Q.

Let S={x : cos–1x=π+sin–1x+sin–1(2x+1)}. Then ∑x∈S(2x–1)2 is equal to __________.          [2025]


Ans.

(5)

We have, cos–1x=π+sin–1x+sin–1(2x+1)

⇒ 2cos–1x–sin–1(2x+1)=3π2

⇒ 2α–β=3π2, where cos–1x=α, sin–1(2x+1)=β

⇒ 2α=3π2+β ⇒ cos 2α=sin β ⇒ 2cos2α–1=sinβ

⇒ 2x2–1=2x+1          [∵ x=cosα and 2x+1=sinβ]

⇒ x2–x–1=0

⇒ x=1±52

⇒ x=1–52          (∵ x=1+52 rejected as cosα≤1)

Now, ∑x∈S(2x–1)2=∑x∈S(4x2+1–4x)=5