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

Options

A. `y=-1=0`
B. `x-1=0`
C. `x+y-1=0`
D. `x-y+1=0`

Answer

A

Explanation

We have,
` y=(2x-1)e^(2(1-x))`
`rArr (dy)/(dx)=2e^(2(1-x))-2(2x-1)e^(2(1-x))`
` rArr (dy)/(dx)=2e^(2(1-x))(2-2x)=4e^(2(1-x))(1-x) `
At points of maximum, we must have
` (dy)/(dx)=0 rArr x=1. `
Now, `(d^(2)y)/(dx^(2))=-8e^(2(1-x))(1-x)-4e^(2(1-x)) `
` rArr ((d^(2)y)/(dx^(2)))_(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 of the tangent at (1,1) is
` y-1=0(x-1) rArr y=1 `