Question Number 94245 by mhmd last updated on 17/May/20
$${find}\:{the}\:{function}\:{f}\left({x}\right)\:{satisfying}\:{the}\:{given}\:{conditions} \\ $$$$\left({i}\right){f}^{'} \left({x}\right)=\mathrm{4}{x}^{\mathrm{2}} −\mathrm{1}\:\:\:,\:{f}\left(\mathrm{0}\right)=\mathrm{3}\:? \\ $$$$\left({ii}\right){f}^{''} \left({x}\right)=\mathrm{12}\:\:,\:{f}^{'} \left(\mathrm{0}\right)=\mathrm{2}\:\:,\:{f}\left(\mathrm{0}\right)=\mathrm{3}\:? \\ $$$$\left({iii}\right){f}^{''} \left({x}\right)=\mathrm{2}{x}\:\:,\:\:{f}^{'} \left(\mathrm{0}\right)=−\mathrm{3}\:\:,\:{f}\left(\mathrm{0}\right)=\mathrm{2}\:? \\ $$$$ \\ $$$${help}\:{me}\:{sir}\:{pleas}\:? \\ $$
Answered by Ar Brandon last updated on 17/May/20
![Q1\ ∫_0 ^1 ∫_0 ^1 (1/(√(1+x^2 −y^2 )))dydx=∫_0 ^1 (1/(√(1+x^2 )))∫_0 ^1 (1/(√(1−((y/(√(1+x^2 ))))^2 )))dydx =∫_0 ^1 [sin^(−1) ((y/(√(1+x^2 ))))]_0 ^1 dx=∫_0 ^1 sin^(−1) ((1/(√(1+x^2 ))))dx u=sin^(−1) ((1/(√(1+x^2 ))))⇒du=−2x∙(1/2)∙(1/((1+x^2 )^(3/2) ))∙(1/( (√(1−(1/(1+x^2 )))))) =−(x/((1+x^2 )^(3/2) ))∙(1/( (√(x^2 /(1+x^2 )))))=−(x/((1+x^2 )^(3/2) ))∙((√(1+x^2 ))/x)=−(1/(1+x^2 ))dx dv=dx⇒v=x ∫_0 ^1 sin^(−1) ((1/(√(1+x^2 ))))dx=[xsin^(−1) ((1/(√(1+x^2 ))))]_0 ^1 +∫_0 ^1 (x/(1+x^2 ))dx =(π/4)+[(1/2)ln(1+x^2 )]_0 ^1 =(π/4)+(1/2)ln(2)](Q94253.png)
$$\mathrm{Q1}\backslash\:\int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} }}\mathrm{dydx}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\sqrt{\mathrm{1}−\left(\frac{\mathrm{y}}{\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }}\right)^{\mathrm{2}} }}\mathrm{dydx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \left[\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{y}}{\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} \mathrm{dx}=\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }}\right)\mathrm{dx} \\ $$$$\mathrm{u}=\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }}\right)\Rightarrow\mathrm{du}=−\mathrm{2x}\centerdot\frac{\mathrm{1}}{\mathrm{2}}\centerdot\frac{\mathrm{1}}{\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} }\centerdot\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }}} \\ $$$$=−\frac{\mathrm{x}}{\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} }\centerdot\frac{\mathrm{1}}{\:\sqrt{\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }}}=−\frac{\mathrm{x}}{\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} }\centerdot\frac{\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }}{\mathrm{x}}=−\frac{\mathrm{1}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx} \\ $$$$\mathrm{dv}=\mathrm{dx}\Rightarrow\mathrm{v}=\mathrm{x} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }}\right)\mathrm{dx}=\left[\mathrm{xsin}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} +\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{x}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx} \\ $$$$=\frac{\pi}{\mathrm{4}}+\left[\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)\right]_{\mathrm{0}} ^{\mathrm{1}} =\frac{\pi}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\left(\mathrm{2}\right) \\ $$$$ \\ $$
Commented by mhmd last updated on 17/May/20
$${thank}\:{you}\:{sir} \\ $$
Commented by Ar Brandon last updated on 17/May/20
��You're welcome. I realised the previous post was deleted and that's why I decided to post it here.
Commented by mhmd last updated on 17/May/20
$${i}\:{dont}\:{know}\:{sir}\:?\:{im}\:{dont}\:{deleted}\: \\ $$
Commented by Tinku Tara last updated on 17/May/20
$$\mathrm{Did}\:\mathrm{the}\:\mathrm{post}\:\mathrm{got}\:\mathrm{deleted}\:\mathrm{after}\:\mathrm{an} \\ $$$$\mathrm{answer}\:\mathrm{or}\:\mathrm{comment}\:\mathrm{added}? \\ $$$$\mathrm{That}\:\mathrm{is}\:\mathrm{disabled}\:\mathrm{from}\:\mathrm{server}. \\ $$
Commented by mhmd last updated on 17/May/20
$${sir}\:{can}\:{help}\:{me}\:{kn}\:{the}\:{question}\:\mathrm{94243}\:? \\ $$
Commented by Ar Brandon last updated on 17/May/20
I think it was deleted before I commented. Because after solving I tried to send but to no avail so I gave up. But my solution was saved. And when I wanted to answer the next question it just appeared so I sent it.