Question

If the term free from \( x \) in the expansion of \( \left(\sqrt{x} - \frac{k}{x^2}\right)^{10} \) is 405, then the value of \( k \) is

• A. $\pm 1$
• B. $\pm 3$
• C. $\pm 4$
• D. $\pm 2$

Hint:

Constant term → power of variable = 0.

Answer 1

The Correct Option is B

Concept: General term and power of $x$

Step 1: General term: \[ T_{r+1} = \binom{10}{r} (\sqrt{x})^{10-r} \left(-\frac{k}{x^2}\right)^r \]

Step 2: Power of $x$: \[ = x^{\frac{10-r}{2}} \cdot x^{-2r} = x^{\frac{10-5r}{2}} \]

Step 3: For constant term: \[ \frac{10-5r}{2} = 0 \Rightarrow r = 2 \]

Step 4: Substitute $r=2$: \[ T_3 = \binom{10}{2} (-k)^2 = 45k^2 \]

Step 5: Equate to 405: \[ 45k^2 = 405 \Rightarrow k^2 = 9 \] \[ k = \pm 3 \] Conclusion:
$k = \pm 3$