The equation of the tangent to the curve y=(2 x-1) e^2(1-x) at the po…

We have,

$y=(2 x-1) e^{2(1-x)}$

$\Rightarrow \frac{d y}{d x}=2 e^{2(1-x)}-2(2 x-1) e^{2(1-x)}$

$\Rightarrow \frac{d y}{d x}=2 e^{2(1-x)}(2-2 x)=4 e^{2(1-x)}(1-x)$

At points of maximum, we must have

$\frac{d y}{d x}=0 \Rightarrow x=1$

Now, $\frac{d^2 y}{d x^2}=-8 e^{2(1-x)}(1-x)-4 e^{2(1-x)}$

$\Rightarrow\left(\frac{d^2 y}{d x^2}\right)_{x=1}=-4<0$

So, y is maximum at x = 1. Clearly, y = 1 for x = 1.

Thus, the point of maximum is (1, 1). The equation of the tangent at (1, 1) is

$y-1=0(x-1) \Rightarrow y=1$