If the coefficients of `x^(7)` in `(x^(2)+(1)/(bx))^(11) and x^(-7)` in `(x-(1)/(bx^(2)))^(11), b ne 0`, are equal, then the value of b is equal to

Options

A. 2
B. `-1`
C. 1
D. `-2`

Answer

To solve the problem, we need to find the value of \( b \) such that the coefficients of \( x^7 \) in the expression \( (x^2 + \frac{1}{bx})^{11} \) and \( x^{-7} \) in the expression \( (x - \frac{1}{bx^2})^{11} \) are equal.

### Step-by-step Solution:

1. **Find the coefficient of \( x^7 \) in \( (x^2 + \frac{1}{bx})^{11} \)**:
- The general term in the binomial expansion is given by:
\[ T_{r+1} = \binom{11}{r} (x^2)^{11-r} \left(\frac{1}{bx}\right)^r \]
- Simplifying this, we get:
\[ T_{r+1} = \binom{11}{r} x^{2(11-r)} \cdot \frac{1}{b^r} x^{-r} = \binom{11}{r} \frac{1}{b^r} x^{22 - 3r} \]
- We need the coefficient of \( x^7 \):
\[ 22 - 3r = 7 \implies 3r = 15 \implies r = 5 \]
- The coefficient of \( x^7 \) is:
\[ \binom{11}{5} \cdot \frac{1}{b^5} \]

2. **Find the coefficient of \( x^{-7} \) in \( (x - \frac{1}{bx^2})^{11} \)**:
- The general term in this expansion is:
\[ T_{r+1} = \binom{11}{r} x^{11-r} \left(-\frac{1}{bx^2}\right)^r \]
- Simplifying this, we get:
\[ T_{r+1} = \binom{11}{r} (-1)^r \frac{1}{b^r} x^{11 - 3r} \]
- We need the coefficient of \( x^{-7} \):
\[ 11 - 3r = -7 \implies 11 + 7 = 3r \implies 3r = 18 \implies r = 6 \]
- The coefficient of \( x^{-7} \) is:
\[ \binom{11}{6} \cdot \left(-\frac{1}{b}\right)^6 = \binom{11}{6} \cdot \frac{(-1)^6}{b^6} = \binom{11}{6} \cdot \frac{1}{b^6} \]

3. **Set the coefficients equal**:
- We have:
\[ \binom{11}{5} \cdot \frac{1}{b^5} = \binom{11}{6} \cdot \frac{1}{b^6} \]
- Since \( \binom{11}{5} = \binom{11}{6} \), we can simplify:
\[ \frac{1}{b^5} = \frac{1}{b^6} \]
- Cross-multiplying gives:
\[ b^6 = b^5 \]
- Dividing both sides by \( b^5 \) (since \( b \neq 0 \)):
\[ b = 1 \]

### Final Answer:
The value of \( b \) is \( 1 \).