Suppose ${\displaystyle X_1,\cdots ,X_n\stackrel{\mathrm{i}.\mathrm{i}.\mathrm{d}}{\sim }\mathrm{Laplace}(\theta )=\frac{1}{2}{\mathrm{e}}^{-|x-\theta |}}$ and we want to test $$H_0:\theta \le 0\text{ vs }{H}_1:\theta >0.$$ Given ${\displaystyle {\theta }_1}$, the LRT is $$\log \frac{p_{{\theta }_1}(X)}{p_0(X)}=\sum _{i=1}^n|X_i|-|X_i-{\theta }_1|={\theta }_1\sum _{i=1}^n{T}_{{\theta }_1}(X_i),$$ where $$T_{\theta }(x)=\{\begin{array}{rl}& -1,x\le 0,\\ & \frac{2x}{\theta }-1,0\le x\le \theta ,\\ & 1,x\ge \theta .\end{array}$$
If take ${\displaystyle {\theta }_1\to 0^{+}}$, ${\displaystyle T_{\theta }(x)}$ approaches to $$T_0(x)=\{\begin{array}{rl}& -1,x\le 0,\\ & +1,x>0.\end{array}$$ Since $$\dot{l}(\theta ;X)=\frac{\mathrm{d}}{\mathrm{d}\theta }\sum _{i=1}^n-|X_i-\theta |=\sum _{i=1}^n{T}_0(X_i),$$ then ${\displaystyle T_0(x)}$ gives the score test.

If we test ${\displaystyle H_0:\theta =0\text{ vs }{H}_1:\theta \ne 0}$ for ${\displaystyle X\sim \mathcal{N}(\theta ,1)}$, we choose $${\varphi }_{{\alpha }_1}(x)=\mathbf{1}\{x<-z_{{\alpha }_1}\}+\mathbf{1}\{x>z_{\alpha -{\alpha }_1}\},\mathrm{\forall }{\alpha }_1\in [0,\alpha ].$$

This is undesirable because the power falls below ${\displaystyle \alpha }$ on part of the alternative: there are alternative values of ${\displaystyle \theta }$ for which our chance of rejecting the null is even less than it would be if the null were true.
Intuitively, we can't find a test that maximizes power on both parts of the alternative.

Consider ${\displaystyle H_0:\theta =1\text{ vs }{H}_1:\theta \ne 1}$ for ${\displaystyle X\sim \mathrm{Exp}(\theta )}$, with c.d.f $$F_{\theta }(t)={\mathbb{P}}_{\theta }(X\le t)=1-{\mathrm{e}}^{-\frac{t}{\theta }}.$$
We set ${\displaystyle c_1=F_1^{-1}(1-\frac{\alpha }{2})=-\log (1-\frac{\alpha }{2})}$, ${\displaystyle c_2=F_1^{-1}\left(\frac{\alpha }{2}\right)=-\log \left(\frac{\alpha }{2}\right)}$. Then the power is $$\begin{array}{rl}{\beta }_{\varphi }(\theta )& ={\mathbb{P}}_{\theta }(X<c_1)+{\mathbb{P}}_{\theta }(X>c_2)\\ & =1-{\mathrm{e}}^{-\frac{c_1}{\theta }}+{\mathrm{e}}^{-\frac{c_2}{\theta }}\\ & =1-{(1-\frac{\alpha }{2})}^{\frac{1}{\theta }}+{\left(\frac{\alpha }{2}\right)}^{\frac{1}{\theta }}.\end{array}$$
The power is below ${\displaystyle \alpha }$ every on ${\displaystyle \theta \in (\frac{1}{2},1)}$, so it is not unbiased.
If we want unbiased test, we have to set derivative of the power to ${\displaystyle 0}$ at ${\displaystyle \theta =1}$. We can solve numerically for $$\{\begin{array}{rl}& c_1({\alpha }_1)=-\log (1-{\alpha }_1),\\ & c_2({\alpha }_2)=-\log ({\alpha }_2)=-\log (\alpha -{\alpha }_1).\end{array}$$

