If the domain of the function $$f(x) = \log_e(4x^2 + 11x + 6) + \sin^{-1}(4x + 3) + \cos^{-1}\left(\frac{10x + 6}{3}\right)$$ is $$(\alpha, \beta]$$, then $$36|\alpha + \beta|$$ is equal to

Find the domain of $$f(x) = \ln(4x^2+11x+6) + \sin^{-1}(4x+3) + \cos^{-1}\left(\dfrac{10x+6}{3}\right)$$, then compute $$36|\alpha+\beta|$$.

Domain constraints.

(i) $$4x^2+11x+6 > 0$$: Factor: $$(4x+3)(x+2) > 0$$. Solutions: $$x < -2$$ or $$x > -\dfrac{3}{4}$$.

(ii) $$-1 \leq 4x+3 \leq 1$$: $$-4 \leq 4x \leq -2$$, so $$-1 \leq x \leq -\dfrac{1}{2}$$.

(iii) $$-1 \leq \dfrac{10x+6}{3} \leq 1$$: $$-3 \leq 10x+6 \leq 3$$, so $$-\dfrac{9}{10} \leq x \leq -\dfrac{3}{10}$$.

From (ii): $$[-1, -1/2]$$

From (iii): $$[-9/10, -3/10]$$

Intersection of (ii) and (iii): $$[-9/10, -1/2]$$

Now intersect with (i): $$x < -2$$ or $$x > -3/4$$.

$$[-9/10, -1/2]$$ intersected with ($$x < -2$$ or $$x > -3/4$$):

$$-9/10 = -0.9$$ and $$-3/4 = -0.75$$. So $$x > -3/4$$ intersected with $$[-9/10, -1/2]$$ gives $$(-3/4, -1/2]$$.

Also check $$x < -2$$: no overlap with $$[-9/10, -1/2]$$.

So domain = $$(-3/4, -1/2] = (\alpha, \beta]$$.

$$\alpha = -3/4, \beta = -1/2$$

$$\alpha + \beta = -3/4 + (-1/2) = -5/4$$

$$36|\alpha + \beta| = 36 \times 5/4 = 45$$

The correct answer is Option D: 45.