二轮做好题目DAY3

本次选取的题目利用到的方法（技巧）:同构、双参数处理、取特值、对称的马尔科夫链

【襄阳四中高三质检(五)】 已知正实数\(x,y\)满足\(\mathrm{e}^{1-2x}=(2x+y)\mathrm{e}^y\),则\(x+\dfrac{2x^2}{y}+\dfrac{y}{x}\)的最小值为\(\underline{\qquad\qquad}.\)

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解. \(\mathrm{e}^{1-2x}=(2x+y)\mathrm{e}^y\implies \mathrm{e}=(2x+y)\mathrm{e}^{2x+y},\)

构造函数\(f(x)=x\mathrm{e}^x\),不难得到\(f(x)\nearrow\),即$$f(1)=f(2x+y)\implies 2x+y=1,$$从而整理所求式

\[\begin{align*} x+\dfrac{2x^2}{y}+\dfrac{y}{x}=&x\left(1+\dfrac{2x}{y}\right)+\dfrac{y}{x}\\ =&\dfrac{x(2x+y)}{y}+\dfrac{y}{x}\\ =&\dfrac{x}{y}+\dfrac{y}{x}\\ \geq& 2. \end{align*}\]

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【江淮十校第一次联考】 已知不等式\(x\leq a\mathrm{e}^x+b\)对任意\(x\in\mathbb{R}\)恒成立, 则\(\dfrac{b}{a}\)的最小值为\(\underline{\qquad\qquad}\).

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解. 整理为:\(a\mathrm{e}^x-x+b\geq 0\), 记\(\varphi(x)=a\mathrm{e}^x-x+b\), 若\(a\leq 0\), 不合题, \(f(x)\)会存在负区间.当\(a>0\), \(\varphi^{\prime}(x)=a\mathrm{e}^x-1\), 有\(\varphi^{\prime}(-\ln a)=0\), 从而$$\varphi(x)_{\min}=\varphi(-\ln a)=1+\ln a+b,$$ 从而
$$1+\ln a+b\geq 0$$
\(\Longrightarrow\)

\[\dfrac{b}{a}>-\dfrac{1+\ln a}{a}\geq -1. \]

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【四川玉林市高三期中考试】 \(\boldsymbol{a}, \boldsymbol{b}\) 不共线, 且满足 \(\boldsymbol{a} \cdot \boldsymbol{b} = {\boldsymbol{a}}^{2} = 4\) , 记 \(\boldsymbol{c} =\) \(\dfrac{3}{4}\boldsymbol{a} + \dfrac{1}{4}\boldsymbol{b}\) , 当 \(\boldsymbol{b}, \boldsymbol{c}\) 的夹角取得最大值时, \(\left| {\boldsymbol{a} - \boldsymbol{b}}\right|\) 的值为
\(\underline{\qquad\qquad}\).

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【法一】 记 \(\boldsymbol{b}, \boldsymbol{c}\) 的夹角为 \(\theta\). 由条件不妨设 \(\boldsymbol{a} = \left( {2, 0}\right) , \boldsymbol{b} = \left( {2, t}\right) \left( {t \geq 0}\right)\) , 则 \(\boldsymbol{c} = \dfrac{3}{4}\boldsymbol{a} + \dfrac{1}{4}\boldsymbol{b} =\)
\(\left( {2, \frac{t}{4}}\right)\) , 所以

\[\cos \theta = \frac{\boldsymbol{b} \cdot \boldsymbol{c}}{\left| \boldsymbol{b}\right| \left| \boldsymbol{c}\right| } = \frac{{16} + {t}^{2}}{\sqrt{{t}^{2} + 4}\sqrt{{t}^{2} + {64}}}, \]

所以

\[{\cos }^{2}\theta = \frac{{t}^{4} + {32}{t}^{2} + {256}}{{t}^{4} + {68}{t}^{2} + {256}} = 1 - \frac{36}{{t}^{2} + {68} + \dfrac{256}{{t}^{2}}}, \]

当且仅当 \({t}^{2} = {16}\) 时取得最小值, 此时 \(t = 4\) , 故 \(\left| {\boldsymbol{a} - \boldsymbol{b}}\right| = \left| \left( {0, - 4}\right) \right| = 4\) .

