Q59 / 160 · AP EAMCET 2024 (23 May Shift 1) · Mathematics

$\begin{aligned} & \frac{4 x^2+5}{(x-2)^4}=\frac{A}{(x-2)}+\frac{B}{(X-2)^2}+\frac{C}{(X-2)^3}+\frac{D}{(X-2)^4} \text { then } \\ & \sqrt{\frac{A}{C}+\frac{B}{C}+\frac{D}{C}}= \end{aligned}$

  • A
    $\frac{\sqrt{29}}{4}$
  • B
    $\frac{\sqrt{23}}{4}$
  • Correct answer
    C
    $\frac{5}{4}$
  • D
    $\frac{4}{5}$

Solution

$\frac{4 x^2+5}{(x-2)^4}=\frac{A}{x-2}+\frac{B}{(x-2)^2}+\frac{C}{(x-2)^3}+\frac{D}{(x-2)^4}$ $\Rightarrow A(x-2)^3+B(x-2)^2+C(x-2)+D=4 x^2+5$ If $x=2 \Rightarrow D=21$ Co-efficient of $x^3, A=0$ Constant term, $4 B-2 C+21=5 \Rightarrow 4 B-2 C=-16$ Co-efficient of $x^2, B=4$ So, $16-2 C=-16 \Rightarrow C=16$ Now, $\sqrt{\frac{A}{C}+\frac{B}{C}+\frac{D}{C}}=\sqrt{\frac{A+B+D}{C}}=\sqrt{\frac{4+21}{16}}=\frac{5}{4}$.

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

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