If sum of the coefficients of `x^(7)` and `x^(4)` in the expansion of `((x^(2))/a-b/x)^(11)` is zero, then

Options

A. ab= 1
B. a= b
C. ab=-1
D. a+ b = 0

Answer

To solve the problem, we need to find the sum of the coefficients of \(x^7\) and \(x^4\) in the expansion of \(\left(\frac{x^2}{a - \frac{b}{x}}\right)^{11}\) and set that sum equal to zero.

### Step-by-step Solution:

1. **Rewrite the Expression**:
The expression can be rewritten as:
\[ \left(\frac{x^2}{a - \frac{b}{x}}\right)^{11} = \left(\frac{x^2}{a - b/x}\right)^{11} = \left(\frac{x^3}{ax - b}\right)^{11} \]

2. **Identify the General Term**:
The general term in the binomial expansion of \((x^2)^{11}\) and \((a - \frac{b}{x})^{11}\) is given by:
\[ T_{r+1} = \binom{11}{r} (x^2)^{11 - r} \left(-\frac{b}{x}\right)^r \]
This simplifies to:
\[ T_{r+1} = \binom{11}{r} (-b)^r x^{2(11 - r) - r} = \binom{11}{r} (-b)^r x^{22 - 3r} \]

3. **Find the Coefficient of \(x^7\)**:
To find the coefficient of \(x^7\), we set the exponent equal to 7:
\[ 22 - 3r = 7 \implies 3r = 15 \implies r = 5 \]
The coefficient for \(x^7\) is:
\[ \binom{11}{5} (-b)^5 \]

4. **Find the Coefficient of \(x^4\)**:
Similarly, for \(x^4\):
\[ 22 - 3r = 4 \implies 3r = 18 \implies r = 6 \]
The coefficient for \(x^4\) is:
\[ \binom{11}{6} (-b)^6 \]

5. **Set Up the Equation**:
The sum of the coefficients of \(x^7\) and \(x^4\) is:
\[ \binom{11}{5} (-b)^5 + \binom{11}{6} (-b)^6 = 0 \]

6. **Factor Out Common Terms**:
Factoring out \((-b)^5\):
\[ (-b)^5 \left(\binom{11}{5} + \binom{11}{6} (-b)\right) = 0 \]
This gives us two cases:
- \((-b)^5 = 0\) (which is not possible since \(b \neq 0\))
- \(\binom{11}{5} + \binom{11}{6} (-b) = 0\)

7. **Solve for \(b\)**:
Using the identity \(\binom{n}{r} = \frac{n}{r} \binom{n-1}{r-1}\):
\[ \binom{11}{6} = \frac{11}{6} \binom{11}{5} \]
Substitute this into the equation:
\[ \binom{11}{5} - \frac{11}{6} b \binom{11}{5} = 0 \]
Dividing by \(\binom{11}{5}\) (assuming it is not zero):
\[ 1 - \frac{11}{6} b = 0 \implies \frac{11}{6} b = 1 \implies b = \frac{6}{11} \]

8. **Find \(a\)**:
Since we have \(ab = 1\):
\[ a \cdot \frac{6}{11} = 1 \implies a = \frac{11}{6} \]

### Conclusion:
Thus, the relationship between \(a\) and \(b\) is:
\[ ab = 1 \]