If only the `4^("th")` term in the expansion of `(2+(3pi)/(8))^(10)` has the greatest numerical value, then the integral values of x are

Options

A. `{-3, -2, 2, 3}`
B. `{-2, -1, 1, 2}`
C. `{-3, 3}`
D. `{-3, -2, -1, 0, 1, 2, 3}`

Answer

To solve the problem, we need to find the integral values of \( x \) such that the fourth term in the expansion of \( (2 + \frac{3\pi}{8})^{10} \) has the greatest numerical value compared to the third and fifth terms.

### Step-by-Step Solution:

1. **Identify the General Term**:
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, \( a = 2 \), \( b = \frac{3\pi}{8} \), and \( n = 10 \).

2. **Find the Fourth, Third, and Fifth Terms**:
- For the fourth term \( T_4 \) (where \( r = 3 \)):
\[ T_4 = \binom{10}{3} \cdot 2^{10-3} \cdot \left(\frac{3\pi}{8}\right)^3 \]
- For the third term \( T_3 \) (where \( r = 2 \)):
\[ T_3 = \binom{10}{2} \cdot 2^{10-2} \cdot \left(\frac{3\pi}{8}\right)^2 \]
- For the fifth term \( T_5 \) (where \( r = 4 \)):
\[ T_5 = \binom{10}{4} \cdot 2^{10-4} \cdot \left(\frac{3\pi}{8}\right)^4 \]

3. **Set Up the Ratios**:
We need to ensure that \( |T_4| > |T_3| \) and \( |T_4| > |T_5| \).
- Calculate the ratio \( \frac{T_4}{T_3} \):
\[ \frac{T_4}{T_3} = \frac{\binom{10}{3} \cdot 2^7 \cdot \left(\frac{3\pi}{8}\right)^3}{\binom{10}{2} \cdot 2^8 \cdot \left(\frac{3\pi}{8}\right)^2} \]
Simplifying this gives:
\[ \frac{T_4}{T_3} = \frac{10 \cdot 9}{3 \cdot 2} \cdot \frac{3\pi}{8} = \frac{15\pi}{8} \]

- Calculate the ratio \( \frac{T_5}{T_4} \):
\[ \frac{T_5}{T_4} = \frac{\binom{10}{4} \cdot 2^6 \cdot \left(\frac{3\pi}{8}\right)^4}{\binom{10}{3} \cdot 2^7 \cdot \left(\frac{3\pi}{8}\right)^3} \]
Simplifying this gives:
\[ \frac{T_5}{T_4} = \frac{10 \cdot 9 \cdot 8}{4 \cdot 3 \cdot 2} \cdot \frac{3\pi}{8} = \frac{21}{64} x \]

4. **Set Up Inequalities**:
From the ratios:
- For \( |T_4| > |T_3| \):
\[ \left|\frac{15\pi}{8}\right| > 1 \implies |x| > \frac{64}{15\pi} \]
- For \( |T_4| > |T_5| \):
\[ \left|\frac{21}{64} x\right| < 1 \implies |x| < \frac{64}{21} \]

5. **Combine the Inequalities**:
We have:
\[ \frac{64}{15\pi} < |x| < \frac{64}{21} \]

6. **Find Integral Values**:
Calculate the approximate values:
- \( \frac{64}{15\pi} \approx 1.36 \)
- \( \frac{64}{21} \approx 3.05 \)

Thus, the integral values of \( x \) that satisfy these inequalities are:
\[ x = -3, -2, 2, 3 \]

### Final Answer:
The integral values of \( x \) are \( -3, -2, 2, 3 \).