Let's determine the values of $a$ and $b$ such that the curve of the function

$$
y=a x^{4}+b x^{3}+1
$$

touches the x-axis at the point $x=1$.
I. Solution: The fact that the curve $y=a x^{4}+b x^{3}+1$ touches the x-axis at $x=1$ means that the equation

$$
a x^{4}+b x^{3}+1=0
$$

has $x=1$ as at least a double root, which means the polynomial is divisible by $(x-1)^{2}$ without a remainder.

Performing the division, we get the quotient $a x^{2}+(2 a+b) x+(3 a+2 b)$, and the remainder is

$$
(4 a+3 b) x-(3 a+3 b-1)
$$

Thus, it must be that

$$
4 a+3 b=0, \quad \text { and } \quad 3 a+2 b-1=0
$$

Solving this system of equations yields $a=3, b=-4$.

Therefore, the function that satisfies the condition is

$$
y=3 x^{4}-4 x^{3}+1
$$

Kovács László (Debrecen, Ref. g. IV. o. t.)

II. Solution: Let the roots of the equation $a x^{4}+b x^{3}+c x^{2}+d x+e=0$, or

$$
x^{4}+\frac{b}{a} x+\frac{c}{a} x+\frac{d}{a} x+\frac{e}{a}=0
$$

be $x_{1}, x_{2}, x_{3}$, and $x_{4}$. The factorized form of the equation is then

$$
\left(x-x_{1}\right)\left(x-x_{2}\right)\left(x-x_{3}\right)\left(x-x_{4}\right)=0
$$

The coefficients of (1) and (2) are equal, so

$$
\begin{gathered}
x_{1}+x_{2}+x_{3}+x_{4}=-\frac{b}{a} \\
x_{1} x_{2}+x_{1} x_{3}+x_{1} x_{4}+x_{2} x_{3}+x_{2} x_{4}+x_{3} x_{4}=\frac{c}{a} \\
x_{1} x_{2} x_{3}+x_{1} x_{2} x_{4}+x_{1} x_{3} x_{4}+x_{2} x_{3} x_{4}=-\frac{d}{a} \\
x_{1} x_{2} x_{3} x_{4}=\frac{e}{a}
\end{gathered}
$$

In this case, $c=d=0, e=1, x_{1}=x_{2}=1$, and thus

$$
\begin{gathered}
2+x_{3}+x_{4}=-\frac{b}{a} \\
1+2\left(x_{3}+x_{4}\right)+x_{3} x_{4}=0 \\
x_{3}+x_{4}+2 x_{3} x_{4}=0 \\
x_{3} x_{4}=\frac{1}{a}
\end{gathered}
$$

Solving this system of equations yields $a=3, b=-4$

Quittner Pál (Bp. I., Petófi g. III. o. t.)

III. Solution: Based on the above, the fourth-degree polynomial $a x^{4}+b x^{3}+1$ can be factored into the product of two quadratic polynomials, one of which is $(x-1)^{2}=x^{2}-2 x+1$, and the other has the first and third coefficients $a$ and 1, respectively, and is thus of the form $a x^{2}+c x+1$. Therefore, the coefficients $a, b$, and $c$ must be determined such that the following identity holds:

$$
\left(x^{2}-2 x+1\right)\left(a x^{2}+c x+1\right)=a x^{4}+b x^{3}+1
$$

The necessary and sufficient condition for this is that the coefficients of equal powers are equal. Expanding the left side,

$$
a x^{4}+(c-2 a) x^{3}+(a-2 c+1) x^{2}+(c-2) x+1=a x^{4}=b x^{3}+1
$$

Thus,

$$
c-2 a=b, \quad a-2 c+1=0, \quad c-2=0
$$

From which we get

$$
c=2, \quad a=2 c-1=3, \quad b=2-6=-4
$$

IV. Solution: Ignoring higher-order tangency, according to the problem, the function $y=a x^{4}+b x^{3}+1$ has an extremum at $x=1$, which is equal to 0:

$$
a+b+1=0, \quad b=-a-1
$$

Then the function $y=a x^{4}+b x^{3}=x^{3}(a x+b)=x^{3}(a x-a-1)$ also has an extremum at $x=1$, and thus the function obtained by multiplying this by the constant $\left(-\frac{a^{3}}{27}\right)$

$$
y-\frac{a x}{3} \cdot \frac{a x}{3} \cdot \frac{a x}{3}(1+a-a x)
$$

also has an extremum.

Here, the factors on the right side are positive near $x=1$ if $a>0$, and their sum is constant: $(a+1)$, so the product has an extremum when the factors are equal, i.e., when

$$
\frac{a x}{3}=1+a-a x
$$

Since the extremum is at $x=1$, $\frac{a}{3}=1$, or $a=3$, and thus $b=-a-1=-4$.

Csiszár Imre (Bp. I., Petőfi g. II. o. t.)[^0]


[^0]:    ${ }^{1}$ Indeed, by the arithmetic and geometric mean inequality, if $s, t, u, v$ are positive, $\sqrt[4]{s t u v}=\sqrt{\sqrt{s t} \sqrt{u v}} \leqq \frac{\sqrt{s t}+\sqrt{u v}}{2} \leqq \frac{\frac{s+t}{2}+\frac{u+v}{2}}{2}=$ $\frac{s+t+u+v}{4}$ and equality holds only if $s=t=u=v$.