Question

The range of $f\left( x \right) = \dfrac{{1 - \tan x}}{{1 + \tan x}}$ is
A. $\left( { - \infty ,\infty } \right)$
B. $\left( { - \infty ,0} \right)$
C. $\left( {0,\infty } \right)$
D. $\left( { - \infty , - 1} \right) \cup \left( { - 1,\infty } \right)$

Answer 1

Hint: In the given question, we need to find the domain and range of the function given by; $f\left( x \right) = \dfrac{{1 - \tan x}}{{1 + \tan x}}$. We use the definition of a range of the function, which states that the range is the possible set of values that the given function can attain for those values of x which are in the domain of the function.

Complete step by step answer:
Let us first consider the given function and denote it by y.
$ \Rightarrow y = f\left( x \right) = \dfrac{{1 - \tan x}}{{1 + \tan x}}$

Let us now simplify the expression and find the value of x in terms of y.
$
   \Rightarrow y = \dfrac{{1 - \tan x}}{{1 + \tan x}} \\
   \Rightarrow y\left( {1 + \tan x} \right) = 1 - \tan x \\
   \Rightarrow y + y\tan x + \tan x = 1 \\
   \Rightarrow \tan x\left( {y + 1} \right) = 1 - y \\
   \Rightarrow \tan x = \dfrac{{1 - y}}{{1 + y}} \\
   \Rightarrow x = {\tan ^{ - 1}}\left( {\dfrac{{1 - y}}{{1 + y}}} \right) \\
 $

Now, we have taken that
$
  y = f\left( x \right) \\
   \Rightarrow x = {f^{ - 1}}\left( y \right) \\
 $
Let us replace x by y, in order to find the range for the given function,
$ \Rightarrow y = {f^{ - 1}}\left( x \right)$

Thus, replacing x and y will also take place in the expression for x.
Thus, using this in the obtained expression of x we get;
\[
   \Rightarrow y = {\tan ^{ - 1}}\left( {\dfrac{{1 - x}}{{1 + x}}} \right) \\
   \Rightarrow {f^{ - 1}}\left( x \right) = {\tan ^{ - 1}}\left( {\dfrac{{1 - x}}{{1 + x}}} \right) \\
 \]

Now, we observe the function and it can be clearly seen that the function is undefined only when $1 + x = 0$ . Thus, we can say that ${f^{ - 1}}\left( x \right)$ is valid only when;
$
  1 + x \ne 0 \\
   \Rightarrow x \ne - 1 \\
 $
Thus, we get that the domain of the function ${f^{ - 1}}\left( x \right)$ is $R - \left\{ { - 1} \right\}$ and range of $f\left( x \right)$ is also, $R - \left\{ { - 1} \right\}$.
 Hence, we will go with option D as range is given by $\left( { - \infty , - 1} \right) \cup \left( { - 1,\infty } \right)$ in which all the real values for \[x\] except at \[x = - 1\]

Note: Whenever we come across such types of problems, the key concept which is the definition of domain and range is to be clear as in this question since we know that the function will be defined when the denominator of tangent function will be non-zero. Using this fact we can easily find the range of the given function. This understanding of basic definitions will help you get the right answer. Also, while determining that the function in terms of \[y\] does the simplification properly. Also remember that \[y = {f^{ - 1}}\left( x \right)\]. $R - \left\{ { - 1} \right\}$ means that \[x = - 1\] is not included in the range or domain of the function.