Question

The maximum value of the function $y=2\tan x-{\tan }^{2}x$ over $\left[0,\frac{\pi }{2}\right]$ is :

• A. 2
• B. $\infty$
• C. 1
• D. 3

Answer 1

Answer: (C)

Given,
$y=2\tan x-{\tan }^{2}x...(i)$
$\therefore \frac{dy}{dx}=2{\sec }^{2}x-2\tan x\cdot {\sec }^{2}x$
$=2{\sec }^{2}x(1-\tan x)...(ii)$
At point of maxima,
$\frac{dy}{dx}=0$
$2{\sec }^{2}x(1-\tan x)=0$ [From Eq. (ii)]
$\therefore x=\frac{\pi }{4},\frac{\pi }{2}$
Here, $x=\frac{\pi }{2}$ is not possible]
$\therefore x=\frac{\pi }{4}$
$[\because x\in \left[0,\frac{\pi }{2}\right]$ (given)
Now, $\frac{d^{2}y}{d{x}^2}=4\sec x\cdot \sec x\cdot \tan x(1-\tan x)+2{\sec }^{2}x\left(0-{\sec }^{2}x\right)$
$=4{\sec }^{2}x\tan x-4{\sec }^{2}x{\tan }^{2}x-2{\sec }^{4}x$
${\therefore \frac{d^{2}y}{d{x}^2}|}_{x-\frac{\pi }{4}}=4{\sec }^2\frac{\pi }{4}\tan \frac{\pi }{4}-4{\sec }^2\frac{\pi }{4}$
${\tan }^2\frac{\pi }{4}-2{\sec }^4\frac{\pi }{4}$
$=4(\sqrt{2}{)}^2\cdot 1-4(\sqrt{2}{)}^2\cdot (1{)}^2-2\cdot (\sqrt{2}{)}^4$
$=8-8-8$
$=-8$ which is negative.
$\therefore$ At $x=\frac{\pi }{4}$, function $y=2\tan x-{\tan }^{2}x$
has maximum value.
$\therefore$ Maximum value of function at point $x=\frac{\pi }{4}$, will be
$[y{]}_{x-\frac{\pi }{4}}=1$