Question

The constant term in the expansion of ${\left(\frac{x}{2}+\frac{1}{x}+\sqrt{2}\right)}^5$ is $\frac{a\sqrt{2}}{2}$, then $a=$

• A. 7
• B. 69
• C. 63
• D. 65

Answer 1

Answer: (C)

${\left(\frac{x}{2}+\frac{1}{x}+\sqrt{2}\right)}^5$
$\Rightarrow {\left(\sqrt{\frac{x}{2}}+\frac{1}{\sqrt{x}}\right)}^{10}={\left(\frac{x+\sqrt{2}}{\sqrt{2}x}\right)}^{10}$
$\Rightarrow \frac{(x+\sqrt{2}{)}^{10}}{32x^5}$
Forms constant term in the expansion
$=\frac{\text{ Coefficient of }{x}^5\text{ in }(x+\sqrt{2}{)}^{10}}{32}$
$\Rightarrow \frac{10^{10}{C}_5\times 4\sqrt{2}}{32}$
$=\frac{a\sqrt{2}}{2}\Rightarrow (a=63)$