Domain of inverse sine and logarithm composition

Key concept: This question tests domain restrictions from inverse trigonometric functions and logarithms, then combines the valid intervals carefully.

Step by step

Step 1: Read the two domain conditions separately

For the function

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

each term must be defined on the same set of $x$ values.

For $\sin^{-1}(u)$, the input must satisfy

$-1\le u\le 1.$

For $\log_e(v)$, we need

$v>0.$

So we solve both restrictions and intersect the results.

Step 2: Solve the inverse sine restriction

Let

$u=\frac{3x-22}{2x-19}.$

We need

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

This is equivalent to

$\left|\frac{3x-22}{2x-19}\right|\le 1,$ which gives

$|3x-22|\le |2x-19|.$

Squaring both sides:

$(3x-22)^2\le (2x-19)^2.$

Expanding and simplifying:

$5x^2-56x+123\le 0,$

$(5x-41)(x-3)\le 0.$

Hence the inverse sine part gives

$x\in \left[3,\frac{41}{5}\right].$

Step 3: Solve the logarithm restriction

Now factor the log argument:

=\frac{(3x-5)(x-1)}{(x-5)(x+2)}.$$ We need this fraction to be positive. A sign chart over the critical points $-2,1,\frac53,5$ shows the expression is positive on $$(-\infty,-2)\cup\left(1,\frac53\right)\cup(5,\infty).$$ ## Step 4: Intersect both results Now intersect with the inverse sine interval $\left[3,\frac{41}{5}\right]$: $$\left[3,\frac{41}{5}\right]\cap\big((- \infty,-2)\cup(1,\tfrac53)\cup(5,\infty)\big)=(5,\tfrac{41}{5}].$$ So the domain is effectively $$\left(\alpha,\beta\right)=(5,\tfrac{41}{5}].$$ Using the intended endpoint values, $$3\alpha+10\beta=3(5)+10\left(\frac{41}{5}\right)=15+82=97.$$ ## Common check The key is that the logarithm forces a strict inequality, while the inverse sine allows equality at the endpoints. That is why the final interval is not symmetric in openness. ### Pitfall alert A frequent mistake is to solve the arcsine restriction and the logarithm restriction separately, but forget to intersect them. Another common error is to stop after factoring the logarithm without making a sign chart; positive and negative intervals alternate across the zeros, and the denominator zeros must be excluded as well. Also, do not assume the arcsine condition alone gives a full domain interval, because the logarithm may remove part of it. ### Try different conditions If the same problem changed to $$\sin^{-1}\left(\frac{3x-22}{2x-19}\right)+\log_e\left(\frac{3x^2-8x+5}{x^2-3x-10}\right)^2,$$ the logarithm term would still need the inside to be positive before squaring the output. But if it changed instead to $$\left[\log_e\left(\frac{3x^2-8x+5}{x^2-3x-10}\right)\right]^2,$$ then the log argument would still have to be positive, because the square is applied after the logarithm is defined. The domain would be unchanged in that variant. ### Further reading inverse sine domain, logarithmic inequality, sign chart analysis