In exponential family, let ${\displaystyle X\sim p_{\theta }(x)={\mathrm{e}}^{\theta T(x)-A(\theta )}h(x)}$, (this is MLR in ${\displaystyle T(X)}$,) then $$\begin{array}{rl}{\dot{\beta }}_{\varphi }({\theta }_0)& =\frac{\mathrm{d}}{\mathrm{d}\theta }{\int \varphi (x){\mathrm{e}}^{\theta T(x)-A(\theta )}h(x)\mathrm{d}\mu (x)|}_{\theta ={\theta }_0}\\ & =\int \varphi (x)(T(x)-\dot{A}({\theta }_0)){\mathrm{e}}^{{\theta }_{0}T(x)-A({\theta }_0)}h(x)\mathrm{d}\mu (x)\\ & ={\mathbb{E}}_{{\theta }_0}[\varphi (X)(T(X)-{\mathbb{E}}_{{\theta }_0}T(X))]\\ & ={\mathrm{Cov}}_{{\theta }_0}(T(X),\varphi (X))\\ & ={\mathbb{E}}_{{\theta }_0}[(\varphi (X)-\alpha )T(X)].\end{array}$$
Set the last expression to ${\displaystyle 0}$, we have $${\mathbb{E}}_{{\theta }_0}T(X)=\frac{{\mathbb{E}}_{{\theta }_0}[\varphi (X)T(X)]}{\alpha }={\mathbb{E}}_{{\theta }_0}[T(X)|\varphi (X)\text{ rejects }{H}_0].$$

WLOG: ${\displaystyle {\theta }_0=0}$. First consider ${\displaystyle \delta =0}$. The proof is similar as proof of NP lemma. For ${\displaystyle \theta \ne 0}$, solve $$\begin{array}{rl}\underset{\varphi }{min}& \int \varphi (x)p_{\theta }(x)\mathrm{d}\mu (x),\\ \mathrm{s}.\mathrm{t}.& \int \varphi (x)p_0(x)\mathrm{d}\mu (x)=\alpha ,\\ & \int \varphi (x)(T(x)-{\nu }_0)p_0(x)\mathrm{d}\mu (x)=0,\end{array}$$ where ${\displaystyle {\nu }_0={\mathbb{E}}_{0}T(X)}$. The Lagrangian is $$\begin{array}{rl}& \int \varphi p_{\theta }\mathrm{d}\mu -{\lambda }_1\int \varphi p_0\mathrm{d}\mu -{\lambda }_2\int \varphi (T-{\nu }_0)p_0\mathrm{d}\mu \\ =& \int \varphi (p_{\theta }-{\lambda }_1{p}_0-{\lambda }_2(T-{\nu }_0)p_0)\mathrm{d}\mu \\ =& \int \varphi (\frac{p_{\theta }}{p_0}-{\lambda }_1-{\lambda }_2(T-{\nu }_0))\mathrm{d}{P}_0.\end{array}$$
Since ${\displaystyle \frac{p_{\theta }}{p_0}(x)={\mathrm{e}}^{\theta T(x)-A(\theta )+A(0)}}$, the test maximizing the Lagrangian is $${\varphi }^{\ast }(x)=\{\begin{array}{rl}& 1,{\mathrm{e}}^{\theta T(x)}>a_0+a_{1}T(x),\\ & 0,{\mathrm{e}}^{\theta T(x)}<a_0+a_{1}T(x),\\ & \text{anything},{\mathrm{e}}^{\theta T(x)}=a_0+a_{1}T(x).\end{array}$$ Here ${\displaystyle a_0=({\lambda }_1-{\lambda }_2{\nu }_0){\mathrm{e}}^{A(0)-A(\theta )}}$, ${\displaystyle a_1={\lambda }_2{\mathrm{e}}^{A(0)-A(\theta )}}$.
${\displaystyle \mathrm{\forall }{c}_1,c_2}$, we can find ${\displaystyle a_1>0,a_0\in \mathbb{R}}$ for which ${\displaystyle {\mathrm{e}}^{\theta t}=a_0+a_{1}t}$ at ${\displaystyle t=c_1,c_2}$, in which case $${\mathrm{e}}^{t\theta }=\{\begin{array}{rl}& >a_0+a_{1}t,t<c_1\text{ or }t>c_2,\\ & <a_0+a_{1}t,\text{otherwise}.\end{array}$$ Then solve ${\displaystyle {\lambda }_1,{\lambda }_2}$.
For any other unbiased test ${\displaystyle \varphi }$ $$\begin{array}{rl}{\beta }_{\varphi }(\theta )& ={\beta }_{\varphi }(\theta )-{\lambda }_1({\beta }_{\varphi }(0)-\alpha )-{\lambda }_2{\dot{\beta }}_{\varphi }(0)\\ & \le {\beta }_{{\varphi }^{\ast }}(\theta )-{\lambda }_1({\beta }_{{\varphi }^{\ast }}(0)-\alpha )-{\lambda }_2{\dot{\beta }}_{{\varphi }^{\ast }}(0)\\ & ={\beta }_{{\varphi }^{\ast }}(\theta ).\end{array}$$
For ${\displaystyle \delta >0}$, replace constraints ${\displaystyle {\beta }_{\varphi }(0)=\alpha ,{\dot{\beta }}_{\varphi }(0)=0}$ by ${\displaystyle {\beta }_{\varphi }(-\delta )={\beta }_{\varphi }(\delta )=\alpha }$.