海边直播目标2017全国初中数学竞赛班课堂测试题解答-The Final

 

1. 设函数 $f(x) = 2^x(ax^2 + bx + c)$ 满足等式 $f(x+1) - f(x) = 2^x\cdot x^2$, 求 $f(1)$.

解答:

由 $f(x) = 2^x(ax^2 + bx + c)$, 可以得出 $$f(x+1) = 2^{x+1}[a(x+1)^2 + b(x+1) + c]= 2\cdot2^x[(ax^2 + bx + c) + 2ax + a+ b]= 2\cdot f(x) + 2\cdot2^x\cdot(2ax + a + b)$$ 因此$$f(x+1) - f(x) = f(x) + 2\cdot2^x(2ax + a + b) = 2^x \cdot x^2$$ 即 $$2^x\cdot(ax^2 + bx + c) + 2\cdot2^x(2ax + a + b) = 2^x \cdot x^2$$ 对比系数可得 $$\begin{cases} a = 1\\ b = -4\\ c = 6 \end{cases}$$ $\therefore f(1) = 2\cdot(a + b + c) = 6$.

 

 

2. 已知函数 $y = |x-1| + |x-3| + \sqrt{1-4x +4x^2}$ 恒表示常数, 求自变量 $x$ 的取值范围.

解答:

易知 $y = |x - 1| + |x - 3| + |2x - 1|$, 若其结果为常数, 则考虑未知项互相抵消即可(存在两种情况但仅有一种情形成立).

因此 $y = 1 - x + 3 - x + 2x - 1 = 3$, 当且仅当 $$\begin{cases}x \le 1\\ x \le 3\\ x\ge \displaystyle{1\over2} \end{cases}$$ 可以得出 $\displaystyle{1\over2} \le x \le 1$.

 

 

3. 若$F\left(\displaystyle{1-x\over 1+x}\right) = x$, 则下列等式正确的是

A. $F(-2-x) = -2 - F(x)$

B. $F(-x) = F\left(\displaystyle{1-x\over 1+x}\right)$

C. $F\left(x^{-1}\right) = F(x)$

D. $F(F(x)) = -x$

解答:

令 $t = \displaystyle{1-x \over 1+x}$, $$\Rightarrow t + tx = 1-x\Rightarrow x = {1-t\over 1+t}\Rightarrow F(t) = {1-t \over 1+t}\Rightarrow F(x) = {1-x \over 1+x}$$ 对于A: $$F(-2 - x) = {1-(-2 - x) \over 1 + (-2 - x)} = {3 + x \over -1 - x}$$ $$-2 - F(x) = -2 - {1-x \over 1 + x} = {-3 - x \over 1 + x} = F(-2 - x)$$ 对于B: $$F(-x) = {1+x \over 1-x} \ne x.$$ 对于C: $$F(x{-1}) = {1 - \displaystyle{1\over x} \over 1 + \displaystyle{1\over x}} = {x - 1 \over x+1} = -F(x).$$ 对于D: $$F(F(x)) = F\left({1 - x \over 1 + x}\right) = x \ne -x.$$

 

4. 若 $f(x+y) = f(x)\cdot f(y)$, 且 $f(1) = 1$, 求 $f(2017)$.

解答:

$$\because f(x + 1) = f(x)\cdot f(1) = f(x), \therefore f(2017) = f(2016) = \cdots = f(1) = 1.$$

 

5. 设 $x$ 是1到12的正整数, $x$ 除以3的余数是 $f(x)$, $y$ 是一位数字的正整数. $y$ 除以4的余数是 $g(y)$. 则当 $f(x) + 3g(y) = 0$ 时, 求 $x+3y$ 的最大值.

解答:

$f(x) = 0$ 时, $x = 3, 6, 9, 12$; $g(y) = 0$ 时, $y = 4, 8$. 由此可得 $$(x + 3y)_\text{max} = 12 + 24 = 36.$$

 

6. 设 $x_1, x_2, x_3, \cdots, x_9$ 均为正整数, 且 $x_1 < x_2 < \cdots < x_9$, $x_1+x_2+\cdots +x_9 = 220$, 则当 $x_1+x_2+\cdots + x_5$ 的值最大时, 求 $y = x_9-x_1$ 的最小值.

解答:

需要求出 $x_9$ 最小且 $x_1$ 最大之取值, 即使得其余7个数尽量平均.

易知9个数的平均数为 $\displaystyle{220\over9} = 24 \displaystyle{4\over9}$.

