The domain of the function f(x)=sin ⁻¹[2 x²-3]+log ₂(log _(1/2)(x²-5 x+5)), where [t] is the greatest integer function, is :

Solution

Maths sir · 4 steps

Treat each part on its own: the GIF bounds x², and the two logs trap x²-5x+5 between 0 and 1.

1. Inverse sine part. sin⁻¹ needs -1 ≤ [2x²-3] ≤ 1. An integer in that range comes from -1 ≤ 2x²-3 < 2, because the GIF stays at 1 until its input reaches 2. This gives 1 ≤ x² < 5/2, so x ∈ (-√(5/2), -1] ∪ [1, √(5/2)).
2. Outer log. log₂ needs a positive input, so log_(1/2)(x²-5x+5) > 0. The base 1/2 is less than 1, so this log is positive only when its input lies strictly between 0 and 1.

   0 < x²-5x+5 < 1

3. Solve both inequalities. x²-5x+5 < 1 gives (x-1)(x-4) < 0, so 1 < x < 4. x²-5x+5 > 0 holds outside the roots (5 ± √5)/2 ≈ 1.38, 3.62. Together: x ∈ (1, (5-√5)/2) ∪ ((5+√5)/2, 4).
4. So. Intersect with the first condition, where √(5/2) ≈ 1.58. Only (1, (5-√5)/2) survives, because 1.38 < 1.58 and the negative interval misses (1, 4) entirely. The answer is (c).

The idea to keep. For logₐ with 0 < a < 1, the log is positive only when its input is between 0 and 1. With [f(x)] = n, use n ≤ f(x) < n+1. Find each condition, then intersect.

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