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If $3 \mathrm{f}(x)-\mathrm{f}\left(\frac{1}{x}\right)=8 \log _2 x^3, x>0$, then $\mathrm{f}(2), \mathrm{f}(4)$ $\mathrm{f}(8)$ are in
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The correct answer is:
A.P.
$$
\begin{aligned}
& 3 f(x)-f\left(\frac{1}{x}\right)=8 \log _2 x^3 \\
& \Rightarrow 3 \mathrm{f}\left(\frac{1}{x}\right)-\mathrm{f}(x)=8 \log _2\left(\frac{1}{x}\right)^3 .
\end{aligned}
$$
From (i) and (ii), we get
$$
\begin{aligned}
& 8 \mathrm{f}(x)=24 \log _2 x^3+8 \log _2\left(\frac{1}{x}\right)^3 \\
& \Rightarrow 8 \mathrm{f}(x)=72 \log _2 x-24 \log _2 x \\
& \Rightarrow 8 \mathrm{f}(x)=48 \log _2 x \\
& \Rightarrow \mathrm{f}(x)=6 \log _2 x
\end{aligned}
$$
$\begin{array}{ll}\therefore \quad & f^{\prime}(2)=6 \log _2 2=6 \\ & f^{\prime}(4)=6 \log _2 4=12 \\ & f^{(}(8)=6 \log _2 8=18 \\ \therefore \quad & f(2), f(4), f(8) \text { are in A.P. }\end{array}$
\begin{aligned}
& 3 f(x)-f\left(\frac{1}{x}\right)=8 \log _2 x^3 \\
& \Rightarrow 3 \mathrm{f}\left(\frac{1}{x}\right)-\mathrm{f}(x)=8 \log _2\left(\frac{1}{x}\right)^3 .
\end{aligned}
$$
From (i) and (ii), we get
$$
\begin{aligned}
& 8 \mathrm{f}(x)=24 \log _2 x^3+8 \log _2\left(\frac{1}{x}\right)^3 \\
& \Rightarrow 8 \mathrm{f}(x)=72 \log _2 x-24 \log _2 x \\
& \Rightarrow 8 \mathrm{f}(x)=48 \log _2 x \\
& \Rightarrow \mathrm{f}(x)=6 \log _2 x
\end{aligned}
$$
$\begin{array}{ll}\therefore \quad & f^{\prime}(2)=6 \log _2 2=6 \\ & f^{\prime}(4)=6 \log _2 4=12 \\ & f^{(}(8)=6 \log _2 8=18 \\ \therefore \quad & f(2), f(4), f(8) \text { are in A.P. }\end{array}$
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