Which of the following graphs are correct for a first order reaction?

For the first order reaction,
$\mathrm{k}=\frac{2.303}{\mathrm{t}}\log \frac{[\mathrm{R}{]}_0}{\left[\mathrm{R}\right]}\Rightarrow \frac{\mathrm{kt}}{2.303}=\log \frac{[\mathrm{R}{]}_0}{\left[\mathrm{R}\right]}$          
    $$\begin{gathered} \log \frac{{\left[\mathrm{R}\right]}_{\mathrm{o}}}{\left[\mathrm{R}\right]}=\left(\frac{\mathrm{k}}{2.303}\right)\mathrm{x} \mathrm{t}+0 \\ \uparrow \uparrow \uparrow \uparrow \\ \mathrm{y} = \mathrm{m} \mathrm{x} +\mathrm{c} \end{gathered}$$
Correct plot of log $\frac{{\left[\mathrm{R}\right]}_{\mathrm{o}}}{\left[\mathrm{R}\right]}$ can be represented by (d)
where, $\mathrm{slope} = \frac{\mathrm{k}}{2.303}$
The time taken for any fraction of the reaction to complete is independent of the initial concentration.
Let, us consider it for half of the reaction to complete.
              $\mathrm{t}=\frac{2303}{\mathrm{k}}\log \frac{\mathrm{a}}{\mathrm{a}-\mathrm{x}}$
For half-life $\left(\mathrm{t}\right)={\mathrm{t}}_{1/2}\text{ and }\mathrm{x}=\frac{\mathrm{a}}{2}$
                   ${\mathrm{t}}_{1/2}=\frac{2.303}{\mathrm{k}}\log \frac{\mathrm{a}}{\mathrm{a}-\frac{\mathrm{a}}{2}}$
           ${\mathrm{t}}_{1/2}=\frac{2.303}{\mathrm{k}}\log 2$
    Half-life time  $\left({\mathrm{t}}_{1/2}\right)=\frac{0.693}{\mathrm{k}}$
${\mathrm{t}}_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$is independent of initial concentration. Hence, correct plot of ${\mathrm{t}}_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$ and ${\left[\mathrm{R}\right]}_{\mathrm{o}}$can be represented by option (a).