Explanation

(a)
$y = (2x - 1)e^{2(1 - x)}$
$\Rightarrow \ \ \ \ \ \ \ \frac{dy}{dx} = 2e^{2(1 - x)} - 2(2x - 1)e^{2(1 - x)}$
$= 2e^{2(1 - x)}(2 - 2x)$
$= 4e^{2(1 - x)}(1 - x)$
Put, $\frac{dy}{dx} = 0 \Rightarrow \ \ x = 1$
Now, $\frac{d^{2}y}{dx^{2}} = - 8e^{2(1 - x)}(1 - x) - 4e^{2(1 - x)}$
$\Rightarrow \ \ \ \ \left( \frac{d^{2}y}{dx^{2}}\ \right)_{x = 1} = - 4 < 0$
So, $y$ is maximum at $x = 1$, when $x = 1,\ y = 1.$
Thus, the point of maximum is (1,1).
The equation of the tangent at (1,1) is
$y - 1 = 0(x - 1) \Rightarrow y = 1$