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JEE Main 202628 JanuaryEvening Shift
Exam
JEE Main
Year
2026
Date
28 January
Shift
Evening Shift
Subject
Mathematics
Chapter
Functions

JEE Main 2026 · 28 January · Evening Shift · Mathematics · Functions

The sum of all the elements in the range of f(x)=Sgn( x)+Sgn(...

The sum of all the elements in the range of f(x)=Sgn⁡(sin⁡x)+Sgn⁡(cos⁡x)+Sgn⁡(tan⁡x)+Sgn⁡(cot⁡x)f(x)=\operatorname{Sgn}(\sin x)+\operatorname{Sgn}(\cos x)+\operatorname{Sgn}(\tan x)+\operatorname{Sgn}(\cot x), x≠nπ2,n∈Z,whereSgn⁡x \neq \frac{\mathrm{n} \pi}{2}, \mathrm{n} \in \mathbf{Z} , where \operatorname{Sgn}(t\mathrm{t})=\left\{1, if t>0−1, if t<0\begin{aligned} 1, & \text{ if } \mathrm{t}>0 \\ -1, & \text{ if } \mathrm{t}<0\end{aligned}\right.$, is :

Options

  1. A2
  2. B0
  3. C4
  4. D-2

Correct answer

A. 2

Step-by-step solution

x∈(0,π/2)⇒y=1+1+1+1=4x \in (0, \pi/2) \Rightarrow y = 1 + 1 + 1 + 1 = 4
x∈(π/2,π)⇒y=1−1−1−1=−2x \in (\pi/2, \pi) \Rightarrow y = 1 - 1 - 1 - 1 = -2
x∈(π,3π/2)⇒y=−1−1+1+1=0x \in (\pi, 3\pi/2) \Rightarrow y = -1 - 1 + 1 + 1 = 0
x∈(3π/2,2π)⇒y=−1+1−1−1=−2x \in (3\pi/2, 2\pi) \Rightarrow y = -1 + 1 - 1 - 1 = -2
∴\therefore Range of y is {−2,0,4}\{-2, 0, 4\}
Required sum =−2+0+4=2= -2 + 0 + 4 = 2
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