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

Question

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

Accepted Answer

A. 1. For $\sin^{-1}(\cdot)$ to be defined, its argument must lie between $-1$ and $1$.
$$
-1\le \frac{5-𝑥}{3+2𝑥}\le 1
$$

2. Handle the left inequality $\frac{5-𝑥}{3+2𝑥}\ge -1$ first by bringing everything to one side.
$$
\frac{2-3𝑥}{3+2𝑥}\le 0 \ \Longrightarrow\ (3𝑥-2)(2𝑥+3)\ge 0
$$
Solving this sign condition (roots at $𝑥=\frac23$ and $𝑥=-\frac32$):
$$
𝑥\in\left(-\infty,-\frac32ight]\cup\left[\frac23,\inftyight)
$$

3. Now handle the right inequality $\frac{5-𝑥}{3+2𝑥}\le 1$ similarly.
$$
\frac{8+𝑥}{3+2𝑥}\ge 0 \ \Longrightarrow\ 𝑥\in(-\infty,-8]\cup\left[-\frac32,\inftyight)
$$

4. Intersecting the results of steps 2 and 3 gives the domain restriction from the inverse sine term.
$$
\left(-\infty,-\frac32ight]\cup\left[\frac23,\inftyight)\ \cap\ \left((-\infty,-8]\cup\left[-\frac32,\inftyight)ight)=(-\infty,-8]\cup\left[\frac23,\inftyight)
$$

5. Next, for $\frac{1}{\log_e(10-𝑥)}$ to be defined, we need $10-𝑥>0$ (so the log itself is defined) **and** $\log_e(10-𝑥)
e 0$ (so we don't divide by zero). This gives:
$$
𝑥<10 \quad 	ext{and} \quad 10-𝑥
e 1 \ \Longrightarrow\ 𝑥
e 9
$$

6. Combining this with the result from step 4 gives the overall domain.
$$
\big((-\infty,-8]\cup[	frac23,\infty)\big)\cap(-\infty,10)-\{9\} = (-\infty,-8]\cup\left[\frac23,10ight)-\{9\}
$$
So $\alpha=-8$, $\beta=\frac23$, $\gamma=10$, $\delta=9$.

7. Now compute the required value.
$$
\alpha+\beta+\gamma+\delta=-8+\frac23+10+9=\frac{35}{3}
$$
$$
6(\alpha+\beta+\gamma+\delta)=6	imes\frac{35}{3}=70
$$

Hence, the answer is **Option A**.