20230228 night

This commit is contained in:
weiye.wang 2023-03-01 06:58:31 +08:00
parent 2e90569b6e
commit ce44476b65
4 changed files with 1139 additions and 66 deletions

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"题号: 010000 , 字段: solution 中已添加数据: (1) $S_{\\triangle ABC}=\\dfrac 12 \\cdot 2\\times 2\\times \\sin 60^\\circ=\\sqrt{3}$, 三棱锥的高$PO=2\\sin 60^\\circ = \\sqrt{3}$, 故$V_{P-ABC}=\\dfrac 13\\cdot S_{\\triangle ABC}\\cdot PO=\\dfrac 13\\times \\sqrt{3}\\times \\sqrt{3}=1$.\\\\\n", "题号: 014204 , 字段: usages 中已添加数据: 20230228\t2023届高三10班\t0.892\n",
"(2) 取$OC$中点$N$, 由于$MN\\parallel BO$, 且$BO\\perp AC$, 故$MN\\perp AC$. 又因为$PO\\perp$平面$ABC$, 故$MN\\perp PO$. 注意到$PO\\cap AC=O$, 于是$MN\\perp$平面$PAC$, 从而$PN$是$PM$在平面$PAC$上的射影. $MN=\\dfrac{\\sqrt{3}}{2}$, $PN=\\dfrac{\\sqrt{13}}{2}$, 故$\\angle MPN=\\arctan \\dfrac{MN}{PN}=\\arctan \\dfrac{\\sqrt{39}}{13}$. 综上, $PM$与平面$PAC$所成角的大小为$\\arctan \\dfrac{\\sqrt{39}}{13}$.\n", "题号: 014217 , 字段: usages 中已添加数据: 20230228\t2023届高三10班\t0.919\n",
"题号: 011301 , 字段: solution 中已添加数据: (1) 由$a^2+c^2-2ac\\cos B=b^2$, 即$c^2-5c-24=0$解得$c=8$或$-3$(负数舍去). 故$c=8$.\\\\\n", "题号: 014206 , 字段: usages 中已添加数据: 20230228\t2023届高三10班\t1.000\n",
"(2) 延长$CM$至$N$, 使得$CM=MN$, 则四边形$ACBN$是平行四边形, 故$S_{\\triangle ABC}=S_{\\triangle ACN}$. $\\triangle ACN$的三边长分别为$5,6,7$, 故$\\cos \\angle CAN=\\dfrac{5^2+7^2-6^2}{2\\times 5\\times 7}=\\dfrac{19}{35}$, 所以$\\sin \\angle CAN=\\dfrac{12\\sqrt{6}}{35}$. 由此, $S_{\\triangle ACN}=\\dfrac 12\\cdot AC\\cdot AN\\cdot \\sin \\angle CAN=6\\sqrt{6}$. 综上, 三角形$ABC$的面积为$6\\sqrt{6}$.\n", "题号: 014207 , 字段: usages 中已添加数据: 20230228\t2023届高三10班\t0.649\n",
"题号: 003691 , 字段: solution 中已添加数据: (1) 前$4$个月的累计投放量为$a_1+a_2+a_3+a_4=20+95+420+430=965$. 前$4$个月的累计损失量为$b_1+b_2+b_3+b_4=6+7+8+9=30$, 因此该地区第$4$个月底的共享单车的保有量为$965-30=935$.\\\\\n", "题号: 014208 , 字段: usages 中已添加数据: 20230228\t2023届高三10班\t0.892\n",
