5. Given the function $f(x)=\left\{\begin{array}{l}2 x^{2}+4 x+1, x<0, \\ \frac{2}{\mathrm{e}^{x}}, x \geqslant 0,\end{array}\right.$ then the number of points on the graph of $y=f(x)(x \in \mathbf{R})$ that are symmetric with respect to the origin $O$ is
A. 0 pairs
B. 1 pair
C. 2 pairs
D. 3 pairs
Consider $y=2 x^{2}+4 x+1$ and its symmetric graph with respect to the origin $y=-2 x^{2}+4 x-1$ intersecting with $y=\frac{2}{\mathrm{e}^{x}}, x \geqslant 0$. Let $g(x)=2 \mathrm{e}^{-x}+2 x^{2}-4 x+1, x \geqslant 0, g^{\prime}(x)=-2 \mathrm{e}^{-x}+4 x-4$, $g^{\prime \prime}(x)=2 \mathrm{e}^{-x}+4>0 \Rightarrow g^{\prime}(x)$ is monotonically increasing, and $g^{\prime}(1)=-2 \mathrm{e}^{-1}<0$, $g^{\prime}(2)=-2 \mathrm{e}^{-2}+4>0$, then there exists $x_{0} \in(1,2)$ such that $g(x)$ is monotonically decreasing on $\left(0, x_{0}\right)$ and monotonically increasing on $\left(x_{0},+\infty\right)$. Since $g(0)=3>0$, $g\left(x_{0}\right)<0$, and $g(x) \to +\infty$ as $x \to +\infty$, $g(x)$ has 2 zeros. Therefore, the answer is $C$.