Statement I: The equation (sin−1x)3+ (cos−1x)3−aπ3=0 has a solution for all a≥132.Statement II: For any x∈R,sin−1x+cos−1x=π2 and 0≤(sin−1x−π4)2≤9π216

Statement I: The equation (sin−1x)3+ (cos−1x)3−aπ3=0 has a solution for all a≥132.

Step 1: We know that sin−1x + cos−1x = π/2. This is a standard result in trigonometry.

Step 2: Cubing both sides, we get (sin−1x)3 + (cos−1x)3 + 3sin−1x * cos−1x * (sin−1x + cos−1x) = (π/2)3.

Step 3: Substituting sin−1x + cos−1x = π/2 in the above equation, we get (sin−1x)3 + (cos−1x)3 = (π/2)3 - 3π/4 = π3/8 - 3π/4.

Step 4: For the equation (sin−1x)3+ (cos−1x)3−aπ3=0 to have a solution for all a≥132, we need π3/8 - 3π/4 ≥ 132π3. This is not true. Hence, Statement I is false.

Statement II: For any x∈R,sin−1x+cos−1x=π2 and 0≤(sin−1x−π4)2≤9π216.

Step 1: We know that sin−1x + cos−1x = π/2. This is a standard result in trigonometry.

Step 2: The second part of the statement is equivalent to -3π/4 ≤ sin−1x - π/4 ≤ 3π/4.

Step 3: Adding π/4 to all parts of the inequality, we get -π/2 ≤ sin−1x ≤ π. This is true for all x∈R. Hence, Statement II is true.