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))`
`because" "(dy)/(dx)=2e^(2(1-x))(2-2x)`
`:." "(dy)/(dx)=0`
`implies" "x=1`
`(d^(2)y)/(dx^(2))=-e^(2(1-x))(3-2x)`
Which is -ve at x = 1.
Hence, y is maixmum at the point (1,1).
Therefore, slope of tangent at the points (1,1) is 0 hence equation of tangent is
`y-1=0(x-1)`
`because" "y-1=0`