Q38 / 75 · JEE Main 2026 (28 January Shift 2) · Mathematics
The sum of all the elements in the range of $f(x)=\operatorname{Sgn}(\sin x)+\operatorname{Sgn}(\cos x)+\operatorname{Sgn}(\tan x)+\operatorname{Sgn}(\cot x)$, $x \neq \frac{\mathrm{n} \pi}{2}, \mathrm{n} \in \mathbf{Z}$, where $\operatorname{Sgn}(\mathrm{t})=\left\{\begin{aligned} 1, & \text { if } \mathrm{t}>0 \\ -1, & \text { if } \mathrm{t}<0\end{aligned}\right.$, is :
- Correct answerA2
- B0
- C4
- D-2
Solution
$x \in (0, \pi/2) \Rightarrow y = 1 + 1 + 1 + 1 = 4$
$x \in (\pi/2, \pi) \Rightarrow y = 1 - 1 - 1 - 1 = -2$
$x \in (\pi, 3\pi/2) \Rightarrow y = -1 - 1 + 1 + 1 = 0$
$x \in (3\pi/2, 2\pi) \Rightarrow y = -1 + 1 - 1 - 1 = -2$
$\therefore$ Range of y is $\{-2, 0, 4\}$
Required sum $= -2 + 0 + 4 = 2$
$x \in (\pi/2, \pi) \Rightarrow y = 1 - 1 - 1 - 1 = -2$
$x \in (\pi, 3\pi/2) \Rightarrow y = -1 - 1 + 1 + 1 = 0$
$x \in (3\pi/2, 2\pi) \Rightarrow y = -1 + 1 - 1 - 1 = -2$
$\therefore$ Range of y is $\{-2, 0, 4\}$
Required sum $= -2 + 0 + 4 = 2$
Hence, the correct answer is option: 'A'.