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

Given quadratic equation has equal roots, thus

$D = 0 \Rightarrow (r(x))^2 = r'(x)\cdot f(x)$

$\frac{f'(x)}{f(x)} = \frac{f''(x)}{f'(x)}$

Integrate

$\ln(f'(x)) = \ln(f(x)) + \ln C \Rightarrow f'(x) = C f(x)$

Put $x = 0$

$1 = C \cdot 2 \Rightarrow C = \frac{1}{2}$

Now $2f'(x) = f(x)$

$\Rightarrow \frac{f'(x)}{f(x)} = 2$

Integrate

$\ln(f(x)) = 2x + d$

$\Rightarrow d = 0$

$\Rightarrow \ln(f(x)) = 2x \Rightarrow f(x) = e^{2x}$

Now let $g(x) = f(\ln x - x) = e^{2(\ln x - x)}$

$\therefore g'(x) = 2e^{2(\ln x - x)}\left(\frac{1}{x} - 1\right) \ge 0$

$\Rightarrow \frac{1 - x}{x} \ge 0$

$\Rightarrow x \in (0,1]$

$\Rightarrow \alpha = 0,; \beta = 1$

$\alpha + \beta = 1$