\mathrm{0}} ^{{a}} \frac{\sqrt{{a}−{x}}}{\:\sqrt{{a}−{x}}+\sqrt{{x}}}{dx}=\int_{\mathrm{0}} ^{{a}} \frac{\sqrt{{x}}}{\:\sqrt{{x}}+\sqrt{{a}−{x}}}{dx}={I} \\ $$$$\mathrm{2}{I}=\int_{\mathrm{0}} ^{{a}} \frac{\sqrt{{a}−{x}}+\sqrt{{x}}}{\:\sqrt{{x}}+\sqrt{{a}−{x}}}{dx}=\int_{\mathrm{0}} ^{{a}} \mathrm{1}{dx}={a} \\ $$$${I}=\frac{{a}}{\mathrm{2}} \\ $$

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