Verified Solution • 2025–26 Session moderate

QJEEM24Ma56c1: JEE Main 2024 Mathematics - Class 11 JEEM24Ma56c1 Solution: If the domain of the function is , then is equal to :

If the domain of the function f(x)=sin⁡−1(x−12x+3)f(x)=\sin ^{-1}\left(\frac{x-1}{2 x+3}\right) is R−(α,β)\mathbf{R}-(\alpha, \beta) , then 12αβ12 \alpha \beta is equal to :

Multiple Choice Options

  • A 40
  • B 36
  • C 24
  • D 32

Step-by-Step Verified Solution

f(x)=sin⁡−1(x−12x+3)amp;−1≤x−12x+3≤1amp;x−12x+3+1≥0x−12x+3−1≤0amp;x−1+2x+32x+3≥0x−1−2x−32x+3≤0amp;3x+22x+3≥0\begin{array}{ll}f(x)=\sin ^{-1}\left(\frac{x-1}{2 x+3}\right) & \\-1 \leq \frac{x-1}{2 x+3} \leq 1 & \frac{x-1}{2 x+3}+1 \geq 0 \\\frac{x-1}{2 x+3}-1 \leq 0 & \frac{x-1+2 x+3}{2 x+3} \geq 0 \\\frac{x-1-2 x-3}{2 x+3} \leq 0 & \frac{3 x+2}{2 x+3} \geq 0\end{array}

JEE Main 2024 (Online) 9th April Morning Shift Mathematics - Functions Question 19 English Explanation 1

amp;−x−42x+3≤0amp;x+42x+3≥0\begin{aligned}& \frac{-x-4}{2 x+3} \leq 0 \\& \frac{x+4}{2 x+3} \geq 0\end{aligned}

JEE Main 2024 (Online) 9th April Morning Shift Mathematics - Functions Question 19 English Explanation 2

amp;x∈(−∞,−4]∪[−23,∞)amp; Domain : R−(−4,−23)amp;α=−4amp;β=−23amp;12×αβ=12×4×23=32\begin{aligned}& x \in(-\infty,-4] \cup\left[\frac{-2}{3}, \infty\right) \\& \text { Domain : } R-\left(-4, \frac{-2}{3}\right) \\& \alpha=-4 \\& \beta=\frac{-2}{3} \\& 12 \times \alpha \beta=12 \times 4 \times \frac{2}{3}=32\end{aligned}

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