Applications Of Derivatives Question 315
Question: $ \frac{x}{1+x,\tan x} $ is maxima at
[UPSEAT 1999]
Options:
A) $ x=\sin x $
B) $ x=\cos x $
C) $ x=\frac{\pi }{3} $
D) $ x=\tan x $
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Answer:
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}{dx^{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 $
therefore $ {{\sec }^{2}}x=\frac{1}{x^{2}} $
therefore $ x^{2}={{\cos }^{2}}x $
therefore $ x=\cos x $
$ \therefore $ $ \frac{d^{2}y}{dx^{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.
BETA
