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Let f(x) = (2x-pie) 3 +2x - cosx. Then show that d[ f -1 (x)] / dx at x=pie/2 is 1/3.
(1) It's "Pi". and Pi is not 22/7 first find df(x) / dx. df(x)/dx = 3(2x - pi)2 * 2 + 2 + sin(x) =6(2x - pi)2 + 2 + sin(x) Let x = pi/2 , we get: 0 + 2 + 1 = 3 So the answer for df-1(x)/dx i.e., 1/[df(x)/dx] is : 1/3 -- 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
(1) It's "Pi". and Pi is not 22/7
first find df(x) / dx. df(x)/dx = 3(2x - pi)2 * 2 + 2 + sin(x)
=6(2x - pi)2 + 2 + sin(x) Let x = pi/2 , we get: 0 + 2 + 1 = 3 So the answer for df-1(x)/dx i.e., 1/[df(x)/dx] is : 1/3
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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
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