2025成年人數(shù)學(xué)數(shù)列單調(diào)性判斷考試題及答案

一、單項(xiàng)選擇題（每題2分，共10題）1.數(shù)列\(zhòng)(a_n=2n+1\)是（）A.遞增數(shù)列B.遞減數(shù)列C.常數(shù)列D.擺動(dòng)數(shù)列2.數(shù)列\(zhòng)(a_n=-3n\)的單調(diào)性為（）A.遞增B.遞減C.先增后減D.先減后增3.數(shù)列\(zhòng)(a_n=5\)是（）A.遞增數(shù)列B.遞減數(shù)列C.常數(shù)列D.擺動(dòng)數(shù)列4.數(shù)列\(zhòng)(a_n=n^2-n\)在\(n\geq1\)時(shí)是（）A.遞增數(shù)列B.遞減數(shù)列C.常數(shù)列D.擺動(dòng)數(shù)列5.數(shù)列\(zhòng)(a_n=(\frac{1}{2})^n\)是（）A.遞增數(shù)列B.遞減數(shù)列C.常數(shù)列D.擺動(dòng)數(shù)列6.若數(shù)列\(zhòng)(a_n\)滿足\(a_{n+1}-a_n>0\)，則\(a_n\)是（）A.遞增數(shù)列B.遞減數(shù)列C.常數(shù)列D.擺動(dòng)數(shù)列7.數(shù)列\(zhòng)(a_n=3-2n\)的單調(diào)性是（）A.遞增B.遞減C.先增后減D.先減后增8.數(shù)列\(zhòng)(a_n=n^2\)在\(n\inN^+\)上是（）A.遞增數(shù)列B.遞減數(shù)列C.常數(shù)列D.擺動(dòng)數(shù)列9.數(shù)列\(zhòng)(a_n=\log_2n\)是（）A.遞增數(shù)列B.遞減數(shù)列C.常數(shù)列D.擺動(dòng)數(shù)列10.數(shù)列\(zhòng)(a_n=(-1)^n\)是（）A.遞增數(shù)列B.遞減數(shù)列C.常數(shù)列D.擺動(dòng)數(shù)列二、多項(xiàng)選擇題（每題2分，共10題）1.以下哪些數(shù)列是遞增數(shù)列（）A.\(a_n=3n\)B.\(a_n=n^2+1\)C.\(a_n=2^n\)D.\(a_n=\frac{1}{n}\)2.下列數(shù)列中是遞減數(shù)列的有（）A.\(a_n=-n+1\)B.\(a_n=(\frac{1}{3})^n\)C.\(a_n=-n^2\)D.\(a_n=\log_{\frac{1}{2}}n\)3.關(guān)于數(shù)列單調(diào)性，正確的有（）A.若\(a_{n+1}>a_n\)恒成立，則數(shù)列遞增B.若\(a_{n+1}<a_n\)恒成立，則數(shù)列遞減C.常數(shù)列不具有單調(diào)性D.擺動(dòng)數(shù)列也有單調(diào)性4.數(shù)列\(zhòng)(a_n=n^2-3n\)（）A.在\(n\geq2\)時(shí)遞增B.在\(n\leq1\)時(shí)遞減C.最小值在\(n=\frac{3}{2}\)處取得D.是遞增數(shù)列5.以下數(shù)列中，有單調(diào)性的是（）A.\(a_n=2n^2-4n\)B.\(a_n=1\)C.\(a_n=(-2)^n\)D.\(a_n=\lnn\)6.下列判斷正確的是（）A.數(shù)列\(zhòng)(a_n=5n-1\)遞增B.數(shù)列\(zhòng)(a_n=4-n\)遞減C.數(shù)列\(zhòng)(a_n=3^n\)遞增D.數(shù)列\(zhòng)(a_n=(\frac{1}{4})^n\)遞減7.若數(shù)列\(zhòng)(a_n\)滿足\(a_{n+1}-a_n=2\)，則（）A.\(a_n\)是遞增數(shù)列B.\(a_n\)是等差數(shù)列C.\(a_n\)可能是常數(shù)列D.\(a_n\)的公差為28.數(shù)列\(zhòng)(a_n=\frac{n}{n+1}\)（）A.遞增B.遞減C.極限為1D.是擺動(dòng)數(shù)列9.以下哪些是判斷數(shù)列單調(diào)性的方法（）A.作差法B.作商法C.圖像法D.歸納法10.數(shù)列\(zhòng)(a_n=n\sin\frac{n\pi}{2}\)（）A.不是遞增數(shù)列B.不是遞減數(shù)列C.