Question

If $\frac{1}{(x-2)(x-3{)}^3}=\frac{A}{x-2}+\frac{B}{x-3}+\frac{C}{(x-3{)}^2}+\frac{D}{(x-3{)}^3}$ then $B=$

• A. $1$
• B. $-1$
• C. $\frac{1}{25}$
• D. $0$

Answer 1

Answer: (A)

$1$

$\frac{1}{(x-2)(x-3{)}^3}=\frac{A}{x-2}+\frac{B}{x-3}+\frac{C}{(x-3{)}^2}+\frac{D}{(x-3{)}^3}$

$\Rightarrow 1=A(x-3{)}^3+B(x-2\left)\right(x^2-6x+9)+C(x-2\left)\right(x-3)+D(x-2)$

Put $x=3\Rightarrow D=1$

Put $x=2\Rightarrow A=-1$

Put $x=0\Rightarrow C-3B=-4$ .....(i)

Put $x=1\Rightarrow -4B+2C=-6$ .....(ii)

On solving, we get $B=1$