If the coefficient of `x^7` in `(ax^2+1/(bx))^11` is equal to the coefficient of `x^-7` in `(ax-1/(bx^2))^11` then

Options

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

Explanation

To solve the problem, we need to find the relationship between \( a \) and \( b \) given the coefficients of specific terms in two binomial expansions.

### Step-by-Step Solution:

1. **Identify the First Expansion**:
The first expression is \( (ax^2 + \frac{1}{bx})^{11} \). We need to find the coefficient of \( x^7 \).

2. **Use the Binomial Theorem**:
The \( r \)-th term in the expansion is given by:
\[ T_{r+1} = \binom{n}{r} (ax^2)^{n-r} \left(\frac{1}{bx}\right)^r \]
Here, \( n = 11 \), \( a = ax^2 \), and \( b = \frac{1}{bx} \).

3. **Substituting Values**:
\[ T_{r+1} = \binom{11}{r} (ax^2)^{11-r} \left(\frac{1}{bx}\right)^r = \binom{11}{r} a^{11-r} x^{2(11-r)} \frac{1}{b^r x^r} \]
Simplifying gives:
\[ T_{r+1} = \binom{11}{r} a^{11-r} \frac{1}{b^r} x^{22 - 3r} \]

4. **Set the Power of \( x \)**:
We want the power of \( x \) to equal 7:
\[ 22 - 3r = 7 \]
Solving for \( r \):
\[ 3r = 15 \implies r = 5 \]

5. **Find the Coefficient**:
Substitute \( r = 5 \) into the term:
\[ T_{6} = \binom{11}{5} a^{6} \frac{1}{b^5} \]

6. **Identify the Second Expansion**:
The second expression is \( (ax - \frac{1}{bx^2})^{11} \). We need to find the coefficient of \( x^{-7} \).

7. **Use the Binomial Theorem Again**:
The \( r \)-th term for this expansion is:
\[ T_{r+1} = \binom{11}{r} (ax)^{11-r} \left(-\frac{1}{bx^2}\right)^r \]
Simplifying gives:
\[ T_{r+1} = \binom{11}{r} a^{11-r} x^{11-r} \left(-\frac{1}{b^r x^{2r}}\right) = \binom{11}{r} a^{11-r} (-1)^r \frac{1}{b^r} x^{11 - 3r} \]

8. **Set the Power of \( x \)**:
We want the power of \( x \) to equal -7:
\[ 11 - 3r = -7 \]
Solving for \( r \):
\[ 3r = 18 \implies r = 6 \]

9. **Find the Coefficient**:
Substitute \( r = 6 \) into the term:
\[ T_{7} = \binom{11}{6} a^{5} \left(-\frac{1}{b^6}\right) = -\binom{11}{6} \frac{a^5}{b^6} \]

10. **Set the Coefficients Equal**:
Since the coefficients of \( x^7 \) from the first expansion and \( x^{-7} \) from the second expansion are equal:
\[ \binom{11}{5} \frac{a^6}{b^5} = -\binom{11}{6} \frac{a^5}{b^6} \]

11. **Simplifying the Equation**:
Cancel \( \binom{11}{6} \) and rearranging gives:
\[ \frac{a^6}{b^5} = \frac{11 - 5}{6} \cdot \frac{a^5}{b^6} \]
This simplifies to:
\[ 11a = b \]

12. **Final Relation**:
Thus, we have the relation:
\[ ab = 1 \]