Let g : (0, ∞) → R be a differentiable function such that ∫ [ (x(cos x - sin x))/(e^x + 1) + (g(x)(e

Q: Wait, why do we differentiate the integral? Shouldn't we try to integrate it first?

That is a very natural instinct! However, in this problem, the integral is not a standard one that can be easily evaluated. It is an integral equation where the function $g(x)$ is hidden inside. By differentiating both sides, we use the Fundamental Theorem of Calculus to 'strip away' the integral sign, which is the standard and most efficient technique for these types of JEE problems.

Q: I got lost in the quotient rule. Is there a trick to avoid mistakes?

The quotient rule is indeed a place where many students make sign errors. The best trick is to write down your $u$ and $v$ clearly before you start. Here, $u = xg(x)$ and $v = e^x + 1$. When you differentiate $u$, remember the product rule: $\frac{d}{dx}(xg(x)) = g(x) + xg'(x)$. Take it one step at a time, and always use brackets to keep your signs organized!

Q: How did the terms cancel out so perfectly? Did I do something wrong?

Not at all! In fact, that perfect cancellation is a hallmark of a well-designed JEE Advanced problem. It is a sign that you are on the right track. When you see complex terms disappearing, it means the structure of the problem was intentionally crafted to simplify. Trust the algebra, and keep going!

Question

Let $g : (0, \infty) 	o \mathbb{R}$ be a differentiable function such that $\int \left( \frac{x(\cos x - \sin x)}{e^x + 1} + \frac{g(x)(e^x + 1 - xe^x)}{(e^x + 1)^2} ight) dx = \frac{xg(x)}{e^x + 1} + c$, for all $x > 0$, where $c$ is an arbitrary constant. Then

Options:

(A) $g$ is decreasing in $(0, \frac{\pi}{4})$

(B) $g'$ is increasing in $(0, \frac{\pi}{4})$

(C) $g + g'$ is increasing in $(0, \frac{\pi}{2})$

(D) $g - g'$ is increasing in $(0, \frac{\pi}{2})$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a massive, intimidating integral equation. It looks like a tangled web of functions, exponentials, and trigonometric terms.

In the world of JEE Advanced, complexity is often just a mask for elegance. Our mission today is to peel back that mask and reveal the simple, beautiful function $g(x)$ hiding underneath.

### The Key to the Kingdom

The first step is to recognize the structure. We are given an equation of the form $\int f(x) dx = F(x) + c$.

When you see an integral on one side and a function on the other, your first instinct should be to use the **Fundamental Theorem of Calculus**. By differentiating both sides with respect to $x$, we can strip away the integral sign.

The left-hand side, which looked so terrifying, simply becomes the integrand:

$$ \frac{x(\cos x - \sin x)}{e^x + 1} + \frac{g(x)(e^x + 1 - xe^x)}{(e^x + 1)^2} $$

This is the power of calculus—it allows us to undo the complexity.

### The Quotient Rule Battle

Now, we turn our attention to the right-hand side: $\frac{xg(x)}{e^x + 1}$. This is a classic quotient rule scenario.

We define $u = xg(x)$ and $v = e^x + 1$. The quotient rule tells us that the derivative is $\frac{v u' - u v'}{v^2}$.

As we differentiate $u = xg(x)$, we must apply the product rule, giving us $g(x) + xg'(x)$. The derivative of the denominator $v = e^x + 1$ is simply $e^x$.

Putting it all together, we get a long expression, but do not let the length discourage you. This is where the magic happens.

### The Great Cancellation

When we equate the differentiated left-hand side and the right-hand side, we notice they share the same denominator, $(e^x + 1)^2$. By multiplying the left-hand side to match this denominator, we can compare the numerators directly.

As we expand and simplify, we see terms like $g(x)(e^x + 1)$ and $-xg(x)e^x$ appearing on both sides. They cancel out!

It is like watching a complex puzzle piece click into place. We are left with:

$$ x(\cos x - \sin x)(e^x + 1) = xg'(x)(e^x + 1) $$

Since $x > 0$, we can divide by $x(e^x + 1)$ to find:

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

### The Final Stretch

With $g'(x)$ in hand, finding $g(x)$ is a simple matter of integration. We integrate $\cos x - \sin x$ to get:

$$ g(x) = \sin x + \cos x + C $$

Finally, we check the options. By defining $\phi(x) = g(x) - g'(x)$, we find $\phi(x) = 2\sin x + C$.

Its derivative, $\phi'(x) = 2\cos x$, is positive in the interval $(0, \frac{\pi}{2})$. This confirms that $g(x) - g'(x)$ is strictly increasing.

You have navigated the complexity and emerged victorious. Remember, the path to the answer is often hidden in the steps you take to simplify the problem. Keep practicing, and you will find that even the most intimidating equations have a soul of pure logic.

Select Answer:

Visualized Solution

Analyze the Integral Equation

Differentiating the LHS

Differentiating the RHS using Quotient Rule

Expanding the RHS Numerator

Equating LHS and RHS

Simplifying the Numerators

Isolating $g^{'} (x)$

Finding $g (x)$ by Integration

Analyzing $g (x) - g^{'} (x)$

Checking Monotonicity

Final Conclusion

The Sigma Insight: Monotonicity

Analyzing the Setup

The Key to the Kingdom

The Quotient Rule Battle

The Great Cancellation

The Final Stretch

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