"(2) 考察不等式$a_n \\geq b_n$在整数范围内的解集. 当$n \\leq 3$时, $5 n^4+15 \\geq n+5$确实成立, 当$n \\geq 4$时, 由$-10 n+470 \\geq n+5$, 解得$n \\leq \\dfrac{465}{11}\\approx 42.27$, 因此第$42$个月底, 保有量达到最大. 当$n \\geq 4$时, $\\{a_n\\}$是公差为$-10$的等差数列, 而$\\{b_n\\}$是公差为$1$的等差数列, 于是到第$42$个月底, 单车保有量为$\\dfrac{38(a_5+a_{42})}{2}+965-\\dfrac{42(b_1+b_{42})}{2}=\\dfrac{38(420+50)}{2}+965-\\dfrac{42(6+47)}{2}=8782$. 而$S_{42}=-4\\times 16+8800=8736<8782$. 因此该月底单车保有量超过了容纳量.\n", "题号: 013024 , 字段: usages 中已添加数据: 20230228\t2023届高三10班\t0.892\n",
"题号: 011450 , 字段: solution 中已添加数据: (1) 设直线$l$的方程为$x=m$, 由$S_{\\triangle OPQ}=\\dfrac 12\\cdot |m|\\cdot |PQ|=\\dfrac{\\sqrt{2}}{2}$得$|PQ|=\\dfrac{\\sqrt{2}}{|m|}$, 于是点$(m,\\dfrac{\\sqrt{2}}{2|m|})$在椭圆上. 代入椭圆$C$的方程, 得$m^2+\\dfrac{2}{8m^2}=1$, 解得$m^2=\\dfrac 12$. 故直线$l$的方程为$x=\\dfrac{\\sqrt{2}}2$或$x=-\\dfrac{\\sqrt{2}}2$.\\\\\n", "题号: 014227 , 字段: usages 中已添加数据: 20230228\t2023届高三10班\t0.784\n",
"(2) 当直线$l$的斜率不存在时, 点$P,Q$关于$x$轴对称, 设$P(x_1,y_1)$, $Q(x_1,-y_1)$, 由$x_1^2+\\dfrac{y_1^2}{2}=1$及$|x_1y_1|=\\dfrac{\\sqrt{2}}{2}$解得$|x_1|=\\dfrac{\\sqrt{2}}2$, $|y_1|=1$, 故此时$x_1^2+x_2^2=1$, $y_1^2+y_2^2=2$.\\\\\n", "题号: 014226 , 字段: usages 中已添加数据: 20230228\t2023届高三10班\t0.946\n",
"当直线$l$的斜率存在时, 设直线$l$的方程为$y=kx+m$, 与椭圆$C$的方程联立, 整理得$(2+k^2)x^2+2kmx+m^2-2=0$, 由$\\dfrac 12|x_1y_2-x_2y_1|=\\dfrac 12|x_1(kx_2+m)-x_2(kx_1+m)|=\\dfrac 12|m||x_1-x_2|=\\dfrac{\\sqrt{2}}{2}$得$\\sqrt{2}(2+k^2)=|m|\\sqrt{2+k^2-m^2}$. 于是$2m^2=2+k^2$, $x_1^2+x_2^2=(x_1+x_2)^2-2x_1x_2=(\\dfrac{-2km}{2+k^2})^2-\\dfrac{2m^2-4}{2+k^2}=1$, $y_1^2+y_2^2=2(1-x_1^2+1-x_2^2)=2$.\\\\\n", "题号: 013010 , 字段: usages 中已添加数据: 20230228\t2023届高三10班\t0.946\n",
"(3) 假设这样的三角形存在, 则三点$D(x_1,y_1)$, $E(x_2,y_2)$, $G(x_3,y_3)$应满足$x_1^2+x_2^2=x_2^2+x_3^2=x_3^2+x_1^2=1$, 从而$x_1^2=x_2^2=x_3^2=\\dfrac 12$, 但椭圆$C$上任取横坐标为$\\pm \\dfrac{\\sqrt{2}}{2}$的三点, 总有两个点与原点共线, 此时三角形不存在, 矛盾. 因此假设不成立, 满足条件的三角形不存在.\n", "题号: 013852 , 字段: usages 中已添加数据: 20230228\t2023届高三10班\t0.730\n",
"题号: 031236 , 字段: solution 中已添加数据: (1) $f'(0)=\\mathrm{e}^0(\\dfrac{1}{1+0}+\\ln (1+0))=1$, 又$f(0)=0$, 故所求切线的方程为$y=x$.\\\\\n", "题号: 013875 , 字段: usages 中已添加数据: 20230228\t2023届高三10班\t0.405\n",
