A particle free to move along the \( x \)-axis has potential energy given by \( U(x)=k\left[1-\e...

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A particle free to move along the \( x \)-axis has potential energy given by \( U(x)=k\left[1-\exp (-x)^{2}\right] \) for \( -\infty \leq x \leq+\infty \), where \( k \) is a positive constant of appropriate dimensions. Then
[IIT-JEE 1999; UPSEAT 2003]
(a) At point away from the origin, the particle is in unstable equilibrium
(b) For any finite non-zero value of \( x \), there is a force directed away from the origin
(c) If its total mechanical energy is \( k / 2 \), it has its minimum kinetic energy at the origin
(d) For small displacements from \( x=0 \), the motion is simple harmonic
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