The coefficient of `x^(9)` in expansion of `(x^(3)+(1)/(2^(log sqrt2(x^(3)/(2)))))^(11)` is equal to

Options

A. `-5`
B. 330
C. 520
D. `5+log_(sqrt2)3`

Answer

To find the coefficient of \( x^9 \) in the expansion of

\[ \left( x^3 + \frac{1}{2^{\log \sqrt{2} \left( \frac{x^3}{2} \right)}} \right)^{11}, \]

we will first simplify the term \( \frac{1}{2^{\log \sqrt{2} \left( \frac{x^3}{2} \right)}} \).

### Step 1: Simplifying the logarithmic term

We start with the logarithmic expression:

\[ \log \sqrt{2} \left( \frac{x^3}{2} \right). \]

Using the property of logarithms, we can separate this into two parts:

\[ \log \sqrt{2} + \log \left( \frac{x^3}{2} \right) = \log \sqrt{2} + \log x^3 - \log 2. \]

Now, we know that \( \log \sqrt{2} = \frac{1}{2} \log 2 \). Thus, we can rewrite the expression:

\[ \frac{1}{2^{\frac{1}{2} \log 2 + \log x^3 - \log 2}}. \]

This simplifies to:

\[ \frac{1}{2^{\frac{1}{2} \log 2 - \log 2 + \log x^3}} = \frac{1}{2^{-\frac{1}{2} \log 2 + \log x^3}} = \frac{1}{\frac{x^3}{\sqrt{2}}} = \frac{\sqrt{2}}{x^3}. \]

### Step 2: Rewriting the expression

Now we can rewrite the original expression:

\[ \left( x^3 + \frac{\sqrt{2}}{x^3} \right)^{11}. \]

### Step 3: Finding the general term

The general term in the binomial expansion of \( (a + b)^n \) is given by:

\[ T_r = \binom{n}{r} a^{n-r} b^r. \]

In our case, \( a = x^3 \) and \( b = \frac{\sqrt{2}}{x^3} \), and \( n = 11 \). Thus, the general term becomes:

\[ T_r = \binom{11}{r} (x^3)^{11-r} \left( \frac{\sqrt{2}}{x^3} \right)^r. \]

This simplifies to:

\[ T_r = \binom{11}{r} x^{3(11-r)} \cdot \frac{(\sqrt{2})^r}{x^{3r}} = \binom{11}{r} (\sqrt{2})^r x^{33 - 6r}. \]

### Step 4: Setting the power of \( x \) to 9

We need the power of \( x \) to equal 9:

\[ 33 - 6r = 9. \]

Solving for \( r \):

\[ 33 - 9 = 6r \implies 24 = 6r \implies r = 4. \]

### Step 5: Finding the coefficient

Now we substitute \( r = 4 \) back into the general term to find the coefficient:

\[ T_4 = \binom{11}{4} (\sqrt{2})^4 x^{9}. \]

Calculating \( \binom{11}{4} \):

\[ \binom{11}{4} = \frac{11!}{4!(11-4)!} = \frac{11 \times 10 \times 9 \times 8}{4 \times 3 \times 2 \times 1} = 330. \]

Since \( (\sqrt{2})^4 = 2^2 = 4 \), the coefficient of \( x^9 \) is:

\[ 330 \times 4 = 1320. \]

### Final Answer

The coefficient of \( x^9 \) in the expansion is **1320**.

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