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7.Binomial Theorem
hard
$\lambda $ ની કઈ કિમત માટે ${x^2}{\left( {\sqrt x + \frac{\lambda }{{{x^2}}}} \right)^{10}}$ ના વિસ્તરણમાં $x^2$ સહગુણક $720$ થાય ?
A
$4$
B
$2\sqrt 2 $
C
$\sqrt 5 $
D
$3$
(JEE MAIN-2019)
Solution
$x^{2}\left(\sqrt{x}+\frac{\lambda}{x^{2}}\right)^{10}$
Consider constant term
$^{10} \mathrm{C}_{\mathrm{r}}(\sqrt{\mathrm{x}})^{10-\mathrm{r}}\left(\frac{\lambda}{\mathrm{x}^{2}}\right)^{r}$
$\frac{10-r}{2}-2 r=0$
$10-5 r=0$
$r=2$
$\Rightarrow^{10} \mathrm{C}_{2} \times \lambda^{2}=720$
$ \Rightarrow \lambda=4$
Standard 11
Mathematics
