Question Number 58639 by rahul 19 last updated on 26/Apr/19
Answered by tanmay last updated on 27/Apr/19
$${F}\left({x}\right)=\int_{\mathrm{0}} ^{{x}} {tf}\left({t}\right){dt} \\ $$$$\frac{{dF}}{{dx}}=\int_{\mathrm{0}} ^{{x}} \frac{\partial}{\partial{x}\:}\left({tf}\left({t}\right){dt}\:+{xf}\left({x}\right)\frac{{dx}}{{dx}}−\mathrm{0}×{f}\left(\mathrm{0}\right)×\frac{{d}\mathrm{0}}{{dx}}\right. \\ $$$$\frac{{dF}}{{dx}}={xf}\left({x}\right) \\ $$$${F}\left({x}^{\mathrm{2}} \right)={x}^{\mathrm{4}} +{x}^{\mathrm{5}} ={x}^{\mathrm{4}} \left(\mathrm{1}+{x}\right)=\left({x}^{\mathrm{2}} \right)^{\mathrm{2}} \left(\mathrm{1}+\sqrt{{x}^{\mathrm{2}} }\:\right) \\ $$$${F}\left({x}\right)={x}^{\mathrm{2}} \left(\mathrm{1}+\sqrt{{x}}\right)={x}^{\mathrm{2}} +{x}^{\frac{\mathrm{5}}{\mathrm{2}}} \\ $$$$\frac{{dF}}{{dx}}=\mathrm{2}{x}+\frac{\mathrm{5}}{\mathrm{2}}{x}^{\frac{\mathrm{3}}{\mathrm{2}}} \\ $$$$\mathrm{2}{x}+\frac{\mathrm{5}}{\mathrm{2}}{x}^{\frac{\mathrm{3}}{\mathrm{2}}} ={xf}\left({x}\right) \\ $$$${f}\left({x}\right)=\mathrm{2}+\frac{\mathrm{5}}{\mathrm{2}}\sqrt{{x}}\: \\ $$$$\left.{a}\right)\:{f}\left(\mathrm{4}\right)=\mathrm{2}+\frac{\mathrm{5}}{\mathrm{2}}×\sqrt{\mathrm{4}}\:=\mathrm{7} \\ $$$$\left.{b}\right){f}\left({x}\right)=\mathrm{2}+\frac{\mathrm{5}}{\mathrm{2}}\sqrt{{x}}\:{continous}\:{function}\:\left[{x}>\mathrm{0}\right] \\ $$$$\left.{c}\right){f}\left({x}\right)=\mathrm{2}+\frac{\mathrm{5}}{\mathrm{2}}\sqrt{{x}}\: \\ $$$$\frac{{df}}{{dx}}=\frac{\mathrm{5}}{\mathrm{2}}×\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}}}=\frac{\mathrm{5}}{\mathrm{4}\sqrt{{x}}}\:{when}\:{x}>\mathrm{0}\:\:\frac{{df}}{{dx}}>\mathrm{0}\:{so}\:{increasing}\:{gunction} \\ $$
Commented by rahul 19 last updated on 27/Apr/19
$${thanks}\:{sir}! \\ $$