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-αdx = e-αx

  Solution is given by, e-αxf(x) = 3e-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.