f(x)=log(pi+ x)/log(e+x) .
For simple calculations we take log on both sides.
We have log f(x) = log log (pi+x) – log log (e+x)
Differntiating both sides with respect to x,
We have f’(x)/f(x) = 1/(pi+x) log (pi+x) - 1/(e+x)log (e+x)
We know e lies between 2 and 3 but pi between 3 and 4. So pi+x>e+x for all x>0
So when x>0 f’(x)/f(x) = 1/ larger number – 1/ smaller number = negative
So decreasing for interval (0, infinity)
Similarly can be proved it is increasing in (-infinity,0]
cos(nx).sin(5x/n) = sin A cos B = ½ [sin (A+b)+Sin (a-b) ] Here a = 5x/n and b=nx
So given function = ½ [ sin (5x/n +nx ) + sin (5x/n –nx)
= ½ [ sin (5x+n^2x)/n + sin 5x-n^2x /n]
Period is given as 3pi. Period of sum of two functions is the period of the largest of the one in the sum
So here larger will be 5x+n^2x /n here coefficient of x = (5+ n^2)/n = 3 pi
2pi/5+n^2/n = 3pi or We solve for n and get answer. Here we do not have real solutions for n. So no
such n exists may be the right answer.
Regards,
Nirmal Singh
Askiitians Faculty