The sum of all the elements in the range of f(x)= Sgn (sin x)+ Sgn (cos x)+ Sgn (tan x)+ Sgn (cot x) , x neq frac n π 2 , n in Z , where Sgn (t)= \ begin array lll 1, & if & t 0 \\ -1 & if & t<0end array . , is

Mathematics · Functions

JEE Main 2026 — 28 January, Evening Shift — Question 15

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

Option A:
$4$
Option B:
$2$
Correct
Option C:
$-2$
Option D:
$0$

Answer: B

Step-by-step solution

$x \in(0, \pi / 2) \Rightarrow y=1+1+1+1=4$

$\mathrm{x} \in(\pi / 2, \pi) \Rightarrow \mathrm{y}=1-1-1-1=-2$

$\mathrm{x} \in(\pi, 3 \pi / 2) \Rightarrow \mathrm{y}=-1-1+1+1=0$

$\mathrm{x} \in(3 \pi / 2,2 \pi) \Rightarrow \mathrm{y}=-1+1-1-1=-2$

∴ Range of y is $\{-2,0,4\}$

Required sum $=-2+0+4=2$

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 28 January, Evening Shift
Subject: Mathematics
Chapter: Functions
Topic: Standard Functions

Given below are two statements : one is labelled as Statement I and the other is labelled as Statement II : Statement I : The function f: R…

Answer: B

Step-by-step solution

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