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If $\small f\left ( \frac{3x-4}{3x+4} \right ) =x+2,\: \: x\neq -\frac{4}{3}$ and

$\small \int f\left ( x \right )dx =A \log \left | 1-x \right |+Bx+C$ then

then
the ordered pair (A, B) is equal to : (where C is a constant of integration)

Option 1)
$\left ( \frac{8}{3} ,\frac{2}{3}\right )$

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

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

Option 4)
$\left ( \frac{8}{3},-\frac{2}{3} \right )$

Answers (1)

As learnt in concept

Rule of integration by Partial fraction -

Linear and non-repeated:

$\frac{P(x)}{Q(x)}=\frac{P(x)}{(x-\alpha _{1})(x-\alpha _{2})\cdot \cdot \cdot (x-\alpha _{n})}$

Let $\frac{P(x)}{Q(x)}=\frac{A}{(x-\alpha _{1})}+\frac{B}{(x-\alpha _{2})}\cdot \cdot \cdot$

Find $A,B$ ...

By comparing $N^{x}$ and $P(x)$

-

$f\left ( \frac{3x-4}{3x+4} \right ) =x+2$

$Put \: \frac{3x-4}{3x+4}=y$

$=> 3xy+4y=3x-4$

$x=\frac{-4\left ( y+1 \right )}{3\left ( y-1 \right )}$

$x=\frac{4\left ( 1+y \right )}{3\left ( 1-y \right )}$

$f\left ( y \right ) =\frac{4\left ( 1+y \right )}{3\left ( 1-y \right )}+2$

$= \frac{10-2y}{3\left ( 1-y \right )}$

$\frac{2}{3}\int \left ( \frac{5-x}{1-x} \right )dx=\frac{2}{3}\int \frac{1-x+4}{\left ( 1-x \right )}dx$

$=\frac{2}{3}x+\frac{8}{3}\:ln\left | 1-x \right |+C$

Option 1)

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

Incorrect option

Option 2)

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

Correct option

Option 3)

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

Incorrect option

Option 4)

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

Incorrect option

Posted by

Vakul

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