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

Given `y=(2x-1)e^(2(1-x))`
`(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 `(dy)/(dx)=0impliesx=1`
Now `(d^(2)y)/(dx^(2))=-8e^(2(1-x))(1-x)-4e^(2(1-x))`
At `(x=1)((d^(2)y)/(dx^(2)))_(x=1) =-4lt0`
So y is maximum at `x=1`
when `x=1` then `y=1`
Thus, the point of maximum is (1,1).
Now, the equation of the tangen at (1,1) is
`y-1=0(x-1)impliesy=1`