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<h1>201710月考复习 高一</h1>
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<span class="date">2017/10/05</span>
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<h5 id="toc_0">一、选择题</h5>
<ol>
<li><p>若集合 \(A = \{ x | x^2 - 1 < 0 \}\), \(B = \{ x | 0 < x < 4 \}\), 则 \(A \cup B = \) <code>( )</code></p>
<p>A. \(\{x| 0<x<1 \}\)<br/>
B. \(\{x| -1<x<1 \}\)<br/>
C. \(\{x| -1<x<4 \}\)<br/>
D. \(\{x| 1<x<4 \}\)</p></li>
<li><p>设集合 \(A = \{ x | 0 \le x \le 3 \}\), \(B = \{ y| y \le 0 \le 1\}\), 则下列对应法则中, 是从 \(A\) 到 \(B\) 的映射的是 <code>( )</code></p>
<p>A. \(x \to y = \sqrt{x}\)<br/>
B. \(x \to y^2 = \dfrac12 x\)<br/>
C. \(x \to y = \dfrac13 x\) <br/>
D. \(x \to y = \dfrac1{2^x - 1}\)</p></li>
<li><p>已知函数 \(f(x) = \begin{cases} \sqrt{x} , & x \ge 2 \\ 3-x ,& x < 2 \end{cases}\), 则 \(f(f(-1)) = \) <code>( )</code></p>
<p>A. \(-1\)<br/>
B. \(0\)<br/>
C. \(1\)<br/>
D. \(2\)</p></li>
<li><p>设 \(A = \{ x \in \mathbb Z | |x| \le 2 \}\), \(B = \{ y | y = x^2 + 1, x \in A \}\), 则 \(B\) 中的元素个数是 <code>( )</code></p>
<p>A. \(5\)<br/>
B. \(4\)<br/>
C. \(3\)<br/>
D. 无数个</p></li>
<li><p>已知 \(P = \{a,b\}\), \(Q = \{ -1,0,1,2 \}\), \(f\) 是从 \(P\) 到 \(Q\) 的映射, 则满足 \(f(a) = 0\) 的映射个数为 <code>( )</code></p>
<p>A. \(3\)<br/>
B. \(4\)<br/>
C. \(6\)<br/>
D. \(8\)</p></li>
<li><p>若函数 \(f(x) = \dfrac{x-1}x\), 则方程 \(f(4x) = x\) 的根是 <code>( )</code></p>
<p>A. \(\dfrac12\)<br/>
B. \(-\dfrac12\)<br/>
C. \(2\)<br/>
D. \(-2\)</p></li>
<li><p>已知 \(f(x) = \begin{cases} x+2, & x \le -1 \\ x^2, & -1 < x < 2 \\ 2x, & x \ge 2 \end{cases}\), 若 \(f(x) = 3\), 则 \(x\) 的值是 <code>( )</code></p>
<p>A. \(1\)<br/>
B. \(\sqrt3\)<br/>
C. \(1\), \(\dfrac32\) 或 \(\pm \sqrt3\)<br/>
D. \(\sqrt3\) 或 \(\dfrac32\)</p></li>
<li><p>设 \(U\) 为全集, \(P\), \(Q\) 为非空集合, 且 \(P \subsetneqq Q \subsetneqq U\), 下面结论中不正确的是 <code>( )</code></p>
<p>A. \((\complement_U P) \cup Q = U\)<br/>
B. \((\complement_U Q) \cap P = \varnothing\)<br/>
C. \(P \cup Q = Q\)<br/>
D. \((\complement_U P) \cap Q = \varnothing\)</p></li>
<li><p>已知函数 \(f(x)\) 是定义在 \(\mathbb R\) 上的奇函数, 它的图像关于直线 \(x = 1\) 对称, 且 \(f(x) = x\) (\(0 < x \le 1\)), 则当 \(x \in (5,7]\) 时, \(y = f(x)\) 的解析式是 <code>( )</code></p>
