Let $\mathrm{S}=\left\{\mathrm{x}: \cos ^{-1} \mathrm{x}=\pi+\sin ^{-1} \mathrm{x}+\sin ^{-1}(2 \mathrm{x}+1)\right\}$. Then $\sum_{\mathrm{x} \in \mathrm{S}}(2 \mathrm{x}-1)^{2}$ is equal to ______.

JEE Main · 2025 · 29 Jan 2025 · Shift 1 · Inverse Trigonometric Functions

Let

S = {x : cos^{- 1} x = π + sin^{- 1} x + sin^{- 1} (2 x + 1)}

.

Then

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

is equal to ______.

Answer: 5

Using

cos^{- 1} x = π /2 - sin^{- 1} x

, the equation

cos^{- 1} x = π + sin^{- 1} x + sin^{- 1} (2 x + 1)

becomes

2 sin^{- 1} x + sin^{- 1} (2 x + 1) = - \frac{π}{2} .

Let

α = sin^{- 1} x

; since the right side is negative and

sin^{- 1} (2 x + 1) \in [- π /2, π /2]

, this forces

α \leq 0

, i.e.

x \leq 0

.

Taking sine after isolating

sin^{- 1} (2 x + 1) = - π /2 - 2 α

2 x + 1 = sin (- \frac{π}{2} - 2 α) = - cos 2 α = - (1 - 2 x^{2}) = 2 x^{2} - 1.

2 x^{2} - 2 x - 2 = 0 ⟹ x^{2} - x - 1 = 0 ⟹ x = \frac{1 \pm 5}{2} .

Only

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

satisfies

x \leq 0

(and lies in the valid domain for both inverse sine terms), so

S

has the single element

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

, which satisfies

x^{2} - x = 1

.

Then

(2 x - 1)^{2} = 4 x^{2} - 4 x + 1 = 4 (x^{2} - x) + 1 = 4 + 1 = 5.

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

JEE Main · 2025 · 29 Jan 2025 · Shift 1 · Integer answer