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

\(\lambda \) ની કઈ કિમત માટે \({x^2}{\left( {\sqrt x  + \frac{\lambda }{{{x^2}}}} \right)^{10}}\) ના વિસ્તરણમાં \(x^2\) સહગુણક \(720\) થાય ?

  1. A \(4\)
  2. B \(2\sqrt 2 \)
  3. C \(\sqrt 5 \)
  4. D \(3\)
Verified Solution

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 \(^{10} \mathrm{C}_{\mathrm{r}}(\sqrt{\mathrm{x}})^{10-\mathrm{r}}\left(\frac{\lambda}{\mathrm{x}^{2}}\right)^{r}\) \(\frac{10-r}{2}-2 r=0\) \(10-5 r=0\) \(r=2\)…
Same subject
Explore more questions on app
From JEE Mains
Explore more questions on app