ExamBro
ExamBro
JEE Mains · Maths · STD 11 - 7. binomial theoram

If the constant term in the binomial expansion of \(\left(\sqrt{x}-\frac{k}{x^{2}}\right)^{10}\) is \(405,\) then \(|k|\) equals 

  1. A \(2\)
  2. B \(1\)
  3. C \(3\)
  4. D \(9\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(3\)

Step-by-step Solution

Detailed explanation

\(\left(\sqrt{x}-\frac{k}{x^{2}}\right)^{10}\) \(T_{r+1}={ }^{10} C_{r}(\sqrt{x})^{10-r}\left(\frac{-k}{x^{2}}\right)^{r}\) \(T_{r+1}={ }^{10} C_{r} \cdot x^{\frac{10-r}{2}} \cdot(-k)^{r} \cdot x^{-2 r}\) \(T_{r+1}={ }^{10} C_{r} x^{\frac{10-5 r}{2}}(-k)^{r}\) Constant term…
Same subject
Explore more questions on app