Question

The range of the function $f(x) = \operatorname{sgn} (\sin x) + \operatorname{sgn} (\cos x) + \operatorname{sgn} (\tan x) + \operatorname{sgn} (\cot x),x \ne \dfrac{{n\pi }}{2}\left( {n \in I} \right)$ is:
[Note: sgn k denotes signum function of k]
A. {-2, 4}
B. {-2, 0, 4}
C. {-4, -2, 0, 4}
D. {0, 2, 4}

Answer 1

Hint: We will first individually calculate the possible values of the given function f(x) in all of the quadrants and thus see what all the values we are getting and thus have our answer.

Complete step by step answer:
We know that we can write the given function f(x) as:-
$ \Rightarrow f(x) = \operatorname{sgn} (\sin x) + \operatorname{sgn} (\cos x) + \operatorname{sgn} (\tan x) + \operatorname{sgn} (\cot x) = \dfrac{{|\sin x|}}{{\sin x}} + \dfrac{{|\cos x|}}{{\cos x}} + \dfrac{{|\tan x|}}{{\tan x}} + \dfrac{{|\cot x|}}{{\cot x}}$
In quadrant one, we know that all the trigonometric functions take up positive values only.
Therefore, the signum function of all the trigonometric values will take positive value.
$ \Rightarrow \dfrac{{|\sin x|}}{{\sin x}} + \dfrac{{|\cos x|}}{{\cos x}} + \dfrac{{|\tan x|}}{{\tan x}} + \dfrac{{|\cot x|}}{{\cot x}} = 1 + 1 + 1 + 1$ in the first quadrant.
Adding this up, we will get: $\dfrac{{|\sin x|}}{{\sin x}} + \dfrac{{|\cos x|}}{{\cos x}} + \dfrac{{|\tan x|}}{{\tan x}} + \dfrac{{|\cot x|}}{{\cot x}} = 4$ in first quadrant. ………(1)
In quadrant two, we know that all the sine / cosine functions take positive values whereas the rest of them are negative.
Therefore, we will get:-
$ \Rightarrow \dfrac{{|\sin x|}}{{\sin x}} + \dfrac{{|\cos x|}}{{\cos x}} + \dfrac{{|\tan x|}}{{\tan x}} + \dfrac{{|\cot x|}}{{\cot x}} = \dfrac{{\sin x}}{{\sin x}} + \dfrac{{( - \cos x)}}{{\cos x}} + \dfrac{{( - \tan x)}}{{\tan x}} + \dfrac{{( - \cot x)}}{{\cot x}}$
$ \Rightarrow \dfrac{{|\sin x|}}{{\sin x}} + \dfrac{{|\cos x|}}{{\cos x}} + \dfrac{{|\tan x|}}{{\tan x}} + \dfrac{{|\cot x|}}{{\cot x}} = 1 - 1 - 1 - 1$ in the second quadrant
Adding this up, we will get: $\operatorname{sgn} (\sin x) + \operatorname{sgn} (\cos x) + \operatorname{sgn} (\tan x) + \operatorname{sgn} (\cot x) = - 2$ in second quadrant. ………….(2)
In quadrant three, we know that all the tangent / cotangent functions take positive value whereas the rest of them are negative.
Therefore, we will get:-
$ \Rightarrow \dfrac{{|\sin x|}}{{\sin x}} + \dfrac{{|\cos x|}}{{\cos x}} + \dfrac{{|\tan x|}}{{\tan x}} + \dfrac{{|\cot x|}}{{\cot x}} = \dfrac{{( - \sin x)}}{{\sin x}} + \dfrac{{( - \cos x)}}{{\cos x}} + \dfrac{{\tan x}}{{\tan x}} + \dfrac{{\cot x}}{{\cot x}}$
$ \Rightarrow \dfrac{{|\sin x|}}{{\sin x}} + \dfrac{{|\cos x|}}{{\cos x}} + \dfrac{{|\tan x|}}{{\tan x}} + \dfrac{{|\cot x|}}{{\cot x}} = - 1 - 1 + 1 + 1$ in the third quadrant
Adding this up, we will get: $\operatorname{sgn} (\sin x) + \operatorname{sgn} (\cos x) + \operatorname{sgn} (\tan x) + \operatorname{sgn} (\cot x) = 0$ in the third quadrant. ………(3)
In quadrant four, we know that all the cosine / secant function takes positive value whereas rest of them are negative.
Therefore, we will get:-
$ \Rightarrow \dfrac{{|\sin x|}}{{\sin x}} + \dfrac{{|\cos x|}}{{\cos x}} + \dfrac{{|\tan x|}}{{\tan x}} + \dfrac{{|\cot x|}}{{\cot x}} = \dfrac{{( - \sin x)}}{{\sin x}} + \dfrac{{\cos x}}{{\cos x}} + \dfrac{{( - \tan x)}}{{\tan x}} + \dfrac{{( - \cot x)}}{{\cot x}}$
$ \Rightarrow \dfrac{{|\sin x|}}{{\sin x}} + \dfrac{{|\cos x|}}{{\cos x}} + \dfrac{{|\tan x|}}{{\tan x}} + \dfrac{{|\cot x|}}{{\cot x}} = - 1 + 1 - 1 - 1$ in the fourth quadrant
Adding this up, we will get: $\operatorname{sgn} (\sin x) + \operatorname{sgn} (\cos x) + \operatorname{sgn} (\tan x) + \operatorname{sgn} (\cot x) = - 2$ in the fourth quadrant. ………(4)
If we combine the equations (1), (2), (3) and (4), we will get:-
The possible values of f(x) can be 4, -2 and 0.

Hence, the correct option is (b) (-2, 0, 4}.

Note: The students must know that the definition of signum function. It is given by:-
$ \Rightarrow \operatorname{sgn} (x) = \dfrac{{|x|}}{x}\forall x \in \mathbb{R}$
After using the definition of the modulus function which is given by the following expression:-
$ \Rightarrow |x| = \left\{ {\begin{array}{*{20}{c}}
  { - x,x < 0} \\
  {0,x = 0} \\
  {x,x > 0}
\end{array}} \right.$
Therefore, after using this definition, we will get the signum function to be equal to:-
$ \Rightarrow \operatorname{sgn} (x) = \left\{ {\begin{array}{*{20}{c}}
  { - 1,x < 0} \\
  {0,x = 0} \\
  {1,x > 0}
\end{array}} \right.$