Q64 / 160 · TS EAMCET 2023 (14 May Shift 2) · Mathematics
If the coefficient of $x^4$ in the expansion of $\frac{x}{(x-1)^2(x-2)}$ is $\frac{m}{n}$ and $|m|,|n|$ are coprimes, then $\sqrt{|m+n|}=$
- A9
- Correct answerB$\sqrt{33}$
- C7
- D$6 \sqrt{2}$
Solution
$\frac{x}{(x-1)^2(x-2)}=x(x-1)^{-2}(x-2)^{-1}$
Coefficient of $x^4=$ Coefficient of $x^3$ in $\frac{1}{(x-1)^2(x-2)}$
$\begin{aligned} & =\frac{1}{-2(1-x)^2\left(1-\frac{x}{2}\right)}=\frac{-1}{2}(1-x)^{-2}\left(1-\frac{x}{2}\right)^{-1} \\ & (1-x)^{-2}=1+2 x+\frac{2 \cdot 3}{2 !} x^2+\frac{2 \cdot 3 \cdot 4}{3 !} x^3+\ldots \ldots \infty \\ & \left(1-\frac{x}{2}\right)^{-1}=1+\frac{x}{2}+\frac{1 \cdot 2}{2 !}\left(\frac{x}{2}\right)^2+\frac{1 \cdot 2 \cdot 3}{3 !}\left(\frac{x}{2}\right)^3+\ldots . . \infty\end{aligned}$
$\therefore$ Coefficient of $x^3=\frac{-1}{2}\left[\frac{2 \cdot 3}{3 ! 8}+\frac{2 \cdot 2}{2 ! 2^2}+\frac{2 \cdot 3}{2 ! 2}+\frac{2 \cdot 3 \cdot 4}{3 !}\right]$
$\begin{aligned} & =-\frac{1}{2}\left[\frac{1}{8}+\frac{1}{2}+\frac{3}{2}+4\right]=\frac{-49}{2 \times 8}=\frac{-49}{16} \\ & m=-49, n=16 \\ & \sqrt{|m+n|}=\sqrt{|-49+16|}=\sqrt{33}\end{aligned}$
Hence, the correct answer is option: 'B'.