Question

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

• A. $a=2$
• B. $b=1$
• C. $a=3$
• D. $b=4$

Answer 1

Answer: (B), (C)

$a=3$
$b=1$

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