If f - x -0- - x - R - f -3-0 and g x - f tan - 2 x -2 tan x -4- 0 lt x ltIdfrac p i2 - then gx is increasing inA- 0- -4B- -6- -3C- 0- -3D- -4- -2

The correct option is D (π4,π2)g'(x)=(f'((tan x 1)^2+3))2(tan x 1)^2x Since f”(x)0,f'(x) is increasing So, f'((tan x 1)^2+3) f'(3)=0 ∀ x∈(0,π4)∪(π4,π2) Al ...

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Byju's Answer

Standard XII

Mathematics

Monotonicity in an Interval

If f '' x >0,...

Question

If $f'' (x) > 0, \forall x \in R, f' (3) = 0$ and $g (x) = f ({tan}^{2} x - 2 tan x + 4), 0 < x < \frac{π}{2}$ , then $g (x)$ is increasing in

(0, \frac{π}{4})

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(\frac{π}{6}, \frac{π}{3})

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(0, \frac{π}{3})

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(\frac{π}{4}, \frac{π}{2})

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Solution

The correct option is D $(\frac{π}{4}, \frac{π}{2})$
$g^{'} (x) = (f^{'} ((tan x - 1)^{2} + 3)) 2 (tan x - 1) {sec}^{2} x$
Since $f^{''} (x) > 0, f^{'} (x)$ is increasing
So,
$f^{'} ((tan x - 1)^{2} + 3) > f^{'} (3) = 0 \forall x \in (0, \frac{π}{4}) \cup (\frac{π}{4}, \frac{π}{2})$
Also, $(tan x - 1) > 0 x \in (\frac{π}{4}, \frac{π}{2})$
So, $g (x)$ is increasing in $(\frac{π}{4}, \frac{π}{2})$ .

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If f''(x)>0,∀ x∈R,f'(3)=0 and g(x)=f(tan2x−2tanx+4),0<x<π2, then g(x) is increasing in

The correct option is D (π4,π2)g′(x)=(f′((tanx−1)2+3))2(tanx−1)sec2x Since f′′(x)>0,f′(x) is increasing So, f′((tanx−1)2+3)>f′(3)=0 ∀ x∈(0,π4)∪(π4,π2) Also, (tanx−1)>0 x∈(π4,π2) So, g(x) is increasing in (π4,π2).

If $f'' (x) > 0, \forall x \in R, f' (3) = 0$ and $g (x) = f ({tan}^{2} x - 2 tan x + 4), 0 < x < \frac{π}{2}$ , then $g (x)$ is increasing in