If $\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}=\overrightarrow{\mathrm{R}} \mathrm{and} 2\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}$ is perpendicular to $\overrightarrow{\mathrm{B}}$ then

Given,

$\overrightarrow{A}+\overrightarrow{B}=\overrightarrow{R}$

We know that,

$\overrightarrow{P}\overrightarrow{Q}=P.Q\cos \theta$

substituting the value,

$$\begin{gathered} (2\overrightarrow{A}+\overrightarrow{B}).\left(\overrightarrow{B}\right)=0 \\ \Rightarrow 2\overrightarrow{A}\overrightarrow{B}+B^2=0 \\ \Rightarrow 2\overrightarrow{A}\overrightarrow{B}=-{\mathit{B}}^{\mathbf{2}} \end{gathered}$$

Now,

$$\begin{gathered} \overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}=\overrightarrow{\mathrm{R}} \\ {\mathrm{R}}^2={\mathrm{A}}^2+{\mathrm{B}}^2+2\overrightarrow{\mathrm{A}}\overrightarrow{\mathrm{B}} \\ \mathrm{on} \mathrm{substituting} \mathrm{the} \mathrm{value} \mathrm{of} 2\overrightarrow{\mathrm{A}}\overrightarrow{\mathrm{B}} \end{gathered}$$

$$\begin{gathered} R^2=A^2+B^2-B^2 \\ \Rightarrow R^2=A^2 \\ \Rightarrow R=A \end{gathered}$$

Hence the correct answer is A=R.