B1. Find all real numbers $x$ for which the value of the function

$$
f(x)=\sqrt{\frac{15 x^{2}+2}{3 x^{2}+5}}
$$

is an integer.

(20 points)
B1. We transform the expression under the root:

$$
f(x)=\sqrt{\frac{15 x^{2}+2}{3 x^{2}+5}}=\sqrt{5-\frac{23}{3 x^{2}+5}}
$$

Since $\frac{23}{3 x^{2}+5} \geq 0$, it follows that $\frac{15 x^{2}+2}{3 x^{2}+5} \leq 5$ and therefore $0 \leq f(x) \leq \sqrt{5}$ for all real numbers $x$. Thus, we have only three possibilities: $f(x)=0$, $f(x)=1$, or $f(x)=2$.

The equation $\sqrt{\frac{15 x^{2}+2}{3 x^{2}+5}}=0$ has no real solutions, as $15 x^{2}+2$ cannot be equal to 0 for any real number $x$. By squaring and eliminating the fraction in the equation $\sqrt{\frac{15 x^{2}+2}{3 x^{2}+5}}=1$, we get $15 x^{2}+2=3 x^{2}+5$. From this, we express $x^{2}=\frac{1}{4}$ or $x= \pm \frac{1}{2}$. Similarly, from the equation

![](https://cdn.mathpix.com/cropped/2024_06_07_cfd2cf88ee04390f062cg-16.jpg?height=117&width=1576&top_left_y=1204&top_left_x=274)
The value of the function $f(x)$ is an integer when $x= \pm \frac{1}{2}$ or $x= \pm \sqrt{6}$.
Conclusion that the radicand is always positive, since $x$ is a real number ..... 1 point
Transforming the radicand to the form $5-\frac{23}{3 x^{2}+5}$ ..... 2 points
Determining that $0 \leq f(x) \leq \sqrt{5}$ for all real numbers $x$ ..... 2 points
Conclusion that $f(x)=0$, $f(x)=1$, or $f(x)=2$ ..... 3 points
Transforming the equation $f(x)=0$ to $15 x^{2}+2=0$ ..... 1 point
Conclusion that the equation $f(x)=0$ has no real solutions ..... 1 point
Transforming the equation $f(x)=1$ to $15 x^{2}+2=3 x^{2}+5$ ..... 1 point
The equation $f(x)=1$ has solutions $x= \pm \frac{1}{2}$ ..... 2 points
Verification of solutions ..... 2 points
Transforming the equation $f(x)=2$ to $15 x^{2}+2=12 x^{2}+20$ ..... 1 point
The equation $f(x)=2$ has solutions $x= \pm \sqrt{6}$ ..... 2 points
Verification of solutions ..... 2 points

2. Alternative method. Let $f(x)=n$, where $n \geq 0$ is an integer. We square the equation to get $\frac{15 x^{2}+2}{3 x^{2}+5}=n^{2}$. Eliminating the fraction and rearranging the equation, we get $\left(15-3 n^{2}\right) x^{2}=5 n^{2}-2$. Since $15-3 n^{2} \neq 0$ for all integers $n$, we can divide the equation by $15-3 n^{2}$ to get

$$
x^{2}=\frac{5 n^{2}-2}{15-3 n^{2}}
$$

Thus, $\frac{5 n^{2}-2}{15-3 n^{2}} \geq 0$ must hold. The number $n=0$ clearly does not satisfy this condition, and for $n \geq 1$, $5 n^{2}-2>0$, so $15-3 n^{2}>0$ must also hold. From this, it follows that $n^{2}<5$ or $n \leq 2$. 

Conclusion that $15-3 n^{2}>0$ ..... 1 point
Conclusion that $n^{2}<5$ or $n \leq 2$ ..... 1 point
Checking the possibility $n=1$ ..... 1 point
Solution $x^{2}=\frac{1}{4}$ or $x= \pm \frac{1}{2}$ ..... 2 points
Verification of solutions ..... 2 points
Checking the possibility $n=2$ ..... 1 point
Solution $x= \pm \sqrt{6}$ ..... 2 points
Verification of solutions ..... 2 points