JEE Main 2026 — Jan 21, Shift 1 — Mathematics solution

JEE Main2026Jan 21, Shift 1Mathematics

If the domain of the function $f (x) = cos^{- 1} (\frac{2 x - 5}{11 - 3 x}) + sin^{- 1} (2 x^{2} - 3 x + 1)$ is the interval $[α, β]$ , then $α + 2 β$ is equal to :

Solution

Both inverse-trig arguments must lie in $[- 1, 1]$ .

From $sin^{- 1}$ : $- 1 \leq 2 x^{2} - 3 x + 1 \leq 1$ . The left inequality $2 x^{2} - 3 x + 2 \geq 0$ holds for all real $x$ (discriminant negative). The right gives $2 x^{2} - 3 x \leq 0 \Rightarrow x \in [0, 3/2]$ .

From $cos^{- 1}$ : $- 1 \leq \frac{2 x - 5}{11 - 3 x} \leq 1$ . Sign analysis on both inequalities yields $x \in (- \infty, \frac{16}{5}] \cup [6, \infty)$ .

Intersection: $x \in [0, 3/2]$ , so $α = 0$ , $β = \frac{3}{2}$ .

$α + 2 β = 0 + 3 = 3$ .

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