If the domain of the function f(x)=cos ⁻¹((2-|x|)/4)+{log ₑ(3-x)}⁻¹ is [-α, β)-{γ}, then α+β+γ is equal to :

Solution

Maths sir · 4 steps

Write one condition for each piece, the cos⁻¹, the log and the reciprocal, then intersect them.

1. Inverse cosine part. cos⁻¹ needs its input in [-1, 1]. So -1 ≤ (2-|x|)/4 ≤ 1, which means -4 ≤ 2-|x| ≤ 4.
2. Solve both sides. The left inequality gives |x| ≤ 6, so -6 ≤ x ≤ 6. The right one gives |x| ≥ -2, which is true for every x.
3. Log and reciprocal. The log needs 3-x > 0, so x < 3. The reciprocal needs logₑ(3-x) ≠ 0, so 3-x ≠ 1, which means x ≠ 2.

   [-6, 6] ∩ (-∞, 3) - {2} = [-6, 3) - {2}

4. So. α = 6, β = 3, γ = 2, and α + β + γ = 11. The answer is (a).

The idea to keep. For a sum of functions, the domain is the intersection of each piece's domain. A reciprocal also removes every point where its denominator is zero, so check where log equals 0.

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