The range of x of which the expansion `( 9 + 25x^2)^(-6//5)` is valid is

Options

A. `(-3//5 , 3//5)`
B. `[-3//5, 3//5]`
C. `(-oo, 3//5)`
D. `(-oo, - 3//5)cup(3//5, oo)`

Answer

To find the range of \( x \) for which the expansion \( (9 + 25x^2)^{-\frac{6}{5}} \) is valid, we will follow these steps:

### Step 1: Identify the form of the expansion
The expression can be rewritten in a form suitable for applying the binomial theorem. We want to express it as:
\[ (1 + u)^{-p} \]
where \( u \) is a function of \( x \).

### Step 2: Factor out the constant
We can factor out 9 from the expression:
\[ (9 + 25x^2)^{-\frac{6}{5}} = 9^{-\frac{6}{5}} \left(1 + \frac{25x^2}{9}\right)^{-\frac{6}{5}} \]

### Step 3: Set up the condition for the binomial expansion
For the binomial expansion \( (1 + u)^{-p} \) to be valid, we need:
\[ |u| < 1 \]
In our case, \( u = \frac{25x^2}{9} \). Therefore, we require:
\[ \left|\frac{25x^2}{9}\right| < 1 \]

### Step 4: Solve the inequality
This leads to:
\[ -\frac{25x^2}{9} < 1 \quad \text{and} \quad \frac{25x^2}{9} < 1 \]
We can simplify the second part:
\[ \frac{25x^2}{9} < 1 \implies 25x^2 < 9 \implies x^2 < \frac{9}{25} \]
Taking the square root of both sides gives:
\[ |x| < \frac{3}{5} \]
Thus, we can write:
\[ - \frac{3}{5} < x < \frac{3}{5} \]

### Step 5: State the final range
The range of \( x \) for which the expansion is valid is:
\[ x \in \left(-\frac{3}{5}, \frac{3}{5}\right) \]