Let f, g :(1, infty) rightarrow mathbb{R} be | vedclass

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(D) Given $f(x) = \frac{2x+3}{5x+2}$ and $g(x) = \frac{2-3x}{1-x}$.We need to find $(f \circ g)(x) = f(g(x))$.$(f \circ g)(x) = f\left(\frac{2-3x}{1-x

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$56$

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(D) Given $f(x) = \frac{2x+3}{5x+2}$ and $g(x) = \frac{2-3x}{1-x}$.We need to find $(f \circ g)(x) = f(g(x))$.$(f \circ g)(x) = f\left(\frac{2-3x}{1-x}\right) = \frac{2\left(\frac{2-3x}{1-x}\right)+3}{5\left(\frac{2-3x}{1-x}\right)+2}$Multiplying numerator and denominator by $(1-x)$:$(f \circ g)(x) = \frac{2(2-3x) + 3(1-x)}{5(2-3x) + 2(1-x)} = \frac{4-6x+3-3x}{10-15x+2-2x} = \frac{7-9x}{12-17x}$.Now,we evaluate the function at the endpoints of the interval $[2, 4]$:$(f \circ g)(2) = \frac{7-9(2)}{12-17(2)} = \frac{7-18}{12-34} = \frac{-11}{-22} = \frac{1}{2}$.$(f \circ g)(4) = \frac{7-9(4)}{12-17(4)} = \frac{7-36}{12-68} = \frac{-29}{-56} = \frac{29}{56}$.Since the function is monotonic in the interval $[2, 4]$,the range is $[\alpha, \beta] = [\frac{1}{2}, \frac{29}{56}]$.Here,$\alpha = \frac{1}{2}$ and $\beta = \frac{29}{56}$.Then $\beta - \alpha = \frac{29}{56} - \frac{1}{2} = \frac{29-28}{56} = \frac{1}{56}$.Therefore,$\frac{1}{\beta-\alpha} = 56$.

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

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