Question ID: #665
Let $f: R\rightarrow R$ 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:
Solution:
Since the quadratic equation in $m$ has two equal roots, its discriminant must be zero.
$$D = (-2f'(x))^2 – 4(f(x))(f”(x)) = 0$$
$$4(f'(x))^2 = 4f(x)f”(x)$$
$$(f'(x))^2 = f(x)f”(x)$$
Rearranging the terms:
$$\frac{f'(x)}{f(x)} = \frac{f”(x)}{f'(x)}$$
Integrating both sides with respect to $x$:
$$\int \frac{f'(x)}{f(x)} dx = \int \frac{f”(x)}{f'(x)} dx$$
$$\ln(f(x)) = \ln(f'(x)) + \ln C$$
$$f(x) = C f'(x) \Rightarrow f'(x) = k f(x)$$
Using the given initial conditions $f(0)=1$ and $f'(0)=2$:
$$2 = k(1) \Rightarrow k = 2$$
So, the differential equation is:
$$\frac{f'(x)}{f(x)} = 2$$
Integrating again:
$$\ln(f(x)) = 2x + C_1$$
$$f(x) = A e^{2x}$$
Using $f(0)=1 \Rightarrow A = 1$. Thus, $f(x) = e^{2x}$.
Now, consider the function $g(x) = f(\ln x – x)$.
Substituting $f(t) = e^{2t}$:
$$g(x) = e^{2(\ln x – x)}$$
For $g(x)$ to be increasing, we need $g'(x) \ge 0$.
$$g'(x) = e^{2(\ln x – x)} \cdot \frac{d}{dx}(2(\ln x – x))$$
$$g'(x) = e^{2(\ln x – x)} \cdot 2\left(\frac{1}{x} – 1\right)$$
Since the exponential term $e^{2(\ln x – x)}$ is always positive, the sign depends only on $\left(\frac{1}{x} – 1\right)$.
$$\frac{1}{x} – 1 \ge 0$$
$$\frac{1-x}{x} \ge 0$$
Since $x$ must be positive for $\ln x$ to be defined ($x > 0$), we multiply by $x$ without flipping the inequality:
$$1 – x \ge 0 \Rightarrow x \le 1$$
Combining with the domain $x > 0$, the interval is $x \in (0, 1]$.
Thus, the largest interval $(\alpha, \beta)$ is $(0, 1)$.
Here $\alpha = 0$ and $\beta = 1$.
Value of $\alpha + \beta$:
$$0 + 1 = 1$$
Ans. 1
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