We have, y=(2x−1)e2(1−x) On differentiating both sides w.r.t. x, we get
dy
dx
=2e2(1−x)−2(2x−1)e2(1−x)
dy
dx
=2e2(1−x)(2−2x)=4e2(1−x)(1−x) At points of maximum, we must have
dy
dx
=0⇒x=1 Now,
d2y
dx2
=−8e2(1−x)(1−x)−4e2(1−x) ⇒[
d2y
dx2
]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) ⇒y=1