"(2) $g(x)=f'(x)=\\mathrm{e}^x(\\dfrac{1}{1+x}+\\ln (1+x))$. $g'(x)=\\mathrm{e}^x(\\ln (1+x)+\\dfrac{2}{1+x}-\\dfrac{1}{(1+x)^2})$, 当$x>0$时, $\\mathrm{e}^x>1>0$, $\\ln(1+x)>0$, 且$\\dfrac{2}{1+x}-\\dfrac{1}{(1+x)^2}=\\dfrac{2x+1}{(1+x)^2}>0$, 所以$g'(x)$在$(0,+\\infty)$上恒正. 因此$g(x)$在$[0,+\\infty)$上是严格增函数.\\\\\n", "题号: 013880 , 字段: usages 中已添加数据: 20230228\t2023届高三10班\t0.378\n",
"(3) 对于固定的常数$t$, 令$h(x)=f(x+t)-f(x)-f(t)$, 则$h(t)=0$, 而$h'(x)=f'(x+t)-f'(x)=g(x+t)-g(x)$. 由(2)可知$g(x+t)>g(x)$, 故$h'(x)$在$(0,+\\infty)$上恒正, 于是有$h(s)>h(0)=0$, 即$f(s+t)>f(s)+f(t)$.\n" "题号: 014221 , 字段: usages 中已添加数据: 20230228\t2023届高三10班\t0.324\n",
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"题号: 014204 , 字段: usages 中已添加数据: 20230228\t2023届高三05班\t0.925\n",
"题号: 014217 , 字段: usages 中已添加数据: 20230228\t2023届高三05班\t1.000\n",
"题号: 014206 , 字段: usages 中已添加数据: 20230228\t2023届高三05班\t0.900\n",
"题号: 014207 , 字段: usages 中已添加数据: 20230228\t2023届高三05班\t0.800\n",
"题号: 014208 , 字段: usages 中已添加数据: 20230228\t2023届高三05班\t0.925\n",
"题号: 013024 , 字段: usages 中已添加数据: 20230228\t2023届高三05班\t0.925\n",
"题号: 014227 , 字段: usages 中已添加数据: 20230228\t2023届高三05班\t0.850\n",
"题号: 014226 , 字段: usages 中已添加数据: 20230228\t2023届高三05班\t0.850\n",
"题号: 013010 , 字段: usages 中已添加数据: 20230228\t2023届高三05班\t0.975\n",
"题号: 013852 , 字段: usages 中已添加数据: 20230228\t2023届高三05班\t0.900\n",
"题号: 013875 , 字段: usages 中已添加数据: 20230228\t2023届高三05班\t0.725\n",
"题号: 013880 , 字段: usages 中已添加数据: 20230228\t2023届高三05班\t0.600\n",
"题号: 014221 , 字段: usages 中已添加数据: 20230228\t2023届高三05班\t0.300\n",
"题号: 014231 , 字段: usages 中已添加数据: 20230228\t2023届高三05班\t0.600\t0.725\n",
"题号: 013879 , 字段: usages 中已添加数据: 20230228\t2023届高三05班\t0.850\t0.750\n",
"题号: 014204 , 字段: usages 中已添加数据: 20230228\t2023届高三06班\t0.850\n",
"题号: 014217 , 字段: usages 中已添加数据: 20230228\t2023届高三06班\t0.950\n",
"题号: 014206 , 字段: usages 中已添加数据: 20230228\t2023届高三06班\t0.925\n",
"题号: 014207 , 字段: usages 中已添加数据: 20230228\t2023届高三06班\t0.875\n",
"题号: 014208 , 字段: usages 中已添加数据: 20230228\t2023届高三06班\t0.950\n",
"题号: 013024 , 字段: usages 中已添加数据: 20230228\t2023届高三06班\t1.000\n",
"题号: 014227 , 字段: usages 中已添加数据: 20230228\t2023届高三06班\t0.800\n",
"题号: 014226 , 字段: usages 中已添加数据: 20230228\t2023届高三06班\t0.925\n",
