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Question Number 131334 by Algoritm last updated on 03/Feb/21

Answered by bluberry508 last updated on 06/Feb/21

$$$$$$letg\left(x\right)={\int }_0^{x}xf{}^{\prime }\left(t\right)-{tf}^{{}^{\prime }}\left(t\right)dt=$$$$x{\int }_0^x{f}^{{}^{\prime }}\left(t\right)dt-{\int }_0^x{tf}^{{}^{\prime }}\left(t\right)dt$$$$$$$$\frac{dg}{dx}={\int }_0^x{f}^{{}^{\prime }}\left(t\right)dt+{xf}^{{}^{\prime }}\left(x\right)-{xf}^{{}^{\prime }}\left(x\right)$$$$={\int }_0^x{f}^{{}^{\prime }}\left(t\right)dt$$$$$$$$differentiatebothsideofeq2times$$$$$$$$f\left(x\right)+{xf}^{{}^{\prime }}\left(x\right)=2x+\frac{dg}{dx}=2x+{\int }_0^x{f}^{{}^{\prime }}\left(t\right)dt$$$$f{}^{{}^{\prime }}\left(x\right)+f^{{}^{\prime }}\left(x\right)+{xf}^{{}^{″}}\left(x\right)=2+f^{{}^{\prime }}\left(x\right)$$$$$$$$then,f{}^{\prime }\left(x\right)+{xf}^{{}^{″}}\left(x\right)=2$$$$andifsubstitute0forx$$$$f{}^{\prime }\left(0\right)=2\mathrm{\&}f\left(0\right)=0$$$$$$$$$$$$\frac{d}{dx}\left({xf}^{{}^{\prime }}\left(x\right)\right)=2$$$$$$$$\therefore {xf}^{{}^{\prime }}\left(x\right)=2x+C_0$$$$$$$$$$$$$$$$howeverf\left(x\right)isdifferentiablefunction$$$$thatisf{}^{\prime }\left(x\right)iscontinuousfunction$$$$$$$$\underset{x\to 0}{\lim }f{}^{\prime }\left(x\right)=\underset{x\to 0}{\lim }\frac{2x+C_0}{x}=f{}^{\prime }\left(0\right)=2$$$$$$$$hence{C}_0=0andf{}^{\prime }\left(x\right)=2$$$$$$$$\therefore f\left(x\right)=2x+C_1$$$$f\left(0\right)=0$$$$$$$$f\left(x\right)=2xf\left(1\right)=2$$$$$$$$$$

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