Question

If the absolute term in the expansion of $( \sqrt {x} - \frac {k }{ x^2})^{10}$ is 405, then k is equal to

• A. $\pm$1
• B. $\pm$2
• C. $\pm$3
• D. none of these.

Answer 1

Answer: (C)

For absolute term $\frac{10-r}{2} - 2r = 0$
$\therefore 10 - r - 4r = 0$
$\therefore 5r = 10$
$\therefore r = 2$
$\therefore $ absolute term $= \,{}^{10}c_{2}k^{2} = 405$
$= \frac{10 \times 9}{1\times 2} k^{2} = 405$
$\Rightarrow k^{2} = \frac{405}{45}= 9$
$\Rightarrow k = \pm 3$.