13. Given the function $f(x)=\mathrm{e}^{x-1}+A x^{2}-1$. If $\forall x \geqslant 1$, we have $f(x) \geqslant A$ holds, find the range of real number $A$.

$$
\begin{array}{l}
f(x) \geqslant A \Rightarrow \mathrm{e}^{x-1}+A x^{2}-A-1 \geqslant 0, \text { let } g(x)=\mathrm{e}^{x-1}+A x^{2}-A-1, x \geqslant 1 . \\
g^{\prime}(x)=\mathrm{e}^{x-1}+2 A x, \quad g^{\prime \prime}(x)=\mathrm{e}^{x-1}+2 A .
\end{array}
$$

Then when $A \geqslant-\frac{1}{2}$, $g^{\prime \prime}(x) \geqslant 0 \Rightarrow g^{\prime}(x)$ is monotonically increasing on $[1,+\infty)$, thus $g^{\prime}(x) \geqslant g^{\prime}(1)=1+2 A \geqslant 0$ $\Rightarrow g(x)$ is monotonically increasing on $[1,+\infty)$ $\Rightarrow g(x) \geqslant g(1)=0$.
Therefore, when $A \geqslant-\frac{1}{2}$, $f(x) \geqslant A$ holds;
When $A<-\frac{1}{2}$, there exists $x_{0}>1$ such that $g^{\prime \prime}\left(x_{0}\right)<0$, then $g^{\prime}(x)$ is monotonically decreasing on $\left(1, x_{0}\right)$, thus on $\left(1, x_{0}\right)$, $g^{\prime}(x)<g^{\prime}(1)=1+2 A<0 \Rightarrow g(x)$ is monotonically decreasing on $\left(1, x_{0}\right)$.

Therefore, on $\left(1, x_{0}\right)$, $g(x)<g(1)=0$, which contradicts the given condition.
In summary, the range of real number $A$ is $\left[-\frac{1}{2},+\infty\right)$.