Question

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

Answer 1

Correct Answer: 1

To solve the given equation, we start with the following expression for $f(x)$:

$$ f(x) = \frac{\sin x}{e^{\pi x}} \frac{(x)^{2023} + 2024x + 2025}{x^2 - x + 3} + \frac{2}{e^{\pi x}} \frac{(x^{2023} + 2024x + 2025)}{x^2 - x + 3} = 0 $$

1. Factorizing the Expression:
We can factor out $\frac{1}{e^{\pi x}} \frac{(x^{2023} + 2024x + 2025)}{x^2 - x + 3}$:

$$ f(x) = \frac{(x^{2023} + 2024x + 2025)}{e^{\pi x}(x^2 - x + 3)} (\sin x + 2) = 0 $$

2. Conditions for Solutions:
The solutions occur when either:

• $\frac{(x^{2023} + 2024x + 2025)}{e^{\pi x}(x^2 - x + 3)} = 0$ or
• $\sin x + 2 = 0$.

3. Analyzing $\frac{(x^{2023} + 2024x + 2025)}{e^{\pi x}(x^2 - x + 3)} = 0$:
Since $e^{\pi x} > 0$ for all $x \in \mathbb{R}$ and $x^2 - x + 3 = (x - \frac{1}{2})^2 + \frac{11}{4} > 0$ for all $x \in \mathbb{R}$, we only need to consider when:

$$ x^{2023} + 2024x + 2025 = 0 $$

4. Analyzing $g(x) = x^{2023} + 2024x + 2025$:
Let $g(x) = x^{2023} + 2024x + 2025$. We find the derivative:

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

Since $g'(x) > 0$ for all $x \in \mathbb{R}$, $g(x)$ is strictly increasing.

5. Finding the Zero of $g(x)$:
We check the values of $g(x)$ at some points:

• At $x = 0$, $g(0) = 2025 > 0$.
• At $x = -1$, $g(-1) = (-1)^{2023} + 2024(-1) + 2025 = -1 - 2024 + 2025 = 0$.

Since $g(x)$ is strictly increasing, $x = -1$ is the only real solution to $g(x) = 0$.

6. Analyzing $\sin x + 2 = 0$:
We need $\sin x = -2$, but $\sin x$ always satisfies $-1 \leq \sin x \leq 1$, so $\sin x = -2$ has no solutions.

7. Conclusion:
The only solution to $f(x) = 0$ is $x = -1$, and therefore the number of solutions is 1.

Final Answer:
The final answer is $\boxed{1}$.

Answer 2

To solve the problem, analyze the given function and find the number of real solutions to \(f(x) = 0\).

Given:
\[ f(x) = \frac{\sin x}{e^{\pi x}} \cdot \frac{x^{2023} + 2024x + 2025}{x^2 - x + 3} + \frac{2}{e^{\pi x}} \cdot \frac{x^{2023} + 2024x + 2025}{x^2 - x + 3} \]

Step 1: Factor \(f(x)\)
Note that: \[ f(x) = \frac{x^{2023} + 2024x + 2025}{x^2 - x + 3} \cdot \frac{\sin x + 2}{e^{\pi x}} \] Since \(e^{\pi x} > 0\) for all real \(x\), the zeroes of \(f(x)\) depend on: \[ \frac{x^{2023} + 2024x + 2025}{x^2 - x + 3} \cdot (\sin x + 2) = 0 \]

Step 2: Analyze each factor
- Denominator \(x^2 - x + 3 > 0\) for all \(x \in \mathbb{R}\) (discriminant \(= 1 - 12 < 0\)) so no zeros come from denominator.
- \(\sin x + 2\) is never zero because \(\sin x \in [-1,1]\), so \(\sin x + 2 \in [1,3]\), always positive → no zeros here.
- Thus zeros come only from numerator polynomial: \[ x^{2023} + 2024x + 2025 = 0 \]

Step 3: Number of real roots of the polynomial \(p(x) = x^{2023} + 2024x + 2025\)
- Degree is odd (2023), so at least one real root exists.
- The function is continuous.
- Since \(x^{2023}\) dominates for large \(|x|\), behavior at infinities:
\(\lim_{x \to \infty} p(x) = +\infty\), \(\lim_{x \to -\infty} p(x) = -\infty\).
- Check \(p(0) = 2025 > 0\).
- Check \(p(-1) = (-1)^{2023} + 2024(-1) + 2025 = -1 - 2024 + 2025 = 0\). So \(x = -1\) is a root.
- Derivative \(p'(x) = 2023 x^{2022} + 2024\) which is always positive since both terms ≥ 0.
- Hence, \(p(x)\) is strictly increasing function (one-to-one).

Step 4: Conclusion on number of roots
- Since \(p(x)\) is strictly increasing and has one root at \(x = -1\), it has exactly one real root.
- Therefore, \(f(x) = 0\) has exactly one real solution.

Final Answer:
\[ \boxed{1} \]