$\text{Let the function}f:R\to R\text{ be defined by}$

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

$\text{Then the number of solutions of}f\left(x\right)=0\text{ in}\mathrm{ R}\text{ is \_\_\_\_\_}$

The correct answer is 1.

Step 1: Simplify and factorize the given function
The given function is defined as:

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

By taking out the common term, we can re-write $f\left(x\right)$ as a product of two expressions:

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

Step 2: Analyze the first term
Consider the term:

$\frac{\sin x-2}{e^{\pi x}}$

1. For all real values of $x$, the range of the sine function is $-1\le \sin x\le 1$.
Subtracting 2 gives $-3\le \sin x-2\le -1$. Thus, $\sin x-2\ne 0$ for any real $x$.
2. The exponential expression $e^{\pi x}>0$ for all real $x$.
Therefore, this factor is strictly negative and never equal to zero for any real $x$.

Step 3: Analyze the rational expression term
Since the first factor is never zero, setting $f\left(x\right)=0$ requires:

$\frac{x^{2023}+2024x+2025}{x^2-x+3}=0$

First, check the denominator $x^2-x+3$:
The discriminant is $D=(-1{)}^2-4(1\left)\right(3)=1-12=-11<0$.
Since the leading coefficient is positive ($1>0$) and the discriminant is negative, $x^2-x+3>0$ for all real numbers $x$. Thus, the denominator is never zero.

Therefore, $f\left(x\right)=0$ reduces strictly to finding the number of real roots of the polynomial numerator:

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

Step 4: Find the number of real roots of $g\left(x\right)$
Let us differentiate $g\left(x\right)$ with respect to $x$:

$g^{\prime }\left(x\right)=2023x^{2022}+2024$

Since $x^{2022}\ge 0$ for all real numbers $x$, we have:

$g^{\prime }\left(x\right)\ge 2024>0$

Since $g^{\prime }\left(x\right)>0$ everywhere, $g\left(x\right)$ is a strictly increasing function on the set of real numbers.
A continuous, strictly increasing function can cross the x-axis at most once.

Furthermore, since $g\left(x\right)$ is a polynomial of odd degree (2023):

$\underset{x\to -\infty }{\lim }g\left(x\right)=-\infty \text{ and }\underset{x\to \infty }{\lim }g\left(x\right)=\infty$

By the Intermediate Value Theorem, $g\left(x\right)$ must cross the x-axis at least once. Combined with strict monotonicity, it crosses the x-axis exactly once.

Thus, $f\left(x\right)=0$ has exactly 1 solution in $R$.