Functions – Range – JEE Main 28 Jan 2026 Shift 2

Solution:

The function is $f(x) = Sgn(\sin x) + Sgn(\cos x) + Sgn(\tan x) + Sgn(\cot x)$.

We analyze the value of $f(x)$ in each quadrant (since $x \ne \frac{n\pi}{2}$, the trigonometric functions are non-zero).

Case 1: $x$ in Quadrant I
All trigonometric functions are positive.
$f(x) = 1 + 1 + 1 + 1 = 4$

Case 2: $x$ in Quadrant II
$\sin x > 0$, $\cos x < 0$, $\tan x < 0$, $\cot x < 0$. $f(x) = 1 + (-1) + (-1) + (-1) = -2$
Case 3: $x$ in Quadrant III
$\sin x < 0$, $\cos x < 0$, $\tan x > 0$, $\cot x > 0$.
$f(x) = -1 + (-1) + 1 + 1 = 0$

Case 4: $x$ in Quadrant IV
$\sin x < 0$, $\cos x > 0$, $\tan x < 0$, $\cot x < 0$. $f(x) = -1 + 1 + (-1) + (-1) = -2$
The range of $f(x)$ is the set of unique values obtained: $\{-2, 0, 4\}$.

Sum of all elements in the range = $-2 + 0 + 4 = 2$.

Ans. (2)