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Assuming $x$ to be so small that $x^2$ and higher powers of $x$ can be neglected, the coefficient of $x$ in $\frac{(1-x)^{1 / 3}+(1-5 x)^2}{(16-x)^{1 / 4}}$ is equal to
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Verified Answer
The correct answer is:
$-\frac{989}{192}$
Given expression
$$
\begin{aligned}
& \frac{(1-x)^{1 / 3}+(1-5 x)^2}{(16-x)^{1 / 4}} \\
= & \frac{1}{2}\left[(1-x)^{1 / 3}\left(1-\frac{x}{16}\right)^{-1 / 4}+(1-5 x)^2\left(1-\frac{x}{16}\right)^{-1 / 4}\right] \\
= & \frac{1}{2}\left[\left(1-\frac{1}{3} x\right)\left(1+\frac{x}{64}\right)+(1-10 x)\left(1+\frac{x}{64}\right)\right]
\end{aligned}
$$
(On ignoring the higher degree terms)
$$
=\frac{1}{2}\left(1-\frac{x}{3}+\frac{x}{64}+1-10 x+\frac{x}{64}\right)
$$
(On ignoring the higher degree terms)
$$
=\frac{1}{2}\left(2-x\left(\frac{1}{3}+10-\frac{1}{32}\right)\right)
$$
$\therefore$ Coefficient of $x=-\frac{32+960-3}{192}=-\frac{989}{192}$
$$
\begin{aligned}
& \frac{(1-x)^{1 / 3}+(1-5 x)^2}{(16-x)^{1 / 4}} \\
= & \frac{1}{2}\left[(1-x)^{1 / 3}\left(1-\frac{x}{16}\right)^{-1 / 4}+(1-5 x)^2\left(1-\frac{x}{16}\right)^{-1 / 4}\right] \\
= & \frac{1}{2}\left[\left(1-\frac{1}{3} x\right)\left(1+\frac{x}{64}\right)+(1-10 x)\left(1+\frac{x}{64}\right)\right]
\end{aligned}
$$
(On ignoring the higher degree terms)
$$
=\frac{1}{2}\left(1-\frac{x}{3}+\frac{x}{64}+1-10 x+\frac{x}{64}\right)
$$
(On ignoring the higher degree terms)
$$
=\frac{1}{2}\left(2-x\left(\frac{1}{3}+10-\frac{1}{32}\right)\right)
$$
$\therefore$ Coefficient of $x=-\frac{32+960-3}{192}=-\frac{989}{192}$
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