Manipal Entrance Test 2015 Math Paper

We have,
f(x)=xex(1−x)
On differentiating both sides w.r.t. x, we get
f′(x)=ex(1−x)+x(1−2x)ex(1−x)
⇒f′(x)=(1+x−2x2)ex(1−x)
⇒f′(x)=−(x−1)(2x+1)ex(1−x)
since, ex(1−x)>0 for all x. Therefore, signs of f′(x) for different values of x are as shown below.Clearly, f(x) is increasing on [−

1

2

,1] and decreasing on (−∞,−

1

2

]∪[1,∞).