Let S = { x : cos⁻¹ x = π + sin⁻¹ x + sin⁻¹ [2x + 1] } . Then Σ_(x ∈ S) (2x - 1)² is equal to _______.

Solution

Maths sir · 5 steps

Swap cos⁻¹ x for π/2-sin⁻¹ x, take sine of both sides, and a quadratic appears.

1. One inverse function. Read the square brackets as ordinary brackets. Put cos⁻¹ x = π/2 - sin⁻¹ x and rearrange: 2 sin⁻¹ x + sin⁻¹(2x+1) = -π/2.
2. Fix the domain. You need -1 ≤ x ≤ 1 and -1 ≤ 2x+1 ≤ 1. Together these give -1 ≤ x ≤ 0, so θ = sin⁻¹ x lies in [-π/2, 0].
3. Take sine. Now sin⁻¹(2x+1) = -π/2 - 2θ, and this angle stays inside [-π/2, π/2], so taking sine loses nothing. That gives 2x+1 = sin(-π/2 - 2θ) = -cos 2θ = 2x² - 1.

   x² - x - 1 = 0 ⇒ x = (1 ± √5)/2

4. Reject one root. (1+√5)/2 ≈ 1.618 is outside [-1, 0]. Only x = (1-√5)/2 ≈ -0.618 survives, so S has one element.
5. So. For this x, 2x - 1 = -√5, so (2x-1)² = 5. The sum is 5.

The common slip: 10. Keeping both roots of x² - x - 1 = 0 adds 5 twice and gives 10. The root (1+√5)/2 is greater than 1, so it is not in the domain.

The idea to keep. Find the domain from every inverse function before you solve. Taking sine or squaring can bring in extra roots, so check each root against that domain.

Answer from the NTA final answer key. Solution by PrepWiser.