If the coefficient of `x^8` in `(ax^2 + (1)/(bx))^13` is equal to the coefficient of `x^(-8)` in `(ax - (1)/(bx^2))^13` , then a and b will satisfy the relation :

Options

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

Answer

To solve the problem, we need to find a relation between \(a\) and \(b\) based on the coefficients of \(x^8\) and \(x^{-8}\) in the given expansions. Let's break down the solution step by step.

### Step 1: Identify the General Term for the First Expansion
The first expression is \((ax^2 + \frac{1}{bx})^{13}\). The general term (T) in the binomial expansion is given by:
\[ T_{r+1} = \binom{n}{r} (ax^2)^r \left(\frac{1}{bx}\right)^{n-r} \]
where \(n = 13\).

### Step 2: Write the General Term
Substituting \(n = 13\):
\[ T_{r+1} = \binom{13}{r} (ax^2)^r \left(\frac{1}{bx}\right)^{13-r} \]
This simplifies to:
\[ T_{r+1} = \binom{13}{r} a^r x^{2r} \cdot \frac{1}{b^{13-r} x^{13-r}} = \binom{13}{r} a^r \frac{x^{2r}}{b^{13-r} x^{13-r}} = \binom{13}{r} a^r \frac{x^{2r}}{b^{13-r} x^{13-r}} = \binom{13}{r} a^r \frac{x^{2r}}{b^{13-r} x^{13-r}} = \binom{13}{r} a^r \frac{x^{2r - (13 - r)}}{b^{13 - r}} = \binom{13}{r} a^r \frac{x^{3r - 13}}{b^{13 - r}} \]

### Step 3: Find the Coefficient of \(x^8\)
To find the coefficient of \(x^8\), we set the exponent equal to 8:
\[ 3r - 13 = 8 \implies 3r = 21 \implies r = 7 \]
Now, substituting \(r = 7\) into the general term:
\[ T_{8} = \binom{13}{7} a^7 \frac{1}{b^{6}} \]

### Step 4: Identify the General Term for the Second Expansion
The second expression is \((ax - \frac{1}{bx^2})^{13}\). The general term is:
\[ T_{r+1} = \binom{13}{r} (ax)^{r} \left(-\frac{1}{bx^2}\right)^{13-r} \]
This simplifies to:
\[ T_{r+1} = \binom{13}{r} a^r x^r \cdot \left(-\frac{1}{b}\right)^{13-r} \cdot \frac{1}{x^{2(13-r)}} = \binom{13}{r} a^r \left(-\frac{1}{b}\right)^{13-r} x^{r - 2(13 - r)} = \binom{13}{r} a^r \left(-\frac{1}{b}\right)^{13-r} x^{3r - 26} \]

### Step 5: Find the Coefficient of \(x^{-8}\)
To find the coefficient of \(x^{-8}\), we set the exponent equal to -8:
\[ 3r - 26 = -8 \implies 3r = 18 \implies r = 6 \]
Now substituting \(r = 6\) into the general term:
\[ T_{7} = \binom{13}{6} a^6 \left(-\frac{1}{b}\right)^{7} \]

### Step 6: Set the Coefficients Equal
According to the problem, the coefficients of \(x^8\) and \(x^{-8}\) are equal:
\[ \binom{13}{7} a^7 \frac{1}{b^6} = \binom{13}{6} a^6 \left(-\frac{1}{b}\right)^{7} \]

### Step 7: Simplify the Equation
This leads to:
\[ \binom{13}{7} a^7 \frac{1}{b^6} = \binom{13}{6} a^6 \frac{-1}{b^7} \]
Cancelling \(a^6\) from both sides gives:
\[ \binom{13}{7} a = -\frac{\binom{13}{6}}{b} \]
Using the identity \(\binom{n}{r} = \binom{n}{n-r}\):
\[ \binom{13}{7} = \binom{13}{6} \implies a = -\frac{1}{b} \]

### Step 8: Final Relation
This implies:
\[ ab + 1 = 0 \]

### Final Answer
Thus, the relation that \(a\) and \(b\) will satisfy is:
\[ ab + 1 = 0 \]