If the absolute term in the expansion of ( √ x - (k / x2))10 is 405, then k is equal to

Question

If the absolute term in the expansion of ( √ x - (k / x2))10 is 405, then k is equal to (A) ±1 (B) ±2 (C) ±3 (D) none of these.

Tardigrade Admin · Accepted Answer

For absolute term (10-r/2) - 2r = 0 ∴ 10 - r - 4r = 0 ∴ 5r = 10 ∴ r = 2 ∴ absolute term = 10c2k2 = 405 = (10 × 9/1× 2) k2 = 405 ⇒ k2 = (405/45)= 9 ⇒ k = ± 3.

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

$\pm$ 1

$\pm$ 2

$\pm$ 3

none of these.

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

Q. If the absolute term in the expansion of (x​−x2k​)10 is 405, then k is equal to

±1

±2

±3