1. 设函数 $f(x) = 2^x(ax^2 + bx + c)$ 满足等式 $f(x+1) - f(x) = 2^xcdot x^2$, 求 $f(1)$.
解答:
由 $f(x) = 2^x(ax^2 + bx + c)$, 可以得出 $$f(x+1) = 2^{x+1}[a(x+1)^2 + b(x+1) + c]= 2cdot2^x[(ax^2 + bx + c) + 2ax + a+ b]= 2cdot f(x) + 2cdot2^xcdot(2ax + a + b)$$ 因此$$f(x+1) - f(x) = f(x) + 2cdot2^x(2ax + a + b) = 2^x cdot x^2$$ 即 $$2^xcdot(ax^2 + bx + c) + 2cdot2^x(2ax + a + b) = 2^x cdot x^2$$ 对比系数可得 $$egin{cases} a = 1\ b = -4\ c = 6 end{cases}$$ $ herefore f(1) = 2cdot(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 $$egin{cases}x le 1\ x le 3\ xge displaystyle{1over2} end{cases}$$ 可以得出 $displaystyle{1over2} le x le 1$.
3. 若$Fleft(displaystyle{1-xover 1+x} ight) = x$, 则下列等式正确的是
A. $F(-2-x) = -2 - F(x)$
B. $F(-x) = Fleft(displaystyle{1-xover 1+x} ight)$
C. $Fleft(x^{-1} ight) = F(x)$
D. $F(F(x)) = -x$
解答:
令 $t = displaystyle{1-x over 1+x}$, $$Rightarrow t + tx = 1-xRightarrow x = {1-tover 1+t}Rightarrow F(t) = {1-t over 1+t}Rightarrow F(x) = {1-x over 1+x}$$ 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: $$F(-x) = {1+x over 1-x} e x.$$ 对于C: $$F(x{-1}) = {1 - displaystyle{1over x} over 1 + displaystyle{1over x}} = {x - 1 over x+1} = -F(x).$$ 对于D: $$F(F(x)) = Fleft({1 - x over 1 + x} ight) = x e -x.$$
4. 若 $f(x+y) = f(x)cdot f(y)$, 且 $f(1) = 1$, 求 $f(2017)$.
解答:
$$ecause f(x + 1) = f(x)cdot f(1) = f(x), herefore f(2017) = f(2016) = cdots = f(1) = 1.$$
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 $$(x + 3y)_ ext{max} = 12 + 24 = 36.$$
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 $displaystyle{220over9} = 24 displaystyle{4over9}$.
若 $x_1 ge 21$, 则 $displaystylesum_{i = 1}^{9}x_i ge 21 + 22 + cdots + 29 = 225$,
若 $x_9 le 28$, 则 $displaystylesum_{i = 1}^{9}x_i le 28 + 27 + cdots + 20 = 180$,
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 $(x_9 - x_1)_ ext{min} = 29 - 20 = 9$.
7. 设 $x, yinmathbf{R}$, 求 $F(x, y) = 5x^2 - 4xy + 4y^2 + 12x + 25$ 之最小值.
解答:
$$F(x, y) = 5left(x^2 - {4over5}xy + {12over5}x ight) + 4y^2 + 25$$ $$= 5left(x - {2over5}y + {6over5} ight)^2 - {4over5}y^2 - {36over5} + {24over5}y + 4y^2 + 25$$ $$= 5left(x - {2over5}y + {6over5} ight)^2 + {16over5}y^2 + {24over5}y + {89over5}$$ $$= 5left(x - {2over5}y + {6over5} ight)^2 + {16over5}left(y + {3over4} ight)^2 + 16 ge 16.$$ 当且仅当 $$egin{cases}x - displaystyle{2over5}y + displaystyle{6over5} = 0\ y + displaystyle{3over4} = 0end{cases} Rightarrow egin{cases}x = -displaystyle{3over2}\ y = -displaystyle{3over4} end{cases}$$ 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: $$y^2 - (2x + 4)y + 2x^2 - 6x + 27 = 0Rightarrow Delta = (2x +4)^2 - 4(2x^2 - 6x + 27) ge 0 Rightarrow x^2 - 10x + 23 le 0Rightarrow 5-sqrt2 le x le 5+sqrt2$$ 即 $x_ ext{min} = 5-sqrt2$, $x_ ext{max} = 5+sqrt2$.
9. 求下列函数之最值: $$y = {3x^2 + 6x + 5 over displaystyle{1over2}x^2 + x + 1}.$$
解答:
判别式法求解: $$3x^2 + 6x + 5 = {1over2}yx^2 + xy + yRightarrow (y-6)x^2 + (2y - 12)x + (2y - 10) = 0$$ 当 $y = 6$ 时代入可知 $x$ 无解, 因此 $y e6$. $$Delta = (2y - 12)^2 - 4(y - 6)(2y - 10) ge 10Rightarrow y^2 - 10y + 24 le 0Rightarrow 4le y le 6$$ 即 $yin [4, 6)$. 当 $x = -1$ 时, $y_ ext{min} = 4$, 无最大值.
另解:
$$y = {6x^2 + 12x + 10over x^2 + 2x + 2}= 6 - {2over x^2 + 2x + 2}$$
因此 $(x+1)^2 + 1 ge 1 Rightarrow displaystyle{2over(x+1)^2 + 1} in (0, 2] Rightarrow y in [4, 6)$.
10. 求函数$y = x + sqrt{-x^2 + 10x - 21}$之最值.
解答:
$$-x^2 + 10x - 21 ge 0Rightarrow x^2 - 10x + 21 le 0Rightarrow 3 le x le 7Rightarrow (y - x)^2 = -x^2 + 10x - 21Rightarrow 2x^2 - (10 +2y)x + y^2 + 21 = 0Rightarrow egin{cases}Delta = (10 + 2y)^2 - 8(y^2 + 21)ge0\ y ge x ge 3 end{cases}Rightarrow 3 le y le 5 + 2sqrt2$$ 上式取等号条件分别为 $x = 3$ 及 $x = 5+sqrt2$.
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