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)

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