Let f (−π2, π2):→ R be given by f(x) = ( log (secx+tanx) ) 3 . Then

Let f \left(\right. \frac{- \pi}{2} , \textrm{ } \frac{\pi}{2} \left.\right) : \rightarrow R be given by f(x) = \left(\left(\right. \textrm{ } log \textrm{ } \left(\right. secx + tanx \left.\right) \textrm{ } \left.\right)\right)^{\textrm{ } 3}. Then

Engineering

Mathematics

Classification of Functions

Even and Odd Functions

Question

Let f $(\frac{- π}{2}, \frac{π}{2}) : \to$ R be given by f(x) = ${(\log (secx + tanx))}^{3}$ .

Then

f(x) is an odd function

f(x) is an even function

f(x) is a one‑one function

f(x) is an onto function

f(x) = – f (– x)

odd function

f '(x) = 3 (log (sec x + tan x))² (sec x tan x + sec²x)

= 3 (log (sec x + tan x))² (sec x) [sec x + tan x]

f '(x) > 0 in x $\in (\frac{- π}{2}, \frac{π}{2})$

say t = sec x + tan x $(\frac{1 + sinx}{cosx}) = \frac{\sin \frac{x}{2} + \cos \frac{x}{2}}{\cos \frac{x}{2} - \sin \frac{x}{2}} = \frac{1 + \tan \frac{x}{2}}{1 - \tan \frac{x}{2}}$

$\Rightarrow T = \tan (\frac{π}{4} + \frac{x}{2})$

$x \in (\frac{- π}{2}, \frac{π}{2}) \Rightarrow x \in (\frac{- π}{4}, \frac{π}{4}) \Rightarrow \frac{π}{4} + \frac{x}{2} \in (0, \frac{π}{2})$

$∴ t \in (0, \infty) \Rightarrow$ y = (log t)³ $\in (- \infty, \infty)$

$∴$ Onto function.

Understanding the Function and Its Properties

Let's analyze the function: f(x) = ${(\log (\sec x + \tan x))}^{3}$ , defined on the interval $(\frac{- π}{2}, \frac{π}{2})$ .

Step 1: Simplify the Expression Inside the Logarithm

First, recall a key trigonometric identity. The expression $\sec x + \tan x$ can be simplified.

Let's find a common denominator and manipulate the expression: $\sec x + \tan x = \frac{1}{\cos x} + \frac{\sin x}{\cos x} = \frac{1 + \sin x}{\cos x}$

This is a standard form. It is known that: $\log (\sec x + \tan x) = \log (\frac{1 + \sin x}{\cos x}) = \log (\frac{1 + \sin x}{\sqrt{1 - \sin^{2} x}})$ However, a more insightful simplification exists. It can be shown that: $\sec x + \tan x = \tan (\frac{π}{4} + \frac{x}{2})$

But the most crucial identity for this problem is: $\log (\sec x + \tan x) = \tanh^{- 1} (\sin x) = \frac{1}{2} \ln (\frac{1 + \sin x}{1 - \sin x})$ For our purpose, the key takeaway is that $g (x) = \log (\sec x + \tan x)$ is an odd function. Let's verify this.

Step 2: Check if g(x) is Odd (and thus f(x) is Odd)

A function is odd if $g (- x) = - g (x)$ .

Let's calculate $g (- x)$ : $g (- x) = \log (\sec (- x) + \tan (- x))$

Since secant is an even function ( $\sec (- x) = \sec (x)$ ) and tangent is an odd function ( $\tan (- x) = - \tan (x)$ ), we substitute: $g (- x) = \log (\sec x + (- \tan x)) = \log (\sec x - \tan x)$

Now, let's rationalize this expression: $\sec x - \tan x = \frac{(\sec x - \tan x) (\sec x + \tan x)}{\sec x + \tan x} = \frac{\sec^{2} x - \tan^{2} x}{\sec x + \tan x}$

Using the Pythagorean Identity $\sec^{2} x - \tan^{2} x = 1$ : $\sec x - \tan x = \frac{1}{\sec x + \tan x}$

Substituting back, we get: $g (- x) = \log (\frac{1}{\sec x + \tan x}) = \log (1) - \log (\sec x + \tan x) = 0 - g (x) = - g (x)$

Conclusion for Step 2: We have proven that $g (x) = \log (\sec x + \tan x)$ is an odd function.

Step 3: Determine the Nature of f(x)

Our original function is $f (x) = {[g (x)]}^{3}$ .

Let's check if f(x) is odd: $f (- x) = {[g (- x)]}^{3</}$

Detailed Solution

Understanding the Function and Its Properties

Step 1: Simplify the Expression Inside the Logarithm

Step 2: Check if g(x) is Odd (and thus f(x) is Odd)

Step 3: Determine the Nature of f(x)

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