91Ó°ÊÓ

To check if f(x) has a local maximum at x=1, we need to analyze the relation and the given conditions, \(f\left(16 x^{2}y\right)=f(-2)-f(4 x y)\), \(f(4)=-255\), and \(f(0)=1\). Notice that, if x=0, the relation is satisfied for all y, since both sides become f(-2). This means that f(-2)=1. Also, if y=0, we have \(f(2x)(1-f\left(\frac{1}{2x}\right))=0 \implies f(2x)=0\) or \(f\left(\frac{1}{2x}\right)=1\). Since the function is an even function, f(2x) cannot be zero. Therefore, f(2x)=1 for all x. However, this doesn't give any information about the value of f(x) at x=1. We don't have sufficient information to check statement A. So we move to the next statement.

Statement B states that \(f(x) f\left(\frac{1}{x}\right) \leq 0\). Note that f(x) is even, so f(x) = f(-x). If we substitute x=-x, the equation won't change. Thus, we can rewrite statement B as \(f(x) f\left(\frac{1}{-x}\right) \leq 0\). Plugging each expression into the given relation: \(f(2 x)\left(1-f\left(\frac{1}{-2 x}\right)\right)+f\left(16(-1) x^{2} y\right)=f(-2)-f(-4 x y) \implies f\left(\frac{1}{-2 x}\right)=1-f\left(16{-1} x^{2}y\right)\) Now, we multiply both expressions together: \(f(x) f\left(\frac{1}{x}\right) = f\left(16 x^{2} y\right) f\left(16{-1} x^{2} y\right) = f\left(16 x^{2} y\right) f\left(-16 x^{2} y\right)\) Since f(x) is an even function, \(f(-x) = f(x)\), so: \(f(x) f\left(\frac{1}{x}\right) = f\left(16 x^{2} y\right) f\left(16 x^{2} y\right) = f\left(16 x^{2} y\right)^{2} \geq 0\) This contradicts statement B. Next, we check statement C.

Statement C says that the range of values of \(\mathrm{k}\) for which \(|\mathrm{f}(\mathrm{x})|=\mathrm{k}-2\) has exactly four distinct solutions is \((2,3)\). We already know that \(f(4)=-255\) and \(f(0)=1\). Plugging x=4 into the given relation: \(f(8)(1-f\left(\frac{1}{8}\right))+f\left(16(-1)16\right)=f(-2)-f(64 y) \implies f\left(\frac{1}{8}\right)=1-\frac{1}{256}=-\frac{255}{256}\) From statement C, we know that \(|\mathrm{f}(\mathrm{x})|=\mathrm{k}-2\). For the values of k in the range (2,3), all the values of |f(x)| should lie in the range (0,1). However, since \(f\left(\frac{1}{8}\right)=-\frac{255}{256}\), statement C doesn't hold true.

Statement D states that \(\int_{ }^{1} \mathrm{f}(\mathrm{x}) \mathrm{d} \mathrm{x}=\frac{3}{4}\). We don't have enough information about the function f(x) to evaluate this integral, so we cannot confirm or refute statement D. We've only refuted statements B and C. Therefore, there isn't enough information available to determine which of the statements A, B, C, or D holds true.