<p>A. \(f(x) = 2 - x\)<br/>
B. \(f(x) = x - 4\)<br/>
C. \(f(x) = 6-x\)<br/>
D. \(f(x) = x - 8\)</p></li>
<li><p>已知函数 \(f(x)\) 是奇函数, 且在 \((0, +\infty)\) 上单调递增, 则以下结论正确的是 <code>( )</code></p>
<p>A. 函数 \(f(|x|)\) 为奇函数, 且在 \((0, +\infty)\) 上单调递增<br/>
B. 函数 \(f(|x|)\) 为偶函数, 且在 \((0, +\infty)\) 上单调递增<br/>
C. 函数 \(|f(x)|\) 为偶函数, 且在 \((-\infty, 0)\) 上单调递增<br/>
D. 函数 \(|f(x)|\) 为奇函数, 且在 \((-\infty, 0)\) 上单调递增</p></li>
</ol>
<h5 id="toc_1">二、填空题</h5>
<ol>
<li><p>已知集合 \(M = \{ (x,y) | x+y=2 \}\), \(N = \{ (x,y) | x-y=4 \}\), 则集合 \(M \cup N =\) <code>________</code>.</p></li>
<li><p>已知函数 \(f(x) = \dfrac{\sqrt{1-2x}}{x^2 - 1}\) 的定义域是 <code>________</code>.</p></li>
<li><p>已知集合 \(A = \{ x | x^2 - 2x - 3 \le 0, x \in \mathbb R \}\), \(B = \{ x | m - 2 \le x \le m + 2 \}\). 若 \(A \subseteq \complement_{\mathbb R} B\), 则实数 \(m\) 的取值的范围是 <code>________</code>.</p></li>
<li><p>已知函数 \(f(x) = -x^2 + 4x\), \(x \in [m,5]\) 的值域是 \([-5,4]\), 则实数 \(m\) 的取值范围是 <code>________</code>.</p></li>
<li><p>已知函数 \(f(x) = ax^3 + bx - 4\), 若 \(f(2) = 6\), 则 \(f(-2) = \) <code>________</code>.</p></li>
<li><p>已知函数 \(f(\sqrt{x} + 1) = x+1\), 则 \(f(2) = \) <code>________</code>.</p></li>
<li><p>已知函数 \(f(x)\) 是定义在 \(\mathbb R\) 上的奇函数, 且当 \(x > 0\) 时, \(f(x) = x-1\), 则不等式 \(f(x) \ge 0\) 的解集为 <code>________</code>.</p></li>
<li><p>某市出租汽车收费标准如下:<br/>
① \(3\) 公里以内收费 \(12\) 元;<br/>
② 超过 \(3\) 公里但不超过 \(15\) 公里, 超过的部分每公里收费 \(2.4\) 元;<br/>
③ 超过 \(15\) 公里, 超过的部分每公里收费 \(3.6\) 元.<br/>
张先生某次乘坐出租车结算时, 发现出租车费恰好平均每公里 \(3\) 元, 则张先生乘坐出租车行驶的里程为 <code>________</code> 公里.</p></li>
<li><p>已知函数 \(f(x) = |x^2 - 2ax + b|\) (\(x \in \mathbb R\)). 给出下列命题:<br/>
① \(f(x)\) 是偶函数;<br/>
② 当 \(f(0) = f(2)\) 时, \(f(x)\) 的图像关于直线 \(x = 1\) 对称;<br/>
③ 若 \(a^2 - b \le 0\), 则 \(f(x)\) 在区间 \([a, +\infty)\) 上是增函数;<br/>
④ \(f(x)\) 有最小值 \(|a^2-b|\);<br/>
⑤ 若方程 \(f(x) = 3\) 恰有 \(3\) 个不相等的实数根, 则 \(a^2 = b+3\).<br/>
其中正确命题的序号是 <code>________</code>.</p></li>
<li><p>德国著名数学家狄利克雷 (Dirichlet) 在数学领域成就显著, 以其名命名的函数 \(f(x) = \begin{cases} 1, & x \in \mathbb Q \\ 0, & x \in \complement_{\mathbb R} \mathbb Q \end{cases}\) 被称为狄利克雷函数. 其中, \(\mathbb R\) 为实数集, \(\mathbb Q\) 为有理数集. 则关于函数 \(f(x)\) 的如下四个命题:<br/>
