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Question Number 110307 by bemath last updated on 28/Aug/20

(1)lim_(x→−∞) ((3−3x)/( (√(x^2 −4x+1)))) ? (2) ∫_0 ^1 arctan (((2x−1)/(1+x−x^2 ))) dx (3)how many integer solution sets exist for the equation x^2 +y^2 =2

$$\left(\mathrm{1}\right)\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\:\frac{\mathrm{3}−\mathrm{3}{x}}{\:\sqrt{{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{1}}}\:? \\ $$$$\left(\mathrm{2}\right)\:\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:\mathrm{arctan}\:\left(\frac{\mathrm{2}{x}−\mathrm{1}}{\mathrm{1}+{x}−{x}^{\mathrm{2}} }\right)\:{dx} \\ $$$$\left(\mathrm{3}\right){how}\:{many}\:{integer}\:{solution}\:{sets} \\ $$$${exist}\:{for}\:{the}\:{equation}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{2} \\ $$

Answered by john santu last updated on 28/Aug/20

★((JS)/≈)★ let x = − t , then t→∞ lim_(t→∞) ((3+3t)/( (√(t^2 +4t+1)))) = lim_(t→∞) ((t(3+(3/t)))/(t(√(1+(4/t)+(1/t^2 )))))= 3

$$\:\:\bigstar\frac{{JS}}{\approx}\bigstar \\ $$$${let}\:{x}\:=\:−\:{t}\:,\:{then}\:{t}\rightarrow\infty \\ $$$$\underset{{t}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{3}+\mathrm{3}{t}}{\:\sqrt{{t}^{\mathrm{2}} +\mathrm{4}{t}+\mathrm{1}}}\:=\:\underset{{t}\rightarrow\infty} {\mathrm{lim}}\:\frac{{t}\left(\mathrm{3}+\frac{\mathrm{3}}{{t}}\right)}{{t}\sqrt{\mathrm{1}+\frac{\mathrm{4}}{{t}}+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }}}=\:\mathrm{3} \\ $$

Answered by Dwaipayan Shikari last updated on 28/Aug/20

∫_0 ^1 tan^(−1) (((x−(1−x))/(1+x(1−x)))) ∫_0 ^1 tan^(−1) x−tan^(−1) (1−x)=∫_0 ^1 tan^(−1) (1−x)−tan^(−1) x=I 2I=0 I=0

$$\int_{\mathrm{0}} ^{\mathrm{1}} {tan}^{−\mathrm{1}} \left(\frac{{x}−\left(\mathrm{1}−{x}\right)}{\mathrm{1}+{x}\left(\mathrm{1}−{x}\right)}\right) \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {tan}^{−\mathrm{1}} {x}−{tan}^{−\mathrm{1}} \left(\mathrm{1}−{x}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} {tan}^{−\mathrm{1}} \left(\mathrm{1}−{x}\right)−{tan}^{−\mathrm{1}} {x}={I} \\ $$$$\mathrm{2}{I}=\mathrm{0} \\ $$$${I}=\mathrm{0} \\ $$

Commented by Dwaipayan Shikari last updated on 28/Aug/20

Is it right?

$${Is}\:{it}\:{right}? \\ $$

Answered by Aziztisffola last updated on 28/Aug/20

(3) (x;y)∈{(1;1);(1;−1);(−1;1);(−1;−1)}

$$\:\left(\mathrm{3}\right)\:\left({x};{y}\right)\in\left\{\left(\mathrm{1};\mathrm{1}\right);\left(\mathrm{1};−\mathrm{1}\right);\left(−\mathrm{1};\mathrm{1}\right);\left(−\mathrm{1};−\mathrm{1}\right)\right\} \\ $$

Answered by Rio Michael last updated on 28/Aug/20

lim_(x→−∞) ((3−3x)/( (√(x^2 −4x+1)))) = lim_(x→−∞) ((3−3x)/(x(√(1−(4/x) + (1/x^2 ))))) = lim_(x→−∞) (((3/x)−3)/( (√(1−(4/x) + (1/x^2 ))))) = −3

$$\:\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\:\frac{\mathrm{3}−\mathrm{3}{x}}{\:\sqrt{{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{1}}}\:=\:\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\:\frac{\mathrm{3}−\mathrm{3}{x}}{{x}\sqrt{\mathrm{1}−\frac{\mathrm{4}}{{x}}\:+\:\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\:\frac{\frac{\mathrm{3}}{{x}}−\mathrm{3}}{\:\sqrt{\mathrm{1}−\frac{\mathrm{4}}{{x}}\:+\:\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}}\:=\:−\mathrm{3}\: \\ $$

