Let y(x) be a solution of the differential equation $\left(1+e^x\right)y^{\mathrm{\prime }}+y{e}^x=1$. 

If $y\left(0\right)=2$ then which of the following statement(s) is (are) true?

we have ,

$\left(1+e^x\right)y^{\mathrm{\prime }}+y{e}^x=1$

$\Rightarrow y^{\mathrm{\prime }}+\frac{e^x}{1+e^x}y=\frac{1}{1+e^x}$             …(i)

This is a linear differential equation with integrating factor 

$\text{ I.F }=e^{\int \frac{e^x}{1+e^x}dx}=e^{\log \left(1+e^x\right)}=1+e^x$

Multiplying both sides of (i) by $I.F. == 1 +e^x$, we obtain

$y^{\mathrm{\prime }}\left(1+e^x\right)+y{e}^x=1$

Integrating both sides with respect to x, we obtain

$y\left(1+e^x\right)=x+C$                       …(ii)

It is given that $y\left(0\right) =2$ i.e.$y =2$ when$x =0.$ Putting $x =0, y =2$ in (ii), we get 

$4=0+C\Rightarrow C=4$

Putting $C =4$ in (ii), we obtain  

$y\left(1+e^x\right)=x+4\Rightarrow y=\frac{x+4}{1+e^x}\Rightarrow y(-4)=0$

Putting $y=\frac{x+4}{1+e^x}$in (i), we obtain 

$y^{\mathrm{\prime }}=\frac{\left(1+e^x\right)-(x+4)e^x}{{\left(1+e^x\right)}^2}$

$\Rightarrow y^{\mathrm{\prime }}(-1)=\frac{\left(1+e^{-1}\right)-3e^{-1}}{{\left(1+e^{-1}\right)}^2}=\frac{e-2}{e{\left(1+e^{-1}\right)}^2}>0$ and $y^{\mathrm{\prime }}\left(0\right)=\frac{1}{2}<0$

$\Rightarrow y\text{'} \left(x\right) =0$ for some $x \in (-1, 0)$

$\Rightarrow y\left(x\right)$ has a critical point in the interval $(-1, 0).$

Hence, options (a) and (c) are true.