6. A. $a \sqrt{1+b^{2}}=\sqrt{2} a \cdot \sqrt{\frac{1+b^{2}}{2}} \leqslant \sqrt{2} \cdot \frac{a^{2}+\frac{1+b^{2}}{2}}{2}=\frac{3 \sqrt{2}}{4}$.
6. Let $a, b \in \mathbf{R}^{+}$, and $a^{2}+\frac{b^{2}}{2}=1, a \sqrt{1+b^{2}}$ has a maximum value of $(\quad)$.
A. $\frac{3 \sqrt{2}}{4}$
B. $2 \sqrt{2}$
C. $\frac{1}{2}$
D. $\sqrt{2}$