If fleft( frac{3x - 4}{3x + 4} right) = x + 2 | vedclass

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(B) Let $t = \frac{3x - 4}{3x + 4}$.Then $3xt + 4t = 3x - 4$,which implies $x(3t - 3) = -4t - 4$,so $x = \frac{4t + 4}{3 - 3t}$.Substituting this into

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$\left( -\frac{8}{3}, \frac{2}{3} \right)$

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(B) Let $t = \frac{3x - 4}{3x + 4}$.Then $3xt + 4t = 3x - 4$,which implies $x(3t - 3) = -4t - 4$,so $x = \frac{4t + 4}{3 - 3t}$.Substituting this into the function: $f(t) = \frac{4t + 4}{3 - 3t} + 2 = \frac{4t + 4 + 6 - 6t}{3 - 3t} = \frac{10 - 2t}{3 - 3t}$.Thus,$f(x) = \frac{10 - 2x}{3 - 3x} = \frac{2x - 10}{3x - 3}$.Now,$\int f(x) dx = \int \frac{2x - 10}{3(x - 1)} dx = \frac{2}{3} \int \frac{x - 1 - 4}{x - 1} dx = \frac{2}{3} \int (1 - \frac{4}{x - 1}) dx$.$= \frac{2}{3} x - \frac{8}{3} \ln |x - 1| + C = -\frac{8}{3} \ln |1 - x| + \frac{2}{3} x + C$.Comparing with $A \log |1 - x| + Bx + C$,we get $A = -\frac{8}{3}$ and $B = \frac{2}{3}$.Therefore,the ordered pair $(A, B) = \left( -\frac{8}{3}, \frac{2}{3} \right)$.

If $f\left( \frac{3x - 4}{3x + 4} \right) = x + 2, x \ne -\frac{4}{3}$,and $\int f(x) dx = A \log |1 - x| + Bx + C$,then the ordered pair $(A, B)$ is equal to: (where $C$ is a constant of integration)

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