"题号: 013010 , 字段: usages 中已添加数据: 20230228\t2023届高三06班\t0.975\n",
"题号: 013852 , 字段: usages 中已添加数据: 20230228\t2023届高三06班\t0.750\n",
"题号: 013875 , 字段: usages 中已添加数据: 20230228\t2023届高三06班\t0.600\n",
"题号: 013880 , 字段: usages 中已添加数据: 20230228\t2023届高三06班\t0.600\n",
"题号: 014221 , 字段: usages 中已添加数据: 20230228\t2023届高三06班\t0.775\n",
"题号: 014231 , 字段: usages 中已添加数据: 20230228\t2023届高三06班\t0.650\t0.950\n",
"题号: 013879 , 字段: usages 中已添加数据: 20230228\t2023届高三06班\t0.725\t0.725\n",
"题号: 014204 , 字段: usages 中已添加数据: 20230228\t2023届高三07班\t0.812\n",
"题号: 014217 , 字段: usages 中已添加数据: 20230228\t2023届高三07班\t0.938\n",
"题号: 014206 , 字段: usages 中已添加数据: 20230228\t2023届高三07班\t0.969\n",
"题号: 014207 , 字段: usages 中已添加数据: 20230228\t2023届高三07班\t0.844\n",
"题号: 014208 , 字段: usages 中已添加数据: 20230228\t2023届高三07班\t0.906\n",
"题号: 013024 , 字段: usages 中已添加数据: 20230228\t2023届高三07班\t0.969\n",
"题号: 014227 , 字段: usages 中已添加数据: 20230228\t2023届高三07班\t0.594\n",
"题号: 014226 , 字段: usages 中已添加数据: 20230228\t2023届高三07班\t0.875\n",
"题号: 013010 , 字段: usages 中已添加数据: 20230228\t2023届高三07班\t0.906\n",
"题号: 013852 , 字段: usages 中已添加数据: 20230228\t2023届高三07班\t0.844\n",
"题号: 013875 , 字段: usages 中已添加数据: 20230228\t2023届高三07班\t0.438\n",
"题号: 013880 , 字段: usages 中已添加数据: 20230228\t2023届高三07班\t0.469\n",
"题号: 014221 , 字段: usages 中已添加数据: 20230228\t2023届高三07班\t0.500\n",
"题号: 014231 , 字段: usages 中已添加数据: 20230228\t2023届高三07班\t0.594\t0.688\n",
"题号: 013879 , 字段: usages 中已添加数据: 20230228\t2023届高三07班\t0.750\t0.781\n",
"题号: 014204 , 字段: usages 中已添加数据: 20230228\t2023届高三08班\t0.852\n",
"题号: 014217 , 字段: usages 中已添加数据: 20230228\t2023届高三08班\t0.926\n",
"题号: 014206 , 字段: usages 中已添加数据: 20230228\t2023届高三08班\t0.963\n",
"题号: 014207 , 字段: usages 中已添加数据: 20230228\t2023届高三08班\t0.926\n",
"题号: 014208 , 字段: usages 中已添加数据: 20230228\t2023届高三08班\t0.963\n",
"题号: 013024 , 字段: usages 中已添加数据: 20230228\t2023届高三08班\t0.963\n",
"题号: 014227 , 字段: usages 中已添加数据: 20230228\t2023届高三08班\t0.852\n",
"题号: 014226 , 字段: usages 中已添加数据: 20230228\t2023届高三08班\t0.815\n",
"题号: 013010 , 字段: usages 中已添加数据: 20230228\t2023届高三08班\t0.963\n",
