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+Cthen

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)
V Vakul

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    

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