Question

The range of a function $f\left( x \right)=sgn \left( \sin x \right)+sgn \left( \cos x \right)+sgn \left( \tan x \right)+sgn \left( \cot x \right)$, $x\ne \dfrac{n\pi }{2}\left( n\in I \right)$ is:
A) \left\\{ -2,4 \right\\}
B) \left\\{ -2,0,4 \right\\}
C) \left\\{ -4,-2,0,4 \right\\}
D) \left\\{ 0,2,4 \right\\}

Answer 1

First, check the value of $\sin x,\cos x,\tan x,\cot x$ in the first quadrant whether it is positive or negative or 0. If the value is greater than 0, substitute 1 in the place of signum of that function. If it is less than 0, substitute -1 in the place and 0 if the value is 0. Then, find the value of the function $f\left( x \right)$. Do the same for the other three quadrants. The value of the function $f\left( x \right)$ is the range.

Formula used: The sign of a real number, also called sgn or signum, is -1 for a negative number (i.e., one with a minus sign "), 0 for the number zero, or +1 for a positive number (i.e., one with a plus sign"). In other words, for any real $x$,
f\left( x \right)=\left\\{ \begin{aligned} & -1\text{for }x<0 \\\ & 0\text{for }x=0 \\\ & 1\text{for }x>0 \\\ \end{aligned} \right.

Complete step-by-step answer:
Given: - $f\left( x \right)=sgn \left( \sin x \right)+sgn \left( \cos x \right)+sgn \left( \tan x \right)+sgn \left( \cot x \right)$ ….. (1)
In the first quadrant, the value of sin, cos, tan, cot is greater than 0. Then,
$sgn \left( \sin x \right)=1,sgn \left( \cos x \right)=1,sgn \left( \tan x \right)=1,sgn \left( \cot x \right)=1$
Substitute these values in the equation (1),
$f\left( x \right)=1+1+1+1$
Add the terms on the right side,
$f\left( x \right)=4$ ….. (2)
In the second quadrant, the value of sin is greater than 0 and the value of cos, tan, cot is less than 0. Then,
$sgn \left( \sin x \right)=1,sgn \left( \cos x \right)=-1,sgn \left( \tan x \right)=-1,sgn \left( \cot x \right)=-1$
Substitute these values in the equation (1),
$f\left( x \right)=1-1-1-1$
Add and subtract the terms on the right side,
$f\left( x \right)=-2$ ….. (3)
In the third quadrant, the value of tan, cot is greater than 0 and the value of sin, cos is less than 0. Then,
$sgn \left( \sin x \right)=-1,sgn \left( \cos x \right)=-1,sgn \left( \tan x \right)=1,sgn \left( \cot x \right)=1$
Substitute these values in the equation (1),
$f\left( x \right)=-1-1+1+1$
Add and subtract the terms on the right side,
$f\left( x \right)=0$ ….. (4)
In the fourth quadrant, the value of cos is greater than 0 and the value of sin, tan, cot is less than 0. Then,
$sgn \left( \sin x \right)=-1,sgn \left( \cos x \right)=1,sgn \left( \tan x \right)=-1,sgn \left( \cot x \right)=-1$
Substitute these values in the equation (1),
$f\left( x \right)=-1+1-1-1$
Add and subtract the terms on the right side,
$f\left( x \right)=-2$ ….. (5)
From the equations (2), (3), (4), and (5),
The range is \left\\{ -2,0,4 \right\\}
Thus, the range of $f\left( x \right)$ is \left\\{ -2,0,4 \right\\}.

Hence, option (B) is the correct answer.

Note: The students must remember the sign formula to solve this type of problem. He/she must also check the value of a and the nature of the quadratic equation to analyze whether the mode function will give a positive sign or negative sign. If he/she forgets any condition the range of the function will be wrong.