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(B) Given the equation: $\int_{0}^{2} (t - \log_{2} a) \, dt = \log_{2} \left( \frac{4}{a^{2}} \right)$.Evaluating the integral on the left side:$\lef

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$a \in R^{+}$

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(B) Given the equation: $\int_{0}^{2} (t - \log_{2} a) \, dt = \log_{2} \left( \frac{4}{a^{2}} \right)$.Evaluating the integral on the left side:$\left[ \frac{t^{2}}{2} - t \log_{2} a \right]_{0}^{2} = \left( \frac{2^{2}}{2} - 2 \log_{2} a \right) - (0 - 0) = 2 - 2 \log_{2} a$.Now,simplify the right side:$\log_{2} \left( \frac{4}{a^{2}} \right) = \log_{2} 4 - \log_{2} a^{2} = 2 - 2 \log_{2} a$.Equating both sides:$2 - 2 \log_{2} a = 2 - 2 \log_{2} a$.This identity holds true for all values of $a$ for which $\log_{2} a$ is defined.Since the logarithm function $\log_{2} a$ is defined only for $a > 0$,the set of values is $a \in R^{+}$.

The set of values of $a$ which satisfy the equation $\int_{0}^{2} (t - \log_{2} a) \, dt = \log_{2} \left( \frac{4}{a^{2}} \right)$ is

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