If the domain of the function f(x)=sec ⁻¹((2 x)/(5 x+3)) is [α, β) U(γ, δ], then |3 α+10(β+γ)+21 δ| is equal to _________.

Solution

Maths sir · 5 steps

For sec⁻¹ the input must stay outside (-1, 1), so split into two inequalities and solve each.

1. Domain condition. sec⁻¹ t is defined only for t ≤ -1 or t ≥ 1. Here t = 2x/(5x+3), and x ≠ -3/5 so the denominator is not zero.
2. Case t ≥ 1. 2x/(5x+3) - 1 ≥ 0 gives (-3x-3)/(5x+3) ≥ 0, that is (x+1)/(5x+3) ≤ 0. The critical points are -1 and -3/5, so -1 ≤ x < -3/5.
3. Case t ≤ −1. 2x/(5x+3) + 1 ≤ 0 gives (7x+3)/(5x+3) ≤ 0. The critical points are -3/5 and -3/7, so -3/5 < x ≤ -3/7.
4. Read off values. The domain is [-1, -3/5) ∪ (-3/5, -3/7]. So α = -1, β = γ = -3/5 and δ = -3/7.

   3α + 10(β+γ) + 21δ = -3 + 10(-6/5) + 21(-3/7) = -3 - 12 - 9 = -24

5. So. Taking the absolute value, |3α + 10(β+γ) + 21δ| = 24.

The idea to keep. Domain of sec⁻¹ or csc⁻¹ means |t| ≥ 1. Bring everything to one side over a common denominator, then use the sign scheme, and never include the point where the denominator is zero.

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