If domain of the function log ₑ((6 x²+5 x+1)/(2 x-1))+cos ⁻¹((2 x²-3 x+4)/(3 x-5)) is (α, β) ∪(γ, δ], then 18(α²+β²+γ²+δ²) is equal to ______________.

Solution

Maths sir · 5 steps

One condition from the log, two from the inverse cosine, then overlap them all on a number line.

1. Log condition. The log needs (6x²+5x+1)/(2x-1) > 0, that is (2x+1)(3x+1)/(2x-1) > 0. The critical points are -1/2, -1/3, 1/2, and the sign chart gives x ∈ (-1/2, -1/3) ∪ (1/2, ∞).
2. Upper limit. cos⁻¹ needs its argument in [-1, 1]. The condition (2x²-3x+4)/(3x-5) ≤ 1 becomes (2x²-6x+9)/(3x-5) ≤ 0. The numerator has discriminant 36 - 72 < 0, so it is always positive. That forces 3x - 5 < 0, so x < 5/3.
3. Lower limit. The condition (2x²-3x+4)/(3x-5) ≥ -1 becomes (2x²-1)/(3x-5) ≥ 0. The critical points are ±1/√2 and 5/3, so x ∈ [-1/√2, 1/√2] ∪ (5/3, ∞). Together with x < 5/3, this leaves [-1/√2, 1/√2].
4. Overlap. Since -1/√2 ≈ -0.707 lies to the left of -1/2, the common part is (-1/2, -1/3) ∪ (1/2, 1/√2]. This gives α = -1/2, β = -1/3, γ = 1/2 and δ = 1/√2.

   18(1/4+1/9+1/4+1/2) = 18 × 40/36 = 20

5. So. 18(α²+β²+γ²+δ²) = 20.

The idea to keep. For cos⁻¹(f) or sin⁻¹(f), solve f ≤ 1 and f ≥ -1 as two separate rational inequalities. A numerator that is always positive hands you the sign of the denominator.

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