Manipal Entrance Test 2015 Math Paper

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