IIT JEE Advanced 2006 Paper 1

We have
f (x) = {

ex	0≤x≤1
2−ex−1	1<x≤2
x−e	2<x≤3

g (x) =

x

∫

0

f (t) dt , x ∊ [1 , 3]

Now, it is obvious from the graph that f(x) has local maxima at x = 1 and local minima at x = 2.
g′(x) = f (x) = 0
That is, ex = 0, which is not possible. Therefore,
2−ex−1] = 0
⇒ ex−1 = 2
⇒ x – 1 = ln2
⇒ x = 1 + ln2
x – e = 0 ⇒ x = e
Maxima at x = 1 + ln2 and minima at x = e.