Verified Solution • 2025–26 Session moderate

QJEEM23Madc42: JEE Main 2023 Mathematics - Class 12 JEEM23Madc42 Solution: Let be the set of all solutions of the equation . Then ∑ x ∈ S…

Let SS be the set of all solutions of the equation cos⁡−1(2x)−2cos⁡−1(1−x2)=π,x∈[−12,12]\cos ^{-1}(2 x)-2 \cos ^{-1}\left(\sqrt{1-x^{2}}\right)=\pi, x \in\left[-\frac{1}{2}, \frac{1}{2}\right] . Then ∑x∈S2sin−1⁡(x2−1) is equal to :

Multiple Choice Options

  • A π−2sin⁡−1(34)\pi-2 \sin ^{-1}\left(\frac{\sqrt{3}}{4}\right)
  • B π−sin⁡−1(34)\pi-\sin ^{-1}\left(\frac{\sqrt{3}}{4}\right)
  • C −2π3\frac{-2 \pi}{3}
  • D None

Step-by-Step Verified Solution

amp;cos⁡−1(2x)=π+2cos⁡−11−x2amp; Since cos⁡−1(2x)∈[0,π]amp; R.H.S. ≥πamp;π+2cos⁡−11−x2=πamp;⇒cos⁡−11−x2=0amp;⇒1−x2=1amp;⇒x=0amp; but at x=0amp;cos⁡−1(2x)=cos⁡−1(0)=π2\begin{aligned}& \cos ^{-1}(2 \mathrm{x})=\pi+2 \cos ^{-1} \sqrt{1-\mathrm{x}^2} \\\\& \text { Since } \cos ^{-1}(2 \mathrm{x}) \in[0, \pi] \\\\& \text { R.H.S. } \geq \pi \\\\& \pi+2 \cos ^{-1} \sqrt{1-\mathrm{x}^2}=\pi \\\\& \Rightarrow \cos ^{-1} \sqrt{1-\mathrm{x}^2}=0 \\\\& \Rightarrow \sqrt{1-\mathrm{x}^2}=1 \\\\& \Rightarrow \mathrm{x}=0 \\\\& \text { but at } \mathrm{x}=0 \\\\& \cos ^{-1}(2 \mathrm{x})=\cos ^{-1}(0)=\frac{\pi}{2}\end{aligned}

∴\therefore No solution possible for given equation.

x∈ϕ\mathrm{x} \in \phi

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