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

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.