"题号: 013852 , 字段: usages 中已添加数据: 20230228\t2023届高三08班\t0.852\n",
"题号: 013875 , 字段: usages 中已添加数据: 20230228\t2023届高三08班\t0.556\n",
"题号: 013880 , 字段: usages 中已添加数据: 20230228\t2023届高三08班\t0.444\n",
"题号: 014221 , 字段: usages 中已添加数据: 20230228\t2023届高三08班\t0.000\n",
"题号: 014231 , 字段: usages 中已添加数据: 20230228\t2023届高三08班\t0.704\t0.815\n",
"题号: 013879 , 字段: usages 中已添加数据: 20230228\t2023届高三08班\t0.815\t0.815\n",
"题号: 014204 , 字段: usages 中已添加数据: 20230228\t2023届高三09班\t0.938\n",
"题号: 014217 , 字段: usages 中已添加数据: 20230228\t2023届高三09班\t0.969\n",
"题号: 014206 , 字段: usages 中已添加数据: 20230228\t2023届高三09班\t1.000\n",
"题号: 014207 , 字段: usages 中已添加数据: 20230228\t2023届高三09班\t0.844\n",
"题号: 014208 , 字段: usages 中已添加数据: 20230228\t2023届高三09班\t0.969\n",
"题号: 013024 , 字段: usages 中已添加数据: 20230228\t2023届高三09班\t1.000\n",
"题号: 014227 , 字段: usages 中已添加数据: 20230228\t2023届高三09班\t0.812\n",
"题号: 014226 , 字段: usages 中已添加数据: 20230228\t2023届高三09班\t0.875\n",
"题号: 013010 , 字段: usages 中已添加数据: 20230228\t2023届高三09班\t0.938\n",
"题号: 013852 , 字段: usages 中已添加数据: 20230228\t2023届高三09班\t0.844\n",
"题号: 013875 , 字段: usages 中已添加数据: 20230228\t2023届高三09班\t0.594\n",
"题号: 013880 , 字段: usages 中已添加数据: 20230228\t2023届高三09班\t0.406\n",
"题号: 014221 , 字段: usages 中已添加数据: 20230228\t2023届高三09班\t0.500\n",
"题号: 014231 , 字段: usages 中已添加数据: 20230228\t2023届高三09班\t0.719\t0.750\n",
"题号: 013879 , 字段: usages 中已添加数据: 20230228\t2023届高三09班\t0.969\t0.781\n"
] ]
} }
], ],

View File

@ -1,26 +1,542 @@
solution usages
014204
20230228 2023届高三10班 0.892
010000 014217
(1) $S_{\triangle ABC}=\dfrac 12 \cdot 2\times 2\times \sin 60^\circ=\sqrt{3}$, 三棱锥的高$PO=2\sin 60^\circ = \sqrt{3}$, 故$V_{P-ABC}=\dfrac 13\cdot S_{\triangle ABC}\cdot PO=\dfrac 13\times \sqrt{3}\times \sqrt{3}=1$.\\ 20230228 2023届高三10班 0.919
(2) 取$OC$中点$N$, 由于$MN\parallel BO$, 且$BO\perp AC$, 故$MN\perp AC$. 又因为$PO\perp$平面$ABC$, 故$MN\perp PO$. 注意到$PO\cap AC=O$, 于是$MN\perp$平面$PAC$, 从而$PN$是$PM$在平面$PAC$上的射影. $MN=\dfrac{\sqrt{3}}{2}$, $PN=\dfrac{\sqrt{13}}{2}$, 故$\angle MPN=\arctan \dfrac{MN}{PN}=\arctan \dfrac{\sqrt{39}}{13}$. 综上, $PM$与平面$PAC$所成角的大小为$\arctan \dfrac{\sqrt{39}}{13}$.
