From 2db568944b7f4462bf4b5401914874ac4778742f Mon Sep 17 00:00:00 2001 From: "weiye.wang" Date: Wed, 10 May 2023 20:29:58 +0800 Subject: [PATCH] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=E6=9C=88=E8=80=83=E8=A7=A3?= =?UTF-8?q?=E7=AD=94=E9=A2=98=E7=AD=94=E6=A1=88=E5=8F=8A=E8=A7=A3=E7=AD=94?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- 工具/文本文件/metadata.txt | 147 ++++++++++++++++++++----------------- 题库0.3/Problems.json | 20 ++--- 2 files changed, 89 insertions(+), 78 deletions(-) diff --git a/工具/文本文件/metadata.txt b/工具/文本文件/metadata.txt index c42f3abc..d5217845 100644 --- a/工具/文本文件/metadata.txt +++ b/工具/文本文件/metadata.txt @@ -1,83 +1,94 @@ -ans -017223 -$\{1\}$ +solution +017239 +(1) 由底面$ABC$为等腰直角三角形且 $AB \perp AC$知$AB\perp$平面$ACC_1A_1$, \\ +从而$BM$在平面$ACC_1A_1$上的投影为$AM$, +故由$BM \perp A_1C$ 知 $AM\perp A_1C $, \\ +结合$AC=2$, $AA_1=4$ 得 $MC=1$, 即$h=1$.\\ +(2) 如图建系：以$A$为原点，分别以$\overrightarrow{AB}$、$\overrightarrow{AC}$、$\overrightarrow{AA_1}$方向为$x$轴、 $y$轴、$z$ 轴正方向建立平面直角坐标系.\\ +$A(0,0,0),B(2,0,0),C(0,2,0),A_1(0,0,4),M(0,2,2)$, +$\overrightarrow{BA_1}=(-2,0,4),\overrightarrow{AB}=(2,0,0),\overrightarrow{AM}=(0,2,2),$\\ +设平面$ABM$的一个法向量为$\overrightarrow{n}=(x,y,z)$, 则$\begin{cases} +2x=0,\\ +2y+2z=0. +\end{cases}$ 取$\overrightarrow{n}=(0,1,-1),$设直线$BA_1$与平面$ABM$所成的角为$\theta$, 则$\sin\theta=|\cos \langle\overrightarrow{BA_1},\overrightarrow{n}\rangle|=\dfrac{|\overrightarrow{BA_1}\cdot\overrightarrow{n}|}{|\overrightarrow{BA_1}|\cdot|\overrightarrow{n}|}=\dfrac{\sqrt{10}}{5}$, 故直线$BA_1$与平面$ABM$所成的角为$\arcsin\dfrac{\sqrt{10}}{5}$. -017224 -$1$ +017240 +(1) 由 $S=\dfrac{1}{2}ac\sin B=\dfrac{\sqrt{3}}{4}ac=\sqrt{3}$知$ac=4$，\\ +由$a^2+c^2-b^2=2ac\cos B$知$a^2+c^2=9$.\\ +结合两式得$(a-c)^2=1$, 故$a-c=\pm 1$\\ +(2) 由$2\cos C (ac\cos B+cb\cos A)=c^2$知$2a\cos B\cos C+2b \cos A\cos C=c$,\\ +又由正弦定理知 $2\cos C(\sin A\cos B+\sin B\cos A)=\sin C$，\\ +$2\cos C \sin (A+B)=2\cos C \sin C=\sin C$,其中$C\in (0,\pi), \sin C>0$,\\ +故$\cos C=\dfrac{1}{2}$, $C\in(0,\pi)$, $C=\dfrac{\pi}{3}$. -017225 -$\dfrac{\pi}{3}$ +017241 +(1) $\dfrac{p}{p+40}=\dfrac{3}{5}$得$p=60$, $q=40$, $x=100$, $y=100$.