A) \[x=\sin x\]
B) \[x=\cos x\]
C) \[x=\frac{\pi }{3}\]
D) \[x=\tan x\]
Correct Answer: B
Solution :
If \[\frac{x}{1+x\tan x}\] is maxima, then its reciprocal \[\frac{1+x\tan x}{x}\] will be minima. Let \[y=\frac{1+x\tan x}{x}\] = \[\frac{1}{x}+\tan x\] \\[\frac{dy}{dx}=-\frac{1}{{{x}^{2}}}+{{\sec }^{2}}x\], \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=\frac{2}{{{x}^{3}}}+2\sec x\sec x\tan x\] On putting \[\frac{dy}{dx}=0\], \[-\frac{1}{{{x}^{2}}}+{{\sec }^{2}}x=0\] Þ \[{{\sec }^{2}}x=\frac{1}{{{x}^{2}}}\] Þ \[{{x}^{2}}={{\cos }^{2}}x\] Þ \[x=\cos x\] \[\therefore \]\[\frac{{{d}^{2}}y}{d{{x}^{2}}}=\frac{2}{{{\cos }^{3}}x}+2{{\sec }^{2}}x\tan x\] = \[2{{\sec }^{2}}x(\sec x+\tan x)\], which is positive. At \[x=\cos x,\] \[\,\frac{1+x\tan x}{x}\] is minimum. So \[\frac{x}{1+x\tan x}\] will be maximum.
You need to login to perform this action.
You will be redirected in
3 sec
