Q59 / 160 · TS EAMCET 2019 (04 May Shift 1) · Mathematics

If $\frac{27 x^2+32 x+16}{\beta x+2)^2(1-x)}=\frac{A}{3 x+2}+\frac{B}{\beta x+2)^2}$ $+\frac{C}{1-x}$, then $A B+B C+C A=$

  • A
    6
  • Correct answer
    B
    12
  • C
    24
  • D
    48

Solution

We have, $$ \begin{aligned} & \frac{27 x^2+32 x+16}{(3 x+2)^2(1-x)}=\frac{A}{3 x+2}+\frac{B}{(3 x+2)^2}+\frac{C}{1-x} \\ & \Rightarrow 27 x^2+32 x+16=A(3 x+2)(1-x)+B \\ & (1-x)+C(3 x+2)^2 \end{aligned} $$ Put $x=1$ both sides, we get $$ \begin{aligned} 27+32+16 & =0+0+25 C \\ C & =3 \end{aligned} $$ Put $x=-2 / 3$ both sides, we get $$ \begin{aligned} 12-\frac{64}{3}+16 & =\frac{5}{3} B \\ B & =4 \end{aligned} $$ Put $x=0$ both sides, we get $$ \begin{aligned} 16 & =2 A+B+4 C \\ 2 A & =16-B-4 C=16-4-12=0 \\ & A=0 \\ \therefore \quad & A B+B C+C A=0+12+0=12 \end{aligned} $$

Hence, the correct answer is option: 'B'.

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