① \(f(f(x)) = 0\);<br/>
② 函数 \(f(x)\) 是偶函数;<br/>
③ 任取一个不为零的有理数 \(T\), \(f(x+T) = f(x)\) 对任意的 \(x \in \mathbb R\) 恒成立;<br/>
④ 存在三个点 \(A (x_1, f(x_1))\), \(B (x_2, f(x_2))\), \(C (x_3, f(x_3))\), 使得 \(\triangle ABC\) 为等边三角形.<br/>
其中, 真命题的序号是 <code>________</code>.</p></li>
</ol>
<h5 id="toc_2">三、解答题</h5>
<ol>
<li><p>已知集合 \(A = \{ -1, a^2 + 1, a^2 - 3 \}\), \(B = \{ a-3, a-1, a+1 \}\), \(A \cap B = \{-2\}\), 求实数 \(a\) 的值.</p></li>
<li><p>设集合 \(A = \{ x | x^2 - 8x + 15 = 0 \}\), \(B = \{ x | ax - 1 = 0\}\).<br/>
(1) 若 \(a = \dfrac15\), 判断集合 \(A\) 与 \(B\) 的关系;<br/>
(2) 若 \(A \cap B = B\), 求实数 \(a\) 组成的集合 \(C\).</p></li>
<li><p>已知函数 \(f(x) = x^2 + ax + b\) 是偶函数, 且 \(f(2) = 0\).<br/>
(1) 求实数 \(a,b\) 的值;<br/>
(2) 若 \(g(x) = f(x) - kx\) 在 \([0,3]\) 上的最小值为 \(-5\), 求实数 \(k\) 的值.</p></li>
<li><p>已知二次函数 \(y = f(x)\) 满足: \(f(0) = 1\), \(f(x+1) - f(x) = 2x\).<br/>
(1) 求 \(f(x)\) 的解析式;<br/>
(2) 当 \(f(x) \ge 3\) 时, 求实数 \(x\) 的取值范围.</p></li>
<li><p>已知函数 \(y = f(x)\) 的定义域为 \(D\), 若存在区间 \([a,b] \subseteq D\), 使得 \(\{ y | y = f(x), x \in [a,b] \} = [a,b]\), 称区间 \([a,b]\) 为函数 \(y=f(x)\) 的“和谐区间”.<br/>
(1) 请直接写出函数 \(f(x) = x^3\) 的所有和谐区间;<br/>
(2) 若 \([0,m]\) (\(m>0\)) 是函数 \(f(x) = \left| \dfrac32 x - 1 \right|\) 的一个和谐区间, 求 \(m\) 的值;<br/>
(3) 求函数 \(f(x) = x^2-2x\) 的所有和谐区间.</p></li>
<li><p>已知函数 \(f(x) = \begin{cases} |x|, & x \in P \\ -x^2+2x, & x \in M \end{cases}\), 其中 \(P,M\) 是非空数集, 且 \(P \cap M = \varnothing\), 设 \(f(P) = \{ y | y = f(x), x \in P\}\), \(f(M) = \{ y | y = f(x), x \in M\}\).<br/>
(1) 若 \(P = (-\infty,0)\), \(M = [0,4]\), 求 \(f(P) \cup f(M)\);<br/>
(2) 是否存在实数 \(a > -3\), 使得 \(P \cup M = [-3, a]\), 且 \(f(P) \cup f(M) = [-3, 2a-3]\)? 若存在, 请求出满足条件的实数 \(a\); 若不存在, 请说明理由;<br/>
(3) 若 \(P \cup M = \mathbb R\), 且 \(0 \in M, 1\in P\), \(f(x)\) 是单调递增函数, 求集合 \(P,M\).</p></li>
</ol>
<h4 id="toc_3">答案</h4>
<h5 id="toc_4">一、选择题</h5>
<table>
<thead>
<tr>
<th>1</th>
<th>2</th>
<th>3</th>
<th>4</th>
<th>5</th>
<th>6</th>
<th>7</th>
<th>8</th>
<th>9</th>
<th>10</th>
</tr>
</thead>
<tbody>
<tr>
<td>C</td>
<td>C</td>
<td>D</td>
<td>C</td>
<td>B</td>
<td>A</td>
<td>B</td>
<td>D</td>