\\ +(2) 原假设$H_0:$ 是否注射此种疫苗与是否感染病毒无关.\\ +$\chi^2=\dfrac{200\times (40\times 40-60\times 60)^2}{100\times 100\times 100\times 100}=8>3.841$\\ +故拒绝原假设，即有$95 \%$的把握认为注射此种疫苗有效.\\ +(3) 抽取$6$只未注射疫苗、$4$只注射疫苗的小白鼠.\\ +$P(X=0)=\dfrac{C_6^0C_4^4}{C_{10}^4}=\dfrac{1}{210}$;\\ +$P(X=1)=\dfrac{C_6^1C_4^3}{C_{10}^4}=\dfrac{4}{35}$;\\ +$P(X=2)=\dfrac{C_6^2C_4^2}{C_{10}^4}=\dfrac{3}{7}$;\\ +$P(X=3)=\dfrac{C_6^3C_4^1}{C_{10}^4}=\dfrac{8}{21}$;\\ +$P(X=4)=\dfrac{C_6^4C_4^0}{C_{10}^4}=\dfrac{1}{14}$.\\ +故$X$的分布为$\begin{pmatrix} +0&1&2&3&4\\ +\dfrac{1}{210}&\dfrac{4}{35}&\dfrac{3}{7}&\dfrac{8}{21}&\dfrac{1}{14} +\end{pmatrix},$\\ +期望$E[X]=0\times \dfrac{1}{210}+1\times \dfrac{4}{35}+2\times \dfrac{3}{7}+3\times \dfrac{8}{21}+4\times \dfrac{1}{14}=\dfrac{12}{5}$. -017226 -$70$ +017242 +(1) 椭圆的离心率$e=\dfrac{1}{2}$;\\ +(2) 证明: 当$x_0=2$时，$y_0=0,$ 过点$P$的椭圆$C$的切线方程为 $x=2$,符合$\dfrac{x_0 x}{4}+\dfrac{y_0 y}{3}=1$，\\ +同理，当$x_0=-2$时，也符合；\\ +当$x_0\neq\pm 2$时，设过点$P$的椭圆$C$的切线方程为$y-y_0=k(x-x_0)$($k$存在),\\ +联立$\begin{cases} +y-y_0=k(x-x_0),\\ + \dfrac{x^2}{4}+\dfrac{y^2}{3}=1 +\end{cases}$得$(3+4k^2)x^2+8k(y-kx_0)x+4(y_0-kx_0)^2-12=0$,\\ +$\Delta=0$得$(x_0^2-4)k^2-2x_0y_0k+y_0^2-3=0$,解得$k=\dfrac{x_0y_0}{x_0^2-4}=\dfrac{x_0y_0}{4(1-\dfrac{y_0^2}{3})-4}=-\dfrac{3x_0}{4y_0}.$\\ +故$y-y_0=-\dfrac{3x_0}{4y_0}(x-x_0)$,即$\dfrac{x_0 x}{4}+\dfrac{y_0 y}{3}=1$.\\ +综上，过点$P$的椭圆$C$的切线方程为$\dfrac{x_0 x}{4}+\dfrac{y_0 y}{3}=1$.\\ +(3) 设$A(x_1,y_1).B(x_2,y_2),x_1\neq x_2,M(4,t)$.\\ +则切线$MA:\dfrac{x_1x}{4}+\dfrac{y_1y}{3}=1,$代入$(4,t)$得$x_1+\dfrac{ty_1}{3}=1$,\\ +同理，$x_1+\dfrac{ty_1}{3}=1$，\\ +故$A(x_1,y_1),B(x_2,y_2)$在直线$x+\dfrac{ty}{3}=1$上，故直线$AB:x=-\dfrac{ty}{3}+1$.