The positive value of `lambda` for which the coefficient of `x^(2)` in the expression `x^(2) (sqrt(x) + (lambda)/(x^(2)))^(10)` is 720 is

Options

A. 3
B. `sqrt(5)`
C. `2 sqrt(2)`
D. 4

Explanation

To solve the problem, we need to find the positive value of \( \lambda \) for which the coefficient of \( x^2 \) in the expression \( x^2 \left( \sqrt{x} + \frac{\lambda}{x^2} \right)^{10} \) is 720.

### Step-by-Step Solution:

1. **Rewrite the Expression**:
\[ x^2 \left( \sqrt{x} + \frac{\lambda}{x^2} \right)^{10} \]
can be rewritten as:
\[ x^2 \left( x^{1/2} + \frac{\lambda}{x^2} \right)^{10} \]

2. **Use the Binomial Theorem**:
According to the Binomial Theorem, the \( r \)-th term in the expansion of \( (a + b)^n \) is given by:
\[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \]
Here, \( a = \sqrt{x} \) and \( b = \frac{\lambda}{x^2} \), and \( n = 10 \).

3. **Identify the General Term**:
The general term \( T_{r+1} \) in our case becomes:
\[ T_{r+1} = \binom{10}{r} \left( \sqrt{x} \right)^{10-r} \left( \frac{\lambda}{x^2} \right)^r \]
Simplifying this gives:
\[ T_{r+1} = \binom{10}{r} \lambda^r x^{(10-r)/2 - 2r} \]
which simplifies to:
\[ T_{r+1} = \binom{10}{r} \lambda^r x^{(10 - r - 4r)/2} = \binom{10}{r} \lambda^r x^{(10 - 5r)/2} \]

4. **Find the Condition for \( x^2 \)**:
We need the exponent of \( x \) to equal 2:
\[ \frac{10 - 5r}{2} = 2 \]
Multiplying through by 2 gives:
\[ 10 - 5r = 4 \implies 5r = 6 \implies r = \frac{6}{5} \]
This is incorrect since \( r \) must be an integer. Let's solve it correctly:
\[ 10 - 5r = 4 \implies 5r = 6 \implies r = 2 \]

5. **Calculate the Coefficient**:
Now substituting \( r = 2 \) into the term:
\[ T_{3} = \binom{10}{2} \lambda^2 x^{(10 - 5 \cdot 2)/2} = \binom{10}{2} \lambda^2 x^0 \]
The coefficient of \( x^2 \) is:
\[ \binom{10}{2} \lambda^2 \]

6. **Set the Coefficient Equal to 720**:
We know:
\[ \binom{10}{2} = \frac{10 \cdot 9}{2 \cdot 1} = 45 \]
Therefore:
\[ 45 \lambda^2 = 720 \]

7. **Solve for \( \lambda^2 \)**:
Dividing both sides by 45:
\[ \lambda^2 = \frac{720}{45} = 16 \]

8. **Find \( \lambda \)**:
Taking the square root:
\[ \lambda = 4 \]

### Final Answer:
The positive value of \( \lambda \) is \( \boxed{4} \).