Given, potential function for the oscillating particle is

$V(x)=\left\{\begin{array}{cl}k(x+a)^2, & x < 0 \\ \frac{k(x-a)^2}{2}, & x > 0 \\ \end{array}\right.$

So, potential energy of the particle (mass $m)$ is

$U(x)=\left\{\begin{array}{cc}\frac{k m(x+a)^2}{2}, & x < 0 \\ \frac{k m(x-a)^2}{2}, & x < 0 \\ \end{array}\right.$

$\frac{d U}{d x}=\left\{\begin{array}{ll}k m(x+a), & x < 0 \\ k m(x-a), & x > 0\end{array}\right.$

If $\frac{d U}{d x}=0$, when $x=\pm a$

Now, $\frac{d^2 U}{d x^2}=k m > 0$

So, particle is in unstable equilibrium at $x=\pm a$.

Hence, particle is unbounded for $-a > x$ and $x > a$.

In region, $-a \leq x \leq a$, time period of particle reduces from a maximum.

So, correct graph is $(b)$.