\\ +联立$\begin{cases} +x=-\dfrac{ty}{3}+1,\\ +\dfrac{x^2}{4}+\dfrac{y^2}{3}=1 +\end{cases}$ +得 +$(4+\dfrac{t^2}{3})y^2-2ty-9=0$, $\Delta=16t^2+144>0$,\\ +$|AB|=\sqrt{1+\dfrac{t^2}{9}}\cdot |y_1-y_2|=\sqrt{1+\dfrac{t^2}{9}}\cdot \dfrac{\sqrt{16t^2+144}}{4+\dfrac{t^2}{3}}$,\\ +$M$到直线$AB$的距离$d=\dfrac{|4+\dfrac{t^2}{3}-1|}{\sqrt{1+\dfrac{t^2}{9}}}=\dfrac{3+\dfrac{t^2}{3}}{\sqrt{1+\dfrac{t^2}{9}}}$,\\ +$\triangle MAB$的面积$S=\dfrac{1}{2}|AB|\cdot d=\dfrac{1}{2}\cdot \sqrt{1+\dfrac{t^2}{9}}\cdot \dfrac{\sqrt{16t^2+144}}{4+\dfrac{t^2}{3}}\cdot \dfrac{3+\dfrac{t^2}{3}}{\sqrt{1+\dfrac{t^2}{9}}}=\dfrac{2(t^2+9)\sqrt{t^2+9}}{t^2+12}$,\\ +令$\lambda=\sqrt{t^2+9}\geq3, S=f(\lambda)=\dfrac{2\lambda^3}{\lambda^2+3},$ +则$f^{'} (\lambda)=\dfrac{2\lambda^4+18\lambda^2}{(\lambda^2+3)^2}>0$,故$f(\lambda)$在$[3,+\infty)$严格增，$f(\lambda)_{\min}=f(3)=\dfrac{9}{2}$,故$\triangle MAB$的面积的最小值为$\dfrac{9}{2},$ 此时$M$的坐标为$(4,0)$. -017227 -$\dfrac{1}{8}$ +017243 +(1) $x_1=\dfrac{1}{2},x_{n+1}=g(x_n)=\dfrac{x_n}{x_{n+1}},$故$x_n>0,\dfrac{1}{x_{n+1}}=\dfrac{x_n+1}{x_n}=\dfrac{1}{x_n}+1$,即$\dfrac{1}{x_{n+1}}-\dfrac{1}{x_{n}}=1$,\\因此数列$\{\dfrac{1}{x_n}\}$是以$2$为首项，$1$为公差的等差数列.\\ +(2) 对任意$x>0$ 均有$f(x)-mg(x)=\ln (x+1)-\dfrac{mx}{x+1}+1>0,$\\ +令$h(x)=\ln (x+1)-\dfrac{mx}{x+1}+1,x>0,$则$h^{'}(x)=\dfrac{x+1-m}{(x+1)^2}$.\\ +当$m\geq4$时，$h(1)=\ln 2-\dfrac{m}{2}+1<2-\dfrac{m}{2}<0,$这与$h(x)>0$对$x>0$恒成立矛盾；\\ +当$m=3$时，$h(x)=\ln (x+1)-\dfrac{3x}{x+1}+1,h^{'}(x)\dfrac{x-2}{(x+1)^2}.h(x)$在$(0,2]$严格减，在$[2,+\infty)$严格增，故$h(x)_{\min}=h(2)=\ln 3-2+1=\ln 3-1>0$,符合题意.\\ +综上，整数$m$的最大值为$3$.\\ +(3) 对任意正整数$t$,取$n=100t,$则 \\$\displaystyle f(n-t)=f(99t)=\ln (1+99t)=\ln \dfrac{1+99t}{99t}+\ln \dfrac{99t}{99t-1}+\cdots +\ln \dfrac{2}{1}=\sum_{k=1}^{99t}\ln (1+\dfrac{1}{k}).$\\ +$\displaystyle n-\sum_{k=1}^{100t}g(k)=\sum_{k=1}^{100t}(1- g(k))=\sum_{k=1}^{100t} \dfrac{1}{1+k}=\sum_{k=1}^{99t} \dfrac{1}{1+k}+\sum_{k=99t+1}^{100t} \dfrac{1}{1+k}.$\\ +$\displaystyle f(n-t)-[ n-\sum_{k=1}^{100t}g(k)]=\sum_{k=1}^{99t} (\ln (1+\dfrac{1}{k})-\dfrac{1}{1+k})-\sum_{k=99t+1}^{100t} \dfrac{1}{1+k}.$\\ +令$H(x)=\ln (x+1)-\dfrac{x}{x+1},x>0$,则$H^{'}(x)=\dfrac{x}{(x+1)^2}>0$恒成立，故$H(x)$在$(0,+\infty)$严格增，$H(x)>H(0)=0$，故$\ln (1+x)>\dfrac{x}{x+1}$对$x>0$恒成立.