$\begin{aligned} & \text { If } \frac{x^4}{(x-a)(x-b)(x-c)}=P(x)+\frac{A}{x-a}+\frac{B}{x-b} \\ &…

1. $a^4+b^4+c^4+a$
2. $a+b+c$
3. $a^4-a-b-c$
4. $a+b+c+a^4$

Solution

If is given that $ \begin{array}{r} \frac{x^4}{(x-a)(x-b)(x-c)}=P(x)+\frac{A}{x-a}+\frac{B}{x-b}+\frac{C}{x-c} \\ \Rightarrow x^4=(x-a)(x-b)(x-c) P(x)+A(x-b)(x-c) \\ +B(x-c)(x-a)+C((x-a)(x-b) \end{array} $ At $x=0, a b c P(0)=b c A+c a B+a b C$
At $x=a, A=\frac{a^4}{(a-b)(a-c)}$, similarly At $x=b, \quad B=\frac{b^4}{(b-c)(b-a)}$ and
So, $ \begin{aligned} & P(0)=\frac{a^3}{(a-b)(a-c)}+\frac{b^3}{(b-c)(b-a)}+\frac{c^3}{(c-a)(c-b)} \\ & =\frac{a^3(c-b)+b^3(a-c)+c^3(b-a)}{(a-b)(b-c)(c-a)} \\ & =\frac{a^3(c-b)+b c\left(c^2-b^2\right)+a\left(b^3-c^3\right)}{(a-b)(b-c)(c-a)} \\ & =\frac{a^3+b c(c+b)-a\left(b^2+c^2+b c\right)}{(a-b(a-c)} \\ & =\frac{a\left(a^2-b^2\right)-c^2(a-b)-b c(a-b)}{(a-b)(a-c)} \\ & =\frac{\left(a^2+b a\right)-c^2-b c}{(a-c)}=a+b+c \end{aligned} $ So, $P(0)+(a-b)(a-c) A=a+b+c+a^4$ Hence, option (d) is correct

Asked in: AP EAMCET 2019 (20 Apr Shift 2)

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