tough problem 1

Quote:

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