QJEEM23Ma3ca1: JEE Main 2023 Mathematics - Class 12 JEEM23Ma3ca1 Solution: Let . If , then is equal to :

Question

Verified Solution • 2025–26 Session moderate

Let

\alpha&gt;0

. If

\int\limits_0^\alpha \frac{x}{\sqrt{x+\alpha}-\sqrt{x}} \mathrm{~d} x=\frac{16+20 \sqrt{2}}{15}

, then

\alpha

is equal to :

Accepted Answer

I=\int\limits_{0}^{\alpha} \frac{x}{\sqrt{x+\alpha}-\sqrt{x}} d x, \alpha&gt;0

\begin{aligned}&amp; =\frac{1}{\alpha} \int\limits_{0}^{\alpha} x(\sqrt{x+\alpha}+\sqrt{x}) d x \end{aligned}

\begin{aligned} = \frac{1}{α} \int_{0}^{α} (x \sqrt{x + α} + x \sqrt{x}) d x \\ = \frac{1}{α} \int_{0}^{α} [(x + α - α) \sqrt{x + α} + x^{\frac{3}{2}}] d x \end{aligned}

\begin{aligned}&amp; =\frac{1}{\alpha}\int\limits_0^\alpha \left[(x+\alpha)^{3 / 2}-\alpha(x+\alpha)^{1 / 2}+x^{3 / 2}\right] d x \\\\&amp; =\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 \\\\&amp; =\frac{1}{\alpha}\left(\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}\right) \\\\&amp; =\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) \\\\&amp; =\frac{\alpha^{3 / 2}}{15}(24 \sqrt{2}-20 \sqrt{2}+10)=\frac{\alpha^{3 / 2}}{15}(4 \sqrt{2}+10)\end{aligned}

Now,

\frac{\alpha^{3 / 2}}{15}(4 \sqrt{2}+10)=\frac{16+20 \sqrt{2}}{15}=\frac{2 \sqrt{2}(4 \sqrt{2}+10)}{15}

\Rightarrow \alpha^{3 / 2}=2 \sqrt{2}=2^{3 / 2}

\Rightarrow \alpha=2

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