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TS EAMCET 04 May 2019 Shift 1 Solved Paper

Section: Mathematics

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Question : 19 of 160

Marks: +1, -0

If

\frac{27x^2+32x+16}{(3x+2)^2(1-x)}

\frac{A}{3x+2}+\frac{B}{(3x+2)^2}+\frac{C}{1-x}

, then AB + BC + CA =

6
12
24
48

Solution:

Consider the expression,

\;\frac{27 x^{2}+32 x+16}{(3 x+2)^{2}(1-x)}=\;\frac{A}{3 x+2}+\;\frac{B}{(3 x+2)^{2}}+\;\frac{C}{1-x}

27 x^{2}+32 x+16 = A(3 x+2)(1-x) + B(1-x) + C(3 x+2)^{2}

Take

x=1

both sides then

27+32+16 = 0+0+25 C

C=3

Take

x = -\;\frac{2}{3}

both sides then

12 - \;\frac{64}{3} + 16 \;\; = \;\frac{5}{3} B

B \;\; = 4

Take

x=0

both sides then

16 \;\; = 2 A + B + 4 C

2 A \;\; = 16 - B - 4 C

\;\; = 16 - 4 - 12

A \;\; = 0

This implies,

AB + BC + CA \;\; = 0 + 12 + 0

\;\; = 12

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