The value of x satisfying ${\sin }^{-1}\left(\sqrt{\frac{3x-1}{25}}\right)+{\sin }^{-1}\left(\sqrt{\frac{3x+1}{25}}\right)=\frac{\pi }{2}$ lies in the interval

${\sin }^{-1}\left(\sqrt{\frac{3x-1}{25}}\right)+{\sin }^{-1}\left(\sqrt{\frac{3x+1}{25}}\right)=\frac{\pi }{2}$

$\Rightarrow {\sin }^{-1}\left(\sqrt{\frac{3x-1}{25}}\right)=\frac{\pi }{2}-{\sin }^{-1}\left(\sqrt{\frac{3x+1}{25}}\right)={\cos }^{-1}\left(\sqrt{\frac{3x+1}{25}}\right)$

$\Rightarrow {\sin }^{-1}\left(\sqrt{\frac{3x-1}{25}}\right)={\sin }^{-1}\left(\sqrt{1-\frac{3x+1}{25}}\right)$

$\Rightarrow {\sin }^{-1}\left(\sqrt{\frac{3x-1}{25}}\right)={\sin }^{-1}\left(\sqrt{\frac{24-3x}{25}}\right)$

$\Rightarrow \frac{3x-1}{25}=\frac{24-3x}{25}$

$\Rightarrow x=\frac{25}{6}$.