2. The graph of the linear function $y=k x+k+1(k>0)$ intersects the $O x$ axis at point $A$, and the $O y$ axis at point $B$ (see the figure). Find the smallest possible value of the area of triangle $A B O$.
Answer: 2.

Solution. The abscissa of point $A$ of intersection with the $O x$ axis: $0=k x+k+1 ; x=$

![](https://cdn.mathpix.com/cropped/2024_05_06_74e32b870097d181f24bg-1.jpg?height=271&width=231&top_left_y=1966&top_left_x=1729)
$-\left(1+\frac{1}{k}\right)$. The ordinate of point $B$ of intersection with the $O y$ axis: $y=k \cdot 0+k+1 ; y=k+1$. Therefore, $S_{A B O}=\frac{1}{2} \cdot O A \cdot O B=\frac{1}{2}(k+1)\left(1+\frac{1}{k}\right)=\frac{1}{2}\left(2+k+\frac{1}{k}\right)$. By the inequality between the arithmetic mean and the geometric mean, $k+\frac{1}{k} \geq 2$, and equality is achieved when $k=1$, so the minimum value of $S_{A B O}$ is also achieved when $k=1$. Thus, the smallest possible area of triangle $A B O$ is 2.

Comment. The correct answer is found based on the consideration of an example, and the minimality is not proven - 1 point. The expression for the area of the triangle in terms of $k$ is found - 5 points. If the logic is correct but errors in transformations prevent obtaining the correct answer - 4 points. The statement that the minimum value of the expression $k+\frac{1}{k}$ is achieved when $k=1$ can be used without proof.