The set of value of 'a' which satisfy the equation `int_0^2(t-log_2a)dt=log_2(4/(a^2))` is

Options

A. `a in R`
B. `a in R^(+)`
C. `a lt2`
D. `a gt2`

Answer

To solve the equation

\[ \int_0^2 \left(t - \log_2 a\right) dt = \log_2\left(\frac{4}{a^2}\right), \]

we will break it down step by step.

### Step 1: Evaluate the Integral

We start by evaluating the left-hand side integral:

\[ \int_0^2 \left(t - \log_2 a\right) dt = \int_0^2 t \, dt - \int_0^2 \log_2 a \, dt. \]

The first integral, \(\int_0^2 t \, dt\), can be computed as follows:

\[ \int_0^2 t \, dt = \left[\frac{t^2}{2}\right]_0^2 = \frac{2^2}{2} - \frac{0^2}{2} = \frac{4}{2} = 2. \]

The second integral, \(\int_0^2 \log_2 a \, dt\), is simply \(\log_2 a\) multiplied by the length of the interval (which is 2):

\[ \int_0^2 \log_2 a \, dt = 2 \log_2 a. \]

Putting it all together, we have:

\[ \int_0^2 \left(t - \log_2 a\right) dt = 2 - 2 \log_2 a. \]

### Step 2: Set the Integral Equal to the Right-Hand Side

Now we set the left-hand side equal to the right-hand side:

\[ 2 - 2 \log_2 a = \log_2\left(\frac{4}{a^2}\right). \]

### Step 3: Simplify the Right-Hand Side

Next, we simplify the right-hand side:

\[ \log_2\left(\frac{4}{a^2}\right) = \log_2 4 - \log_2 a^2 = 2 - 2 \log_2 a. \]

### Step 4: Equate Both Sides

Now we have:

\[ 2 - 2 \log_2 a = 2 - 2 \log_2 a. \]

This equation is always true for any value of \(a\) that makes \(\log_2 a\) defined, which means \(a\) must be positive.

### Step 5: Conclusion

Thus, the set of values of \(a\) that satisfy the equation is:

\[ a > 0. \]