JEE Main 2026 Jan 22 Shift 1, Mathematics Q14: Invertible And Composite Functions

JEE Main2026Jan 22, Shift 1Mathematics

If the domain of the function $f (x) = sin^{- 1} (\frac{5 - x}{3 + 2 x}) + \frac{1}{lo g _{e} ( 10 - x )}$ is $(- \infty, α] \cup [β, γ) - {δ}$ , then $6 (α + β + γ + δ)$ is equal to

A
70
B
66
C
67
D
68

Solution

Three constraints:

1) $sin^{- 1}$ domain: $\frac{5 - x}{3 + 2 x} \leq 1$ , i.e. $(5 - x)^{2} \leq (2 x + 3)^{2}$ . Factor as a difference of squares: $[(2 x + 3) - (5 - x)] [(2 x + 3) + (5 - x)] \geq 0$ , i.e. $(3 x - 2) (x + 8) \geq 0$ , giving $x \leq - 8$ or $x \geq 2/3$ .

2) $lo g_{e} (10 - x)$ defined: $10 - x > 0$ , so $x < 10$ .

3) Denominator $lo g_{e} (10 - x) \neq = 0$ : $10 - x \neq = 1$ , so $x \neq = 9$ .

Combining: $x \in (- \infty, - 8] \cup [2/3, 10) - {9}$ .

So $α = - 8, β = 2/3, γ = 10, δ = 9$ . Then $6 (α + β + γ + δ) = 6 (- 8 + 2/3 + 10 + 9) = 6 \cdot \frac{35}{3} = 70$ .

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JEE Main 2026 Jan 22 Shift 1, Mathematics Q14: Invertible And Composite Functions

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