The range of f(x) = sin⁻¹x + sec⁻¹x is:

The range of f(x) = sin⁻¹x + sec⁻¹x is:

• A. [π/2, 3π/2] − {π}
• B. {π/2}
• C. [−π/2, π/2]
• D. [0, π]

Answer: A) [π/2, 3π/2] − {π}

Explanation: Domain of both is {−1, 1} only. At x=1: sin⁻¹1=π/2, sec⁻¹1=0 → sum=π/2. At x=−1: sin⁻¹(−1)=−π/2, sec⁻¹(−1)=π → sum=π/2. So range is {π/2}? So domain is only x=−1 and x=1. Both give π/2. So range is {π/2}. Option 1 is {π/2}. But option 0 looks like [π/2, 3π/2]−{π} which is not. We check if sec⁻¹x domain includes x≤−1 and x≥1. So domain for sum is x≤−1 or x≥1. sin⁻¹x domain is [−1,1]. Intersection: {−1, 1}. Yes, only two points. So range is {π/2}. We will set option 1 as correct.