62f(x)=cos x - 1
g(x) = ax+b^2
What i can see is f(g(x))=g(f(x))
Cant see anythign beyond this right now [2] except the feeling that the answer could probably be infinity!
cant think! (may be bcos of the fever :P)
341Since it is an identity, it should be true for x=0. On substituting this, we get b^2 = \cos b^2 -1 \le 0
This immediately gives us b = 0.
Hence we have for all x, a(\cos x -1) = \cos(ax) -1 \Rightarrow \cos (ax) - a \cos x + a-1=0
Differentiating, we get a (\sin (ax) - \sin x) =0
a=0 is a solution. Else we must have \sin (ax) - \sin x =0 for all x.
WLOG a>0. Now if we choose x such that 0<x<\frac{\pi}{2a}, then we have both x and ax lying in the interval \left(0, \frac{\pi}{2} \right)
For such x, \sin ax = \sin x \Rightarrow a=1 as sin x is strictly monotonic in that interval
Hence the only solutions are (0,0) and (1,0).
1ur solution is not visible ... plz post it again!!
341
Hari Shankar
·2010-07-29 06:29:10
Nishant sir, I dont know what the problem is. The solution is visible to me though.
1Yes PROPHET sir ... the solution is visible !!!
341
Hari Shankar
·2010-07-29 06:40:56
maybe the latex parts were not visible?
1
ARKA(REEK)
·2010-07-29 06:44:17
The LATEX parts r very much visible .... Sir ... !!!
62
Lokesh Verma
·2010-07-29 08:04:58
visible is ok.. but did u understand it ;)
btw yeah i missed out on a sitter :P making things so much more complex :(
