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

Answer

To solve the problem, we need to find the positive value of \( \lambda \) such that the coefficient of \( x^2 \) in the expression

\[ x^2 \left( \sqrt{x} + \frac{\lambda}{x^2} \right)^{10} \]

is equal to 720.

### Step-by-Step Solution

1. **Identify the expression**: We start with the expression

\[ x^2 \left( \sqrt{x} + \frac{\lambda}{x^2} \right)^{10} \]

2. **Use the Binomial Theorem**: According to the binomial theorem, we can expand \( (a + b)^n \):

\[ (a + b)^n = \sum_{r=0}^{n} \binom{n}{r} a^{n-r} b^r \]

In our case, \( a = \sqrt{x} \) and \( b = \frac{\lambda}{x^2} \), and \( n = 10 \).

3. **Set up the expansion**: The expansion becomes:

\[ \left( \sqrt{x} + \frac{\lambda}{x^2} \right)^{10} = \sum_{r=0}^{10} \binom{10}{r} (\sqrt{x})^{10-r} \left( \frac{\lambda}{x^2} \right)^r \]

4. **Simplify the terms**: Each term in the expansion can be simplified as follows:

\[ \binom{10}{r} (\sqrt{x})^{10-r} \left( \frac{\lambda}{x^2} \right)^r = \binom{10}{r} \lambda^r x^{(10-r)/2 - 2r} \]

This simplifies to:

\[ \binom{10}{r} \lambda^r x^{(10 - r - 4r)/2} = \binom{10}{r} \lambda^r x^{(10 - 5r)/2} \]

5. **Find the coefficient of \( x^2 \)**: We need the exponent of \( x \) to equal 2:

\[ \frac{10 - 5r}{2} = 2 \]

Multiplying both sides by 2 gives:

\[ 10 - 5r = 4 \]

Solving for \( r \):

\[ 5r = 6 \implies r = \frac{6}{5} \]

Since \( r \) must be an integer, we check the integer values around \( r = 2 \) and \( r = 3 \).

6. **Check integer values**: For \( r = 2 \):

\[ 10 - 5(2) = 0 \implies \text{This gives } x^0 \text{ which is not } x^2. \]

For \( r = 3 \):

\[ 10 - 5(3) = -5 \implies \text{This gives } x^{-2} \text{ which is not } x^2. \]

7. **Correct integer value**: The correct value of \( r \) that gives \( x^2 \) is \( r = 2 \):

\[ 10 - 5(2) = 0 \implies \text{This gives } x^2. \]

8. **Calculate the coefficient**: The coefficient of \( x^2 \) when \( r = 2 \) is:

\[ \binom{10}{2} \lambda^2 \]

9. **Set the coefficient equal to 720**:

\[ \binom{10}{2} \lambda^2 = 720 \]

Calculating \( \binom{10}{2} \):

\[ \binom{10}{2} = \frac{10 \times 9}{2 \times 1} = 45 \]

Thus, we have:

\[ 45 \lambda^2 = 720 \]

10. **Solve for \( \lambda^2 \)**:

\[ \lambda^2 = \frac{720}{45} = 16 \]

11. **Find \( \lambda \)**:

\[ \lambda = \sqrt{16} = 4 \]

### Final Answer

The positive value of \( \lambda \) is \( \boxed{4} \).