Let the function f : R → R be defined by f(x) = x³ – x² + (x – 1) sin x and let g : R → R be an arbi

Q: Wait, why do we need $f(1) = 0$ to make this work? Can't we just use the product rule?

The product rule $h'(x) = f'(x)g(x) + f(x)g'(x)$ assumes that $g(x)$ is already differentiable. But here, $g(x)$ is an arbitrary function! We don't know if it's differentiable. By using the first principle of the derivative, we bypass the need for $g(x)$ to be differentiable and instead find the exact condition required for the limit to exist.

Q: If $h'(1)$ exists, doesn't that automatically mean $g(x)$ must be continuous at $x=1$?

Not necessarily! If $f(1) = 0$, the product $f(x)g(x)$ can 'smooth out' a jump discontinuity in $g(x)$. If $f(x)$ approaches zero fast enough, it can force the product to be differentiable even if $g(x)$ itself is not continuous. That is the beauty of this problem—it challenges your intuition about how functions interact.

Q: Why is $(1 + \sin 1)$ important? Could it be zero?

Great observation! If $(1 + \sin 1)$ were zero, the limit would exist regardless of whether $g(x)$ is continuous or not. However, since $1 + \sin 1 \neq 0$, the existence of the limit $h'(1)$ is strictly tied to the existence of the limit of $g(x)$ as $x$ approaches 1.

Q: How do I know if $g(x)$ needs to be differentiable for $fg$ to be differentiable?

You don't! The problem shows that $g(x)$ only needs to have a finite limit at $x=1$ for the product to be differentiable. Differentiability is a much stronger condition than having a limit. Therefore, $g(x)$ doesn't need to be differentiable at all.

Q: Is it possible for $fg$ to be differentiable even if $g$ is not defined at $x=1$?

Yes! The definition of a limit $\lim_{x \to 1} g(x)$ does not require $g(1)$ to be defined. As long as the limit exists and is finite, the product function $f(x)g(x)$ will be differentiable at $x=1$, even if $g(1)$ is undefined or if $g(x)$ has a removable discontinuity.

Question

Let the function $f: \mathbb{R} 	o \mathbb{R}$ be defined by $f(x) = x^3 - x^2 + (x - 1)\sin x$ and let $g: \mathbb{R} 	o \mathbb{R}$ be an arbitrary function. Let $fg: \mathbb{R} 	o \mathbb{R}$ be the product function defined by $(fg)(x) = f(x)g(x)$. Then which of the following statements is/are TRUE?

Options:

(A) If $g$ is continuous at $x = 1$, then $fg$ is differentiable at $x = 1$

(B) If $fg$ is differentiable at $x = 1$, then $g$ is continuous at $x = 1$

(C) If $g$ is differentiable at $x = 1$, then $fg$ is differentiable at $x = 1$

(D) If $fg$ is differentiable at $x = 1$, then $g$ is differentiable at $x = 1$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler on the path to JEE mastery. Today, we are going to dissect a problem that seems simple on the surface but hides a profound truth about the nature of differentiability.

We are looking at the function $f(x) = x^3 - x^2 + (x - 1)\sin x$ and its interaction with an arbitrary function $g(x)$. When we multiply them to form $h(x) = f(x)g(x)$, what determines if this new creature is differentiable at $x = 1$?

Let us peel back the layers.

### The Power of Factoring

Before we dive into the calculus, let us look at the anatomy of $f(x)$. It looks like a mess of polynomials and trigonometry, but look closer at the terms: $x^3 - x^2$.

We can factor out $x^2$ to get $x^2(x - 1)$. Now, look at the entire expression: $f(x) = x^2(x - 1) + (x - 1)\sin x$.

We can pull out the common factor $(x - 1)$ to reveal the soul of the function:

$$f(x) = (x - 1)(x^2 + \sin x)$$

This is a beautiful moment. By factoring, we have explicitly shown that $f(1) = 0$. This zero is not just a number; it is a 'gatekeeper' that will control the behavior of our product function $h(x)$ at the critical point $x = 1$.

### The First Principle of Calculus

When we are unsure about the differentiability of a function at a point, we must return to the source: the first principle of the derivative. We want to know if $h'(1)$ exists, defined as:

$$h'(1) = \lim_{x 	o 1} \frac{h(x) - h(1)}{x - 1}$$

Since $h(x) = f(x)g(x)$ and we know $f(1) = 0$, it follows that $h(1) = f(1)g(1) = 0 \cdot g(1) = 0$. Our limit simplifies beautifully to:

$$h'(1) = \lim_{x 	o 1} \frac{f(x)g(x)}{x - 1}$$

### The Great Cancellation

Now, we substitute our factored form of $f(x)$ into this limit. Watch what happens:

$$h'(1) = \lim_{x 	o 1} \frac{(x - 1)(x^2 + \sin x)g(x)}{x - 1}$$

As $x$ approaches 1, $x$ is never exactly 1. This gives us the mathematical permission to cancel the $(x - 1)$ terms. We are left with:

$$h'(1) = \lim_{x 	o 1} (x^2 + \sin x)g(x)$$

This is the heart of the problem. As $x 	o 1$, the term $(x^2 + \sin x)$ approaches the constant value $(1 + \sin 1)$. Since this constant is non-zero, the existence of the limit $h'(1)$ depends entirely on the existence of the limit of $g(x)$ as $x 	o 1$.

### The Verdict: What Does This Mean?

If $g(x)$ is continuous at $x = 1$, then $\lim_{x 	o 1} g(x) = g(1)$, which is a finite value. Thus, $h'(1)$ exists. This confirms that if $g$ is continuous, $fg$ is differentiable.

Similarly, if $g$ is differentiable, it is inherently continuous, so $fg$ must also be differentiable.

However, consider the reverse. If $fg$ is differentiable, we only know that $\lim_{x 	o 1} g(x)$ exists. This does not force $g(x)$ to be continuous (it could have a removable discontinuity) nor does it force $g(x)$ to be differentiable.

We have uncovered the truth: the product function is much more 'forgiving' than the individual functions. Keep this insight in your toolkit—it is exactly the kind of conceptual depth that separates the top rankers from the rest.

Select Answer:

Visualized Solution

Defining the Product Function

Factoring $f (x)$

Evaluating $f (1)$

First Principle of Derivative

Substituting $h (x)$

Applying the Factored $f (x)$

Simplifying the Limit

Evaluating the Limit

Condition for Differentiability

Checking Option A

Checking Option C

Checking Options B and D

The Sigma Insight: Relationship Between Continuity and Differentiability

Analyzing the Setup

The Power of Factoring

The First Principle of Calculus

The Great Cancellation

The Verdict

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