腾讯课堂目标2017高中数学联赛基础班-2作业题解答-8

 

课程链接：目标2017高中数学联赛基础班-2（赵胤授课）

 

1. 当 $x$ 为何值时, $x+5$, $x + 2$, $x - 1$, $x - 4$ 的积不大于 $-80$?

解答: $$(x + 5)(x + 2)(x - 1)(x - 4) \le-80 \Rightarrow (x^2 + x - 20)(x^2 + x - 2) + 80 \le 0$$ $$\Rightarrow (x^2 + x)^2 - 22(x^2 + x) + 120 \le 0 \Rightarrow (x^2 + x - 10)(x^2 + x - 12) \le 0$$ $$\Rightarrow \left(x - {-1 + \sqrt{41} \over 2}\right)\left(x - {-1 - \sqrt{41} \over 2}\right)(x + 4)(x - 3) \le 0$$ $$\Rightarrow x \in \left[-4, {-1 - \sqrt{41} \over 2}\right] \cup \left[{-1 + \sqrt{41} \over 2}, 3\right].$$

 

2. 求 $k$ 的取值范围, 使得下列不等式恒成立 $${2x^2 + 2kx + k \over 4x^2 + 6x + 3} < 1.$$ 解答: $${2x^2 + 2kx + k \over 4x^2 + 6x + 3} < 1 \Rightarrow 2x^2 + 2kx + k < 4x^2 + 6x + 3$$ $$2x^2 + (6 - 2k)x + (3 - k) > 0 \Rightarrow \Delta = (6 - 2k)^2 - 8(3 - k) < 0$$ $$\Rightarrow 4k^2 - 16k + 12 < 0 \Rightarrow k^2 - 4k + 3 < 0$$ $$ \Rightarrow k\in (1, 3).$$

 

3. 解关于 $x$ 的不等式: $$x^{\log_ax} > {x^{9\over2} \over a^2}$$ 其中 $a > 0$, $a \ne 1$.

解答:

令 $\log_ax = t$, 则 $$x^t > {x^{9\over2} \over a^2} \Rightarrow x^{t - {9\over2}} > a^{-2}$$ $a > 1$ 时, $$\log_ax^{t - {9\over2}} > -2 \Rightarrow t\left(t - {9\over2}\right) > -2$$ $$\Rightarrow 2t^2 - 9t + 4 > 0 \Rightarrow t < {1\over2},\ t > 4$$ $$\Rightarrow \log_ax < {1\over2},\ \log_ax > 4$$ $$\Rightarrow x \in\left(0, a^{1\over2}\right)\cup\left(a^4, +\infty\right).$$ $0 < a < 1$ 时, $$\log_ax^{t - {9\over2}} < -2 \Rightarrow t\left(t - {9\over2}\right) < -2$$ $$\Rightarrow 2t^2 - 9t + 4 < 0 \Rightarrow {1\over2} < t < 4$$ $$\Rightarrow {1\over2} < \log_ax < 4$$ $$\Rightarrow x \in\left(a^4, a^{1\over2}\right).$$ 综上, 不等式的解集为 $$\begin{cases} x \in\left(0, a^{1\over2}\right)\cup\left(a^4, +\infty\right),\ (a > 1)\\ x \in\left(a^4, a^{1\over2}\right),\ (0 < a < 1)\end{cases}$$

 

4. 解不等式: $${1\over 2^x - 1} > {1 \over 1 - 2^{x - 1}}.$$ 解答:

令 $2^x = t$, 则 $${1\over t-1} > {1\over 1 - {1\over2}t} \Rightarrow {1 \over t - 1} > {2 \over 2 - t} \Rightarrow {1\over t - 1} + {2 \over t - 2} > 0$$ $$\Rightarrow {3t-4 \over (t - 1)(t - 2)} > 0 \Rightarrow (3t - 4)(t - 1)(t - 2) > 0 \Rightarrow 1 < t < {4\over3},\ t > 2$$ $$\Rightarrow 1 < 2^x < {4\over3},\ 2^x > 2 \Rightarrow x\in\left(0, 2 - \log_23\right) \cup (1, +\infty).$$

 

5. 设 $a$ 是正常数, 解不等式: $$2^{3x} - 2^x < a(2^x - 2^{-x}).$$ 解答:

