Question

Which of the following pairs of functions is/ are identical
(This questions has multiple correct options)
A) $f\left( x \right) = \tan \left( {{{\tan }^{ - 1}}x} \right)$ and $g\left( x \right) = \cot \left( {{{\cot }^{ - 1}}x} \right)$
B) $f\left( x \right) = \operatorname{sgn} \left( x \right)$ and $g\left( x \right) = \operatorname{sgn} \left( {\operatorname{sgn} \left( x \right)} \right)$
C) $f\left( x \right) = {\cot ^2}x \cdot {\cos ^2}x$ and $g\left( x \right) = {\cot ^2}x - {\cos ^2}x$
D) $f\left( x \right) = {e^{\ln {{\sec }^{ - 1}}x}}$ and $g\left( x \right) = {\sec ^{ - 1}}x$

Answer 1

We will consider that all the four pairs are well defined functions. In order to solve this question, we will use the concept of identical functions i.e., Two functions are said to identical functions when:

1. $f\left( x \right) = g\left( x \right)$
2. Domain of both the functions is same
3. The range of both functions is the same.
   So, we will check each option one by one and get the required result.

Complete step by step answer:
We have to identify which pair is an identical function,
Now, let’s start from option (A)
Given: $f\left( x \right) = \tan \left( {{{\tan }^{ - 1}}x} \right)$ and $g\left( x \right) = \cot \left( {{{\cot }^{ - 1}}x} \right)$
We know that
$\tan \left( {{{\tan }^{ - 1}}x} \right) = x$ and $\cot \left( {{{\cot }^{ - 1}}x} \right) = x$
Therefore, we have
$f\left( x \right) = x$ and $g\left( x \right) = x$
Thus, the first condition is satisfied.
Now domain of $f\left( x \right)$ is $x \in R$ and also domain of $g\left( x \right)$ is $x \in R$
And the range of both the functions is also the same as we get the same result from both the functions.
Thus, the second and third condition is also satisfied.
Hence, option (A) satisfies all the conditions
Which means $f\left( x \right)$ and $g\left( x \right)$ are identical functions.

Now let’s consider option (B)
Given: $f\left( x \right) = \operatorname{sgn} \left( x \right)$ and $g\left( x \right) = \operatorname{sgn} \left( {\operatorname{sgn} \left( x \right)} \right)$
Here, domain of $f\left( x \right)$ is $x \in R$ and also domain of $g\left( x \right)$ is $x \in R$
Thus, domain of $f\left( x \right)$ and $g\left( x \right)$ are same
Now we know that signum function is defined as
\operatorname{sgn} \left( x \right)=\left\\{ \begin{matrix} 1,\text{ if }x\text{> 0} \\\ 0,\text{ if }x\text{= 0 } \\\ -1,\text{ if }x\text{< 0 } \\\ \end{matrix} \right.
Therefore, the range of $f\left( x \right)$ is \left\\{ { - 1,0,1} \right\\}
And also, the range of $g\left( x \right)$ is \left\\{ { - 1,0,1} \right\\}
Thus, the range of $f\left( x \right)$ and $g\left( x \right)$ are the same.
And we know if range of two functions is same then, $f\left( x \right) = g\left( x \right)$
Hence, option (B) satisfies all the conditions
Which means $f\left( x \right)$ and $g\left( x \right)$ are identical functions.

Now consider option (C)
Given: $f\left( x \right) = {\cot ^2}x \cdot {\cos ^2}x$ and $g\left( x \right) = {\cot ^2}x - {\cos ^2}x$
Here, domain of $f\left( x \right)$ is $x \in R - n\pi$
and domain of $g\left( x \right)$ is $x \in R - n\pi$
Therefore, domain of $f\left( x \right)$ and $g\left( x \right)$ are same
Now, $g\left( x \right) = {\cot ^2}x - {\cos ^2}x$
$\Rightarrow g\left( x \right) = \dfrac{{{{\cos }^2}x}}{{{{\sin }^2}x}} - {\cos ^2}x$
On taking common, we get
$\Rightarrow g\left( x \right) = {\cos ^2}x\left( {\dfrac{1}{{{{\sin }^2}x}} - 1} \right)$
On taking L.C.M we get
$\Rightarrow g\left( x \right) = {\cos ^2}x\left( {\dfrac{{1 - {{\sin }^2}x}}{{{{\sin }^2}x}}} \right)$
$\Rightarrow g\left( x \right) = {\cos ^2}x\left( {\dfrac{{{{\cos }^2}x}}{{{{\sin }^2}x}}} \right)$
$\Rightarrow g\left( x \right) = {\cos ^2}x \cdot {\cot ^2}x = f\left( x \right)$
Therefore, $f\left( x \right) = g\left( x \right)$
And range of both the function is also same as we get the same result from both the functions.
Hence, option (C) satisfies all the conditions
Which means $f\left( x \right)$ and $g\left( x \right)$ are identical functions.

Now let’s consider option (D)
Given: $f\left( x \right) = {e^{\ln {{\sec }^{ - 1}}x}}$ and $g\left( x \right) = {\sec ^{ - 1}}x$
We know that domain of ${\sec ^{ - 1}}x$ is $x \in \left( { - \infty , - 1} \right] \cup \left[ {1,\infty } \right)$
Now if we see $f\left( x \right) = {e^{\ln {{\sec }^{ - 1}}x}}$
Now we know range of ${\sec ^{ - 1}}x$ is $\left[ {0,\pi } \right]$
So, if we see, $1 \notin$ domain of $f\left( x \right)$
Therefore, domain of $f\left( x \right)$ is not equal to domain of $g\left( x \right)$
Hence, $f\left( x \right)$ and $g\left( x \right)$ are not identical functions.
Therefore, the correct answer is Option (A), (B) and (C).

Note:
In order to solve these types of questions, always remember there are no direct connections between any of the pairs, so we have to check for each option. Here we used different tactics to show the identical functions. So, the basic knowledge of concepts is must. Also, recall the concept of domain, range, and identical function to get a good grasp.