In this I could not understand how they have written th first equation in the solution.
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Lookup integral of inverse functions. – dxiv May 04 '17 at 18:53
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I think the author(s) of the exam assumed that $f$ was not only invertible but also continuous. The formula for integrating an inverse function is not valid for discontinuous invertible functions (and yes, such functions exist). Maybe there's some way to infer from the formula in the problem statement that $f$ is continuous, but it's not obvious to me. – David K May 04 '17 at 19:41
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Put $$F (x)=\int_0^x f(t)dt+\int_{f (0)}^{f (x)}f^{-1}(t)dt. $$
then by FTC, $$F'(x)=f (x)+f^{-1}(f (x))f'(x) $$ $$=f (x)+xf'(x) =(xf (x))'$$ thus $$F (x)=xf (x)+C $$
but $F (0)=0$.
then $$F (x)=xf (x)-0$$
hamam_Abdallah
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It is given that, \begin{align*} f(x)+xf(x)&=1+\int_1^{f(x)}f^{-1}(t)\ dt\\ \implies f'(x)+xf'(x)+f(x)&=f^{-1}(f(x))\cdot f'(x)\hspace{25pt}\text{ by differentiating both side. }\\ \implies f'(x)+xf'(x)+f(x)&=xf'(x)\\ \implies \dfrac{d}{dx}f(x)+f(x)&=0\\ \implies \dfrac{d\,f(x)}{f(x)}&=-dx\\ \implies \int\dfrac{d\,f(x)}{f(x)}&=-\int dx+k\hspace{25pt}k=\text{integration const. }\\ \implies \ln\left|f(x)\right|&=-x+k\\ \implies f(x)&=ce^{-x}\hspace{25pt}e^k=c=\text{const. }\\ \implies f(x)&=e^{-x}\hspace{25pt}\text{ as, }f(0)=1\implies c=1 \end{align*} Now, \begin{align*} g(x)&=f(|x|)\\ \implies g(x)&=e^{-|x|}\\ \implies g(x^-)&=e^x\hspace{25pt}\text{ i.e when, }x<0\\ \implies g'(x^-)&=e^x\\ \implies g'(0^-)&=1. \end{align*}
k.Vijay
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