Let f be a differentiable function defined on [0, π/2] such that f(x) 0 and f(x) + ∫₀ˣ f(t)√(1 - (

Q: Wait, why can we just differentiate the integral? Is that always allowed?

Great question! We use the Leibniz Integral Rule. Since the function $f(x)$ is given as differentiable, the integrand $f(t)\sqrt{1-(\ln f(t))^2}$ is continuous, allowing us to apply the Fundamental Theorem of Calculus to differentiate the integral $\int_{0}^{x} g(t) dt$ simply as $g(x)$.

Q: Why did we set $f(0) = e$? Where did that come from?

Look closely at the original equation: $f(x) + \int_{0}^{x} f(t) \sqrt{1 - (\ln f(t))^2} dt = e$. If we plug in $x = 0$, the integral becomes $\int_{0}^{0} \dots dt$, which is identically zero. Thus, $f(0) + 0 = e$, giving us our vital initial condition.

Q: I got stuck at $\sin^{-1}(\ln f(x)) = -x + C$. How do I know if I should use $\sin$ or $\cos$?

The integral $\int \frac{dt}{\sqrt{1-t^2}}$ is the standard derivative of $\sin^{-1}(t)$. While it is also equal to $-\cos^{-1}(t) + C'$, using $\sin^{-1}$ is usually cleaner. As long as you determine your constant $C$ correctly using the initial condition, both paths will lead to the same final result!

Q: Is it possible that $\ln f(x)$ could be greater than 1?

The term $\sqrt{1 - (\ln f(x))^2}$ in the original equation forces the domain of the square root to be $1 - (\ln f(x))^2 \ge 0$, which means $(\ln f(x))^2 \le 1$. This implies $-1 \le \ln f(x) \le 1$. So, no, it cannot exceed 1!

Q: Why did we square the final expression? Is there a trick?

The question asks for $(6 \ln f(\pi/6))^2$. Often in JEE Advanced, the final expression is designed to eliminate radicals or fractions, turning a messy irrational number into a clean integer. It's a sign you're on the right track!

Question

Let $f$ be a differentiable function defined on $[0, \pi/2]$ such that $f(x) > 0$ and $f(x) + \int_{0}^{x} f(t) \sqrt{1 - (\log_e f(t))^2} dt = e, \forall x \in [0, \pi/2]$. Then $(6 \log_e f(\pi/6))^2$ is equal to ______.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler on the path to JEE excellence. Today, we are going to dissect a problem that at first glance looks like a formidable wall of symbols.

We have a function $f(x)$ trapped inside an integral, and we are asked to find a specific value related to its behavior at $x = \pi/6$. Do not be intimidated.

In the world of advanced calculus, every integral equation is just a differential equation in disguise, waiting for the right key to unlock it.

### The Power of Differentiation

We start with the given equation:

$$f(x) + \int_{0}^{x} f(t) \sqrt{1 - (\ln f(t))^2} dt = e$$

Imagine you are standing before this equation. The integral is the obstacle. How do we remove it?

We use the most powerful tool in our arsenal: the Leibniz Rule. By differentiating both sides with respect to $x$, we invoke the Fundamental Theorem of Calculus.

The derivative of the integral $\int_{0}^{x} g(t) dt$ is simply $g(x)$. Thus, differentiating the entire equation yields:

$$f'(x) + f(x) \sqrt{1 - (\ln f(x))^2} = 0$$

Look at that! The integral has vanished, and we are left with a beautiful, clean differential equation. We have successfully translated the problem from the language of accumulation to the language of rates of change.

### Separation and Substitution

Now, let us treat $f(x)$ as $y$. Our equation becomes $\frac{dy}{dx} = -y \sqrt{1 - (\ln y)^2}$. This is a separable differential equation.

We want all the $y$ terms on one side and the $x$ terms on the other:

$$\frac{dy}{y \sqrt{1 - (\ln y)^2}} = -dx$$

This looks like a standard integral, but the $\ln y$ inside the square root is a slight nuisance. Let us perform a substitution. Let $t = \ln y$. Then, $dt = \frac{1}{y} dy$.

Suddenly, the complexity collapses. The left side transforms into the classic integral of $\frac{1}{\sqrt{1-t^2}}$, which is the derivative of $\sin^{-1}(t)$. We are left with:

$$\sin^{-1}(t) = -x + C$$

Substituting back $t = \ln f(x)$, we find the general solution: $\sin^{-1}(\ln f(x)) = -x + C$.

### The Boundary Condition

We are almost there, but we have an unknown constant $C$. To find it, we return to our original equation.

By setting $x = 0$, the integral term becomes zero, leaving us with $f(0) = e$. Plugging this into our general solution:

$$\sin^{-1}(\ln f(0)) = -0 + C \Rightarrow \sin^{-1}(\ln e) = C \Rightarrow \sin^{-1}(1) = C$$

Since $\sin^{-1}(1) = \pi/2$, we have found our constant: $C = \pi/2$. Our specific solution is now fully defined:

$$\sin^{-1}(\ln f(x)) = \frac{\pi}{2} - x$$

### The Final Victory

We need to evaluate this at $x = \pi/6$. Substituting the value:

$$\sin^{-1}(\ln f(\pi/6)) = \frac{\pi}{2} - \frac{\pi}{6} = \frac{\pi}{3}$$

Taking the sine of both sides, we get $\ln f(\pi/6) = \sin(\pi/3) = \frac{\sqrt{3}}{2}$. The final step is to calculate $(6 \ln f(\pi/6))^2$.

Substituting our value:

$$(6 \cdot \frac{\sqrt{3}}{2})^2 = (3\sqrt{3})^2 = 9 \cdot 3 = 27$$

And there it is. The complexity dissolves into a simple, elegant integer. You have navigated the integral, mastered the differential equation, and arrived at the truth.

The final answer is **27**.

Animated Solution for Mathematics - Differential Equations: Let $f$ be a differentiable function defined on $[0, π /2]$ such that $f (x) > 0$ and $f (x) + \int_{0}^{x} f (t) 1 - (lo g_{e} f (t))^{2} d t = e, \forall x \in [0, π /2]$ . Then $(6 lo g_{e} f (π /6))^{2}$ is equal to ______.

Enter Numerical Value:

Visualized Solution

Analyze the Integral Equation

The Leibniz Rule

Differentiating the Equation

Formulate the Differential Equation

Variable Separation

Integrating Both Sides

Solving the Integral

The General Solution

Finding the Initial Condition

Calculate Constant $C$

The Specific Solution

Evaluate at $x = \frac{π}{6}$

Final Calculation

The Sigma Insight: Variable Separable Method

Analyzing the Setup

The Power of Differentiation

Separation and Substitution

The Boundary Condition

The Final Victory

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