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TS EAMCET 18-July-2022 Shift 1 Solved Paper

Section: Mathematics

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Question : 19 of 160

Marks: +1, -0

If

\;\frac{x^{2}-3x+2}{(x-4)(x-3)^{2}} =\;\frac{A}{x-4} +\;\frac{B}{x-3} +\;\frac{C}{(x-3)^{2}}

A+B+C=

1
0
-1
5

Solution: 👈: Video Solution

Given,

\;\frac{x^{2}-3x+2}{(x-4)(x-3)^{2}}\;=\;\frac{A}{x-4}+\;\frac{B}{x-3}+\;\frac{C}{(x-3)^{2}}

x^{2}-3x+2\;=A(x-3)^{2}

\;+B(x-4)(x-3)+C(x-4)\ldots\;\text{(i)}\;

This is an identity so, true for every value of

x

.

On putting

x=3

in Eq. (i), we have

C=-2

On putting

x=4

in Eq. (i), we have

A=6

On putting

x=0

in Eq. (i) and

A=6

and

C:=-2

, we have

B=-5

Now,

A+B+C=6-5-2=-1

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