Let f: R - R be a continuous odd function, which vanishes exactly at one point and f(1) = 1/2. Supp

Q: Wait, why is the integral of an odd function over $[-1, 1]$ always zero? Can't it be non-zero?

That is a fantastic question! Think of the graph. Because the function is odd, the area from $-1$ to $0$ is exactly the negative of the area from $0$ to $1$. When you sum them up, they cancel out perfectly. It is not just a rule; it is a geometric necessity of symmetry.

Q: Why did we use L'Hôpital's Rule? Is there another way to solve this?

L'Hôpital's Rule is the most direct path because we have an indeterminate form $\frac{0}{0}$. While you could theoretically use Taylor series expansions, they would be overkill here. The Leibniz Rule allows us to bypass the integration entirely and look at the behavior of the functions at the point of interest.

Q: How do we know $f(1/2)$ is positive 7 and not negative 7?

We are given that $f(1) = 1/2$, which is positive. Since the function only crosses the x-axis at $x=0$ (it vanishes at exactly one point), it cannot change sign for $x > 0$. Therefore, if it is positive at $x=1$, it must be positive for all $x > 0$. Thus, $f(1/2)$ must be positive.

Question

Let $f: R 	o R$ be a continuous odd function, which vanishes exactly at one point and $f(1) = 1/2$. Suppose that $F(x) = \int_{-1}^{x} f(t) dt$ for all $x \in [-1, 2]$ and $G(x) = \int_{-1}^{x} t |f(f(t))| dt$ for all $x \in [-1, 2]$. If $\lim_{x 	o 1} \frac{F(x)}{G(x)} = \frac{1}{14}$, then the value of $f(1/2)$ is

Sigma Percentile · Accepted Answer

### The Symphony of Symmetry: Unlocking the Integral

Welcome, future engineers. Today, we are going to dissect a problem that, at first glance, looks like a terrifying labyrinth of integrals and limits. But I want you to take a deep breath.

In JEE Advanced, the most complex-looking problems are often hiding a beautiful, elegant secret. Our goal today is not just to solve for $f(1/2)$, but to understand the 'why' behind every step.

### Phase 1: The Power of the Odd Function

We start with a function $f(x)$ that is continuous and odd. Geometrically, it means the graph is perfectly symmetric about the origin. Algebraically, this gives us the identity $f(-x) = -f(x)$.

Now, consider the root. The problem tells us it vanishes at exactly one point. If we plug $x=0$ into our odd function identity, we get $f(0) = -f(0)$, which forces $f(0) = 0$.

This confirms that the origin is the only place where our function kisses the x-axis. This is our anchor.

### Phase 2: The Integral Trap

We are given two functions, $F(x) = \int_{-1}^{x} f(t) dt$ and $G(x) = \int_{-1}^{x} t |f(f(t))| dt$. We need to evaluate the limit of their ratio as $x 	o 1$. If we try to plug in $x=1$ immediately, we get $\frac{F(1)}{G(1)}$.

Let's look at $F(1) = \int_{-1}^{1} f(t) dt$. Because $f(t)$ is an odd function, the area below the x-axis from $-1$ to $0$ is the exact negative of the area above the x-axis from $0$ to $1$. They cancel out, so $F(1) = 0$.

Now, what about $G(1)$? We have the integrand $h(t) = t |f(f(t))|$. Let's test its symmetry by checking $h(-t)$:

$$h(-t) = (-t) |f(f(-t))| = (-t) |f(-f(t))| = -t |f(f(t))| = -h(t)$$

Since the integrand is an odd function, integrating over the symmetric interval $[-1, 1]$ gives us zero. Thus, $G(1) = 0$.

### Phase 3: The Calculus Rescue

We have arrived at the indeterminate form $\frac{0}{0}$. This is the moment to call upon L'Hôpital's Rule. To differentiate these integrals, we use the Leibniz Rule.

The derivative of $F(x)$ is simply $f(x)$, and the derivative of $G(x)$ is $x |f(f(x))|$. Our limit now becomes:

$$\lim_{x 	o 1} \frac{f(x)}{x |f(f(x))|} = \frac{1}{14}$$

This is where the problem simplifies beautifully. We substitute $x=1$ into the expression:

$$\frac{f(1)}{1 \cdot |f(f(1))|} = \frac{1}{14}$$

### Phase 4: The Final Reveal

We know $f(1) = 1/2$. Let's plug that in:

$$\frac{1/2}{|f(1/2)|} = \frac{1}{14}$$

Cross-multiplying, we find $|f(1/2)| = 7$. Now, we must decide: is it positive or negative?

We established earlier that the function only crosses the x-axis at $x=0$. Since $f(1) = 1/2$ (a positive value), the function must remain positive for all $x > 0$. Therefore, $f(1/2)$ cannot be negative.

It must be **$7$**.

Enter Numerical Value:

Visualized Solution

Properties of $f (x)$

The Single Root of $f (x)$

The Given Value

Evaluating $F (1)$

Integral of an Odd Function

Analyzing $G (x)$

Symmetry of $h (t) = t ∣ f (f (t)) ∣$

Simplifying $h (- t)$

Evaluating $G (1)$

The $\frac{0}{0}$ Indeterminate Form

Differentiating Integrals (Leibniz Rule)

Applying L'Hôpital's Rule

Substituting $x = 1$

Solving for $f (\frac{1}{2})$

Final Value of $f (\frac{1}{2})$

The Sigma Insight: Newton-Leibniz & Reduction Formulas

The Symphony of Symmetry

Phase 1

Phase 2

Phase 3

Phase 4

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Enter Numerical Value:

Visualized Solution

Properties of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

The Single Root of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

The Given Value

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } F(1)

Integral of an Odd Function

Analyzing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } G(x)

Symmetry of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } h(t)=t∣f(f(t))∣

Simplifying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } h(−t)

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } G(1)

The .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 00​ Indeterminate Form