\\ +因此$\ln (1+\dfrac{1}{k})>\dfrac{\dfrac{1}{k}}{1+\dfrac{1}{k}}=\dfrac{1}{1+k}$,\\ +$\displaystyle f(n-t)-[ n-\sum_{k=1}^{100t}g(k)]=\sum_{k=1}^{99t} (\ln (1+\dfrac{1}{k})-\dfrac{1}{1+k})-\sum_{k=99t+1}^{100t} \dfrac{1}{1+k}$\\ +$\displaystyle>\sum_{k=2}^{99t} (\ln (1+\dfrac{1}{k})-\dfrac{1}{1+k})+(\ln 2-\dfrac{1}{2})- \dfrac{t}{1+99t+1}>\ln 2-\dfrac{1}{2}-\dfrac{1}{99+\dfrac{2}{t}}>0.1-\dfrac{1}{99}>0.$\\ +因此不存在正整数$t$使得对任意$n \in \mathbf{N}$, $n \geq t$, 都有$\displaystyle f(n-t)0$,\\\\\n故$\\cos C=\\dfrac{1}{2}$, $C\\in(0,\\pi)$, $C=\\dfrac{\\pi}{3}$.", "duration": -1, "usages": [], "origin": "2023届高三下学期月考2试题18", @@ -441407,8 +441407,8 @@ "objs": [], "tags": [], "genre": "解答题", - "ans": "", - "solution": "", + "ans": "(1) $p=60$, $q=40$, $x=100$, $y=100$; (2) $\\chi^2=8$, 有$95 \\%$的把握认为注射此种疫苗有效; (3) 分布为$\\begin{pmatrix}0&1&2&3&4\\\\\\dfrac{1}{210}&\\dfrac{4}{35}&\\dfrac{3}{7}&\\dfrac{8}{21}&\\dfrac{1}{14}\\end{pmatrix}$, $E[X]=\\dfrac{12}{5}$", + "solution": "(1) $\\dfrac{p}{p+40}=\\dfrac{3}{5}$得$p=60$, $q=40$, $x=100$, $y=100$.\\\\\n(2) 原假设$H_0:$ 是否注射此种疫苗与是否感染病毒无关.\\\\\n$\\chi^2=\\dfrac{200\\times (40\\times 40-60\\times 60)^2}{100\\times 100\\times 100\\times 100}=8>3.841$\\\\\n故拒绝原假设，即有$95 \\%$的把握认为注射此种疫苗有效.\\\\\n(3) 抽取$6$只未注射疫苗、$4$只注射疫苗的小白鼠.\\\\\n$P(X=0)=\\dfrac{C_6^0C_4^4}{C_{10}^4}=\\dfrac{1}{210}$;\\\\\n$P(X=1)=\\dfrac{C_6^1C_4^3}{C_{10}^4}=\\dfrac{4}{35}$;\\\\\n$P(X=2)=\\dfrac{C_6^2C_4^2}{C_{10}^4}=\\dfrac{3}{7}$;\\\\\n$P(X=3)=\\dfrac{C_6^3C_4^1}{C_{10}^4}=\\dfrac{8}{21}$;\\\\\n$P(X=4)=\\dfrac{C_6^4C_4^0}{C_{10}^4}=\\dfrac{1}{14}$.\\\\\n故$X$的分布为$\\begin{pmatrix}\n0&1&2&3&4\\\\\n\\dfrac{1}{210}&\\dfrac{4}{35}&\\dfrac{3}{7}&\\dfrac{8}{21}&\\dfrac{1}{14}\n\\end{pmatrix},$\\\\\n期望$E[X]=0\\times \\dfrac{1}{210}+1\\times \\dfrac{4}{35}+2\\times \\dfrac{3}{7}+3\\times \\dfrac{8}{21}+4\\times \\dfrac{1}{14}=\\dfrac{12}{5}$.", "duration": -1, "usages": [], "origin": "2023届高三下学期月考2试题19", @@ -441427,8 +441427,8 @@ "objs": [], "tags": [], "genre": "解答题", - "ans": "", - "solution": "", + "ans": "(1) $\\dfrac 12$; (2) 证明略; (3) $\\dfrac 