alleryMaximum value of x² + (arcsin(x))² for x in [-1, 1]:
AAt x = 1: 1 + (π/2)² ≈ 3.47
B0
CCannot determine without calculus
DAlways 1
Answer & Solution
Correct answer: A. At x = 1: 1 + (π/2)² ≈ 3.47
f(x) = x² + (arcsin(x))² on [-1, 1].
f'(x) = 2x + 2 arcsin(x) × 1/sqrt(1-x²).
At x = 1: f = 1 + (π/2)² ≈ 1 + 2.47 = 3.47. By symmetry same value at x = -1. So max ≈ 3.47.
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