5. Let $f(x)=x^{2}-\pi x, \alpha=\arcsin \frac{1}{3}, \beta=\arctan \frac{5}{4}, \gamma=\arccos \left(-\frac{1}{3}\right), \delta=\operatorname{arccot}\left(-\frac{5}{4}\right)$, then ( ).
A. $f(\alpha)>f(\beta)>f(\delta)>f(\gamma)$
B. $f(\alpha)>f(\delta)>f(\beta)>f(\gamma)$
C. $f(\delta)>f(\alpha)>f(\beta)>f(\gamma)$
D. $f(\delta)>f(\alpha)>f(\gamma)>f(\beta)$
5. B

According to the problem, the graph of $f(x)$ is symmetric about the line $x=\frac{\pi}{2}$, and it is monotonically decreasing in $\left(-\infty, \frac{\pi}{2}\right)$, and monotonically increasing in $\left(\frac{\pi}{2},+\infty\right)$. Therefore, when $\left|x_{1}-\frac{\pi}{2}\right|>\left|x_{2}-\frac{\pi}{2}\right|$, we have $f\left(x_{1}\right)>f\left(x_{2}\right)$. It is also easy to see that $0f(\delta)>f(\beta)>f(\gamma)$.
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