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function f(x)=log(pi+ x)/log(e+x) is increasing in?


6 years ago

							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>0So when x>0 f’(x)/f(x) = 1/ larger number – 1/ smaller number = negativeSo 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=nxSo 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 sumSo here larger will be 5x+n^2x /n here coefficient of x = (5+ n^2)/n = 3 pi2pi/5+n^2/n = 3pi or We solve for n and get answer. Here we do not have real solutions for n. So nosuch n exists may be the right answer.Regards,Nirmal SinghAskiitians Faculty

6 years ago
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