Q59 / 160 · TS EAMCET 2020 (14 Sep Shift 2) · Mathematics
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
- Correct answerA3
- B2
- C-2
- D4
Solution
We have,
$\begin{aligned}
\frac{x^5-5}{x^5+x^2} & =f(x)+\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x+1} \\
\frac{x^5-5}{x^3+x^2} & =x^2-x+1+\frac{-x^2-5}{x^3+x^2} \\
\therefore \quad \frac{-x^2-5}{x^3+x^2} & =\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x+1} \\
-x^2-5 & =A x(x+1)+B(x+1)+C\left(x^2\right) \\
-x^2-5 & =(A+C) x^2+(A+B) x+B \\
A+C & =-1, A+B=0, B=-5
\end{aligned}$
Solving we get, $A=5, B=-5, C=-6$
$\begin{aligned}
& \therefore \quad f(x)=x^2-x+1 \quad \therefore \quad f(K)=K^2-K+1 \\
& \text { Given, } f(K)+A+B+C=1 \\
& K^2-K+1+5-5-6=1 \Rightarrow K^2-K-6=0 \\
& \Rightarrow \quad(K+2)(K-3)=0 \Rightarrow K=3,-2
\end{aligned}$
Given, $f(K)+A+B+C=1$
$\begin{aligned}
K^2-K+1+5-5-6 & =1 \Rightarrow K^2-K-6=0 \\
\Rightarrow \quad(K+2)(K-3) & =0 \Rightarrow K=3,-2
\end{aligned}$
Largest value of $K=3$
Hence, the correct answer is option: 'A'.