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

Options

A. 1
B. 2
C. 3
D. 4

Answer

To solve the problem, we need to find the values of \( a \) and \( b \) such that the coefficient of \( x^7 \) in the expansion of \( [ax^2 + \frac{1}{bx}]^{11} \) is equal to the coefficient of \( x^{-7} \) in the expansion of \( [ax - \frac{1}{bx^2}]^{11} \).

### Step 1: Find the coefficient of \( x^7 \) in \( [ax^2 + \frac{1}{bx}]^{11} \)

The general term in the binomial expansion is given by:
\[ T_{r+1} = \binom{n}{r} (ax^2)^{n-r} \left(\frac{1}{bx}\right)^r \]
For our case, \( n = 11 \), so:
\[ T_{r+1} = \binom{11}{r} (ax^2)^{11-r} \left(\frac{1}{bx}\right)^r \]
This simplifies to:
\[ T_{r+1} = \binom{11}{r} a^{11-r} x^{2(11-r)} \cdot \frac{1}{b^r} x^{-r} = \binom{11}{r} a^{11-r} \frac{1}{b^r} x^{22 - 3r} \]
We need the coefficient of \( x^7 \):
\[ 22 - 3r = 7 \implies 3r = 15 \implies r = 5 \]
Now substituting \( r = 5 \) into the term:
\[ T_{6} = \binom{11}{5} a^{6} \frac{1}{b^{5}} x^{7} \]
Thus, the coefficient of \( x^7 \) is:
\[ \text{Coefficient}_1 = \binom{11}{5} a^6 \frac{1}{b^5} \]

### Step 2: Find the coefficient of \( x^{-7} \) in \( [ax - \frac{1}{bx^2}]^{11} \)

The general term for this expansion is:
\[ T_{r+1} = \binom{11}{r} (ax)^{11-r} \left(-\frac{1}{bx^2}\right)^r \]
This simplifies to:
\[ T_{r+1} = \binom{11}{r} a^{11-r} x^{11-r} \left(-\frac{1}{b}\right)^r x^{-2r} = \binom{11}{r} a^{11-r} \left(-\frac{1}{b}\right)^r x^{11 - 3r} \]
We need the coefficient of \( x^{-7} \):
\[ 11 - 3r = -7 \implies 3r = 18 \implies r = 6 \]
Now substituting \( r = 6 \) into the term:
\[ T_{7} = \binom{11}{6} a^{5} \left(-\frac{1}{b}\right)^{6} x^{-7} \]
Thus, the coefficient of \( x^{-7} \) is:
\[ \text{Coefficient}_2 = \binom{11}{6} a^5 \left(-\frac{1}{b^6}\right) \]

### Step 3: Set the coefficients equal

According to the problem, these coefficients are equal:
\[ \binom{11}{5} a^6 \frac{1}{b^5} = \binom{11}{6} a^5 \left(-\frac{1}{b^6}\right) \]
We can simplify this using the property \( \binom{n}{r} = \binom{n}{n-r} \):
\[ \binom{11}{5} = \binom{11}{6} \]
So we have:
\[ a^6 \frac{1}{b^5} = a^5 \left(-\frac{1}{b^6}\right) \]
Cross-multiplying gives:
\[ a^6 b^6 = -a^5 b^5 \]
Dividing both sides by \( a^5 b^5 \) (assuming \( a \neq 0 \) and \( b \neq 0 \)):
\[ ab = -1 \]

### Conclusion

Thus, the value of \( ab \) is:
\[ \boxed{-1} \]