The graph of the function $\mathrm{y}=\mathrm{f}\left(\mathrm{x}\right)$ passing through the point (0, 1) and satisfying the differential equation $\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{ycos} \mathrm{x}=\cos \mathrm{x}$ is such that

$\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{ycos} \mathrm{x}=\cos \mathrm{x} (\mathrm{linear})$
I.F. $={\mathrm{e}}^{\int \cos \mathrm{xdx}}={\mathrm{e}}^{\sin \mathrm{x}}$
Thus. solution is $$\begin{gathered} {\mathrm{ye}}^{\sin \mathrm{x}}=\int {\mathrm{e}}^{\sin \mathrm{x}}\cos \mathrm{xdx} \\ y{e}^{\sin x}={\mathrm{e}}^{\sin \mathrm{x}}+\mathrm{c} \end{gathered}$$
When $\mathrm{x}=0,\mathrm{y}=1$,then $\mathrm{c}=0$
Thus, $\mathrm{y}=1$. Hence, options(a), (b), (d) are true.