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

The Correct Option is 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 6 k=3-48+81=36 $
$\Rightarrow k=6 $
$\therefore k+A-B+C=6+\frac{1}{2}+16+\frac{81}{2} $
$=6+16+41=63$