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(D) The relationship between force $F(x)$ and potential energy $U(x)$ is given by $F(x) = -\frac{dU}{dx}$.Integrating this,we get $U(x) = -\int F(x) d

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(D) The relationship between force $F(x)$ and potential energy $U(x)$ is given by $F(x) = -\frac{dU}{dx}$.Integrating this,we get $U(x) = -\int F(x) dx$.Substituting $F(x) = -kx + ax^3$,we have $U(x) = -\int (-kx + ax^3) dx = \frac{kx^2}{2} - \frac{ax^4}{4} + C$.Assuming $U(0) = 0$,we get $C = 0$,so $U(x) = \frac{kx^2}{2} - \frac{ax^4}{4} = \frac{x^2}{4}(2k - ax^2)$.For $x \ge 0$,$U(x) = 0$ at $x = 0$ and $x = \sqrt{\frac{2k}{a}}$.Between $x = 0$ and $x = \sqrt{\frac{2k}{a}}$,$U(x)$ is positive. For $x > \sqrt{\frac{2k}{a}}$,$U(x)$ becomes negative.The graph starts at the origin,rises to a maximum,and then decreases,crossing the $x-$axis at $x = \sqrt{\frac{2k}{a}}$. This matches Graph $D$.

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