Question

Let $ f:\left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right) \to R $ be given by $ f(x) = {(\log (\sec x + \tan x))^3} $ . Then
 A. $ f(x) $ is an odd function.
 B. $ f(x) $ is one to one function.
 C. $ f(x) $ is an onto function.
 D. $ f(x) $ is an even function.

Answer 1

Hint: To solve the above question, first we will check if the function is odd or even function i.e. $ f( - x) = - f(x) $ or not. Then we will check if the function is one-one or not followed by checking onto or not. We might get multiple answers to the question here.

Complete step-by-step answer:
We have, $ f(x) = {(\log (\sec x + \tan x))^3} $
To check $ f(x) $ is even or odd function-
Differentiating the above function we get,
 $ f( - x) = {\{ \log (\sec ( - x) + \tan ( - x))\} ^3} $
We know that, $ \tan ( - x) = - \tan x $ .
Putting this value in the above equation we get,
 $ f( - x) = {\{ \log (\sec x - \tan x)\} ^3} $
Multiplying and dividing $ \sec x + \tan x $ with the trigonometric function of right hand side,
 $ f( - x) = {\left\{ {\log \dfrac{{(\sec x - \tan x)(\sec x + \tan x)}}{{(\sec x + \tan x)}}} \right\}^3} $
As we know, $ (a + b)(a - b) = {a^2} - {b^2} $ , hence
 $ f( - x) = {\left\{ {\log \dfrac{{({{\sec }^2}x - {{\tan }^2}x)}}{{(\sec x + \tan x)}}} \right\}^3} $
Again we know that, $ {\sec ^2}x - {\tan ^2}x = 1 $
 $ f( - x) = {\left\{ {\log \dfrac{1}{{(\sec x + \tan x)}}} \right\}^3} $
As we know, $ \log {a^{ - 1}} = - \log a $ ,
 $ f( - x) = {\left\{ { - \log (\sec x + \tan x)} \right\}^3} $
Again we know that, $ ( - {a^3}) = - ({a^3}) $
 $ f( - x) = - {\left\{ {\log (\sec x + \tan x)} \right\}^3} $
 $ f( - x) = - f(x) $
Thus, $ f(x) $ is an odd function.
To check if $ f(x) $ is one-one function or not-
Let $ g(x) = \sec x + \tan x $ $ \forall x \in \left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right) $
Differentiating it we get,
 $ g'(x) = \sec x(\sec x + \tan x) $
Both the terms of the right hand side are positive at a given interval.
i.e. $ g'(x) > 0\forall x \in \left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right) $
Hence, $ g(x) $ is a one-one function.
Thus $ {\{ \log \left( {g(x)} \right)\} ^3} $ is a one-one function.
Thus $ f(x) $ is a one-one function.
To check if $ f(x) $ is onto function or not-
For $ f(x) $ to be onto we have to prove,
 $ \left( {g(x)} \right) \in (0,\infty )\forall x \in \left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right) $
We know that,
 $ g(x) = \sec x + \tan x = \dfrac{{1 + \sin x}}{{\cos x}} = \dfrac{{1 - ( - \sin x)}}{{\cos x}} $
Expanding it we get,
 $ g(x) = \dfrac{{1 - (\cos \dfrac{\pi }{2} + x)}}{{\sin (\dfrac{\pi }{2} + x)}} $
Converting the above equation into $ \dfrac{{1 - \cos 2A}}{{\sin 2A}} $ form we get,
 $ g(x) = \dfrac{{1 - (\cos 2(\dfrac{\pi }{4} + \dfrac{x}{2})}}{{\sin 2(\dfrac{\pi }{4} + \dfrac{x}{2})}} $
Putting the formula $ \dfrac{{1 - \cos 2A}}{{\sin 2A}} = \tan A $ we get,
 $ g(x) = \tan (\dfrac{\pi }{4} + \dfrac{x}{2}) $
Hence, $ - \dfrac{\pi }{2} < x < \dfrac{\pi }{2} $
Dividing 2 we get,
 $ - \dfrac{\pi }{4} < \dfrac{x}{2} < \dfrac{\pi }{4} $
Adding $ \dfrac{\pi }{4} $ we get,
 $ \therefore $ $ o < \dfrac{x}{2} + \dfrac{\pi }{4} < \dfrac{\pi }{2} $
Taking tan,
 $ \tan o < \tan \left( {\dfrac{x}{2} + \dfrac{\pi }{4}} \right) < \tan \dfrac{\pi }{2} $
Putting trigonometric values we get,
 $ o < \tan \left( {\dfrac{x}{2} + \dfrac{\pi }{4}} \right) < \infty $
 $ \therefore $ $ \left( {g(x)} \right) \in (0,\infty )\forall x \in \left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right) $
Hence, $ g(x) $ is onto function.
Thus $ {\{ \log \left( {g(x)} \right)\} ^3} $ is onto function.
Thus $ f(x) $ is onto function.

So, the correct answer is “Option A”, “Option B” and “Option C”.

Note: A function is called one to one if no two elements in the domain of that function correspond to the same element in the range of that function which means each x in the domain has exactly one image in the range. This is also called injective function.
The mapping of f is said to be onto if every element of the Y is the f-image of at least one element of x. It is also called a surjective function. A function f from a set X to set Y is surjective, if for every element y in the codomain Y of f, there is at least one element x in the domain X of f such that f(x)=y.
If a function is both one to one and onto, then it is called a bijective function.
Even functions and odd functions are functions which satisfy particular symmetry relations, with respect to taking additive inverse. If for a function $ f( - x) = - f(x) $ , then it is odd, if $ f( - x) = f(x) $ , then the function is even.
You might think it's done after proving it is an odd function, but you have to check each possibility or options one by one.