<td>C</td>
<td>B</td>
</tr>
</tbody>
</table>
<h5 id="toc_5">二、填空题</h5>
<ol>
<li>\(\{ (3,-1) \}\).</li>
<li>\((-\infty, -1) \cup \left( -1, \dfrac12 \right]\).</li>
<li>\((-\infty, -3) \cup (5, +\infty)\).</li>
<li>\([-1,2]\).</li>
<li>\(-14\).</li>
<li>\(2\).</li>
<li>\([-1,0] \cup [1, +\infty)\).</li>
<li>\(8\) 或 \(22\).</li>
<li>③⑤.</li>
<li>②③④.</li>
</ol>
<h5 id="toc_6">三、解答题</h5>
<ol>
<li>\(a = -1\). (\(a = 1\) 时， \(A \cap B = \{ 2, -2\}\) 不合题意)</li>
<li>(1) \(A = \{3,5\}\), \(B = \{ 5\}\), \(A \supsetneqq B\).<br/>
(2) \(A \cup B = B \Rightarrow B \subset A\). \(B = \{3\}\) 时, \(a = \dfrac13\); \(B = \{5\}\) 时, \(a = \dfrac15\); \(B = \varnothing\) 时, \(a = 0\). 故所求的 \(C = \left\{ 0, \dfrac13, \dfrac15 \right\}\).</li>
<li>(1) \(a = 0, b = 4\).<br/>
(2) \(g(x) = x^2 - kx +4\) 在 \([0,3]\) 上的最小值为 \(-5\), 按对称轴位置讨论. \(k \le 0\) 时 \(g(0) = -5\) 不成立; \(k \ge 6\) 时, \(g(3) = -5\) 可得 \(k = 6\); \(0 < k < 6\) 时, \(g \left(\dfrac{k}2\right) = -5\) 可得 \(k^2 = 36\) 不成立. 故 \(k = 6\).</li>
<li>(1) 设 \(f(x) = ax^2 + bx +1\), \(f(x+1) - f(x) = 2ax + a + b = 2x \Rightarrow a = 1, b = -1\). \(f(x) = x^2 - x + 1\).<br/>
(2) \(x^2 - x + 1 \ge 3\) 解得 \(x \le -1\) 或 \(x \ge 2\).</li>
<li>(1) \([-1,0]\), \([0,1]\), \([-1,1]\).<br/>
(2) \(x \in [0,m]\) 时, \(f(x) \in [0,m]\). \(f(x)\) 能取到的最小值是 \(0\), 故其唯一的零点 \(\dfrac23 \in [0,m]\), \(m \ge \dfrac23\). \(f(x)\) 能取到的最大值是 \(m\), 最大值可能在 \(x = 0\) 或 \(x = m\) 处取得, 即有 \(f(1) = m \ge f(m)\) 及 \(f(m) = m \ge f(1)\) 这两种情况, 可得 \(m = 1\) 或 \(2\).<br/>
(3) 二次函数 \(f(x)\) 顶点为 \((1, -1)\), 设 \([a,b]\) (\(a<b\)) 是它的和谐区间.<br/>
若 \(b \le 1\), 则 \(f(a) = b, f(b) = a\), 可解得 \(a = \dfrac{1 - \sqrt5}2\), \(b = \dfrac{1+\sqrt5}2 > 1\), 这种情况不成立;<br/>
若 \(a \ge 1\), 则 \(f(a) = a, f(b) = b\), 解得 \(a = 0 < 1\), \(b = 3\), 这种情况也不成立;<br/>
若 \(a < 1 < b\), 则 \(a = f(1) = -1\), \(b = \max \{f(a), f(b)\}\), 有两种情况: \(b = f(a) \ge f(b)\) 或 \(b = f(b) \ge f(a)\). 而此时 \(f(a) = f(-1) = 3\), 两种情况得到的都是 \(b = 3\). 故所求的和谐区间只有一个, \([-1,3]\).</li>
<li>(1) \(f(P) = (0,+\infty)\), \(f(M) = [-8,1]\), \(f(P) \cup f(M) = [-8, +\infty)\).<br/>
(2) <mark>待补充</mark></li>
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