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

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 1: Write the General Term
The general term in the binomial expansion of \( (a + b)^n \) is given by:
\[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \]
In our case, \( a = px^{\frac{1}{12}} \), \( b = qx^{-\frac{1}{9}} \), and \( n = 14 \). Therefore, the general term becomes:
\[ T_{r+1} = \binom{14}{r} (px^{\frac{1}{12}})^{14-r} (qx^{-\frac{1}{9}})^r \]

### Step 2: Simplify the General Term
Expanding this, we have:
\[ T_{r+1} = \binom{14}{r} p^{14-r} q^r x^{\frac{14-r}{12} - \frac{r}{9}} \]

### Step 3: Find the Power of \( x \)
To find the term independent of \( x \), we need the exponent of \( x \) to be zero:
\[ \frac{14 - r}{12} - \frac{r}{9} = 0 \]

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

### Step 5: Substitute \( r \) Back into the General Term
Now we substitute \( r = 6 \) back into the general term:
\[ T_{7} = \binom{14}{6} p^{14-6} q^6 = \binom{14}{6} p^8 q^6 \]

### Step 6: Use the Given Condition
We are given that \( p^4 + q^3 = 2 \). We will use the method of Lagrange multipliers or the AM-GM inequality to find the maximum value of \( p^8 q^6 \) under this constraint.

### Step 7: Apply AM-GM Inequality
By the AM-GM inequality:
\[ \frac{p^4 + q^3}{2} \geq \sqrt{p^4 q^3} \]
Given \( p^4 + q^3 = 2 \):
\[ 1 \geq \sqrt{p^4 q^3} \implies p^4 q^3 \leq 1 \]

### Step 8: Find Maximum of \( p^8 q^6 \)
We can express \( p^8 q^6 \) in terms of \( p^4 q^3 \):
\[ p^8 q^6 = (p^4 q^3)^{\frac{8}{4}} (p^4 q^3)^{\frac{6}{3}} = (p^4 q^3)^2 (p^4 q^3)^2 \leq 1^2 = 1 \]

### Step 9: Final Calculation
Thus, the maximum value of the term independent of \( x \) is:
\[ \binom{14}{6} \cdot 1 = 14C6 \]

### Conclusion
The maximum value of the term independent of \( x \) in the expansion is:
\[ \boxed{14C6} \]