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UPSEE 2017 Solved Paper

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Question : 140 of 150

Marks: +1, -0

The maximum value of the function

y=2\tan x-\tan^{2}x

\left[0,\frac{\pi}{2}\right]

is

∞
1
3
2

Solution:

Given,

y=2\tan x-\tan^{2}x

.......(i)

∴

\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}{dx^{2}}=4\sec x\cdot\sec x\cdot\tan x(1-\tan x)

+2\sec^{2}x(0-\sec^{2}x)

=4\sec^{2}x\tan x-4\sec^{2}x\tan^{2}x-2\sec^{4}x

\therefore \left.\frac{d^{2}y}{dx^{2}}\right|_{x=\frac{\pi}{2}} = 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}\cdot1-4(\sqrt{2})^{2}\cdot(1)^{2}-2\cdot(\sqrt{2})^{4}

=8-8-8

=-8

\therefore

which is negative.

x=\frac{\pi}{4}

At

y=2\tan x-\tan^{2}x

, function

\therefore

has maximum value.

x=\frac{\pi}{4}

Maximum value of function at point

[y]_{x=\frac{\pi}{4}}=1

, will be

[y]_{x={\pi}/{4}}=1

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