Let α be a non-zero real number. Suppose f: R → R is a differentiable function such that f(0)=2 and lim(x →-∞) f(x)=1. If f′(x)=α f(x)+3, for all x ∈ R, then f(-log ₑ 2) is equal to :

Solution

Maths sir · 4 steps

This is a linear equation: solve it, let the limit fix α, then let f(0)=2 fix the constant.

1. General solution. Rewrite as f' - α f = 3 and multiply by the integrating factor e^(-α x). This gives (f e^(-α x))' = 3e^(-α x). Integrating both sides gives the general solution below.

   f(x) = C e^(α x) - 3/α

2. Use the limit. For f to settle at a finite value, Ce^(α x) must die out, which leaves -3/α = 1, so α = -3. With α = -3, e⁻³ˣ → 0 only as x → +∞. So the limit condition only works at +∞, and the key follows that reading. At -∞ as printed, no non-zero α fits.
3. Find C. Now f(x) = Ce⁻³ˣ + 1, and f(0) = C + 1 = 2 gives C = 1. So f(x) = e⁻³ˣ + 1.
4. So. f(-logₑ 2) = e^(3 logₑ 2) + 1 = 2³ + 1 = 9, which is option (b).

The idea to keep. For f' = α f + k, the solution is Ce^(α x) - k/α. A limit condition kills the exponential and hands you α, and the initial value then gives C.

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