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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:
Option 1)
$2\sqrt2$
Option 2)
$\sqrt{5}$
Option 3)
3
Option 4)
4

Answers (1)

best_answer

General Term in the expansion of (x+a)^n -

$T_{r+1}= ^{n}c_{r}\cdot x^{n-r}\cdot a^{r}$

- wherein

Where $r\geqslant 0 \, and \, r\leqslant n$

$r= 0,1,2,----n$

From the concept

$x^{2}[_{r}^{10}\textrm{C}(\sqrt x)^{10-r}\cdot (\frac{\lambda }{x^{2}})^{r}]$

$x^{2}[_{r}^{10}\textrm{C}(x)^\frac{{10-r}}{2}\cdot \lambda ^{r}(x)^{-2r}]$

$x^{2}[_{r}^{10}\textrm{C}(x)^\frac{{10-5r}}{2}\cdot \lambda ^{r}]$

Coefficient of $x^{2}$ So, r = 2

Hence,

$_{2}^{10}\textrm{C}\lambda ^{2}=720$

$\lambda ^{2}=16$

$\lambda =\pm 4$

Option 1)

$2\sqrt2$

Option 2)

$\sqrt{5}$

Option 3)

3

Option 4)
4

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