011301 014206
(1) 由$a^2+c^2-2ac\cos B=b^2$, 即$c^2-5c-24=0$解得$c=8$或$-3$(负数舍去). 故$c=8$.\\ 20230228 2023届高三10班 1.000
(2) 延长$CM$至$N$, 使得$CM=MN$, 则四边形$ACBN$是平行四边形, 故$S_{\triangle ABC}=S_{\triangle ACN}$. $\triangle ACN$的三边长分别为$5,6,7$, 故$\cos \angle CAN=\dfrac{5^2+7^2-6^2}{2\times 5\times 7}=\dfrac{19}{35}$, 所以$\sin \angle CAN=\dfrac{12\sqrt{6}}{35}$. 由此, $S_{\triangle ACN}=\dfrac 12\cdot AC\cdot AN\cdot \sin \angle CAN=6\sqrt{6}$. 综上, 三角形$ABC$的面积为$6\sqrt{6}$.
003691 014207
(1) 前$4$个月的累计投放量为$a_1+a_2+a_3+a_4=20+95+420+430=965$. 前$4$个月的累计损失量为$b_1+b_2+b_3+b_4=6+7+8+9=30$, 因此该地区第$4$个月底的共享单车的保有量为$965-30=935$.\\ 20230228 2023届高三10班 0.649
(2) 考察不等式$a_n \geq b_n$在整数范围内的解集. 当$n \leq 3$时, $5 n^4+15 \geq n+5$确实成立, 当$n \geq 4$时, 由$-10 n+470 \geq n+5$, 解得$n \leq \dfrac{465}{11}\approx 42.27$, 因此第$42$个月底, 保有量达到最大. 当$n \geq 4$时, $\{a_n\}$是公差为$-10$的等差数列, 而$\{b_n\}$是公差为$1$的等差数列, 于是到第$42$个月底, 单车保有量为$\dfrac{38(a_5+a_{42})}{2}+965-\dfrac{42(b_1+b_{42})}{2}=\dfrac{38(420+50)}{2}+965-\dfrac{42(6+47)}{2}=8782$. 而$S_{42}=-4\times 16+8800=8736<8782$. 因此该月底单车保有量超过了容纳量.
014208
20230228 2023届高三10班 0.892
011450 013024
(1) 设直线$l$的方程为$x=m$, 由$S_{\triangle OPQ}=\dfrac 12\cdot |m|\cdot |PQ|=\dfrac{\sqrt{2}}{2}$得$|PQ|=\dfrac{\sqrt{2}}{|m|}$, 于是点$(m,\dfrac{\sqrt{2}}{2|m|})$在椭圆上. 代入椭圆$C$的方程, 得$m^2+\dfrac{2}{8m^2}=1$, 解得$m^2=\dfrac 12$. 故直线$l$的方程为$x=\dfrac{\sqrt{2}}2$或$x=-\dfrac{\sqrt{2}}2$.\\ 20230228 2023届高三10班 0.892
(2) 当直线$l$的斜率不存在时, 点$P,Q$关于$x$轴对称, 设$P(x_1,y_1)$, $Q(x_1,-y_1)$, 由$x_1^2+\dfrac{y_1^2}{2}=1$及$|x_1y_1|=\dfrac{\sqrt{2}}{2}$解得$|x_1|=\dfrac{\sqrt{2}}2$, $|y_1|=1$, 故此时$x_1^2+x_2^2=1$, $y_1^2+y_2^2=2$.\\
当直线$l$的斜率存在时, 设直线$l$的方程为$y=kx+m$, 与椭圆$C$的方程联立, 整理得$(2+k^2)x^2+2kmx+m^2-2=0$, 由$\dfrac 12|x_1y_2-x_2y_1|=\dfrac 12|x_1(kx_2+m)-x_2(kx_1+m)|=\dfrac 12|m||x_1-x_2|=\dfrac{\sqrt{2}}{2}$得$\sqrt{2}(2+k^2)=|m|\sqrt{2+k^2-m^2}$. 于是$2m^2=2+k^2$, $x_1^2+x_2^2=(x_1+x_2)^2-2x_1x_2=(\dfrac{-2km}{2+k^2})^2-\dfrac{2m^2-4}{2+k^2}=1$, $y_1^2+y_2^2=2(1-x_1^2+1-x_2^2)=2$.\\ 014227
(3) 假设这样的三角形存在, 则三点$D(x_1,y_1)$, $E(x_2,y_2)$, $G(x_3,y_3)$应满足$x_1^2+x_2^2=x_2^2+x_3^2=x_3^2+x_1^2=1$, 从而$x_1^2=x_2^2=x_3^2=\dfrac 12$, 但椭圆$C$上任取横坐标为$\pm \dfrac{\sqrt{2}}{2}$的三点, 总有两个点与原点共线, 此时三角形不存在, 矛盾. 因此假设不成立, 满足条件的三角形不存在. 20230228 2023届高三10班 0.784
