If f'(x) > 0 x R, f'(3)=0 and g(x)=f(² x-2 x+4), 0

Correct Answer is : (π4,π2)≤ft(π/4, π/2)(4π,2π)

Correct Answer is : (π4,π2)≤ft(π/4, π/2)(4π,2π)

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CGPET 2014 Solved Paper

Section: Mathematics

Question : 113 of 150

Marks: +1, -0

f'(x) > 0 \forall x \in R, f'(3)=0

g(x)=f(\tan^2 x-2\tan x+4), 0 < x < \frac{\pi}{2},

g(x)

Solution:

Given,

g(x)=f(\tan^2 x-2\tan x+4)

On differentiating w.r.t.

x

, we get

g'(x)=f'(\tan^2 x-2.\,\tan x+4)

\times (2\tan x \sec^2 x-2\sec^2 x)

= f'(\tan^2 x-2\tan x+4) 2 \sec^2 x(\tan x-1)

\because \; f'(x) > 0 \forall x \in \left(0, \frac{\pi}{2}\right)

Also,

2 \sec^2 x > 0 \forall x \in \left(0, \frac{\pi}{2}\right)

[ \because x \in \left(0, \frac{\pi}{2}\right) .

, given

]

But

\tan x-1 > 0 \forall x \in \left(\frac{\pi}{4}, \frac{\pi}{2}\right)

\therefore \; g'(x) > 0 \forall x \in \left(\frac{\pi}{4}, \frac{\pi}{2}\right)

Hence,

g(x)

is increasing in

\left(\frac{\pi}{4}, \frac{\pi}{2}\right)

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