A force $\overline{F}=k(y^2\widehat{i}+x\widehat{j})$ , where k is a positive constant, acts on a particle when it is at position (x, y). First, the particle is taken through path OAB and the work done by the force is $W_{OAB}.$ Then the particle is taken on through path OCB and the work done by the force is $W_{OCB}.$ Then

 $W_{OA}=0\left(\because Y=0\text{\hspace{0.17em}along\hspace{0.17em}}OA\text{\hspace{0.17em}and\hspace{0.17em}}\overrightarrow{F.}\overrightarrow{X}\to kx\widehat{j}.x\widehat{i}=0\right)$
 $W_{AB}=\underset{0}{\overset{a}{\int }}k(y^2\widehat{i}+a\widehat{j}).dy\widehat{j}=k\underset{0}{\overset{a}{\int }}ady=k{a}^2;$
 $\therefore W_{OAB}=k{a}^2$
$\Rightarrow \left(a\right)$ is correct
$W_{OC{B}^{:}}$  path  $OC\Rightarrow x=0$
 $\Rightarrow F.ds=k\left(y^{2}i\right).dyj=0$
Path  $CB\Rightarrow y=a\Rightarrow F=k(a^2\widehat{i}+x\widehat{j}).dx\widehat{i}$
 $\therefore w_{C\mathbf{B}}=\underset{0}{\overset{a}{\int }}\overline{F}.\overline{ds}=\underset{0}{\overset{a}{\int }}k(a^2\widehat{i}+x\widehat{j}).dx\widehat{i}$
 $=k\underset{0}{\overset{a}{\int }}{a}^{2}dx=k{a}^3$
Since the work done is path dependent, the force is non-conservative. 
(d) is correct
$\therefore$  (a) and (d) are correct