【法二】 如图, 作 \(\overrightarrow{OA} = \boldsymbol{a}, \overrightarrow{OB} = \boldsymbol{b}, \overrightarrow{OC} = \boldsymbol{c}, \boldsymbol{b}\) 与 \(c\) 的夹角为 \(\angle {BOC}\) , 不妨设 \({BC} = {3AC} = {3x}\) ,

\[\begin{align*}\tan \angle {BOC} =& \tan \left( {\angle {BOA} - \angle {COA}}\right)\\=& \dfrac{\tan \angle {BOA} - \tan \angle {COA}}{1 + \tan \angle {BOA} \cdot \tan \angle {COA}}\\=& \dfrac{\dfrac{4x}{2} - \dfrac{x}{2}}{1 + \dfrac{4x}{2} \cdot \dfrac{x}{2}}\\=& \dfrac{\dfrac{3}{2}x}{1 + {x}^{2}} \\=& \dfrac{3}{2} \cdot \dfrac{1}{\dfrac{1}{x} + x}, \end{align*}\]

当 \(x = 1\) 时, \(\tan \angle {BOC}\) 最大, 即 \(\angle {BOC}\) 最大, 此时 \(\left| {\boldsymbol{a} - \boldsymbol{b}}\right| = \left| \overrightarrow{BA}\right| = {4x} = 4\) .

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【湖北云学联盟高三下学期2月阶段训练】 在篮球训练场上,教练甲指导三名学员 \(A,B,C\) 进行传球训练,训练开始时,篮球在教练甲手中. 由甲开始传球,他每次等可能地将篮球传给学员 \(A,B,C\) 其中一人,学员接球后,将篮球传出,传给教练甲的概率为 \(\dfrac{1}{4}\) ,传给另外两学员的概率相等, 篮球在四人之间传递.设 \({P}_n\) 表示经过 \(n\) 次传球后篮球在 \(A\) 手中的概率,求 \({P}_n=\underline{\qquad\qquad}.\)

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解.
设 \({Q}_n\) 表示经过 \(n\) 次传球后篮球在教练甲手中的概率,
由题得\({Q}_{1} = 0,{Q}_{2} = \dfrac{1}{4}\) ,
且 \({Q}_n = \dfrac{1}{4}\left( {1 - {Q}_{n - 1}}\right)\) ,
即

\[{Q}_n - \dfrac{1}{5} = - \dfrac{1}{4}\left( {{Q}_{n - 1} - \dfrac{1}{5}}\right), \]

则数列 \(\left\{ {{Q}_n - \dfrac{1}{5}}\right\}\) 是首项为 \(- \dfrac{1}{5}\) ,公比为 \(- \dfrac{1}{4}\) 的等比数列,
从而得到$$ {Q}_n - \dfrac{1}{5} = - \dfrac{1}{5} \times {\left( -\dfrac{1}{4}\right) }^{n - 1},$$ 即 $${Q}_n = \dfrac{1}{5} - \dfrac{1}{5} \times {\left( -\dfrac{1}{4}\right) }^{n - 1},$$
又传给学员 \(A,B,C\) 的概率相等,所以

\[ \begin{align*}{P}_n =&\dfrac{1 - {Q}_n}{3} \\=& \dfrac{1 - \left\lbrack {\dfrac{1}{5} - \dfrac{1}{5} \times {\left( -\dfrac{1}{4}\right) }^{n - 1}}\right\rbrack }{3} \\=& \dfrac{4}{15} + \dfrac{1}{15} \times {\left( -\dfrac{1}{4}\right) }^{n - 1}. \end{align*}\]