Let f : (0 , pie) is defined to R by f(x) = max { sin2x, cos2x}. then show that the set of points of which f(x) has a local maxiam or minima is equlas to { pie/8 , pie/4 , 5 pie/8 } .

From the figure , the curve f(x) is shown in black bold line

the local minima is at : Pi/8 , 5Pi/8

and local maxima at : Pi/4

---

Please feel free to post as many doubts on our disucssion forum as you can. If you find any question difficult to understand - post it here and
we will get you the answer and detailed solution very quickly.We are all IITians and here to help you in your IIT JEE preparation. All the best.

Regards,
Naga Ramesh
IIT Kgp - 2005 batch

period of cos2x,sin2x is pie

we will redefine the function given

now in(0,pie/8) cos2x.>sin2x , in (pie/8 to 5pie/8) sin2x>cos2x and in ( 5pie/8 to pie)again cos2x>sin2x (by comparing graph)

sof(x) = cos2x.  (0,pie/8)

               sin2x    (pie/8 to 5pie/8)

               cos2x   ( 5pie/8 to pie)

        now u can get it by drawing the graph

note : pie is not the answer because pie is the last point of the fonction so it is the global maxima not the local maxima.