Mathematics

If $$\displaystyle \frac{ax+b}{(3x+4)^2} =\frac{1}{3x+4}-\frac{3}{(3x+4)^2}$$ then


ANSWER

$$b=1$$

$$a=3$$


SOLUTION
Given, $$\displaystyle \frac{ax+b}{(3x+4)^2} =\frac{1}{3x+4}-\frac{3}{(3x+4)^2}$$       ....(1)
Resolving into partial fractions
$$\displaystyle \frac { ax+b }{ (3x+4)^{ 2 } } =\frac { A }{ 3x+4 } +\frac { B }{ (3x+4)^{ 2 } } $$      .....(2)
$$\displaystyle \frac { ax+b }{ (3x+4)^{ 2 } } =\frac { A(3x+4)+B }{ (3x+4)^{ 2 } } $$
$$\Rightarrow ax+b=A(3x+4)+B$$
$$\Rightarrow a=3A,b=4A+B$$
$$\Rightarrow \displaystyle A=\frac{a}{3}, B=b-\frac{4a}{3}$$
Put these values in (2)
$$\displaystyle \frac { ax+b }{ (3x+4)^{ 2 } } =\frac { a }{ 3(3x+4) } +\frac { 3b-4a }{ 3(3x+4)^{ 2 } } $$
Comparing this with (1),
$$\displaystyle \frac{a}{3}=1,\frac { 3b-4a }{ 3}=-3$$
$$\Rightarrow a=3,b=1$$
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