SATHEE: Application of Derivatives 2 Question 7

Application of Derivatives 2 Question 7

####7. If the function $g : (- \infty, \infty) \to - \frac{π}{2}, \frac{π}{2}$ is given by $g (u) = 2 \tan^{- 1} (e^{u}) - \frac{π}{2}$ . Then, $g$ is

(2008, 3M)

(a) even and is strictly increasing in $(0, \infty)$

(b) odd and is strictly decreasing in $(- \infty, \infty)$

(d) neither even nor odd but is strictly increasing in $(- \infty, \infty)$

Show Answer

Answer:

(a)

Solution:

Given, $g (u) = 2 \tan^{- 1} (e^{u}) - \frac{π}{2}$ for $u \in (- \infty, \infty)$

$\begin{aligned} g (- u) = 2 \tan^{- 1} (e^{- u}) - \frac{π}{2} = 2 (\cot^{- 1} (e^{u})) - \frac{π}{2} \\ = 2 \frac{π}{2} - \tan^{- 1} (e^{u}) - \frac{π}{2} \\ = π / 2 - 2 \tan^{- 1} (e^{u}) = - g (u) \\ ∴ g (- u) = - g (u) \\ \Rightarrow g (u) is an odd function. \end{aligned}$

We have, $g (u) = 2 \tan^{- 1} (e^{u}) - π / 2$

$\begin{aligned} g^{'} (u) = \frac{2 e^{u}}{1 + e^{2 u}} \\ g^{'} (u) > 0, \forall x \in R \end{aligned}$

$[∵ e^{u} > 0]$

So, $g^{'} (u)$ is increasing for all $x \in R$ .

Application of Derivatives 2 Question 7

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Application of Derivatives 2 Question 7

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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