1. ## tough problem 1

please see problem attached to this post

2. Originally Posted by anshu
pls see problem attached 2 dis post
The global minimum of $f(x)$ occurs at an end point of the interval if the calculus like minimum (the solution of $f'(x)=0$) does not lie in the interval $(0,1).$

Now find the solution of $f'(u)=0$ (in terms of $a$) and use the fact that $f(u)=3$ (which gives a quadratic in $a$) and solve.

Now how many values of $a$ do you have and do the corresponding values of $u$ lie in $(0,1)$ ?

CB