JEE Main 2025 29 January Morning Shift
Exam JEE Main
Year 2025
Date 29 January
Shift Morning Shift
Subject Mathematics
Chapter Inverse Trigonometric Functions JEE Main 2025 · 29 January · Morning Shift · Mathematics · Inverse Trigonometric Functions
Let (S=\ x: ^ -1 x=π+ ^ -1 x+ ^ -1 (2 x+1)\ ). Then (∑_ x ∈ ~S (2...
Let (\mathrm{S}=\left\{x: \cos ^{-1} x=\pi+\sin ^{-1} x+\sin ^{-1}(2 x+1)\right\}). Then (\sum_{x \in \mathrm{~S}}(2 x-1)^2) is equal to ______.
Step-by-step solution
cos − 1 x = π + sin − 1 x + sin − 1 ( 2 x + 1 ) 2 cos − 1 x − sin − 1 ( 2 x + 1 ) = 3 π 2 2 α − β = 3 π 2 where cos − 1 x = α , sin − 1 ( 2 x + 1 ) = β 2 α = 3 π 2 + β cos 2 α = sin β 2 cos 2 α − 1 = sin β 2 x 2 − 1 = 2 x + 1 x 2 − x − 1 = 0 \begin{aligned} & \cos ^{-1} x=\pi+\sin ^{-1} x+\sin ^{-1}(2 x+1) \\ & 2 \cos ^{-1} x-\sin ^{-1}(2 x+1)=\frac{3 \pi}{2} \\ & 2 \alpha-\beta=\frac{3 \pi}{2} \text{ 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 \\ & x^2-x-1=0\end{aligned} cos − 1 x = π + sin − 1 x + sin − 1 ( 2 x + 1 ) 2 cos − 1 x − sin − 1 ( 2 x + 1 ) = 2 3 π 2 α − β = 2 3 π where cos − 1 x = α , sin − 1 ( 2 x + 1 ) = β 2 α = 2 3 π + β cos 2 α = sin β 2 cos 2 α − 1 = sin β 2 x 2 − 1 = 2 x + 1 x 2 − x − 1 = 0 ⇒ n = 1 ± 5 2 = [ n = 1 + 5 2 rejected n = 1 − 5 2 ∴ 4 x 2 − 4 x = 4 ( 2 x − 1 ) 2 = 5 \begin{aligned} & \Rightarrow \mathrm{n}=\frac{1 \pm \sqrt{5}}{2}=\left[\begin{array}{c}\mathrm{n}=\frac{1+\sqrt{5}}{2} \text{ rejected } \\ \mathrm{n}=\frac{1-\sqrt{5}}{2}\end{array}\right. \\ & \therefore 4 \mathrm{x}^2-4 \mathrm{x}=4 \\ & (2 \mathrm{x}-1)^2=5\end{aligned} ⇒ n = 2 1 ± 5 = [ n = 2 1 + 5 rejected n = 2 1 − 5 ∴ 4 x 2 − 4 x = 4 ( 2 x − 1 ) 2 = 5
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