Question

Let $\alpha >0$. If $\underset{0}{\overset{\alpha }{\int }}\frac{x}{\sqrt{x+\alpha }-\sqrt{x}}dx=\frac{16+20\sqrt{2}}{15}$, then $\alpha$ is equal to :

• A. 2
• B. $2\sqrt{2}$
• C. 4
• D. $\sqrt{2}$

Answer 1

Answer: (A)

After rationalising
$\underset{0}{\overset{\alpha }{\int }}\frac{x}{\alpha }(\sqrt{x+\alpha }+\sqrt{x})$
$\underset{0}{\overset{\alpha }{\int }}\frac{1}{\alpha }\left[(x+\alpha {)}^{3/2}-\alpha (x+\alpha {)}^{1/2}+x^{3/2}\right]$
${\frac{1}{\alpha }\left[\frac{2}{5}(x+\alpha {)}^{5/2}-\alpha \frac{2}{3}(x+\alpha {)}^{3/2}+\frac{2}{5}{x}^{5/2}\right]|}_0^{\alpha }$
$=\frac{1}{\alpha }(\frac{5}{2}(2\alpha {)}^{5/2}-\frac{2\alpha }{3}(2\alpha {)}^{3/2}+\frac{2}{5}{\alpha }^{5/2}-\frac{2}{5}{\alpha }^{5/2}+\frac{2}{3}{\alpha }^{5/2}$
$=\frac{1}{\alpha }\left(\frac{2^{7/2}{\alpha }^{5/2}}{5}\frac{2^{5/2}{\alpha }^{5/2}}{3}+\frac{2}{3}{\alpha }^{5/2}\right)$
$={\alpha }^{3/2}\left(\frac{2^{7/2}}{5}-\frac{2^{5/2}}{3}+\frac{2}{3}\right)$
$=\frac{{\alpha }^{3/2}}{15}(24\sqrt{2}-20\sqrt{2}+10)=\frac{{\alpha }^{3/2}}{15}(4\sqrt{2}+10)$
Now,
$\frac{{\alpha }^{3/2}}{15}(4\sqrt{2}+10)=\frac{16+20\sqrt{2}}{15}$
$\Rightarrow \alpha =2$