92$", + "solution": "(1) 椭圆的离心率$e=\\dfrac{1}{2}$;\\\\\n(2) 证明: 当$x_0=2$时，$y_0=0,$ 过点$P$的椭圆$C$的切线方程为 $x=2$,符合$\\dfrac{x_0 x}{4}+\\dfrac{y_0 y}{3}=1$，\\\\\n同理，当$x_0=-2$时，也符合；\\\\\n当$x_0\\neq\\pm 2$时，设过点$P$的椭圆$C$的切线方程为$y-y_0=k(x-x_0)$($k$存在),\\\\\n联立$\\begin{cases}\ny-y_0=k(x-x_0),\\\\\n \\dfrac{x^2}{4}+\\dfrac{y^2}{3}=1\n\\end{cases}$得$(3+4k^2)x^2+8k(y-kx_0)x+4(y_0-kx_0)^2-12=0$,\\\\\n$\\Delta=0$得$(x_0^2-4)k^2-2x_0y_0k+y_0^2-3=0$,解得$k=\\dfrac{x_0y_0}{x_0^2-4}=\\dfrac{x_0y_0}{4(1-\\dfrac{y_0^2}{3})-4}=-\\dfrac{3x_0}{4y_0}.$\\\\\n故$y-y_0=-\\dfrac{3x_0}{4y_0}(x-x_0)$,即$\\dfrac{x_0 x}{4}+\\dfrac{y_0 y}{3}=1$.\\\\\n综上，过点$P$的椭圆$C$的切线方程为$\\dfrac{x_0 x}{4}+\\dfrac{y_0 y}{3}=1$.\\\\\n(3) 设$A(x_1,y_1).B(x_2,y_2),x_1\\neq x_2,M(4,t)$.\\\\\n则切线$MA:\\dfrac{x_1x}{4}+\\dfrac{y_1y}{3}=1,$代入$(4,t)$得$x_1+\\dfrac{ty_1}{3}=1$,\\\\\n同理，$x_1+\\dfrac{ty_1}{3}=1$，\\\\\n故$A(x_1,y_1),B(x_2,y_2)$在直线$x+\\dfrac{ty}{3}=1$上，故直线$AB:x=-\\dfrac{ty}{3}+1$.\\\\\n联立$\\begin{cases}\nx=-\\dfrac{ty}{3}+1,\\\\\n\\dfrac{x^2}{4}+\\dfrac{y^2}{3}=1\n\\end{cases}$\n得\n$(4+\\dfrac{t^2}{3})y^2-2ty-9=0$, $\\Delta=16t^2+144>0$,\\\\\n$|AB|=\\sqrt{1+\\dfrac{t^2}{9}}\\cdot |y_1-y_2|=\\sqrt{1+\\dfrac{t^2}{9}}\\cdot \\dfrac{\\sqrt{16t^2+144}}{4+\\dfrac{t^2}{3}}$,\\\\\n$M$到直线$AB$的距离$d=\\dfrac{|4+\\dfrac{t^2}{3}-1|}{\\sqrt{1+\\dfrac{t^2}{9}}}=\\dfrac{3+\\dfrac{t^2}{3}}{\\sqrt{1+\\dfrac{t^2}{9}}}$,\\\\\n$\\triangle MAB$的面积$S=\\dfrac{1}{2}|AB|\\cdot d=\\dfrac{1}{2}\\cdot \\sqrt{1+\\dfrac{t^2}{9}}\\cdot \\dfrac{\\sqrt{16t^2+144}}{4+\\dfrac{t^2}{3}}\\cdot \\dfrac{3+\\dfrac{t^2}{3}}{\\sqrt{1+\\dfrac{t^2}{9}}}=\\dfrac{2(t^2+9)\\sqrt{t^2+9}}{t^2+12}$,\\\\\n令$\\lambda=\\sqrt{t^2+9}\\geq3, S=f(\\lambda)=\\dfrac{2\\lambda^3}{\\lambda^2+3},$\n则$f^{'} (\\lambda)=\\dfrac{2\\lambda^4+18\\lambda^2}{(\\lambda^2+3)^2}>0$,故$f(\\lambda)$在$[3,+\\infty)$严格增，$f(\\lambda)_{\\min}=f(3)=\\dfrac{9}{2}$,故$\\triangle MAB$的面积的最小值为$\\dfrac{9}{2},$ 此时$M$的坐标为$(4,0)$.", "duration": -1, "usages": [], "origin": "2023届高三下学期月考2试题20", @@ -441447,8 +441447,8 @@ "objs": [], "tags": [], "genre": "解答题", - "ans": "", - "solution": "", + "ans": "(1) 证明略; (2) $3$; (3) 不存在, 证明略", + "solution": "(1) $x_1=\\dfrac{1}{2},x_{n+1}=g(x_n)=\\dfrac{x_n}{x_{n+1}},$故$x_n>0,\\dfrac{1}{x_{n+1}}=\\dfrac{x_n+1}{x_n}=\\dfrac{1}{x_n}+1$,即$\\dfrac{1}{x_{n+1}}-\\dfrac{1}{x_{n}}=1$,\\\\因此数列$\\{\\dfrac{1}{x_n}\\}$是以$2$为首项，$1$为公差的等差数列.\\\\\n(2) 对任意$x>0$ 均有$f(x)-mg(x)=\\ln (x+1)-\\dfrac{mx}{x+1}+1>0,$\\\\\n令$h(x)=\\ln (x+1)-\\dfrac{mx}{x+1}+1,x>0,$则$h^{'}(x)=\\dfrac{x+1-m}{(x+1)^2}$.\\\\\n当$m\\geq4$时，$h(1)=\\ln 2-\\dfrac{m}{2}+1<2-\\dfrac{m}{2}<0,$这与$h(x)>0$对$x>0$恒成立矛盾；\\\\\n当$m=3$时，$h(x)=\\ln (x+1)-\\dfrac{3x}{x+1}+1,h^{'}(x)\\dfrac{x-2}{(x+1)^2}.h(x)$在$(0,2]$严格减，在$[2,+\\infty)$严格增，故$h(x)_{\\min}=h(2)=\\ln 3-2+1=\\ln 3-1>0$,符合题意.\\\\\n综上，整数$m$的最大值为$3$.\\\\\n(3) 对任意正整数$t$,取$n=100t,$则 \\\\$\\displaystyle f(n-t)=f(99t)=\\ln (1+99t)=\\ln \\dfrac{1+99t}{99t}+\\ln \\dfrac{99t}{99t-1}+\\cdots +\\ln \\dfrac{2}{1}=\\sum_{k=1}^{99t}\\ln (1+\\dfrac{1}{k}).$\\\\\n$\\displaystyle n-\\sum_{k=1}^{100t}g(k)=\\sum_{k=1}^{100t}(1- g(k))=\\sum_{k=1}^{100t} \\dfrac{1}{1+k}=\\sum_{k=1}^{99t} \\dfrac{1}{1+k}+\\sum_{k=99t+1}^{100t} \\dfrac{1}{1+k}.$\\\\\n$\\displaystyle f(n-t)-[ n-\\sum_{k=1}^{100t}g(k)]=\\sum_{k=1}^{99t} (\\ln (1+\\dfrac{1}{k})-\\dfrac{1}{1+k})-\\sum_{k=99t+1}^{100t} \\dfrac{1}{1+k}.$\\\\\n令$H(x)=\\ln (x+1)-\\dfrac{x}{x+1},x>0$,则$H^{'}(x)=\\dfrac{x}{(x+1)^2}>0$恒成立，故$H(x)$在$(0,+\\infty)$严格增，$H(x)>H(0)=0$，故$\\ln (1+x)>\\dfrac{x}{x+1}$对$x>0$恒成立.\\\\\n因此$\\ln (1+\\dfrac{1}{k})>\\dfrac{\\dfrac{1}{k}}{1+\\dfrac{1}{k}}=\\dfrac{1}{1+k}$,\\\\\n$\\displaystyle f(n-t)-[ n-\\sum_{k=1}^{100t}g(k)]=\\sum_{k=1}^{99t} (\\ln (1+\\dfrac{1}{k})-\\dfrac{1}{1+k})-\\sum_{k=99t+1}^{100t} \\dfrac{1}{1+k}$\\\\\n$\\displaystyle>\\sum_{k=2}^{99t} (\\ln (1+\\dfrac{1}{k})-\\dfrac{1}{1+k})+(\\ln 2-\\dfrac{1}{2})- \\dfrac{t}{1+99t+1}>\\ln 2-\\dfrac{1}{2}-\\dfrac{1}{99+\\dfrac{2}{t}}>0.1-\\dfrac{1}{99}>0.$\\\\\n因此不存在正整数$t$使得对任意$n \\in \\mathbf{N}$, $n \\geq t$, 都有$\\displaystyle f(n-t)