JEE Mains · Maths · STD 11 - 7. binomial theoram
The positive value of \(\lambda \) for which the co-efficient of \(x^2\) in the expression \({x^2}{\left( {\sqrt x + \frac{\lambda }{{{x^2}}}} \right)^{10}}\) is \(720\) is
- A \(4\)
- B \(2\sqrt 2 \)
- C \(\sqrt 5 \)
- D \(3\)
Answer & Solution
Correct Answer
(A) \(4\)
Step-by-step Solution
Detailed explanation
\(x^{2}\left(\sqrt{x}+\frac{\lambda}{x^{2}}\right)^{10}\) Consider constant term \(^{10} \mathrm{C}_{\mathrm{r}}(\sqrt{\mathrm{x}})^{10-\mathrm{r}}\left(\frac{\lambda}{\mathrm{x}^{2}}\right)^{r}\) \(\frac{10-r}{2}-2 r=0\) \(10-5 r=0\) \(r=2\)…
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