Question 42 of 42advanced🔧 ApplyMCQ1 mark

sin⁻¹(1-x) - 2sin⁻¹x = π/2, then x is equal to

Exercise: Miscellaneous Exercise on Chapter 2 | Q: 14 | (Page 14)
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Explanation

Option C is correct because only x = 0 satisfies the given equation. When solving sin⁻¹(1-x) - 2sin⁻¹x = π/2, we substitute sin⁻¹x = θ, giving sin⁻¹(1-sinθ) = π/2 + 2θ. For this equality to hold, the RHS must lie in the range of sin⁻¹, which is [-π/2, π/2]. This requires -π/2 ≤\leq π/2 + 2θ ≤\leq π/2, giving θ ∈ [-π/2, 0]. Solving algebraically yields sinθ = 0 or sinθ = 1/2. While sinθ = 1/2 gives θ = π/6, this violates the constraint θ ≤\leq 0. Only θ = 0 is valid, giving x = sin(0) = 0. Verification: sin⁻¹(1) - 2sin⁻¹(0) = π/2 - 0 = π/2 ✓. For x = 1/2: sin⁻¹(1/2) - 2sin⁻¹(1/2) = π/6 - π/3 = -π/6 ≠\neq π/2.

Solution Steps

  1. Step 1: Let sin⁻¹x = θ, so x = sinθ where θ ∈ [-π/2, π/2]

  2. Step 2: Substitute into the equation: sin⁻¹(1-sinθ) = π/2 + 2θ

  3. Step 3: Apply constraint: RHS must be in range [-π/2, π/2], so -π/2 ≤\leq π/2 + 2θ ≤\leq π/2, giving θ ∈ [-π/2, 0]

  4. Step 4: Take sine of both sides: 1-sinθ = sin(π/2 + 2θ) = cos(2θ) = 1 - 2sin²θ

  5. Step 5: Simplify: sinθ = 2sin²θ, giving sinθ(1-2sinθ) = 0

  6. Step 6: Solutions: sinθ = 0 or sinθ = 1/2. With θ ∈ [-π/2, 0], only θ = 0 is valid

  7. Step 7: Therefore x = sin(0) = 0 is the only solution

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