Let `1 le x le 256` and M be the maximum value of `(log_(2)x)^(4)+16(log_(2)x)^(2)log_(2)((16)/(x))`. The sum of the digits of M is :

Options

A. 9
B. 11
C. 13
D. 15

Answer

To solve the problem, we need to find the maximum value of the expression:

\[ M = (\log_2 x)^4 + 16 (\log_2 x)^2 \log_2 \left(\frac{16}{x}\right) \]

where \(1 \leq x \leq 256\).

### Step 1: Rewrite the expression
We can rewrite \(\log_2 \left(\frac{16}{x}\right)\) using the properties of logarithms:

\[ \log_2 \left(\frac{16}{x}\right) = \log_2 16 - \log_2 x = 4 - \log_2 x \]

Substituting this back into \(M\):

\[ M = (\log_2 x)^4 + 16 (\log_2 x)^2 (4 - \log_2 x) \]

### Step 2: Simplify the expression
Let \(t = \log_2 x\). Then we can rewrite \(M\) as:

\[ M = t^4 + 16 t^2 (4 - t) \]

Expanding this gives:

\[ M = t^4 + 64 t^2 - 16 t^3 \]

### Step 3: Rearrange the expression
Rearranging the expression, we have:

\[ M = t^4 - 16 t^3 + 64 t^2 \]

### Step 4: Find the critical points
To find the maximum value, we take the derivative of \(M\) with respect to \(t\) and set it to zero:

\[ \frac{dM}{dt} = 4t^3 - 48t^2 + 128t \]

Setting the derivative equal to zero:

\[ 4t^3 - 48t^2 + 128t = 0 \]

Factoring out \(4t\):

\[ 4t(t^2 - 12t + 32) = 0 \]

This gives us \(t = 0\) or solving the quadratic \(t^2 - 12t + 32 = 0\).

### Step 5: Solve the quadratic equation
Using the quadratic formula:

\[ t = \frac{12 \pm \sqrt{(-12)^2 - 4 \cdot 1 \cdot 32}}{2 \cdot 1} = \frac{12 \pm \sqrt{144 - 128}}{2} = \frac{12 \pm \sqrt{16}}{2} = \frac{12 \pm 4}{2} \]

Thus, we have:

\[ t = \frac{16}{2} = 8 \quad \text{and} \quad t = \frac{8}{2} = 4 \]

### Step 6: Evaluate \(M\) at critical points
We need to evaluate \(M\) at \(t = 0\), \(t = 4\), and \(t = 8\).

1. For \(t = 0\):
\[ M(0) = 0^4 - 16 \cdot 0^3 + 64 \cdot 0^2 = 0 \]

2. For \(t = 4\):
\[ M(4) = 4^4 - 16 \cdot 4^3 + 64 \cdot 4^2 = 256 - 1024 + 1024 = 256 \]

3. For \(t = 8\):
\[ M(8) = 8^4 - 16 \cdot 8^3 + 64 \cdot 8^2 = 4096 - 8192 + 4096 = 0 \]

### Step 7: Determine the maximum value
The maximum value of \(M\) occurs at \(t = 4\) and is:

\[ M = 256 \]

### Step 8: Find the sum of the digits of \(M\)
Now, we find the sum of the digits of \(256\):

\[ 2 + 5 + 6 = 13 \]

### Final Answer
The sum of the digits of \(M\) is:

\[ \boxed{13} \]