JEE Main 2026 Apr 4 Shift 1, Mathematics Q1: Inverse Trigonometric Functions

JEE Main2026Apr 4, Shift 1Mathematics

Let $[\cdot]$ denote the greatest integer function. If the domain of the function $f (x) = cos^{- 1} (\frac{4 x + 2 [ x ]}{3})$ is $[α, β]$ , then $12 (α + β)$ is equal to :

A
$6$
B
$8$
C
$9$
D
$4$

Solution

The domain requires the argument of $cos^{- 1}$ to lie in $[- 1, 1]$ :

$- 1 \leq \frac{4 x + 2 [ x ]}{3} \leq 1 \Rightarrow - 3 - 4 x \leq 2 [x] \leq 3 - 4 x .$

Treat this as two linear bounds on $2 [x]$ and intersect with the step function $y = 2 [x]$ . The lower bound $y = - 3 - 4 x$ and upper bound $y = 3 - 4 x$ together with the staircase $2 [x]$ admit valid $x$ exactly on $[- \frac{1}{4}, \frac{3}{4}]$ .

Hence $α = - \frac{1}{4}$ , $β = \frac{3}{4}$ , and

$12 (α + β) = 12 \cdot \frac{1}{2} = 6.$

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JEE Main 2026 Apr 4 Shift 1, Mathematics Q1: Inverse Trigonometric Functions

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