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,
$\frac{x^{5}-5}{x^{5}+x^{2}}=f(x)+\frac{A}{x}+\frac{B}{x^{2}}+\frac{C}{x+1}$
$\frac{x^{5}-5}{x^{3}+x^{2}}=x^{2}-x+1+\frac{-x^{2}-5}{x^{3}+x^{2}}$
$\therefore \frac{-x^{2}-5}{x^{3}+x^{2}}=\frac{A}{x}+\frac{B}{x^{2}}+\frac{C}{x+1}$
$-x^{2}-5=A x(x+1)+B(x+1)+C\left(x^{2}\right)$
$-x^{2}-5=(A+C) x^{2}+(A+B) x+ B$
$A+C=-1, A+B=0, B=-5$
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 $K=3$