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
- A \(2\)
- B \(1\)
- C \(3\)
- D \(9\)
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…
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