Let f : R - R be a twice differentiable function such that the quadratic equation f(x)m² - 2f'(x)m

Q: Wait, why do we set the discriminant to zero? Does 'equal roots' always mean that?

Absolutely! For any quadratic equation $am^2 + bm + c = 0$, the roots are given by the quadratic formula. If the discriminant $D = b^2 - 4ac$ is zero, the $\pm \sqrt{D}$ term vanishes, leaving you with a single, repeated root of $-b/2a$. It is the algebraic way of saying the function has a unique solution.

Q: I got stuck at $(f'(x))^2 = f(x)f''(x)$. How do I integrate this?

That is a classic differential form! Notice that if you divide both sides by $f(x)f'(x)$, you get $\frac{f'(x)}{f(x)} = \frac{f''(x)}{f'(x)}$. Both sides are now perfect derivatives: the left is the derivative of $\ln(f(x))$ and the right is the derivative of $\ln(f'(x))$. Integrating both sides gives you $\ln(f'(x)) = \ln(f(x)) + C$.

Q: Why is the domain $x > 0$ for the function $f(\ln x - x)$?

Great catch! Always look at the 'inner' functions first. The term $\ln x$ is only defined for $x > 0$. Even if the rest of the function looks perfectly fine, the entire expression is tethered to the domain of its most restrictive component. Never forget to check your domain constraints before you start differentiating!

Q: Do I really need to worry about the exponential term when checking where $g'(x) > 0$?

Mathematically, you should always consider it, but physically/intuitively, you can breathe easy. The term $e^{2(\ln x - x)}$ is an exponential function, and $e^u$ is strictly positive for all real $u$. Since it is always positive, it doesn't change the sign of the inequality. It's like multiplying by a positive constant—it scales the slope, but it doesn't flip the direction of growth.

Q: Is there a shortcut to finding $f(x)$ once I have the differential equation?

You could recognize that $(f'(x))^2 = f(x)f''(x)$ is the condition for $f(x)$ to be an exponential function of the form $Ae^{kx}$. If you see this structure, you can jump straight to the general solution. However, deriving it step-by-step as we did is the safest way to ensure you don't miss any constants!

Question

Let $f : \mathbb{R} 	o \mathbb{R}$ 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 $x \in \mathbb{R}$. 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 .........

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a mathematical mystery. You are given a quadratic equation $f(x)m^2 - 2f'(x)m + f''(x) = 0$, and told that for every value of $x$, this equation yields two equal roots.

At first glance, this looks like a standard algebra problem, but the coefficients are functions of $x$. This is the beauty of JEE Advanced problems; they hide a deep, structural truth behind a simple algebraic facade.

### The Discriminant: The Key to the Lock

When a quadratic equation has equal roots, the discriminant must be zero. The discriminant, $D = b^2 - 4ac$, is the gatekeeper of roots.

By setting $D = (-2f'(x))^2 - 4(f(x))(f''(x)) = 0$, we force the function to maintain a perfect balance. Simplifying this, we arrive at the elegant relation:

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

This equation is a statement about the growth of $f(x)$. It tells us that the square of the rate of change is proportional to the function itself multiplied by its curvature. To solve this, we rearrange it into a form that invites integration:

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

### The Integration Journey

Now, we integrate both sides. The left side is the integral of the logarithmic derivative of $f'(x)$, and the right side is the logarithmic derivative of $f(x)$.

This leads us to $\ln|f'(x)| = \ln|f(x)| + \ln|k|$, which simplifies beautifully to $f'(x) = k \cdot f(x)$. This is the hallmark of exponential growth. Solving this first-order differential equation, we find the general form:

$$f(x) = Ae^{kx}$$

Using our initial conditions, $f(0) = 1$ and $f'(0) = 2$, we find that $A = 1$ and $k = 2$. Thus, the function that has been hiding in plain sight is none other than $f(x) = e^{2x}$.

### The Final Ascent: Analyzing the Composite Function

We must now investigate the behavior of $g(x) = f(\ln x - x)$. Substituting our newly discovered $f(x)$, we get $g(x) = e^{2(\ln x - x)}$.

To find where this function is increasing, we need to find the interval where its derivative $g'(x)$ is positive. Applying the chain rule, we calculate:

$$g'(x) = e^{2(\ln x - x)} \cdot \frac{d}{dx}[2(\ln x - x)] = 2 e^{2(\ln x - x)} \cdot \left( \frac{1}{x} - 1 \right)$$

Since the exponential term $2 e^{2(\ln x - x)}$ is always positive, the sign of $g'(x)$ depends entirely on the term $\left( \frac{1}{x} - 1 \right)$.

For $g(x)$ to be increasing, we require:

$$\frac{1-x}{x} > 0$$

Given the domain constraint $x > 0$, the denominator is always positive, leaving us with the simple inequality $1 - x > 0$, or $x < 1$. Combining this with our domain, we find the interval is $(0, 1)$.

### The Conclusion

We have arrived at our interval $(\alpha, \beta) = (0, 1)$. Thus, $\alpha = 0$ and $\beta = 1$.

The sum $\alpha + \beta$ is $0 + 1 = 1$. It is a simple integer, but the path to get there required us to bridge the gap between algebra, differential equations, and calculus. The final result is **1**.

Enter Numerical Value:

Visualized Solution

The Quadratic Equation in $m$

Condition for Equal Roots

Applying the Discriminant

Simplifying the Equation

Rearranging the Differential Equation

Integrating Both Sides

Finding the Function $f (x)$

Applying Initial Conditions

The Exact Function $f (x)$

Defining the Composite Function $g (x)$

Differentiating $g (x)$

Calculating $g^{'} (x)$

Condition for Increasing Function

Solving for the Interval

Finding $α + β$

The Sigma Insight: Variable Separable Method

Analyzing the Setup

The Discriminant

The Integration Journey

The Final Ascent

The Conclusion

Similar Questions

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