STD 12 - 6. Application of derivatives — Mathematics STD 12 Science — Question

Let $y = \frac{{1 + x\tan x}}{x}  =  \frac{1}{x} + \tan x$

$\therefore$ $\frac{{dy}}{{dx}} = - \frac{1}{{{x^2}}} + {\sec ^2}x$, $\frac{{{d^2}y}}{{d{x^2}}} = \frac{2}{{{x^3}}} + 2\sec x\sec x\tan x$

On putting $\frac{{dy}}{{dx}} = 0$, $ - \frac{1}{{{x^2}}} + {\sec ^2}x = 0$

==> ${\sec ^2}x = \frac{1}{{{x^2}}}$ ==> ${x^2} = {\cos ^2}x$ ==> $x = \cos x$

$\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.