Let f and g be real valued functions defined on interval (-1, 1) such that g''(x) is continuous, g(0

Q: Why can't I just use L'Hopital's Rule immediately? It seems faster.

While L'Hopital's Rule is a powerful tool in your arsenal, it can be a trap here. Differentiating $g(x) \cot x$ involves the second derivative of $g(x)$ and the derivative of $\cot x$, which leads to a messy expression involving $\csc^2 x$. The algebraic approach we used is more elegant because it reveals the fundamental structure of the derivative definition, which is exactly what JEE examiners want you to see.

Q: How did you know to split the fraction like that? It felt like a leap of faith.

That is a great observation! It wasn't a leap of faith; it was a strategic move. We looked at the numerator $f'(x) - f'(0) - g'(x) \sin x$ and realized that $f'(x) - f'(0)$ is the perfect setup for the definition of the second derivative $f''(0)$. By splitting the fraction, we isolated that term, allowing the definition to emerge naturally. Always look for the 'derivative definition' structure when you see $f(x) - f(a)$ in a limit!

Q: Is Statement-2 really the 'explanation' for Statement-1? It feels like they are just two separate facts.

They are deeply connected! Think of Statement-2 as the 'key' that unlocks the door to Statement-1. In our proof, we could not have simplified the limit to $f''(0)$ without replacing $g(0)$ with $f'(0)$. Because the logic of the first statement relies entirely on the truth of the second, Statement-2 is indeed the correct explanation.

Question

Let $f$ and $g$ be real valued functions defined on interval $(-1, 1)$ such that $g''(x)$ is continuous, $g(0) 
eq 0, g'(0) = 0, g''(0) 
eq 0$, and $f(x) = g(x) \sin x$. \ STATEMENT-1: $\lim_{x 	o 0} [g(x) \cot x - g(0) \csc x] = f''(0)$ and \ STATEMENT-2: $f'(0) = g(0)$.

Options:

(A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1

(B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1

(C) Statement-1 is True, Statement-2 is False

(D) Statement-1 is False, Statement-2 is True

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler on the path to JEE Advanced mastery. Today, we are not just solving a problem; we are peeling back the layers of a mathematical onion.

We have two functions, $f(x)$ and $g(x)$, locked in a relationship defined by $f(x) = g(x) \sin x$. At first glance, this looks like a standard calculus problem, but beneath the surface lies a beautiful interplay between the definition of a derivative and the algebraic manipulation of limits.

### The Foundation (Statement-2)

Before we tackle the intimidating limit in Statement-1, let us ground ourselves with Statement-2. We are asked to verify if $f'(0) = g(0)$.

We know that $f(x) = g(x) \sin x$. To find the derivative, we must invoke the Product Rule:

$$f'(x) = \frac{d}{dx}[g(x)] \cdot \sin x + g(x) \cdot \frac{d}{dx}[\sin x]$$

This yields the expression:

$$f'(x) = g'(x) \sin x + g(x) \cos x$$

Now, we evaluate this at $x = 0$. We know that $\sin(0) = 0$ and $\cos(0) = 1$. Substituting these values, the first term, $g'(0) \sin(0)$, vanishes into thin air.

We are left with:

$$f'(0) = g'(0) \cdot 0 + g(0) \cdot 1 = g(0)$$

Just like that, Statement-2 is proven true. This equality, **$f'(0) = g(0)$**, is the bridge we will cross to solve the complex limit in Statement-1.

### The Limit Transformation (Statement-1)

Now, let us face the beast: $\lim_{x 	o 0} [g(x) \cot x - g(0) \csc x]$. The secret to conquering such expressions is to return to the basics: sines and cosines.

Let us rewrite the expression:

$$L = \lim_{x 	o 0} \left[ \frac{g(x) \cos x}{\sin x} - \frac{g(0)}{\sin x} ight] = \lim_{x 	o 0} \frac{g(x) \cos x - g(0)}{\sin x}$$

Now, look closely at the numerator. Recall our derivative expression from Phase 1: $f'(x) = g'(x) \sin x + g(x) \cos x$. If we rearrange this, we find:

$$g(x) \cos x = f'(x) - g'(x) \sin x$$

This is the "Aha!" moment. We substitute this into our limit, along with the identity $g(0) = f'(0)$:

$$L = \lim_{x 	o 0} \frac{[f'(x) - g'(x) \sin x] - f'(0)}{\sin x}$$

### The Elegant Cancellation

We are in the home stretch. Let us rearrange the terms in the numerator to group the derivative terms together:

$$L = \lim_{x 	o 0} \frac{f'(x) - f'(0) - g'(x) \sin x}{\sin x}$$

Now, we split this into two separate limits:

$$L = \lim_{x 	o 0} \frac{f'(x) - f'(0)}{\sin x} - \lim_{x 	o 0} \frac{g'(x) \sin x}{\sin x}$$

In the second term, the $\sin x$ terms cancel out beautifully, leaving us with $\lim_{x 	o 0} g'(x)$. In the first term, we multiply and divide by $x$ to reveal the definition of the second derivative:

$$L = \lim_{x 	o 0} \left[ \frac{f'(x) - f'(0)}{x} \cdot \frac{x}{\sin x} ight] - g'(0)$$

We know that $\lim_{x 	o 0} \frac{x}{\sin x} = 1$, and $\lim_{x 	o 0} \frac{f'(x) - f'(0)}{x}$ is the definition of $f''(0)$. Thus, our limit simplifies to:

$$L = f''(0) \cdot 1 - g'(0)$$

Since the problem explicitly states that $g'(0) = 0$, we are left with **$L = f''(0)$**.

### Conclusion: The Logical Symphony

We have proven that Statement-1 is true. The entire derivation relied on the identity $f'(0) = g(0)$, which we established in Statement-2.

Therefore, Statement-2 is not just a true statement; it is the logical foundation—the explanation—for Statement-1. In the world of JEE Advanced, problems like this are about recognizing the hidden architecture of mathematics. You have successfully navigated the product rule, trigonometric identities, and the definition of the derivative.

Select Answer:

Visualized Solution

Understanding the Given Conditions

Finding the First Derivative $f^{'} (x)$

Evaluating $f^{'} (0)$ and Verifying Statement-2

Setting up the Limit in Statement-1

The Logical Bridge: Using Statement-2

Algebraic Manipulation of the Limit

Evaluating the Limit via Derivative Definition

Final Conclusion and Logical Link

The Sigma Insight: Higher Order Derivatives

Analyzing the Setup

The Foundation (Statement-2)

The Limit Transformation (Statement-1)

The Elegant Cancellation

Conclusion

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