Question

Find the value of $x$ for which $y=\frac{x}{1+x\tan x}$ is max. $0\le x\le \frac{\pi }{2}.$

• A. $x=\cos x$
• B. $x=\sin x$
• C. $x=\sec x$
• D. $x=cosecx$

Answer 1

Answer: (A)

$x=\cos x$

$y=\frac{x}{1+x\tan x}$
y will be maximum when its reciprocal is minimum.
So reciprocal of y is $\frac{1}{y}=\frac{1+x\tan x}{x}=\frac{1}{x}+\tan x$
Let $z=\frac{1}{x}+\tan x$
$\therefore \frac{dz}{dx}=-\frac{1}{x^2}+{\sec }^{2}x$
For maximum or minimum,
$\frac{dz}{dx}=0$
$\therefore \frac{1}{x^2}={\sec }^{2}x$
$\therefore x=\cos x$
Now, $\frac{d^{2}z}{d{x}^2}=\frac{2}{x^3}+2\sec x\sec x\tan x=+ive$ ($\because 0\le x\le \frac{\pi }{2}$)
Hence $z$ is minimum when $x=\cos x$
So, $y$ is maximum at $x=\cos x.$