If `T_(9)=495` in the binomial expansion of `((1)/(x^(2))+(x)/(2)log_(2)x)^(12)` then x is equal to

Options

A. 1
B. an integer `gt` 1
C. fraction
D. none of these

Answer

To solve the problem, we need to find the value of \( x \) given that \( T_9 = 495 \) in the binomial expansion of

\[ \left( \frac{1}{x^2} + \frac{x}{2} \log_2 x \right)^{12}. \]

### Step-by-Step Solution:

1. **Identify the General Term in Binomial Expansion**:
The general term \( T_{r+1} \) in the binomial expansion of \( (a + b)^n \) is given by:
\[ T_{r+1} = \binom{n}{r} a^{n-r} b^r. \]
Here, \( a = \frac{1}{x^2} \), \( b = \frac{x}{2} \log_2 x \), and \( n = 12 \).

2. **Find the 9th Term**:
Since we need \( T_9 \), we have \( r = 8 \) (because \( T_{r+1} \) corresponds to \( r \)).
Thus,
\[ T_9 = \binom{12}{8} \left( \frac{1}{x^2} \right)^{12-8} \left( \frac{x}{2} \log_2 x \right)^8. \]

3. **Substitute Values**:
Substituting the values, we get:
\[ T_9 = \binom{12}{8} \left( \frac{1}{x^2} \right)^4 \left( \frac{x}{2} \log_2 x \right)^8. \]
This simplifies to:
\[ T_9 = \binom{12}{8} \cdot \frac{1}{x^8} \cdot \frac{x^8}{2^8} (\log_2 x)^8. \]

4. **Simplify the Expression**:
The \( x^8 \) terms cancel out:
\[ T_9 = \binom{12}{8} \cdot \frac{(\log_2 x)^8}{2^8}. \]
We know \( T_9 = 495 \), so:
\[ 495 = \binom{12}{8} \cdot \frac{(\log_2 x)^8}{2^8}. \]

5. **Calculate \( \binom{12}{8} \)**:
Using the formula for combinations:
\[ \binom{12}{8} = \frac{12!}{8! \cdot (12-8)!} = \frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2 \times 1} = 495. \]
Thus, we have:
\[ 495 = 495 \cdot \frac{(\log_2 x)^8}{256}. \]

6. **Set Up the Equation**:
Dividing both sides by 495:
\[ 1 = \frac{(\log_2 x)^8}{256}. \]
Multiplying both sides by 256 gives:
\[ (\log_2 x)^8 = 256. \]

7. **Solve for \( \log_2 x \)**:
Taking the eighth root of both sides:
\[ \log_2 x = 2. \]

8. **Find \( x \)**:
Converting from logarithmic form:
\[ x = 2^2 = 4. \]

### Final Answer:
Thus, the value of \( x \) is \( 4 \).