gaokao 2010 Q14

gaokao · China · shanghai-arts Sequences and series, recurrence and convergence Convergence proof and limit determination

14. The three lines $l _ { 1 } : x + y - 1 = 0 , l _ { 2 } : n x + y - n = 0 , l _ { 3 } : x + n y - n = 0 \left( n \in \mathbf { N } ^ { * } , n \geq 2 \right)$ form a triangle with area denoted as $S _ { n }$ .
Then $\lim _ { n \rightarrow \infty } S _ { n } =$ $\_\_\_\_$.

II. Multiple Choice (Total Score: 20 points, 5 points each)

(5 points) Let $y = f ( x )$ be a function whose graph is a continuous curve on the interval $(0,1]$, and $0 \leqslant f ( x ) \leqslant 1$ always holds. The area $S$ enclosed by the curve $y = f ( x )$ and the lines $x = 0 , x = 1 , y = 0$ can be estimated using the Monte Carlo method. First, generate two groups of $N$ uniformly distributed random numbers on the interval $(0,1]$: $x _ { 1 } , x _ { 2 } , \ldots , x _ { N }$ and $y _ { 1 } , y _ { 2 } , \ldots , y _ { N }$, obtaining $N$ points $( x_i , y_i )$ $(i = 1,2 \ldots , N)$. Then count the number $N _ { 1 }$ of points satisfying $y _ { i } \leqslant f ( x_i )$ $(i = 1,2 \ldots , N)$. By the Monte Carlo method, the approximate value of $S$ is $\_\_\_\_$.

14. The three lines $l _ { 1 } : x + y - 1 = 0 , l _ { 2 } : n x + y - n = 0 , l _ { 3 } : x + n y - n = 0 \left( n \in \mathbf { N } ^ { * } , n \geq 2 \right)$ form a triangle with area denoted as $S _ { n }$ .

Then $\lim _ { n \rightarrow \infty } S _ { n } =$ $\_\_\_\_$.

\section*{II. Multiple Choice (Total Score: 20 points, 5 points each)}

Paper Questions

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