Q33 / 75 · JEE Main 2026 (22 January Shift 1) · Mathematics
If the domain of the function $f(x)=\sin ^{-1}\left(\frac{5-x}{3+2 x}\right)+\frac{1}{\log _{e}(10-x)}$ is $(-\infty, \alpha] \cup[\beta, \gamma)-\{\delta\}$, then $6(\alpha+\beta+\gamma+\delta)$ is equal to
- A67
- B66
- Correct answerC70
- D68
Solution
$-1 \leq \dfrac{5-x}{2x+3} \leq$ & $10 - x > 0$, $10 - x \neq 1$
$\left| \dfrac{5-x}{2x+3} \right| \geq 1$ & $x < 10$ & $x \neq 9$
$(5-x)^2 - (2x+3)^2 \leq 0$ & $x < 10$ & $4x \neq 9$
$(x+8)(3x-2) \geq 0$ & $x < 10$ & $x \neq 9$
$\Rightarrow (-\infty, -8] \cup \left[\dfrac{2}{3}, 10\right) - \{9\}$
$\Rightarrow (\alpha + \beta + \gamma + \delta) = 6\left(-8 + \dfrac{2}{3} + 10 + 9\right)$
$= 70$
$\left| \dfrac{5-x}{2x+3} \right| \geq 1$ & $x < 10$ & $x \neq 9$
$(5-x)^2 - (2x+3)^2 \leq 0$ & $x < 10$ & $4x \neq 9$
$(x+8)(3x-2) \geq 0$ & $x < 10$ & $x \neq 9$
$\Rightarrow (-\infty, -8] \cup \left[\dfrac{2}{3}, 10\right) - \{9\}$
$\Rightarrow (\alpha + \beta + \gamma + \delta) = 6\left(-8 + \dfrac{2}{3} + 10 + 9\right)$
$= 70$
Hence, the correct answer is option: 'C'.