JEE Main 2025 Jan 29 Shift 1, Mathematics Q25: Inverse Trigonometric Functions

Given,

${\cos }^{-1}x=\pi +{\sin }^{-1}x+{\sin }^{-1}(2x+1)$

Since $2x+1\in [-1,1]$,

$x\in [-1,0]$

Also,

${\cos }^{-1}x=\frac{\pi }{2}-{\sin }^{-1}x$

$\therefore \frac{\pi }{2}-{\sin }^{-1}x=\pi +{\sin }^{-1}x+{\sin }^{-1}(2x+1)$

$\Rightarrow {\sin }^{-1}(2x+1)=-\frac{\pi }{2}-2{\sin }^{-1}x$

Taking sine on both sides,

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

Using $\cos 2\theta =1-2{\sin }^2\theta$,

$2x+1=-(1-2x^2)=2x^2-1$

$\Rightarrow 2x^2-2x-2=0$

$\Rightarrow x^2-x-1=0$

$\Rightarrow x=\frac{1\pm \sqrt{5}}{2}$

Since $x\in [-1,0]$,

$x=\frac{1-\sqrt{5}}{2}$

$2x-1=2\left(\frac{1-\sqrt{5}}{2}\right)-1=-\sqrt{5}$

$\therefore (2x-1{)}^2=5$

$\boxed{\sum _{x\in S}(2x-1{)}^2=5}$