Let $$f, g : (1, \infty) \to \mathbb{R}$$ be defined as $$f(x) = \dfrac{2x + 3}{5x + 2}$$ and $$g(x) = \dfrac{2 - 3x}{1 - x}$$. If the range of the function $$f \circ g : [2, 4] \to \mathbb{R}$$ is $$[\alpha, \beta]$$, then $$\dfrac{1}{\beta - \alpha}$$ is equal to

We first simplify the composition $$\bigl(f \circ g\bigr)(x) = f\!\bigl(g(x)\bigr)$$.

Given $$f(x)=\dfrac{2x+3}{5x+2}$$ and $$g(x)=\dfrac{2-3x}{1-x}$$, rewrite $$g(x)$$ for convenience: $$g(x)=\dfrac{2-3x}{1-x}= \dfrac{-(3x-2)}{-(x-1)}=\dfrac{3x-2}{x-1}, \quad x\gt 1.$$

Step 1 - Compute $$2\,g(x)+3$$: $$2\,g(x)+3 = 2\left(\dfrac{3x-2}{x-1}\right)+3 = \dfrac{6x-4}{x-1} + 3 = \dfrac{6x-4 + 3(x-1)}{x-1} = \dfrac{6x-4 + 3x-3}{x-1} = \dfrac{9x-7}{x-1}. \quad -(1)$$

Step 2 - Compute $$5\,g(x)+2$$: $$5\,g(x)+2 = 5\left(\dfrac{3x-2}{x-1}\right)+2 = \dfrac{15x-10}{x-1} + 2 = \dfrac{15x-10 + 2(x-1)}{x-1} = \dfrac{15x-10 + 2x-2}{x-1} = \dfrac{17x-12}{x-1}. \quad -(2)$$

Step 3 - Form the quotient using $$(1)$$ and $$(2)$$: $$\bigl(f \circ g\bigr)(x)=\dfrac{2\,g(x)+3}{5\,g(x)+2} =\dfrac{\dfrac{9x-7}{x-1}}{\dfrac{17x-12}{x-1}} =\dfrac{9x-7}{17x-12}. \quad -(3)$$

Therefore, for every $$x\gt 1$$, $$h(x)=\bigl(f \circ g\bigr)(x)=\dfrac{9x-7}{17x-12}.$$

Step 4 - Check monotonicity on $$x\in(1,\infty)$$.

Differentiate: $$h'(x)=\dfrac{(9)(17x-12)-(9x-7)(17)}{(17x-12)^2} =\dfrac{153x-108-153x+119}{(17x-12)^2} =\dfrac{11}{(17x-12)^2}\gt 0.$$

Since $$h'(x)\gt 0$$ for all $$x\gt 1$$, $$h(x)$$ is strictly increasing. Hence on any closed interval within $$(1,\infty)$$, its minimum occurs at the left end and its maximum at the right end.

Step 5 - Evaluate $$h(x)$$ at the endpoints of the given domain $$[2,4]$$.

At $$x=2$$: $$h(2)=\dfrac{9(2)-7}{17(2)-12} =\dfrac{18-7}{34-12} =\dfrac{11}{22} =\dfrac12.$$

At $$x=4$$: $$h(4)=\dfrac{9(4)-7}{17(4)-12} =\dfrac{36-7}{68-12} =\dfrac{29}{56}.$$

Step 6 - State the range.

Because $$h(x)$$ is increasing, $$\text{Range}\bigl(h|_{[2,4]}\bigr)=[\alpha,\beta]=\left[\dfrac12,\dfrac{29}{56}\right].$$

Step 7 - Compute $$\dfrac1{\beta-\alpha}$$.

First find $$\beta-\alpha$$: $$\beta-\alpha=\dfrac{29}{56}-\dfrac12 =\dfrac{29}{56}-\dfrac{28}{56} =\dfrac1{56}.$$

Therefore, $$\dfrac1{\beta-\alpha}=\dfrac1{\,\tfrac1{56}\,}=56.$$

The required value equals $$56$$. Hence the correct option is Option D.