1. Given function

$f(x)=\sin^{-1}\left(\frac{3x-22}{2x-19}\right)+\ln\left(\frac{3x^2-8x+5}{x^2-3x-10}\right)$

We need the domain of this function.

2. Condition from the inverse sine term

For $\sin^{-1}(u)$ to be defined, we need $-1\le u\le 1.$

So, $-1\le \frac{3x-22}{2x-19}\le 1, \qquad x\ne \frac{19}{2}.$

We solve both inequalities.

(i) Solve

$\frac{3x-22}{2x-19}\le 1$

$\frac{3x-22}{2x-19}-1\le 0$ $\frac{3x-22-(2x-19)}{2x-19}\le 0$ $\frac{x-3}{2x-19}\le 0.$

Critical points: $x=3,\; x=\frac{19}{2}$.

Sign analysis gives $x\in [3,\tfrac{19}{2}).$

(ii) Solve

$\frac{3x-22}{2x-19}\ge -1$

$\frac{3x-22}{2x-19}+1\ge 0$ $\frac{3x-22+2x-19}{2x-19}\ge 0$ $\frac{5x-41}{2x-19}\ge 0.$

Critical points: $x=\frac{41}{5},\; x=\frac{19}{2}$.

Sign analysis gives $x\in (-\infty,\tfrac{41}{5}]\cup(\tfrac{19}{2},\infty).$

Now intersect both conditions:

$[3,\tfrac{19}{2})\cap\left[(-\infty,\tfrac{41}{5}]\cup(\tfrac{19}{2},\infty)\right]=[3,\tfrac{41}{5}].$

So, from the $\sin^{-1}$ part, $x\in [3,\tfrac{41}{5}].$

3. Condition from the logarithm term

For $\ln\left(\frac{3x^2-8x+5}{x^2-3x-10}\right)$ to be defined, we need $\frac{3x^2-8x+5}{x^2-3x-10}>0.$

Factorize:

$3x^2-8x+5=(3x-5)(x-1),$ $x^2-3x-10=(x-5)(x+2).$

Thus, $\frac{(3x-5)(x-1)}{(x-5)(x+2)}>0.$

Critical points are $x=-2,\;1,\;\frac53,\;5.$

Now do sign analysis:

• For $x<-2$: positive
• For $-2<x<1$: negative
• For $1<x<\frac53$: positive
• For $\frac53<x<5$: negative
• For $x>5$: positive

Hence the logarithm domain is $(-\infty,-2)\cup(1,\tfrac53)\cup(5,\infty).$

4. Intersect both domain conditions

From $\sin^{-1}$: $[3,\tfrac{41}{5}]$

From $\ln$: $(-\infty,-2)\cup(1,\tfrac53)\cup(5,\infty)$

Their intersection is $[3,\tfrac{41}{5}]\cap\left[(-\infty,-2)\cup(1,\tfrac53)\cup(5,\infty)\right]=(5,\tfrac{41}{5}].$

So the domain is $(\alpha,\beta]=(5,\tfrac{41}{5}].$

Thus, $\alpha=5,\qquad \beta=\frac{41}{5}.$

5. Compute the required value

$3\alpha+10\beta=3(5)+10\left(\frac{41}{5}\right)=15+82=97.$

6. Compare with options

The value is $97,$ which corresponds to Option C.