014226
20230228 2023届高三10班 0.946
013010
20230228 2023届高三10班 0.946
013852
20230228 2023届高三10班 0.730
013875
20230228 2023届高三10班 0.405
013880
20230228 2023届高三10班 0.378
014221
20230228 2023届高三10班 0.324
014231
20230228 2023届高三10班 0.432 0.351
013879
20230228 2023届高三10班 0.757 0.135
014204
20230228 2023届高三11班 0.809
014217
20230228 2023届高三11班 0.905
014206
20230228 2023届高三11班 0.952
014207
20230228 2023届高三11班 0.905
014208
20230228 2023届高三11班 0.952
013024
20230228 2023届高三11班 0.952
014227
20230228 2023届高三11班 0.809
014226
20230228 2023届高三11班 0.905
013010
20230228 2023届高三11班 0.905
013852
20230228 2023届高三11班 0.667
013875
20230228 2023届高三11班 0.333
013880
20230228 2023届高三11班 0.048
014221
20230228 2023届高三11班 0.619
014231
20230228 2023届高三11班 0.571 0.571
013879
20230228 2023届高三11班 0.857 0.619
014204
20230228 2023届高三12班 0.826
014217
20230228 2023届高三12班 0.913
014206
20230228 2023届高三12班 1.000
014207
20230228 2023届高三12班 0.870
014208
20230228 2023届高三12班 0.870
013024
20230228 2023届高三12班 0.913
014227
20230228 2023届高三12班 0.783
014226
20230228 2023届高三12班 0.783
013010
20230228 2023届高三12班 0.957
013852
20230228 2023届高三12班 0.739
013875
20230228 2023届高三12班 0.739
013880
20230228 2023届高三12班 0.435
014221
20230228 2023届高三12班 0.696
014231
20230228 2023届高三12班 0.652 0.739
013879
20230228 2023届高三12班 0.957 0.870
014204
20230228 2023届高三01班 0.967
014217
20230228 2023届高三01班 0.967
014206
20230228 2023届高三01班 1.000
014207
20230228 2023届高三01班 0.933
014208
20230228 2023届高三01班 0.867
013024
20230228 2023届高三01班 1.000
014227
20230228 2023届高三01班 0.767
014226
20230228 2023届高三01班 0.933
013010
20230228 2023届高三01班 0.967
013852
20230228 2023届高三01班 0.900
013875
20230228 2023届高三01班 0.767
013880
20230228 2023届高三01班 0.900
014221
20230228 2023届高三01班 0.700
014231
20230228 2023届高三01班 0.800 0.933
013879
20230228 2023届高三01班 0.833 0.900
014204
20230228 2023届高三02班 1.000
014217
20230228 2023届高三02班 0.933
014206
20230228 2023届高三02班 0.967
014207
20230228 2023届高三02班 0.933
014208
20230228 2023届高三02班 0.900
013024
20230228 2023届高三02班 0.967
014227
20230228 2023届高三02班 0.833
014226
20230228 2023届高三02班 1.000
013010
20230228 2023届高三02班 0.900
