Question

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

• A. 3
• B. 2
• C. -2
• D. 4

Answer 1

Answer: (A)

We have,
$\frac{x^5-5}{x^5+x^2}=f(x)+\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x+1}$
$\frac{x^5-5}{x^3+x^2}=x^2-x+1+\frac{-x^2-5}{x^3+x^2}$
$\therefore \frac{-x^2-5}{x^3+x^2}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x+1}$
$-x^2-5=Ax(x+1)+B(x+1)+C\left(x^2\right)$
$-x^2-5=(A+C)x^2+(A+B)x+B$
$A+C=-1,A+B=0,B=-5$
Solving we get, $A=5,B=-5,C=-6$
$\therefore f(x)=x^2-x+1$
$\therefore f(K)=K^2-K+1$
Given, $f(K)+A+B+C=1$
$K^2-K+1+5-5-6=1$
$\Rightarrow K^2-K-6=0$
$\Rightarrow (K+2)(K-3)=0$
$\Rightarrow K=3,-2$
Largest value of $K=3$