1. Find the domain of $\log_5(18x-x^2-77)$

For a logarithm to be defined, its argument must be positive: $18x-x^2-77>0$ Rewriting, $-x^2+18x-77>0$ $x^2-18x+77<0$ Factorize: $x^2-18x+77=(x-7)(x-11)$ So, $(x-7)(x-11)<0$ This holds for $7<x<11$ Hence, $\alpha=7,\quad \beta=11$

2. Find the domain of $\log_{(x-1)}\left(\dfrac{2x^2+3x-2}{x^2-3x-4}\right)$

For $\log_a b$ to be defined, we need:

1. $a>0$
2. $a\neq 1$
3. $b>0$

Here base is $x-1$, so: $x-1>0 \Rightarrow x>1$ and $x-1\neq 1 \Rightarrow x\neq 2$

Now simplify the argument: $\frac{2x^2+3x-2}{x^2-3x-4}$ Factor numerator and denominator: $2x^2+3x-2=(2x-1)(x+2)$ $x^2-3x-4=(x-4)(x+1)$ So we need $\frac{(2x-1)(x+2)}{(x-4)(x+1)}>0$ with also $x\neq 4,-1$.

Critical points are: $-2,\,-1,\,\frac12,\,4$

But since $x>1$, only the interval $(1,\infty)$ matters, excluding $x=2,4$.

Test sign of $\frac{(2x-1)(x+2)}{(x-4)(x+1)}$ for $x>1$:

• For $1<x<4$, take $x=3$: $\frac{(6-1)(5)}{(-1)(4)}<0$ So negative.

• For $x>4$, take $x=5$: $\frac{(10-1)(7)}{(1)(6)}>0$ So positive.

Thus argument $>0$ only for $x>4$ This already satisfies $x>1$ and excludes $x=2$ automatically.

Hence, $\gamma=4,\quad \delta=\infty$ So the domain is $(4,\infty)$.

3. Compute the required value

We need: $\alpha^2+\beta^2+\gamma^2$ Substitute values: $=7^2+11^2+4^2$ $=49+121+16$ $=186$

4. Compare with options

$186$ matches Option A.

5. Verification with stored answer

Stored correct answer: A

Our derived answer: A

So the answer agrees with the stored correct answer.