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TS EAMCET 2017 Code A Solved Paper

Section: Mathematics

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Question : 18 of 160

Marks: +1, -0

if

\frac{x^2+5}{(x^2+1)(x-2)}=\frac{A}{x-2}+\frac{Bx+C}{x^2+1}

A+B+C=

$-1$
$\frac{2}{5}$
$\frac{-3}{5}$
$0$

Solution:

Solve the following

\;\;\frac{x^{2}+5}{(x^{2}+1)(x-2)}=\;\;\frac{A}{x-2}+\;\;\frac{Bx+C}{x^{2}+1}

x^{2}+5=A(x^{2}+1)+(B x+C)(x-2)

x^{2}+5=A x^{2}+A+B x^{2}-2 B x+C x-2 C

Equate coefficient of

x^{2}, x

and constant terms

A+B=1

-2 B + C =0

A -2 C =5

After solving the above three equations, we get

A =\;\;\frac{9}{5}

B =-\;\;\frac{4}{5}

C=-\;\;\frac{8}{5}

The value of

A+B+C

is,

A+B+C =\;\;\frac{9-4-8}{5}

=\;\;\frac{-3}{5}

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