3. Let $f(x)=\frac{b x+1}{2 x+a}$, where $a, b$ are constants, and $a b \neq 2$. If $f(x) \cdot f\left(\frac{1}{x}\right)=k$ is a constant, then $k=$
Solution: $f(x) \cdot f\left(\frac{1}{x}\right)=\frac{b x+1}{2 x+a} \cdot \frac{b+x}{2+a x}=\frac{b x^{2}+\left(b^{2}+1\right) x+b}{2 a x^{2}+\left(a^{2}+4\right) x+2 a}=k$,
If for $\forall x \neq 0$ we have $\frac{b x^{2}+\left(b^{2}+1\right) x+b}{2 a x^{2}+\left(a^{2}+4\right) x+2 a}=k$ is a constant,
then $\frac{b}{2 a}=\frac{b^{2}+1}{a^{2}+4}$, which means $(a b-2)(a-2 b)=0$,
Since $a b \neq 2$, we have $a=2 b$,
Thus, $f(x) \cdot f\left(\frac{1}{x}\right)=\frac{b x^{2}+\left(b^{2}+1\right) x+b}{2 a x^{2}+\left(a^{2}+4\right) x+2 a}=\frac{b}{2 a}=\frac{1}{4}$, i.e., $k=\frac{1}{4}$.