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If $a>0$ and the coefficient of $x^2$ in the expansion of $\left(a x^3+\frac{c}{x}\right)^6$ is 60, then $a c^2=$
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Verified Answer
The correct answer is:
2
General term,
$$
\begin{aligned}
& T_{r+1}={ }^6 C_r\left(a x^3\right)^{6-r}\left(\frac{c}{x}\right)^r \\
& ={ }^6 C_r a^{6-r} c^r x^{18-3 r-r} \\
& ={ }^6 C_r a^{6-r} c^r x^{18-4 r}
\end{aligned}
$$
According to the question,
$$
\begin{aligned}
18-4 r & =2 \Rightarrow 4 r=16 \\
r & =4 .
\end{aligned}
$$
So, ${ }^6 C_4 a^{6-4} c^4=60 \Rightarrow 15 a^2 c^4=60$
$$
\begin{array}{rlrl}
& \Rightarrow & a^2 c^4=4 \Rightarrow a c^2= \pm 2 \\
\because & a>0 & \\
\therefore & a c^2=2
\end{array}
$$
$$
\begin{aligned}
& T_{r+1}={ }^6 C_r\left(a x^3\right)^{6-r}\left(\frac{c}{x}\right)^r \\
& ={ }^6 C_r a^{6-r} c^r x^{18-3 r-r} \\
& ={ }^6 C_r a^{6-r} c^r x^{18-4 r}
\end{aligned}
$$
According to the question,
$$
\begin{aligned}
18-4 r & =2 \Rightarrow 4 r=16 \\
r & =4 .
\end{aligned}
$$
So, ${ }^6 C_4 a^{6-4} c^4=60 \Rightarrow 15 a^2 c^4=60$
$$
\begin{array}{rlrl}
& \Rightarrow & a^2 c^4=4 \Rightarrow a c^2= \pm 2 \\
\because & a>0 & \\
\therefore & a c^2=2
\end{array}
$$
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