Direction: In the given question, read the given statement and compare the two given quantities on its basis.
`1 gt a gt 0 gt b`
Quantity I. Value of `((a+b)^2-a^2-b^2)/((a+b)^2-(a-b)^2)`
Quantity II. `1/(2(ab^3+ab))`

Options

A. Quantity I `lt` Quantity II
B. Quantity I `ge` Quantity II
C. Quantity I `le` Quantity II
D. Quantity I `gt` Quantity II

Answer

To solve the given problem, we need to analyze and simplify both quantities step by step.

### Given:
1. \( 1 > a > 0 > b \)

### Quantity I:
We need to evaluate:
\[ \frac{(a+b)^2 - a^2 - b^2}{(a+b)^2 - (a-b)^2} \]

#### Step 1: Simplify the numerator
The numerator is:
\[ (a+b)^2 - a^2 - b^2 \]
Using the identity \( (x+y)^2 = x^2 + y^2 + 2xy \):
\[ = (a^2 + 2ab + b^2) - a^2 - b^2 = 2ab \]

#### Step 2: Simplify the denominator
The denominator is:
\[ (a+b)^2 - (a-b)^2 \]
Using the identity \( (x+y)^2 - (x-y)^2 = 4xy \):
\[ = (a+b)^2 - (a^2 - 2ab + b^2) = (a^2 + 2ab + b^2) - (a^2 - 2ab + b^2) = 4ab \]

#### Step 3: Combine the results
Now we can substitute the simplified numerator and denominator back into the expression:
\[ \frac{2ab}{4ab} = \frac{1}{2} \]

### Quantity II:
We need to evaluate:
\[ \frac{1}{2(ab^3 + ab)} \]

#### Step 4: Analyze the expression
Given that \( b < 0 \), we can see that \( ab \) will be positive (since \( a > 0 \) and \( b < 0 \)), but \( b^3 \) will be negative (since the cube of a negative number is negative). Therefore, \( ab^3 \) will also be negative.

#### Step 5: Combine terms in Quantity II
Thus:
\[ ab^3 + ab < 0 \quad \text{(since \( ab^3 \) is negative and dominates)} \]
This means that:
\[ 2(ab^3 + ab) < 0 \implies \frac{1}{2(ab^3 + ab)} < 0 \]

### Final Comparison:
- Quantity I: \( \frac{1}{2} \) (positive)
- Quantity II: \( \frac{1}{2(ab^3 + ab)} \) (negative)

Since a positive number is always greater than a negative number, we conclude that:
\[ \text{Quantity I} > \text{Quantity II} \]

### Answer:
**Quantity I is greater than Quantity II.**