The range of x of which the expansion of `(2-3x^2)^(-11/2)`is valid is

Options

A. `(-sqrt(2/3), 2/3)`
B. `(-2/3,sqrt(2/3))`
C. `(-sqrt(2/3),sqrt(2/3))`
D. `(-oo,sqrt(2/3)) cup (sqrt(2/3) , oo)`

Answer

To find the range of \( x \) for which the expansion of \( (2 - 3x^2)^{-11/2} \) is valid, we can follow these steps:

### Step 1: Identify the form of the binomial expansion
The binomial expansion is valid for expressions of the form \( (1 + u)^n \) where \( |u| < 1 \). Therefore, we need to rewrite \( (2 - 3x^2)^{-11/2} \) in a suitable form.

### Step 2: Factor out the constant
We can factor out the constant from the expression:
\[ (2 - 3x^2)^{-11/2} = 2^{-11/2} \left(1 - \frac{3}{2} x^2\right)^{-11/2} \]
This allows us to identify \( u = -\frac{3}{2} x^2 \).

### Step 3: Set up the inequality for the binomial expansion
For the binomial expansion to be valid, we need:
\[ |u| < 1 \implies \left| -\frac{3}{2} x^2 \right| < 1 \]
This simplifies to:
\[ \frac{3}{2} x^2 < 1 \]

### Step 4: Solve the inequality
Now we can solve the inequality:
\[ x^2 < \frac{2}{3} \]
Taking the square root of both sides gives us:
\[ |x| < \sqrt{\frac{2}{3}} \]
This means:
\[ -\sqrt{\frac{2}{3}} < x < \sqrt{\frac{2}{3}} \]

### Step 5: Write the final answer
Thus, the range of \( x \) for which the expansion is valid is:
\[ x \in \left(-\sqrt{\frac{2}{3}}, \sqrt{\frac{2}{3}}\right) \]