A potential is given by $V\left(x\right)=k(x+a{)}^2/2$ for $x<0$ and $V\left(x\right)=k(x-a{)}^2/2$ for $x>0$. The schematic variation of oscillation period $T$ for a particle performing periodic motion in this potential as a function of its energy $E$ is

 Given, potential function for the oscillating particle is

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

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

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

$\frac{dU}{dx}=\left\{\begin{array}{ll}km(x+a),& x<0\\ km(x-a),& x>0\end{array}\right.$

If $\frac{dU}{dx}=0,$ when $x=\pm a$

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

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

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

In region, $-a\le x\le a$, time period of particle reduces from a maximum.
So, correct graph is (b)..