Question

If $\frac{x^4}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}=x+k+\frac{A}{x-1}+\frac{B}{x-2}+\frac{C}{x-3}$, then $k+A-B+C=$

• A. $104$
• B. $52$
• C. $63$
• D. $\frac{127}{2}$

Answer 1

Answer: (C)

Given,
$\frac{x^4}{(x-1)(x-2)(x-3)}$
$=x+k+\frac{A}{x-1}+\frac{B}{x-2}+\frac{C}{x-3}$
$\Rightarrow x^4=x(x-1)(x-2)(x-3)+k(x-1)$
$(x-2)(x-3)+A(x-2)(x-3)+B(x-3)$
$(x-1)+C(x-1)(x-2)$
on putting $x=1$, we get
$A=\frac{1}{2}$
on putting $x=2$, we get
$B=\frac{16}{-1}=-16$
on putting $x=3$, we get
$C=\frac{81}{2}$
on putting $x=0$, we get
$0=k(-1)(-2)(-3)+\frac{1}{2}(-2)(-3)-16$
$(-3)(-1)+\frac{81}{2}(-1)(-2)$
$\Rightarrow 6k=3-48+81=36$
$\Rightarrow k=6$
$\therefore k+A-B+C=6+\frac{1}{2}+16+\frac{81}{2}$
$=6+16+41=63$