Let f: R R be a twice differentiable function such that the quadratic equation f(x) m ²-2 f^ (x) m +f^ (x)=0 in m, has two equal roots for every x R . If f(0)=1, f^ (0)=2 , and ( , ) is the largest interval in which the function f ( _ e x-x

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Let f: R R be a twice differentiable function such that the quadratic equation f(x) m ²-2 f^ (x) m +f^ (x)=0 in m, has two equal roots for every x R . If f(0)=1, f^ (0)=2 , and ( , ) is the largest interval in which the function f ( _ e x-x ) is increasing, then + is equal to _ _ _ _ .

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Correct answer: 0

For equal roots in f(x)m^2 - 2f'(x)m + f''(x) = 0 , discriminant = 0: [f'(x)]^2 = f(x)f''(x) This gives f''(x) f'(x) = f'(x) f(x) Integrating: f'(x) = kf(x) , so f(x) = Ae^ kx Using f(0) = 1 and f'(0) = 2 : A = 1 , k = 2 f(x) = e^ 2x For g(x) = f( x - x) : g'(x) = 2e^ 2( x - x) ( 1 x - 1 ) g'(x) > 0 when 1-x x > 0 , i.e., 0 ( , ) = (0, 1) + = 1

Let f: R R be a twice differentiable function such that the quadratic equation f(x) m ²-2 f^ (x) m +f^ (x)=0 in m, has two equal roots for every x R . If f(0)=1, f^ (0)=2 , and ( , ) is the largest interval in which the function f ( _ e x-x ) is increasing, then + is equal to _ _ _ _ .

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