Condition for Equal Roots

The quadratic equation $f(x)m^2 - 2f'(x)m + f''(x) = 0$ has two equal roots. This implies the discriminant ($\Delta$) is zero.

$ \Delta = (-2f'(x))^2 - 4(f(x))(f''(x)) = 0 $

Simplifying, we get $ 4(f'(x))^2 = 4f(x)f''(x) $, which leads to $ (f'(x))^2 = f(x)f''(x) $.

Solving the Differential Equation

Rearranging the equation $ (f'(x))^2 = f(x)f''(x) $, we get $ \frac{f''(x)}{f'(x)} = \frac{f'(x)}{f(x)} $ (assuming $f(x) \neq 0$ and $f'(x) \neq 0$).

Integrating both sides with respect to $x$:

• $ \int \frac{f''(x)}{f'(x)} dx = \int \frac{f'(x)}{f(x)} dx $
• $ \log|f'(x)| = \log|f(x)| + C_1 $
• Exponentiating both sides gives $ f'(x) = e^{C_1} f(x) $. Let $C = e^{C_1}$ (a constant).
• So, $ f'(x) = C f(x) $.

Integrating $f'(x) = C f(x)$ again:

• $ \int \frac{f'(x)}{f(x)} dx = \int C dx $
• $ \log|f(x)| = Cx + C_2 $
• Exponentiating gives $ f(x) = e^{Cx + C_2} = e^{C_2} e^{Cx} $. Let $A = e^{C_2}$ (a constant).
• Thus, $ f(x) = A e^{Cx} $.

Determining Function Constants

Use the given initial conditions $f(0) = 1$ and $f'(0) = 2$ to find the constants A and C.

• Using $f(0) = 1$: $ A e^{C \cdot 0} = 1 \implies A = 1 $.
• The function is $ f(x) = e^{Cx} $.
• The derivative is $ f'(x) = C e^{Cx} $.
• Using $f'(0) = 2$: $ C e^{C \cdot 0} = 2 \implies C = 2 $.

Therefore, the function is $ f(x) = e^{2x} $.

Finding the Interval of Increase

Consider the function $g(x) = f(\log_e x - x)$. We need to find the interval $(\alpha, \beta)$ where $g(x)$ is increasing, meaning $g'(x) > 0$.

• Substitute $f(x)$: $ g(x) = e^{2(\log_e x - x)} $.
• Simplify $g(x)$: $ g(x) = e^{2\log_e x} \cdot e^{-2x} = e^{\log_e x^2} \cdot e^{-2x} = x^2 e^{-2x} $.
• Calculate the derivative $g'(x)$ using the product rule: $ g'(x) = \frac{d}{dx}(x^2 e^{-2x}) = (2x)e^{-2x} + x^2(-2e^{-2x}) $
• Factor the derivative: $ g'(x) = 2x e^{-2x} (1 - x) $.

For $g(x)$ to be increasing, $g'(x) > 0$.

• $ 2x e^{-2x} (1 - x) > 0 $.
• Since $x$ is the argument of $\log_e x$, we must have $x > 0$. Also, $e^{-2x}$ is always positive.
• Thus, the inequality simplifies to $ 2x(1 - x) > 0 $.
• Since $x > 0$, we must have $1 - x > 0$, which implies $x < 1$.

The function $g(x)$ is increasing on the interval $ (0, 1) $. Therefore, $(\alpha, \beta) = (0, 1)$.

Calculating $\alpha + \beta$

Given $\alpha = 0$ and $\beta = 1$, the sum is:

$ \alpha + \beta = 0 + 1 = 1 $