Question

Which of the following pairs of functions are identical
This questions has multiple correct options
A. \[f\left( x \right) = {\log _x}e\] and \[g\left( x \right) = \dfrac{1}{{{{\log }_e}x}}\]
B. \[f\left( x \right) = \operatorname{sgn} \left( {{x^2} + 1} \right)\] and \[g\left( x \right) = {\sin ^2}x + {\cos ^2}x\]
C. \[f\left( x \right) = {\sec ^2}x - {\tan ^2}x\] and \[g\left( x \right) = \cos e{c^2}x - {\cot ^2}x\]
D. \[f\left( x \right) = \dfrac{1}{{|x|}}\] and \[g\left( x \right) = \sqrt {{x^{ - 2}}} \]

Answer 1

Hint: Here 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 if:
- domain of \[f\left( x \right)\] = domain of \[g\left( x \right)\]
- range of \[f\left( x \right)\] = domain of \[g\left( x \right)\]
- \[f\left( x \right) = g\left( x \right)\]
So, we will check each option one by one and get the required result.

Complete step by step answer:
Here, we have to identify which pair is an identical function, as we know that two functions are identical when:
- domain of \[f\left( x \right)\] = domain of \[g\left( x \right)\]
- range of \[f\left( x \right)\] = domain of \[g\left( x \right)\]
- \[f\left( x \right) = g\left( x \right)\]

Now let’s start from option (A)
Given: \[f\left( x \right) = {\log _x}e\] and \[g\left( x \right) = \dfrac{1}{{{{\log }_e}x}}\]
Here, domain of \[f\left( x \right)\] is \[x \in \left( {0,\infty } \right) - \left\{ 1 \right\}\]
Also, the domain of \[g\left( x \right)\] is \[x \in \left( {0,\infty } \right) - \left\{ 1 \right\}\]
Since, log function is defined for \[x > 0\]
\[ \Rightarrow {\log _e}x \ne 0\]
\[ \Rightarrow x \ne 1\]
Hence, domain of \[g\left( x \right)\] is \[x \in \left( {0,\infty } \right) - \left\{ 1 \right\}\]
Now, we know that
\[{\log _a}b = \dfrac{1}{{{{\log }_b}a}}\]
So, \[f\left( x \right) = {\log _x}e = \dfrac{1}{{{{\log }_e}x}} = g\left( x \right)\]
Thus, the function is the same, and the domain is also the same. Therefore, the range will also be the same.Thus, all the conditions are satisfied. Hence, option (A) is an identical function.

Now let’s consider option (B)
Given: \[f\left( x \right) = \operatorname{sgn} \left( {{x^2} + 1} \right)\] and \[g\left( x \right) = {\sin ^2}x + {\cos ^2}x\]
As we know that signum function is defined as
\[\operatorname{sgn}(x)=\left\{ \begin{align}
& 1,x>0 \\
& 0, x=0 \\
& -1, x<0\\
\end{align} \right.\]
Domain of signum function is \[x \in R\]
Therefore, domain of \[f\left( x \right)\] is \[x \in R\]
Now, we can see \[{x^2} + 1 > 0\]
\[ \Rightarrow \operatorname{sgn} \left( {{x^2} + 1} \right) = 1{\text{ }}\forall x \in R\]
Therefore, range of \[f\left( x \right)\] is \[1\]
Now, for \[g\left( x \right)\]
domain of \[g\left( x \right)\] is \[x \in R\]
and we know \[{\sin ^2}x + {\cos ^2}x = 1{\text{ }}\forall x \in R\]
Therefore, range of \[g\left( x \right)\] is \[1\]
So, \[f\left( x \right)\] and \[g\left( x \right)\] are identical functions as
domain of \[f\left( x \right)\] = domain of \[g\left( x \right)\]
range of \[f\left( x \right)\] = domain of \[g\left( x \right)\]
and \[\forall x{\text{ }}f\left( x \right) = g\left( x \right)\]
Hence, option (B) represents an identical function.

Now let’s consider option (C)
Given: \[f\left( x \right) = {\sec ^2}x - {\tan ^2}x\] and \[g\left( x \right) = \cos e{c^2}x - {\cot ^2}x\]
Here, domain of \[f\left( x \right)\] is \[R - \left\{ {\left( {2n + 1} \right)\dfrac{\pi }{2}} \right\}\]
And domain of \[g\left( x \right)\] is \[R - \left\{ {n\pi } \right\}\]
As, \[f\left( x \right)\] and \[g\left( x \right)\] have different domains.Hence, they are not identical.

Now let’s consider option (D)
Given: \[f\left( x \right) = \dfrac{1}{{|x|}}\] and \[g\left( x \right) = \sqrt {{x^{ - 2}}} \]
Here, domain of \[f\left( x \right)\] is \[R - \left\{ 0 \right\}\]
Also, \[g\left( x \right) = \sqrt {{x^{ - 2}}} = \sqrt {\dfrac{1}{{{x^2}}}} \]
So, domain of \[g\left( x \right)\] is \[R - \left\{ 0 \right\}\]
Now if we see
\[g\left( x \right) = \sqrt {\dfrac{1}{{{x^2}}}} = \dfrac{1}{{|x|}}\] because \[{x^2} \geqslant 0\]
\[ \Rightarrow f\left( x \right) = g\left( x \right)\]
Which means range of both the functions is also same
Therefore, \[f\left( x \right)\] and \[g\left( x \right)\] are identical functions. Hence, option (D) represents an identical function.

So, the correct answer is option (A), (B) and (D).

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.