令 $2^x = t$, 则 $$t^3 - t < a\left(t - {1\over t}\right) \Rightarrow t^4 - t^2 < a(t^2 - 1) \Rightarrow (t^2 - a)(t - 1)(t + 1) < 0$$ $a = 1$ 时, $$(t+1)^2(t-1)^2 < 0 \Rightarrow t\in\phi.$$ $a > 1$ 时, $$(t + \sqrt{a})(t - \sqrt{a})(t + 1)(t - 1) < 0 \Rightarrow -\sqrt{a} < t < -1,\ 1 < t < \sqrt{a}$$ $$1 < 2^x < \sqrt{a} \Rightarrow 0 < x < {1\over2}\log_2a.$$ $0 < a < 1$ 时, $$(t + \sqrt{a})(t - \sqrt{a})(t + 1)(t - 1) < 0 \Rightarrow -1 < t < -\sqrt{a},\ \sqrt{a} < t < 1$$ $$\sqrt{a} < 2^x < 1 \Rightarrow {1\over2}\log_2a < x < 0.$$ 综上, $$\begin{cases}x\in\phi,\ (a = 1)\\ 0 < x < {1\over2}\log_2a,\ (a > 1)\\ {1\over2}\log_2a < x < 0,\ (0 < a < 1)\end{cases}$$

 

6. 在直角坐标平面上求满足不等式 $$2 - x^2 - y^2 - \sqrt{(1 - x^2)^2 + (1 + y^2)^2} > 0$$ 的点集 $(x, y)$.

解答: $$\left(1 - x^2\right) + \left(1 - y^2\right) > \sqrt{(1 - x^2)^2 + (1 + y^2)^2} \ge 0$$ $$\Rightarrow \left[\left(1 - x^2\right) + \left(1 - y^2\right)\right]^2 > (1 - x^2)^2 + (1 + y^2)^2$$ $$\Rightarrow \begin{cases}\left(1 - x^2\right) + \left(1 - y^2\right) > 0\\ \left(1 - x^2\right)\left(1 - y^2\right) > 0 \end{cases}$$ $$\Rightarrow \begin{cases}1 - x^2 > 0\\ 1 - y^2 > 0 \end{cases} \Rightarrow \begin{cases}|x| < 1\\ |y| < 1 \end{cases}$$ 即满足题意的点集 $(x, y)$ 是位于正方形 $-1 < x < 1$, $-1 < y < 1$ 之内点.

 

 

7. 解不等式: $$10^{\lg^2x} + x^{\lg{1\over x}} + {1\over \sqrt{y - y^2}} \le 2\sqrt2(\sin\pi z + \cos\pi z).$$ 解答:

与例题类似处理方法. $$\begin{cases}10^{\lg^2x} + x^{\lg{1\over x}} = x^{\lg x} + x^{-\lg x} \ge 2\\ y-y^2 = y(1 - y) \le \left({y + 1-y \over 2}\right)^2 = {1\over 4}\Rightarrow {1\over\sqrt{y - y^2}} \ge 2\\ 2\sqrt2(\sin\pi z + \cos\pi z) \le 2\sqrt2 \cdot \sqrt2 = 4\end{cases}$$ 因此上述不等式均取等号. $$\begin{cases}x = 1\\ y = {1\over2}\\ z = 2k + {1\over4},\ (k\in\mathbf{Z})\end{cases}$$

 

8. 解关于 $x$ 的不等式: $$\sqrt{a^2 - 2x^2} > x + a.$$ 解答:

分类讨论.

$x + a < 0$ 时, $$\begin{cases}x + a <0\\ a^2 - 2x^2 \ge 0\Rightarrow x^2 \le {1\over2}a^2 \end{cases}$$ 若 $a = 0$, 则无解.

若 $a > 0$, 则 $$\begin{cases}x < -a\\ -{\sqrt2 \over 2}a \le x \le {\sqrt2\over2}a\end{cases} \Rightarrow x\in\phi$$ 若 $a < 0$, 则 $$\begin{cases}x < -a\\ {\sqrt2 \over 2}a \le x \le -{\sqrt2\over2}a\end{cases} \Rightarrow x\in\left[{\sqrt2\over2}a, -{\sqrt2\over2}a\right]$$ $x + a \ge 0$ 时, $$\begin{cases}x + a \ge 0\\ a^2 - 2x^2 \ge 0\\ a^2 - 2x^2 > (x + a)^2 \end{cases} \Rightarrow \begin{cases}x \ge -a\\ x^2 \le {1\over2}a^2\\ 3x^2+2ax < 0 \Rightarrow x(3x + 2a) < 0\end{cases}$$ 若 $a = 0$, 则无解.

若 $a > 0$, 则 $$\begin{cases}x \ge -a\\ -{\sqrt2 \over 2}a \le x \le {\sqrt2\over2}a\\ -{2\over3}a < x < 0\end{cases} \Rightarrow x \in \left(-{2\over3}a, 0\right)$$ 若 $a < 0$, 则 $$\begin{cases}x \ge -a\\ {\sqrt2 \over 2}a \le x \le -{\sqrt2\over2}a\\ 0 < x < -{2\over3}a\end{cases} \Rightarrow x \in \phi$$ 综上, $$\begin{cases} x \in \left(-{2\over3}a, 0\right),\ (a > 0)\\ x\in\left[{\sqrt2\over2}a, -{\sqrt2\over2}a\right],\ (a < 0)\\ x\in\phi,\ (a = 0)\end{cases}$$