The equation of the tangent to the curve $y=(2 x-1) e^{2(1-x)}$ at the point of its maximum, is

We have, $y=(2 x-1) e^{2(1-x)}$ On differentiating both sides w.r.t. $x$, we get $\begin{aligned} \frac{d y}{d x} & =2 e^{2(1-x)}-2(2 x-1) e^{2(1-x)} \\ \Rightarrow \quad \frac{d y}{d x} & =2 e^{2(1-x)}(2-2 x)=4 e^{2(1-x)}(1-x)\end{aligned}$ At points of maximum, we must have $\frac{d y}{d x}=0 \Rightarrow x=1$ Now, $\frac{d^2 y}{d x^2}=-8 e^{2(1-x)}(1-x)-4 e^{2(1-x)}$ $\Rightarrow \quad\left[\frac{d^2 y}{d x^2}\right]_{x=1}=-4 \lt 0$ So, $y$ is maximum at $x=1$. Clearly, $y=1$ for $x=1$. Thus, the point of maximum is $(1,1)$. The equation is the tangent at $(1,1)$ is $y-1=0(x-1)$ $\Rightarrow \quad y=1$