If the absolute term (independent of x ) in the expansion of `(sqrtx-k//x^2)^10` is 405 then `k=`

Options

A. `pm 3^(1//4)`
B. `pm 4^(1//3)`
C. `pm 2`
D. `pm 3`

Answer

To solve the problem, we need to find the value of \( k \) such that the absolute term (independent of \( x \)) in the expansion of \( \left( \sqrt{x} - \frac{k}{x^2} \right)^{10} \) is equal to 405.

### Step-by-Step Solution:

1. **Identify the Terms in the Expansion**:
The expression can be rewritten as \( \left( \sqrt{x} - kx^{-2} \right)^{10} \). Here, we can identify \( a = \sqrt{x} \) and \( b = -\frac{k}{x^2} \).

2. **Use the Binomial Theorem**:
According to the Binomial Theorem, the general term \( T_r \) in the expansion of \( (a + b)^n \) is given by:
\[ T_r = \binom{n}{r} a^{n-r} b^r \]
For our case:
\[ T_r = \binom{10}{r} (\sqrt{x})^{10-r} \left(-\frac{k}{x^2}\right)^r \]

3. **Simplify the General Term**:
\[ T_r = \binom{10}{r} (\sqrt{x})^{10-r} \left(-1\right)^r \frac{k^r}{x^{2r}} = \binom{10}{r} (-1)^r k^r x^{\frac{10-r}{2} - 2r} \]
Simplifying the exponent of \( x \):
\[ T_r = \binom{10}{r} (-1)^r k^r x^{\frac{10 - r - 4r}{2}} = \binom{10}{r} (-1)^r k^r x^{\frac{10 - 5r}{2}} \]

4. **Find the Condition for the Absolute Term**:
For the term to be independent of \( x \), the exponent of \( x \) must be zero:
\[ \frac{10 - 5r}{2} = 0 \implies 10 - 5r = 0 \implies 5r = 10 \implies r = 2 \]

5. **Substitute \( r \) back into the General Term**:
Now substitute \( r = 2 \) into the general term:
\[ T_2 = \binom{10}{2} (-1)^2 k^2 x^{0} = \binom{10}{2} k^2 \]
Calculate \( \binom{10}{2} \):
\[ \binom{10}{2} = \frac{10 \times 9}{2 \times 1} = 45 \]
Thus, the term becomes:
\[ T_2 = 45 k^2 \]

6. **Set the Term Equal to 405**:
According to the problem, this term equals 405:
\[ 45 k^2 = 405 \]

7. **Solve for \( k^2 \)**:
\[ k^2 = \frac{405}{45} = 9 \]

8. **Find \( k \)**:
Taking the square root of both sides:
\[ k = \pm 3 \]

### Final Answer:
Thus, the value of \( k \) is \( \pm 3 \).