For a differentiable function f: ℝ → ℝ, suppose f′(x)=3 f(x)+α, where α ∈ ℝ, f(0)=1 and lim x →-∞ f(x)=7. Then 9 f(-log ₑ 3) is equal to _________.

Solution

Maths sir · 4 steps

Solve the linear equation; the limit at minus infinity gives α, and f(0) fixes the constant.

1. Integrating factor. Write it as f' - 3f = α. The integrating factor is e⁻³ˣ, so d/dx(f e⁻³ˣ) = α e⁻³ˣ.

   f(x) = Ce³ˣ - α/3

2. Use the limit. As x → -∞, e³ˣ → 0, so f(x) → -α/3. This equals 7, so α = -21 and f(x) = Ce³ˣ + 7.
3. Use f(0) = 1. C + 7 = 1 gives C = -6, so f(x) = 7 - 6e³ˣ.
4. So. At x = -logₑ 3, e³ˣ = e^(-3 logₑ 3) = 1/27. Then f = 7 - 6/27 = 61/9, and 9f(-logₑ 3) = 61.

The common slip: 45. Taking e³ˣ at x = -logₑ 3 as 1/3 instead of 1/27 gives f = 5 and an answer of 45, not 61.

The idea to keep. For y' = ky + c, the solution is Ceᵏˣ - c/k. A limit where eᵏˣ dies off reads out the constant part straight away; the initial value then fixes C.

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