Which of the following is a partial fraction of {-x²+6x+13}{(3x+5)(x²+4x+4)}=

Correct Answer is : 23x+5+−1x+2+3(x+2)22/3x+5+-1/x+2+{3}{(x+2)²}3x+52+x+2−1+(x+2)23

Correct Answer is : 23x+5+−1x+2+3(x+2)22/3x+5+-1/x+2+{3}{(x+2)²}3x+52+x+2−1+(x+2)23

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AP EAPCET 19-Aug-2021 Shift 2 Paper

Section: Mathematics

Question : 19 of 160

Marks: +1, -0

\frac{-x^{2}+6x+13}{(3x+5)(x^{2}+4x+4)}=

Solution:

\frac{-x^{2}+6x+13}{(3x+5)(x^{2}+4x+4)}=\frac{-x^{2}+6x+13}{(3x+5)(x+2)^{2}}

\Rightarrow \frac{-x^{2}+6x+13}{(3x+5)(x+2)^{2}}=\frac{A}{3x+5}+\frac{B}{x+2}+\frac{C}{(x+2)^{2}}

\Rightarrow -x^{2}+6x+13=A(x+2)^{2}+B(3x+5)(x+2)+C(3x+5)

...(i)

It is an identity

∴ True for every value of

x

Put

x=-2

in Eq. (i),

-(-2)^{2}+6(-2)+13=A\cdot 0+B\cdot 0+C(3(-2)+5)

\Rightarrow -4-12+13=-C

\Rightarrow C=3

Put

x=\frac{-5}{3}

in Eq. (i), we get

-\left(\frac{-5}{3}\right)^{2}+6\left(\frac{-5}{3}\right)+13

=A\left(\frac{-5}{2}+2\right)^{2}+B\cdot 0+C\cdot 0

\Rightarrow \frac{-25}{9}-10+13=A\cdot \frac{1}{9}

\frac{-25}{9}+3=\frac{A}{9}

\Rightarrow A=2

Put

x=0

in Eq. (i), we get

13=4A+10B+5C

\Rightarrow 13=8+10B+15

\Rightarrow 10B=13-23

B=-1

\therefore \frac{-x^{2}+6x+13}{(3x+5)(x^{2}+4x+4)}

=\frac{2}{3x+5}+\frac{-1}{x+2}+\frac{3}{(x+2)^{2}}

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