Aspire Faculty ID #17948 · Topic: JEE Main 2026 (22 January Morning Shift) · Just now
JEE Main 2026 (22 January Morning Shift)

If the domain of the function $f(x) = \sin^{-1}\left(\frac{5 - x}{3 + 2x}\right) + \log\left(\frac{1}{10 - x}\right)$ is $(-\infty, \alpha] \cup [\beta, \gamma) - {8}$, then $6(\alpha + \beta + \gamma + \delta)$ is equal to

Solution

$-1 \le \frac{5-x}{2x+3} \le 1,; 10-x > 0,; 10-x \ne 1$

$\left|\frac{5-x}{2x+3}\right| \le 1$ & $x < 10$ & $x \ne 9$

$(5-x)^2 - (2x+3)^2 \le 0$

$(x+8)(3x-2) \ge 0$ & $x < 10$ & $x \ne 9$

$\Rightarrow (-\infty,-8] \cup \left[\frac{2}{3},10\right) - {9}$

$\Rightarrow (\alpha + \beta + \gamma + \delta) = 6\left(-8 + \frac{2}{3} + 10 + 9\right)$

$= 70$

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