Question

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 \(g(x) = f(\log_e x - x)\) is increasing, then \(\alpha+\beta\) is equal to:

Hint:

The differential equation \((f'(x))^2 = f(x)f''(x)\) is a classic form.
It can be quickly identified as \(\left(\frac{f'(x)}{f(x)}\right)' = 0\), which implies \(\frac{f'(x)}{f(x)}\) is a constant.
This leads directly to an exponential solution \(f(x) = Ke^{Cx}\), saving time on integration steps.

Answer 1

Correct Answer: 2

Step 1: Understanding the Question:
The problem has two main parts.
First, we must determine the function \(f(x)\) using the given information that a quadratic equation involving \(f(x)\) and its derivatives has equal roots.
Second, we must find the largest interval on which a related composite function, \(g(x)\), is increasing, and then find the sum of the interval's endpoints.
Step 2: Key Formula or Approach:
1. A quadratic equation \(Am^2+Bm+C=0\) has equal roots if its discriminant is zero: \(B^2-4AC=0\).
2. Solving the resulting differential equation to find \(f(x)\).
3. A function \(g(x)\) is increasing where its derivative \(g'(x)\) is positive.
4. The Chain Rule for differentiation: If \(g(x) = f(h(x))\), then \(g'(x) = f'(h(x)) \cdot h'(x)\).
Step 3: Detailed Explanation:
Part 1: Finding the function f(x)
The given quadratic equation in \(m\) is \(f(x)m^2 - 2f'(x)m + f''(x) = 0\).
For this equation to have equal roots, its discriminant must be zero.
\[ (-2f'(x))^2 - 4(f(x))(f''(x)) = 0
\] \[ 4(f'(x))^2 - 4f(x)f''(x) = 0
\] \[ (f'(x))^2 = f(x)f''(x)
\] This can be rewritten, for \(f'(x) \neq 0\) and \(f(x) \neq 0\), as:
\[ \frac{f''(x)}{f'(x)} = \frac{f'(x)}{f(x)}
\] Integrating both sides with respect to \(x\):
\[ \int \frac{f''(x)}{f'(x)} dx = \int \frac{f'(x)}{f(x)} dx
\] \[ \ln|f'(x)| = \ln|f(x)| + C_1
\] \[ f'(x) = e^{C_1} f(x) = C f(x)
\] This is a first-order linear differential equation. We use the initial conditions \(f(0)=1\) and \(f'(0)=2\) to find the constant \(C\).
\[ f'(0) = C \cdot f(0) \implies 2 = C \cdot 1 \implies C=2.
\] So, the differential equation is \(f'(x) = 2f(x)\), or \(\frac{dy}{dx} = 2y\).
The solution to this standard differential equation is \(f(x) = K e^{2x}\).
Using the initial condition \(f(0)=1\):
\[ 1 = K e^{2(0)} \implies K=1.
\] Therefore, the function is \(f(x) = e^{2x}\). Consequently, its derivative is \(f'(x) = 2e^{2x}\).
Part 2: Finding the interval of increase for g(x)
The function is \(g(x) = f(\log_e x - x)\). The domain requires \(\log_e x\) to be defined, so \(x>0\).
To find where \(g(x)\) is increasing, we need to find where its derivative \(g'(x)>0\).
Using the chain rule, \(g'(x) = f'(\log_e x - x) \cdot \frac{d}{dx}(\log_e x - x)\).
We know \(f'(u) = 2e^{2u}\). Let \(u = \log_e x - x\).
\[ g'(x) = 2e^{2(\log_e x - x)} \cdot \left(\frac{1}{x} - 1\right)
\] We need to solve \(g'(x)>0\).
\[ 2e^{2\log_e x - 2x} \cdot \left(\frac{1-x}{x}\right)>0
\] The exponential term \(2e^{2\log_e x - 2x}\) is always positive.
So, the sign of \(g'(x)\) depends entirely on the sign of the term \(\left(\frac{1-x}{x}\right)\).
We need to solve the inequality:
\[ \frac{1-x}{x}>0
\] Since the domain of \(g(x)\) is \(x>0\), the denominator \(x\) is positive.
Thus, the inequality simplifies to the numerator being positive:
\[ 1-x>0 \implies x<1.
\] Combining this result with the domain restriction \(x>0\), we find that \(g(x)\) is increasing on the interval \((0, 1)\).
This is the largest such interval, so we have \((\alpha, \beta) = (0, 1)\).
The question asks for the value of \(\alpha + \beta\).
\[ \alpha + \beta = 0 + 1 = 1.
\] Step 4: Final Answer:
The value of \(\alpha+\beta\) is 1.