If the domain of the function sin ⁻¹((3 x-22)/(2 x-19))+log ₑ((3 x²-8 x+5)/(x²-3 x-10)) is (α, β], then 3 α+10 β is equal to:

Solution

Maths sir · 5 steps

Find the window allowed by sin⁻¹ first, then let the log condition cut it down to (α, β].

1. Inverse sine condition. We need -1 ≤ (3x-22)/(2x-19) ≤ 1. Split it into two inequalities and bring everything to one side. Do not cross-multiply, because 2x-19 can be negative.
2. Solve both halves. Upper half: (3x-22)/(2x-19) - 1 = (x-3)/(2x-19) ≤ 0, so 3 ≤ x < 19/2. Lower half: (3x-22)/(2x-19) + 1 = (5x-41)/(2x-19) ≥ 0, so x ≤ 41/5 or x > 19/2. Together these give x ∈ [3, 41/5].
3. Log condition. Factorise the log's argument and ask for it to be strictly positive. A sign chart with critical points -2, 1, 5/3, 5 gives x ∈ (-∞, -2) ∪ (1, 5/3) ∪ (5, ∞).

   (3x²-8x+5)/(x²-3x-10) = (3x-5)(x-1)/(x-5)(x+2) > 0

4. Intersect. Inside [3, 41/5], only the piece (5, ∞) survives. So the domain is (5, 41/5], which gives α = 5 and β = 41/5.
5. So. 3α + 10β = 15 + 82 = 97, which is option (c).

The idea to keep. For sin⁻¹(f), write -1 ≤ f ≤ 1 as two separate inequalities, move everything to one side and use sign charts. Never cross-multiply when the denominator can change sign.

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