First slide

Monotonicity

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

Moderate

A

$(0, \frac{π}{4})$
B

$(\frac{π}{6}, \frac{π}{3})$
C

$(0, \frac{π}{3})$
D

$(\frac{π}{4}, \frac{π}{2})$

Solution

$g^{'} (x) = (f^{'} ((\tan x - 1)^{2} + 3)) 2 (\tan x - 1) \sec^{2} x$
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, $g (x) is increasing in (\frac{π}{4}, \frac{π}{2})$ .

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App