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

### Related Formula For a quadratic equation Am^2 + Bm + C = 0$Am^2 + Bm + C = 0$ having equal roots, Discriminant D = 0 Rightarrow B^2 - 4AC = 0$D = 0 \Rightarrow B^2 - 4AC = 0$. ### Core Logic Given quadratic equation in m$m$: f(x)m^2 - 2f'(x)m + f''(x) = 0$f(x)m^2 - 2f'(x)m + f''(x) = 0$ has equal roots. D = 0 Rightarrow (-2f'(x))^2 - 4(f(x))(f''(x)) = 0$$D = 0 \Rightarrow (-2f'(x))^2 - 4(f(x))(f''(x)) = 0$$ 4(f'(x))^2 = 4f(x)f''(x) Rightarrow (f'(x))^2 = f(x)f''(x)$$4(f'(x))^2 = 4f(x)f''(x) \Rightarrow (f'(x))^2 = f(x)f''(x)$$ ### Step 1: Solve the Differential Equation Rewrite the DE: fracf''(x)f'(x) = fracf'(x)f(x)$\frac{f''(x)}{f'(x)} = \frac{f'(x)}{f(x)}$ Integrate both sides: int fracf''(x)f'(x) dx = int fracf'(x)f(x) dx$$\int \frac{f''(x)}{f'(x)} dx = \int \frac{f'(x)}{f(x)} dx$$ ln|f'(x)| = ln|f(x)| + ln|c| Rightarrow f'(x) = c cdot f(x)$$\ln|f'(x)| = \ln|f(x)| + \ln|c| \Rightarrow f'(x) = c \cdot f(x)$$ Using given f(0) = 1$f(0) = 1$ and f'(0) = 2$f'(0) = 2$: f'(0) = c cdot f(0) Rightarrow 2 = c(1) Rightarrow c = 2$$f'(0) = c \cdot f(0) \Rightarrow 2 = c(1) \Rightarrow c = 2$$ Now we have f'(x) = 2f(x) Rightarrow fracf'(x)f(x) = 2$f'(x) = 2f(x) \Rightarrow \frac{f'(x)}{f(x)} = 2$. Integrate again: ln|f(x)| = 2x + d$$\ln|f(x)| = 2x + d$$ Use f(0) = 1 Rightarrow ln(1) = 0 + d Rightarrow d = 0$f(0) = 1 \Rightarrow \ln(1) = 0 + d \Rightarrow d = 0$. So, ln f(x) = 2x Rightarrow f(x) = e^2x$\ln f(x) = 2x \Rightarrow f(x) = e^{2x}$. ### Step 2: Investigate increasing interval Let g(x) = f(ln x - x) = e^2(ln x - x)$g(x) = f(\ln x - x) = e^{2(\ln x - x)}$. For g(x)$g(x)$ to be increasing, g'(x) geq 0$g'(x) \geq 0$. g'(x) = 2e^2(ln x - x) cdot fracddx(ln x - x)$$g'(x) = 2e^{2(\ln x - x)} \cdot \frac{d}{dx}(\ln x - x)$$ g'(x) = 2e^2(ln x - x) left(frac1x - 1right)$$g'(x) = 2e^{2(\ln x - x)} \left(\frac{1}{x} - 1\right)$$ Since exponential is always positive, g'(x) geq 0 Rightarrow frac1x - 1 geq 0$g'(x) \geq 0 \Rightarrow \frac{1}{x} - 1 \geq 0$. frac1 - xx geq 0$$\frac{1 - x}{x} \geq 0$$ The critical points are x=0, x=1$x=0, x=1$. Based on domain of ln x$\ln x$, x > 0$x > 0$. Sign scheme yields positive derivative in x in (0, 1]$x \in (0, 1]$. Therefore, (alpha, beta) = (0, 1) Rightarrow alpha = 0, beta = 1$(\alpha, \beta) = (0, 1) \Rightarrow \alpha = 0, \beta = 1$. ### Step 3: Final Output alpha + beta = 0 + 1 = 1$$\alpha + \beta = 0 + 1 = 1$$ ### Pattern Recognition The relation (f')^2 = f cdot f''$(f')^2 = f \cdot f''$ is a classical indicator of exponential functions (f = Ce^kx$f = Ce^{kx}$). Solving via double logarithmic integration collapses the differential equation almost instantaneously. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Differential Equations Class 12 Maths: Applications of Derivatives