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JEE Main 202622 JanuaryMorning Shift
Exam
JEE Main
Year
2026
Date
22 January
Shift
Morning Shift
Subject
Mathematics
Chapter
Functions

JEE Main 2026 · 22 January · Morning Shift · Mathematics · Functions

If the domain of the function f(x)= ^ -1 ((5-x)/(3+2 x))+ 1 _ e...

If the domain of the function f(x)=sin⁡−1(5−x3+2x)+1log⁡e(10−x)f(x)=\sin ^{-1}\left(\frac{5-x}{3+2 x}\right)+\frac{1}{\log _{e}(10-x)} is (−∞,α]∪[β,γ)−δ(-\infty, \alpha] \cup[\beta, \gamma)-\delta, then 6(α+β+γ+δ)6(\alpha+\beta+\gamma+\delta) is equal to

Options

  1. A67
  2. B66
  3. C70
  4. D68

Correct answer

C. 70

Step-by-step solution

-1 \leq \dfrac{5-x}{2x+3} \leq&10 - x > 0,,10 - x \neq 1<br>∣5−x2x+3∣≥1<br>\left| \dfrac{5-x}{2x+3} \right| \geq 1 & x<10x < 10 & x≠9x \neq 9
(5−x)2−(2x+3)2≤0(5-x)^2 - (2x+3)^2 \leq 0 & x<10x < 10 & 4x≠94x \neq 9
(x+8)(3x−2)≥0(x+8)(3x-2) \geq 0 & x<10x < 10 & x≠9x \neq 9
⇒(−∞,−8]∪[23,10)−{9}\Rightarrow (-\infty, -8] \cup \left[\dfrac{2}{3}, 10\right) - \{9\}
⇒(α+β+γ+δ)=6(−8+23+10+9)\Rightarrow (\alpha + \beta + \gamma + \delta) = 6\left(-8 + \dfrac{2}{3} + 10 + 9\right)
=70= 70
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