013852
20230228 2023届高三02班 0.733
013875
20230228 2023届高三02班 0.467
013880
20230228 2023届高三02班 0.700
014221
20230228 2023届高三02班 0.467
014231
20230228 2023届高三02班 0.700 0.833
013879
20230228 2023届高三02班 0.833 0.333
014204
20230228 2023届高三03班 0.897
014217
20230228 2023届高三03班 0.897
014206
20230228 2023届高三03班 1.000
014207
20230228 2023届高三03班 0.793
014208
20230228 2023届高三03班 0.931
013024
20230228 2023届高三03班 0.931
014227
20230228 2023届高三03班 0.828
014226
20230228 2023届高三03班 0.966
013010
20230228 2023届高三03班 0.966
013852
20230228 2023届高三03班 0.862
013875
20230228 2023届高三03班 0.690
013880
20230228 2023届高三03班 0.724
014221
20230228 2023届高三03班 0.448
014231
20230228 2023届高三03班 0.655 0.759
013879
20230228 2023届高三03班 0.828 0.759
014204
20230228 2023届高三04班 0.923
014217
20230228 2023届高三04班 0.923
014206
20230228 2023届高三04班 0.923
014207
20230228 2023届高三04班 0.846
014208
20230228 2023届高三04班 1.000
013024
20230228 2023届高三04班 0.962
014227
20230228 2023届高三04班 0.923
014226
20230228 2023届高三04班 0.923
013010
20230228 2023届高三04班 0.808
013852
20230228 2023届高三04班 0.808
013875
20230228 2023届高三04班 0.577
013880
20230228 2023届高三04班 0.808
014221
20230228 2023届高三04班 0.769
014231
20230228 2023届高三04班 0.615 0.615
013879
20230228 2023届高三04班 0.808 0.654
014204
20230228 2023届高三05班 0.925
014217
20230228 2023届高三05班 1.000
014206
20230228 2023届高三05班 0.900
014207
20230228 2023届高三05班 0.800
014208
20230228 2023届高三05班 0.925
013024
20230228 2023届高三05班 0.925
014227
20230228 2023届高三05班 0.850
014226
20230228 2023届高三05班 0.850
013010
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031236
(1) $f'(0)=\mathrm{e}^0(\dfrac{1}{1+0}+\ln (1+0))=1$, 又$f(0)=0$, 故所求切线的方程为$y=x$.\\
(2) $g(x)=f'(x)=\mathrm{e}^x(\dfrac{1}{1+x}+\ln (1+x))$. $g'(x)=\mathrm{e}^x(\ln (1+x)+\dfrac{2}{1+x}-\dfrac{1}{(1+x)^2})$, 当$x>0$时, $\mathrm{e}^x>1>0$, $\ln(1+x)>0$, 且$\dfrac{2}{1+x}-\dfrac{1}{(1+x)^2}=\dfrac{2x+1}{(1+x)^2}>0$, 所以$g'(x)$在$(0,+\infty)$上恒正. 因此$g(x)$在$[0,+\infty)$上是严格增函数.\\
(3) 对于固定的常数$t$, 令$h(x)=f(x+t)-f(x)-f(t)$, 则$h(t)=0$, 而$h'(x)=f'(x+t)-f'(x)=g(x+t)-g(x)$. 由(2)可知$g(x+t)>g(x)$, 故$h'(x)$在$(0,+\infty)$上恒正, 于是有$h(s)>h(0)=0$, 即$f(s+t)>f(s)+f(t)$.

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