若 $x_1 \ge 21$, 则 $\displaystyle\sum_{i = 1}^{9}x_i \ge 21 + 22 + \cdots + 29 = 225$,

若 $x_9 \le 28$, 则 $\displaystyle\sum_{i = 1}^{9}x_i \le 28 + 27 + \cdots + 20 = 180$,

因此可取 $x_1 = 20, x_2 = 21, x_3 = 22, x_4 = 23, x_5 = 24, x_6 = 26, x_7 = 27, x_8 = 28, x_9 = 29$.

此时 $(x_9 - x_1)_\text{min} = 29 - 20 = 9$.

 

 

7. 设 $x, y\in\mathbf{R}$, 求 $F(x, y) = 5x^2 - 4xy + 4y^2 + 12x + 25$ 之最小值.

解答:

$$F(x, y) = 5\left(x^2 - {4\over5}xy + {12\over5}x\right) + 4y^2 + 25$$ $$= 5\left(x - {2\over5}y + {6\over5}\right)^2 - {4\over5}y^2 - {36\over5} + {24\over5}y + 4y^2 + 25$$ $$= 5\left(x - {2\over5}y + {6\over5}\right)^2 + {16\over5}y^2 + {24\over5}y + {89\over5}$$ $$= 5\left(x - {2\over5}y + {6\over5}\right)^2 + {16\over5}\left(y + {3\over4}\right)^2 + 16 \ge 16.$$ 当且仅当 $$\begin{cases}x - \displaystyle{2\over5}y + \displaystyle{6\over5} = 0\\ y + \displaystyle{3\over4} = 0\end{cases} \Rightarrow \begin{cases}x = -\displaystyle{3\over2}\\ y = -\displaystyle{3\over4} \end{cases}$$ 时上式取等号. 即 $F(x, y)$ 最小值为16.

另解:

$$F(x, y) = (x^2 - 4xy + 4y^2) + 4x^2 + 12x + 25 = (x-2y)^2 + (2x+3)^2 + 16 \ge 16.$$

 

8. 已知$2x^2 - 2xy + y^2 - 6x - 4y + 27 = 0$, 求实数$x$之最值.

解答:

视$y$为主元: $$y^2 - (2x + 4)y + 2x^2 - 6x + 27 = 0\Rightarrow \Delta = (2x +4)^2 - 4(2x^2 - 6x + 27) \ge 0 \Rightarrow x^2 - 10x + 23 \le 0\Rightarrow 5-\sqrt2 \le x \le 5+\sqrt2$$ 即 $x_\text{min} = 5-\sqrt2$, $x_\text{max} = 5+\sqrt2$.

 

 

9. 求下列函数之最值: $$y = {3x^2 + 6x + 5 \over \displaystyle{1\over2}x^2 + x + 1}.$$

解答:

判别式法求解: $$3x^2 + 6x + 5 = {1\over2}yx^2 + xy + y\Rightarrow (y-6)x^2 + (2y - 12)x + (2y - 10) = 0$$ 当 $y = 6$ 时代入可知 $x$ 无解, 因此 $y\ne6$. $$\Delta = (2y - 12)^2 - 4(y - 6)(2y - 10) \ge 10\Rightarrow y^2 - 10y + 24 \le 0\Rightarrow 4\le y \le 6$$ 即 $y\in [4, 6)$. 当 $x = -1$ 时, $y_\text{min} = 4$, 无最大值.

另解:

$$y = {6x^2 + 12x + 10\over x^2 + 2x + 2}= 6 - {2\over x^2 + 2x + 2}$$

因此 $(x+1)^2 + 1 \ge 1 \Rightarrow \displaystyle{2\over(x+1)^2 + 1} \in (0, 2] \Rightarrow y \in [4, 6)$.

 

 

10. 求函数$y = x + \sqrt{-x^2 + 10x - 21}$之最值.

解答:

$$-x^2 + 10x - 21 \ge 0\Rightarrow x^2 - 10x + 21 \le 0\Rightarrow 3 \le x \le 7\Rightarrow (y - x)^2 = -x^2 + 10x - 21\Rightarrow 2x^2 - (10 +2y)x + y^2 + 21 = 0\Rightarrow \begin{cases}\Delta = (10 + 2y)^2 - 8(y^2 + 21)\ge0\\ y \ge x \ge 3 \end{cases}\Rightarrow 3 \le y \le 5 + 2\sqrt2$$ 上式取等号条件分别为 $x = 3$ 及 $x = 5+\sqrt2$.

 

 

 

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