GATE Computer Science (CS) 2019 Solved Papers

(a) $\left\{\mathit{w}{\mathit{w}}^{\mathit{R}}\mid \mathit{w}\in {\left\{\mathit{a},\mathit{b}\right\}}^{*}\right\}$ is a CFL
(b) $\left\{\mathit{w}{\mathit{a}}^{\mathit{n}}{\mathit{b}}^{\mathit{n}}{\mathit{w}}^{\mathit{R}}\mid \mathit{w}\in {\left\{\mathit{a},\mathit{b}\right\}}^{*},\mathit{n}\ge 0\right\}$ is a CFL, since we can first push w, then a's, b's pop with a's and ${\mathit{w}}^{\mathit{R}}$ pops with the $\mathit{w}$. So PDA can accept the language.
(c) $\left\{{\mathit{w}\mathit{a}}^{\mathit{n}}{\mathit{w}}^{\mathit{R}}{\mathit{b}}^{\mathit{n}}\mid \mathit{w}\in {\left\{\mathit{a},\mathit{b}\right\}}^{*},\mathit{n}\ge 0\right\}$ is a not CFL because after pushing w, we need to push a's into stack which will stop the $\mathit{w}$ from being matched with ${\mathit{w}}^{\mathit{R}}.$ If we don't push a's after $\mathit{w}$, than later we cannot match with bn. So this language is not acceptable by a PDA and hence not a CFL.
(d) $\left\{{\mathit{a}}^{\mathit{n}}{\mathit{b}}^{\mathit{i}}\mid \mathit{i}\in \left\{\mathit{n},3\mathit{n},5\mathit{n}\right\},\mathit{n}\ge 0\right\}$ $={\mathit{a}}^{\mathit{n}}{\mathit{b}}^{\mathit{n}}\cup {\mathit{a}}^{\mathit{n}}{\mathit{b}}^{3\mathit{n}}\cup {\mathit{a}}^{\mathit{n}}{\mathit{b}}^{5\mathit{n}}$ is $\mathit{C}\mathit{F}\mathit{L}$ since each of the three parts is a CFL and closure under union guarantees that result also is a CFL.