Q59 / 160 · TS EAMCET 2022 (20 Jul Shift 1) · Mathematics

$\text { If } \frac{2 x^2-3 x+5}{(x-7)^3}=\frac{\mathrm{A}}{x-7}+\frac{\mathrm{B}}{(x-7)^2}+\frac{\mathrm{C}}{(x-7)^3}$ then $2 \mathrm{~A}-3 \mathrm{~B}+\mathrm{C}=$

  • A
    0
  • B
    27
  • Correct answer
    C
    11
  • D
    15

Solution

Given $\frac{2 x^2-3 x+5}{(x-7)^3}=\frac{A}{(x-7)}+\frac{B}{(x-7)^2}+\frac{C}{(x-7)^3}$ $\begin{aligned} \frac{2 x^2-3 x+5}{(x-7)^3} & =\frac{A(x-7)^2+B(x-7)+C}{(x-7)^3} \\ 2 x^2-3 x+5 & =A\left(x^2+49-14 x\right)+B x-7 B+C \\ 2 x^2-3 x+5 & =A x^2+x(B-14 A)+(49 A-7 B+C) \end{aligned}$ Compare each coefficients. $\begin{aligned} & \Rightarrow \mathrm{A}=2, \\ & \Rightarrow \mathrm{B}-14 \mathrm{~A}=-3 \\ & \mathrm{~B}-14 \times 2=-3 \\ & \mathrm{~B}=-3+28=25 \\ & \Rightarrow 49 \mathrm{~A}-7 \mathrm{~B}+\mathrm{C}=5 \\ & 49 \times 2-7 \times 25+\mathrm{C}=5 \\ & 98-175+\mathrm{C}=5 \\ & \mathrm{C}-77=5 \\ & \mathrm{C}=77+5=82 \end{aligned}$ Now, $2 \mathrm{~A}-3 \mathrm{~B}+\mathrm{C}=2 \times 2-3 \times 25+82$ $=4-75+82=86-75=11$ Therefore, option (c) is correct.

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

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