Question

Let $\alpha>0$ If $\int\limits_0^\alpha \frac{x}{\sqrt{x+\alpha}-\sqrt{x}} d x=\frac{16+20 \sqrt{2}}{15}$, then $\alpha$ is equal to :

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

Answer 1

Answer: (A)

2

Answer 2

After rationalising
$\int\limits_{0}^{a}\frac{x}{\alpha}\left(\sqrt{x+\alpha}+\sqrt{x}\right)dx$
$=\int\limits_{0}^{a}\frac{1}{\alpha}[(x+\alpha)^{3/2}-\alpha(x+\alpha)^{1/2}+x^{3/2}]dx$
$=\frac{1}{\alpha}\left[\frac{2}{5}(x+\alpha)^{5/2}-\alpha\frac{2}{3}(x+\alpha)^{3/2}+\frac{2}{5}x^{5/2}\right]|^a_{0}$
$=\frac{1}\alpha{}[\frac{2}{5}(2\alpha)^{5/2}-\frac{2\alpha}{3}(2\alpha)^{3/2}+\frac{2}{5}\alpha^{5/2}-\frac{2}{5}\alpha^{5/2}+\frac{2}{3}\alpha^{5/2}]|^{a}_0$
$=\frac{1}{\alpha}[\frac{2^{7/2}\alpha^{5/2}}{5}-\frac{2^{5/2}\alpha^{5/2}}{3}+\frac{2}{3}\alpha^{5/2}]$
$=\alpha^{3/2}[\frac{2^{7/2}}{5}-\frac{2^{5/2}}{3}+\frac{2}{3}]$
$=\frac{\alpha^{3/2}}{15}(24\sqrt2-20\sqrt2+10)$
$=\frac{\alpha^{3/2}}{15}(4\sqrt2+10)$
Now,
$\frac{\alpha^{3/2}}{15}(4\sqrt2+10)=\frac{16+20\sqrt2}{15}$
$⇒\alpha=2$
So, the correct option is (A) : 2