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TS EAMCET 14-Sep-2020 Shift 2 Solved Paper

Section: Mathematics

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Question : 19 of 160

Marks: +1, -0

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,\text{ is}

3
2
-2
4

Solution:

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(x^{2})

-x^{2}-5\;\; =(A+C) x^{2}+(A+B) x+B

A+C\;\; =-1,\quad A+B=0,\quad B=-5

Solving we get,

A=5,\quad B=-5,\quad 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

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