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(B) Given that $f\left(3x + \frac{1}{2x}\right) = 9x^2 + \frac{1}{4x^2}$.We can rewrite the expression as $f\left(3x + \frac{1}{2x}\right) = (3x)^2 +

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$\pm 1$

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(B) Given that $f\left(3x + \frac{1}{2x}\right) = 9x^2 + \frac{1}{4x^2}$.We can rewrite the expression as $f\left(3x + \frac{1}{2x}\right) = (3x)^2 + \left(\frac{1}{2x}\right)^2 + 2(3x)\left(\frac{1}{2x}\right) - 3$.This simplifies to $f\left(3x + \frac{1}{2x}\right) = \left(3x + \frac{1}{2x}\right)^2 - 3$.Therefore, the general form of the function is $f(t) = t^2 - 3$.Given $f\left(x + \frac{1}{x}\right) = 1$, we substitute into the function: $\left(x + \frac{1}{x}\right)^2 - 3 = 1$.This gives $\left(x + \frac{1}{x}\right)^2 = 4$, so $x + \frac{1}{x} = \pm 2$.Case $1$: $x + \frac{1}{x} = 2 \Rightarrow x^2 - 2x + 1 = 0 \Rightarrow (x - 1)^2 = 0 \Rightarrow x = 1$.Case $2$: $x + \frac{1}{x} = -2 \Rightarrow x^2 + 2x + 1 = 0 \Rightarrow (x + 1)^2 = 0 \Rightarrow x = -1$.Thus, $x = \pm 1$.

$f$ is a real-valued function satisfying the relation $f\left(3x + \frac{1}{2x}\right) = 9x^2 + \frac{1}{4x^2}$. If $f\left(x + \frac{1}{x}\right) = 1$, then $x =$

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