Q.

Let α be a non-zero real number. Suppose f : R → R is a differentiable function such that f(0) = 2 and limx → -∞f(x) = 1. If f'(x) = αf(x) + 3, for all x∈ R, then f(- loge 2) is equal to ________.            [2024]

1 7  
2 9  
3 3  
4 5  

Ans.

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We have, f'(x) = αf(x) + 3  ⇒  f'(x) - αf(x) = 3

Now, Let α > 0, IF = e-α∫dx = e-αx

∴  Solution is given by, e-αxf(x) = 3∫e-αxdx + C

⇒  f(x) = -3α + Ceαx

Now, let α > 0, we have limx →-∞f(x) = 1

⇒  -3α + C limx →-∞eαx = 1

Let α < 0, limx →∞f(x) = -3α + C limx →-∞eαx = 1

⇒  -3α + ∞ = 1 which is not possible

∴  Value of α does not exist.