If frac{x^5-5}{x^3+x^2}=f(x)+frac{A}{x}+frac{ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Performing polynomial long division on $\frac{x^5-5}{x^3+x^2}$:$\frac{x^5-5}{x^3+x^2} = x^2 - x + 1 + \frac{-x^2-5}{x^3+x^2}$Thus, $f(x) = x^2 - x

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(A) Performing polynomial long division on $\frac{x^5-5}{x^3+x^2}$:$\frac{x^5-5}{x^3+x^2} = x^2 - x + 1 + \frac{-x^2-5}{x^3+x^2}$Thus, $f(x) = x^2 - x + 1$ and $\frac{-x^2-5}{x^2(x+1)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x+1}$.Multiplying by $x^2(x+1)$, we get:$-x^2-5 = Ax(x+1) + B(x+1) + Cx^2$$-x^2-5 = (A+C)x^2 + (A+B)x + B$Comparing coefficients:$B = -5$, $A+B = 0 \Rightarrow A = 5$, $A+C = -1 \Rightarrow C = -6$.Given $f(K) + A + B + C = 1$:$(K^2 - K + 1) + 5 - 5 - 6 = 1$$K^2 - K - 6 = 0$$(K-3)(K+2) = 0$$K = 3$ or $K = -2$.The larger value of $K$ is $3$.

If $\frac{x^5-5}{x^3+x^2}=f(x)+\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x+1}$, then the larger value of $K$ for which $f(K)+A+B+C=1$ is:

Similar Questions

If $\frac{3x}{(x-a)(x-b)} = \frac{2}{x-a} + \frac{1}{x-b}$,then $a:b$ is equal to

$\frac{x^2 + 13x + 15}{(2x + 3)(x + 3)^2} = $

The coefficient of $x^5$ in the expansion of $\frac{x^2 + 1}{(x^2 + 4)(x - 2)}$ is

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module