If the domain of the function f(x) = sin^{-1} | vedclass

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(A) For the function $f(x) = \sin^{-1}\left(\frac{5-x}{2x+3}\right) + \frac{1}{\log_e(10-x)}$ to be defined:$1$. The argument of $\sin^{-1}$ must be i

Vedclass · Accepted Answer

$70$

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(A) For the function $f(x) = \sin^{-1}\left(\frac{5-x}{2x+3}ight) + \frac{1}{\log_e(10-x)}$ to be defined:$1$. The argument of $\sin^{-1}$ must be in $[-1, 1]$, so $-1 \leq \frac{5-x}{2x+3} \leq 1$.$2$. The argument of $\log_e$ must be positive and not equal to $1$, so $10-x > 0$ and $10-x 
eq 1$, which implies $x $3$. Solving $-1 \leq \frac{5-x}{2x+3} \leq 1$:$\left|\frac{5-x}{2x+3}ight| \leq 1 \implies (5-x)^2 \leq (2x+3)^2$$(5-x)^2 - (2x+3)^2 \leq 0$$(5-x-2x-3)(5-x+2x+3) \leq 0$$(2-3x)(x+8) \leq 0 \implies (3x-2)(x+8) \geq 0$This gives $x \in (-\infty, -8] \cup [\frac{2}{3}, \infty)$.$4$. Combining with $x $5$. Comparing with $(-\infty, \alpha] \cup [\beta, \gamma) - \{\delta\}$, we get $\alpha = -8$, $\beta = \frac{2}{3}$, $\gamma = 10$, and $\delta = 9$.$6$. The value $6(\alpha + \beta + \gamma + \delta) = 6(-8 + \frac{2}{3} + 10 + 9) = 6(\frac{35}{3}) = 70$.

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

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