If `p^(4)+q^(3)=2(p gt 0, q gt 0)`, then the maximum value of term independent of `x` in the expansion of `(px^((1)/(12))+qx^(-(1)/(9)))^(14)` is (a) `"^(14)C_(4)` (b) `"^(14)C_(6)` (c) `"^(14)C_(7)` (d) `"^(14)C_(12)`

Options

A. `"^(14)C_(4)`
B. `"^(14)C_(6)`
C. `"^(14)C_(7)`
D. `"^(14)C_(12)`

Explanation

To find the maximum value of the term independent of \( x \) in the expansion of \( (px^{\frac{1}{12}} + qx^{-\frac{1}{9}})^{14} \), we will follow these steps:

### Step-by-Step Solution:

1. **Identify the General Term**:
The general term \( T_r \) in the expansion of \( (a + b)^n \) is given by:
\[ T_r = \binom{n}{r} a^{n-r} b^r \]
Here, \( a = px^{\frac{1}{12}} \), \( b = qx^{-\frac{1}{9}} \), and \( n = 14 \). Thus, the general term becomes:
\[ T_r = \binom{14}{r} (px^{\frac{1}{12}})^{14-r} (qx^{-\frac{1}{9}})^r \]

2. **Simplify the General Term**:
\[ T_r = \binom{14}{r} p^{14-r} q^r x^{\frac{14-r}{12} - \frac{r}{9}} \]

3. **Find the Power of \( x \)**:
We need the power of \( x \) to be zero for the term to be independent of \( x \):
\[ \frac{14 - r}{12} - \frac{r}{9} = 0 \]

4. **Solve for \( r \)**:
To solve the equation, find a common denominator (which is 36):
\[ \frac{3(14 - r)}{36} - \frac{4r}{36} = 0 \]
This simplifies to:
\[ 42 - 3r - 4r = 0 \implies 42 - 7r = 0 \implies 7r = 42 \implies r = 6 \]

5. **Substitute \( r \) Back into the General Term**:
Substitute \( r = 6 \) into the general term:
\[ T_6 = \binom{14}{6} p^{14-6} q^6 = \binom{14}{6} p^8 q^6 \]

6. **Maximize the Value of \( p^8 q^6 \)**:
We are given that \( p^4 + q^3 = 2 \). To maximize \( p^8 q^6 \), we can apply the AM-GM inequality:
\[ \frac{p^4 + q^3}{2} \geq \sqrt{p^4 q^3} \]
From \( p^4 + q^3 = 2 \):
\[ 1 \geq \sqrt{p^4 q^3} \implies p^4 q^3 \leq 1 \]
Squaring both sides gives:
\[ (p^4 q^3)^2 \leq 1 \implies p^8 q^6 \leq 1 \]

7. **Final Result**:
Thus, the maximum value of the term independent of \( x \) is:
\[ T_6 \leq \binom{14}{6} \cdot 1 = \binom{14}{6} \]

### Conclusion:
The maximum value of the term independent of \( x \) in the expansion is \( \binom{14}{6} \).