If `ab ne 0` and the sum of the coefficient of `x^7 and x^4` in the expansion of `((x^2)/(a)-(b)/(x))^11` zero, then

Options

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

Answer

To solve the problem, we need to find the coefficients of \(x^7\) and \(x^4\) in the expansion of \(\left(\frac{x^2}{a} - \frac{b}{x}\right)^{11}\) and set their sum to zero. Let's go through the steps systematically.

### Step 1: Identify the general term in the expansion
The general term in the binomial expansion of \(\left(\frac{x^2}{a} - \frac{b}{x}\right)^{11}\) can be expressed as:
\[ T_{r+1} = \binom{11}{r} \left(\frac{x^2}{a}\right)^{11-r} \left(-\frac{b}{x}\right)^r \]
This simplifies to:
\[ T_{r+1} = \binom{11}{r} \frac{(-b)^r}{a^{11-r}} x^{2(11-r) - r} = \binom{11}{r} \frac{(-b)^r}{a^{11-r}} x^{22 - 3r} \]

### Step 2: Find the coefficient of \(x^7\)
To find the coefficient of \(x^7\), we set the exponent equal to 7:
\[ 22 - 3r = 7 \implies 3r = 15 \implies r = 5 \]
Now, substituting \(r = 5\) into the general term:
\[ \text{Coefficient of } x^7 = \binom{11}{5} \frac{(-b)^5}{a^{11-5}} = \binom{11}{5} \frac{(-b)^5}{a^6} \]

### Step 3: Find the coefficient of \(x^4\)
Next, we find the coefficient of \(x^4\) by setting the exponent equal to 4:
\[ 22 - 3m = 4 \implies 3m = 18 \implies m = 6 \]
Substituting \(m = 6\) into the general term:
\[ \text{Coefficient of } x^4 = \binom{11}{6} \frac{(-b)^6}{a^{11-6}} = \binom{11}{6} \frac{(-b)^6}{a^5} \]

### Step 4: Set the sum of coefficients to zero
According to the problem, the sum of the coefficients of \(x^7\) and \(x^4\) is zero:
\[ \binom{11}{5} \frac{(-b)^5}{a^6} + \binom{11}{6} \frac{(-b)^6}{a^5} = 0 \]

### Step 5: Factor out common terms
Factoring out the common terms:
\[ \binom{11}{5} \frac{(-b)^5}{a^6} + \binom{11}{6} \frac{(-b)^6}{a^5} = 0 \]
This can be rewritten as:
\[ \frac{(-b)^5}{a^6} \left(\binom{11}{5} + \frac{(-b)}{a} \binom{11}{6}\right) = 0 \]
Since \(ab \neq 0\), we can conclude:
\[ \binom{11}{5} + \frac{(-b)}{a} \binom{11}{6} = 0 \]

### Step 6: Solve for the ratio of \(b\) and \(a\)
Rearranging gives:
\[ \frac{(-b)}{a} = -\frac{\binom{11}{5}}{\binom{11}{6}} = -\frac{11-5}{6} = -\frac{6}{6} = -1 \]
Thus:
\[ \frac{b}{a} = 1 \implies b = a \]

### Conclusion
The solution to the problem is:
\[ \boxed{a = b} \]