JEE Main 2025 — 29 January, Morning Shift — Question 69

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

$2 \cos ^{-1} x-\sin ^{-1}(2 x+1)=\frac{3 \pi}{2}$

$2 \alpha-\beta=\frac{3 \pi}{2}$

where $\cos ^{-1} x=\alpha, \sin ^{-1}(2 x+1)=\beta$

$2 \alpha=\frac{3 \pi}{2}+\beta$

$\cos 2 \alpha=\sin \beta$

$2 \cos ^{2} \alpha-1=\sin \beta$

$2 x^{2}-1=2 x+1$

$\mathrm{x}^{2}-\mathrm{x}-1=0$

$\Rightarrow \mathrm{x}=\frac{1 \pm \sqrt{5}}{2}=\left[\begin{array}{c}\mathrm{x}=\frac{1+\sqrt{5}}{2} \text { rejected } \\ \mathrm{x}=\frac{1-\sqrt{5}}{2}\end{array}\right.$

$\therefore 4 x^{2}-4 x=4$

$\sum_{x \in S}(2 x-1)^{2}=5$