Let $f (x) = 2 x + 1$ . Then the number of real values of $x$ for which the three unequal numbers $f (x), f (2 x), f (4 x)$ are in G.P is ..............

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Solution

$f (x) = 2 x + 1, f (2 x) = 4 x + 1, f (4 x) = 8 x + 1$
From the given conditions, $(4 x + 1)^{2} = (2 x + 1) \times (8 x + 1)$
$\Rightarrow 16 x^{2} + 8 x + 1 = 16 x^{2} + 10 x + 1$
$\Rightarrow 2 x = 0$
Thus, the only real value possible is 0 but then $x, 2 x, 4 x$ become the same.
Hence no real $x$ possible!

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