Let the function f: R → R be defined by f(x) = [sin x (x²⁰²³ + 2024x + 2025)] / [eˣ (x²-x+3)] + [2(x

Q: Wait, why can't we just use energy conservation or some other physics principle here? This looks like a physics problem.

I love the enthusiasm! However, this is a pure mathematical analysis problem. In JEE, sometimes the notation is designed to look like a complex physical system to test your ability to strip away the 'noise' and focus on the underlying function behavior. We aren't dealing with energy here; we are dealing with the roots of a function, which is a fundamental concept in calculus.

Q: How do I know for sure that $x^2 - x + 3$ is always positive? Can't it be zero at some point?

That is a fantastic question. To check if a quadratic $ax^2 + bx + c$ is always positive, we look at two things: the leading coefficient $a$ and the discriminant $\Delta = b^2 - 4ac$. Here, $a = 1 > 0$ (so it opens upwards) and $\Delta = (-1)^2 - 4(1)(3) = 1 - 12 = -11$. Since the discriminant is negative, the parabola never touches the x-axis. Because it opens upwards, it must stay strictly above the x-axis for all real $x$.

Q: Why does $g'(x) > 0$ mean there is only one root? Couldn't it have multiple roots if it's increasing?

Think of it this way: if a function is strictly increasing, it is constantly moving 'upwards'. If it were to cross the x-axis at two different points, say $x_1$ and $x_2$ (where $x_1 < x_2$), then by the Mean Value Theorem, there would have to be a point between them where the function turns or stays flat. But if the derivative is strictly positive, the function never turns or flattens. It can only cross the x-axis exactly once. It is a one-way street to infinity!

Question

Let the function $f: \mathbb{R} 	o \mathbb{R}$ be defined by
$f(x) = \frac{\sin x (x^{2023} + 2024x + 2025)}{e^x (x^2-x+3)} + \frac{2(x^{2023} + 2024x + 2025)}{e^x (x^2-x+3)}$.
Then the number of solutions of $f(x) = 0$ in $\mathbb{R}$ is ________.

Sigma Percentile · Accepted Answer

### The Illusion of Complexity: A Journey into Function Analysis

Welcome, future engineer. Today, we are going to dismantle a problem that, at first glance, looks like a nightmare. You see a function $f(x)$ filled with high-degree polynomials, exponential terms, and trigonometric functions.

It is designed to make you panic. But here is the secret of the JEE Advanced: the most complex-looking problems often have the most elegant, simple solutions. Let us peel back the layers of this 'monster' together.

### Phase 1: The Art of Deconstruction

Look at the function again:

$$f(x) = \frac{\sin x (x^{2023} + 2024x + 2025)}{e^x (x^2-x+3)} + \frac{2(x^{2023} + 2024x + 2025)}{e^x (x^2-x+3)}$$

When you see a sum of two fractions, your first instinct should be to look for commonality. Both terms share the exact same numerator polynomial, $(x^{2023} + 2024x + 2025)$, and the exact same denominator, $e^x (x^2-x+3)$.

This is not a coincidence; it is an invitation. Let us factor out these common terms:

$$f(x) = \frac{(\sin x + 2)(x^{2023} + 2024x + 2025)}{e^x (x^2-x+3)}$$

Suddenly, the 'monster' has been tamed. We have a product of three distinct parts. For $f(x)$ to be zero, the numerator must be zero, and the denominator must be non-zero.

### Phase 2: The 'Safe' Zones

We need to determine if any of these factors can ever be zero. Let us start with the trigonometric part: $(\sin x + 2)$.

We know that for any real number $x$, the value of $\sin x$ is trapped between $-1$ and $1$. Therefore, $\sin x + 2$ must be between $1$ and $3$. It is strictly positive and can never be zero.

Now, look at the denominator: $e^x (x^2 - x + 3)$. The exponential term $e^x$ is always positive for all real $x$.

What about the quadratic $x^2 - x + 3$? The discriminant is calculated as:

$$\Delta = (-1)^2 - 4(1)(3) = -11$$

Since the discriminant is negative and the leading coefficient is positive, this parabola never touches the x-axis. It is always positive.

This is a massive realization. Since the denominator and the $(\sin x + 2)$ factor are always positive, they cannot contribute to making $f(x) = 0$. They are essentially 'noise'.

### Phase 3: The Polynomial Battle

We are left with the core equation:

$$g(x) = x^{2023} + 2024x + 2025 = 0$$

We need to find the number of real roots for this polynomial. Let us find the derivative, $g'(x)$:

$$g'(x) = 2023x^{2022} + 2024$$

Look closely at this derivative. The term $x^{2022}$ has an even exponent, which means it is always greater than or equal to zero. When we multiply it by $2023$ and add $2024$, the result is strictly greater than zero for all real $x$.

This means $g(x)$ is a **strictly increasing function**.

### Phase 4: The Final Revelation

Think about what a strictly increasing function looks like. It starts from negative infinity and climbs steadily toward positive infinity.

Because it is an odd-degree polynomial (degree $2023$), it must cross the x-axis at least once. But because it is strictly increasing, it can never turn back down to cross the axis a second time. Therefore, it must cross exactly once.

We have proven there is exactly one root. Now, let us test a simple integer. If we plug in $x = -1$:

$$g(-1) = (-1)^{2023} + 2024(-1) + 2025 = -1 - 2024 + 2025 = 0$$

It works! The root is $x = -1$.

We have successfully navigated the complexity, simplified the expression, analyzed the behavior of the function, and found the unique solution. The number of solutions is **exactly 1**.

Enter Numerical Value:

Visualized Solution

Problem Setup

Factoring $f (x)$

Analyzing $sin x + 2$

Analyzing the Denominator

The Core Equation

Defining $g (x)$

Monotonicity of $g (x)$

Roots of Odd Polynomial

Finding the Root

Final Conclusion

The Sigma Insight: Monotonicity

The Illusion of Complexity

Phase 1

Phase 2

Phase 3

Phase 4

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