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

(\frac{8}{3}, \frac{2}{3})

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(- \frac{8}{3}, \frac{2}{3})

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(- \frac{8}{3}, - \frac{2}{3})

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(\frac{8}{3}, - \frac{2}{3})

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Solution

The correct option is B $(- \frac{8}{3}, \frac{2}{3})$
$\int f (x) = A log (1 - x) + B x + C$
Differentiate w.r.t $x$
$f (x) = \frac{- A}{1 - x} + B$

$f (\frac{3 x - 4}{3 x + 4}) = \frac{- A}{1 - (\frac{3 x - 4}{3 x + 4})} + B$

$x + 2 = \frac{- A (3 x + 4)}{8} + B$

$\Rightarrow x + 2 = \frac{- 3 A x}{8} - \frac{4 A}{8} + B$

$\Rightarrow \frac{- 3 A}{8} = 1$

$\Rightarrow A = \frac{- 8}{3}$

$\Rightarrow \frac{- 4 A}{8} + B = 2$

$\Rightarrow B = 2 + \frac{A}{2}$

$\Rightarrow B = 2 - \frac{4}{3} = \frac{2}{3}$

Hence,
$(A, B) \equiv (- \frac{8}{3}, \frac{2}{3})$

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