If the domain of the function f ( x ) = sin - 1 ( 5 - x 3 + 2 x ) + 1 log e ( 10 - x ) is , then 6 is equal to [2026]

Question

If the domain of the function $f (x)$ $= \sin^{- 1} (\frac{5 - x}{3 + 2 x}) + \frac{1}{\log_{e} (10 - x)}$ is $(- \infty, α] \cup [β, γ) - {δ},$ then $6 (α + β + γ + δ)$ is equal to [2026]

Smart Achievers · Accepted Answer

(4)

$- 1 \leq \frac{5 - x}{2 x + 3} \leq 1 & 10 - x > 0, 10 - x \neq 1$

$| \frac{5 - x}{2 x + 3} |$ $\geq 1 & x < 10 & x \neq 9$

${(5 - x)}^{2} - {(2 x + 3)}^{2} \leq 0 & x < 10 & 4 x \neq 9$

$(x + 8) (3 x - 2) \geq 0 & x < 10 & x \neq 9$

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

$\Rightarrow (α + β + γ + δ) = 6 (- 8 + \frac{2}{3} + 10 + 9)$

$= 70$

Solution

Q.
If the domain of the function $f (x)$ $= \sin^{- 1} (\frac{5 - x}{3 + 2 x}) + \frac{1}{\log_{e} (10 - x)}$ is $(- \infty, α] \cup [β, γ) - {δ},$ then $6 (α + β + γ + δ)$ is equal to [2026]

1 68

2 67

3 66

4 70

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