Commented by bemath last updated on 28/Aug/20

wrong sir. x→−∞ it should be (√(x^2 )) = −x not x

$${wrong}\:{sir}.\:{x}\rightarrow−\infty\:{it}\:{should}\:{be}\: \\ $$$$\sqrt{{x}^{\mathrm{2}} \:}\:=\:−{x}\:{not}\:{x}\: \\ $$

Commented by Rio Michael last updated on 28/Aug/20

correct. thanks for the correction

$$\mathrm{correct}.\:\mathrm{thanks}\:\mathrm{for}\:\mathrm{the}\:\mathrm{correction} \\ $$

Answered by 1549442205PVT last updated on 28/Aug/20

(1)lim_(x→−∞) ((3−3x)/( (√(x^2 −4x+1)))) =lim_(x→−∞) ((3−3x)/( (−x ) (√(1−(4/x)+(1/x^2 ))))) =lim_(x→−∞) ((((−3)/x)+3)/( (√(1−(4/x)+(1/x^2 )))))=((0+3)/( (√(1−0+0))))=3 3)Solve x^2 +y^2 =2 in Z. x^2 =2−y^2 ≤2 but since x∈Z,∣x∣≤1 ⇒x∈−1,1⇒y∈{±1,±1} The solution set of the given eq. has four elements S={(−1,−1),(−1,1),(1,−1),(1,1)}

$$\left(\mathrm{1}\right)\underset{\mathrm{x}\rightarrow−\infty} {\mathrm{lim}}\:\frac{\mathrm{3}−\mathrm{3}{x}}{\:\sqrt{{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{1}}}\:=\underset{\mathrm{x}\rightarrow−\infty} {\mathrm{lim}}\:\frac{\mathrm{3}−\mathrm{3x}}{\:\left(−\mathrm{x}\:\right)\:\:\sqrt{\mathrm{1}−\frac{\mathrm{4}}{\mathrm{x}}+\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }}} \\ $$$$=\underset{\mathrm{x}\rightarrow−\infty} {\mathrm{lim}}\:\frac{\frac{−\mathrm{3}}{\mathrm{x}}+\mathrm{3}}{\:\sqrt{\mathrm{1}−\frac{\mathrm{4}}{\mathrm{x}}+\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }}}=\frac{\mathrm{0}+\mathrm{3}}{\:\sqrt{\mathrm{1}−\mathrm{0}+\mathrm{0}}}=\mathrm{3} \\ $$$$\left.\mathrm{3}\right)\mathrm{Solve}\:\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} =\mathrm{2}\:\mathrm{in}\:\mathbb{Z}. \\ $$$$\mathrm{x}^{\mathrm{2}} =\mathrm{2}−\mathrm{y}^{\mathrm{2}} \leqslant\mathrm{2}\:\mathrm{but}\:\mathrm{since}\:\mathrm{x}\in\mathbb{Z},\mid\mathrm{x}\mid\leqslant\mathrm{1} \\ $$$$\Rightarrow\mathrm{x}\in−\mathrm{1},\mathrm{1}\Rightarrow\mathrm{y}\in\left\{\pm\mathrm{1},\pm\mathrm{1}\right\} \\ $$$$\:\mathrm{The}\:\mathrm{solution}\:\mathrm{set}\:\mathrm{of}\:\mathrm{the}\:\mathrm{given}\:\mathrm{eq}. \\ $$$$\mathrm{has}\:\mathrm{four}\:\mathrm{elements} \\ $$$$\mathrm{S}=\left\{\left(−\mathrm{1},−\mathrm{1}\right),\left(−\mathrm{1},\mathrm{1}\right),\left(\mathrm{1},−\mathrm{1}\right),\left(\mathrm{1},\mathrm{1}\right)\right\} \\ $$

Answered by john santu last updated on 28/Aug/20

(2)∫ tan^(−1) (((2x−1)/(1+x−x^2 ))) dx = I setting x = u−(1/2) I=∫_(−1/2) ^(1/2) tan^(−1) (((2u)/((5/4)−u^2 ))) du since the resulting integrand is a continous odd function , so we got I = 0.

$$\left(\mathrm{2}\right)\int\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{2}{x}−\mathrm{1}}{\mathrm{1}+{x}−{x}^{\mathrm{2}} }\right)\:{dx}\:=\:{I} \\ $$$${setting}\:{x}\:=\:{u}−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${I}=\underset{−\mathrm{1}/\mathrm{2}} {\overset{\mathrm{1}/\mathrm{2}} {\int}}\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{2}{u}}{\frac{\mathrm{5}}{\mathrm{4}}−{u}^{\mathrm{2}} }\right)\:{du}\: \\ $$$${since}\:{the}\:{resulting}\:{integrand}\:{is} \\ $$$${a}\:{continous}\:{odd}\:{function}\:,\:{so}\:{we}\:{got} \\ $$$${I}\:=\:\mathrm{0}. \\ $$

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