是擺動(dòng)數(shù)列D.有單調(diào)性三、判斷題（每題2分，共10題）1.數(shù)列\(zhòng)(a_n=4\)是遞增數(shù)列。（）2.若\(a_{n+1}-a_n<0\)，則數(shù)列\(zhòng)(a_n\)遞減。（）3.數(shù)列\(zhòng)(a_n=n^2+1\)在\(n\inN^+\)上遞增。（）4.擺動(dòng)數(shù)列一定沒有單調(diào)性。（）5.數(shù)列\(zhòng)(a_n=\frac{1}{n+1}\)是遞減數(shù)列。（）6.常數(shù)列既不是遞增數(shù)列也不是遞減數(shù)列。（）7.若數(shù)列\(zhòng)(a_n\)滿足\(a_{n+1}=2a_n\)，則\(a_n\)遞增。（）8.數(shù)列\(zhòng)(a_n=\log_3n\)在\(n\geq1\)時(shí)遞增。（）9.數(shù)列\(zhòng)(a_n=(-1)^nn\)是遞增數(shù)列。（）10.用圖像法可以判斷數(shù)列的單調(diào)性。（）四、簡答題（每題5分，共4題）1.簡述判斷數(shù)列單調(diào)性的作差法。2.已知數(shù)列\(zhòng)(a_n=n^2-5n\)，判斷其單調(diào)性。3.如何用定義判斷數(shù)列\(zhòng)(a_n\)是遞增數(shù)列？4.對于數(shù)列\(zhòng)(a_n=2^n-n\)，判斷其單調(diào)性。五、討論題（每題5分，共4題）1.討論數(shù)列\(zhòng)(a_n=\frac{n}{n^2+1}\)的單調(diào)性及最值情況。2.探討數(shù)列單調(diào)性在實(shí)際生活中的應(yīng)用實(shí)例。3.當(dāng)數(shù)列\(zhòng)(a_n\)的通項(xiàng)公式較復(fù)雜時(shí)，如何選擇合適方法判斷單調(diào)性？4.若數(shù)列\(zhòng)(a_n\)和\(b_n\)都是遞增數(shù)列，那么數(shù)列\(zhòng)(a_n+b_n\)和\(a_n\cdotb_n\)的單調(diào)性如何？答案一、單項(xiàng)選擇題1.A2.B3.C4.A5.B6.A7.B8.A9.A10.D二、多項(xiàng)選擇題1.ABC2.ABCD3.AB4.AB5.AD6.ABCD7.ABD8.AC9.ABC10.ABC三、判斷題1.×2.√3.√4.×5.√6.√7.×8.√9.×10.√四、簡答題1.作差法：對于數(shù)列\(zhòng)(a_n\)，計(jì)算\(a_{n+1}-a_n\)，若\(a_{n+1}-a_n>0\)，則數(shù)列遞增；若\(a_{n+1}-a_n<0\)，則數(shù)列遞減；若\(a_{n+1}-a_n=0\)，則數(shù)列為常數(shù)列。2.\(a_{n+1}-a_n=(n+1)^2-5(n+1)-(n^2-5n)=2n-4\)，當(dāng)\(n\geq2\)時(shí)，\(a_{n+1}-a_n\geq0\)，數(shù)列遞增；當(dāng)\(n=1\)時(shí)，\(a_{n+1}-a_n<0\)，數(shù)列遞減。3.若對于任意的\(n\inN^+\)，都有\(zhòng)(a_{n+1}-a_n>0\)，即\(a_{n+1}>a_n\)，那么數(shù)列\(zhòng)(a_n\)是遞增數(shù)列。4.\(a_{n+1}-a_n=2^{n+1}-(n+1)-(2^n-n)=2^n-1>0\)，所以數(shù)列\(zhòng)(a_n\)遞增。五、討論題1.\(a_{n+1}-a_n=\frac{n+1}{(n+1)^2+1}-\frac{n}{n^2+1}=\frac{1-n^2-n}{[(n+1)^2+1](n^2+1)}\)。當(dāng)\(n=1\)時(shí)，\(a_2-a_1>0\)；\(n\geq2\)時(shí)，\(a_{n+1}-a_n<0\)。所以\(a_1<a_2\)，\(a_2>a_3>a_4>\cdots\)，最大值為\(a_2=\frac{2}{5}\)。2.例如在經(jīng)濟(jì)領(lǐng)域，分析企業(yè)利潤隨時(shí)間變化的數(shù)列，判斷其單調(diào)性可了解企業(yè)發(fā)展趨勢；在人口增長問題中，通過數(shù)列單調(diào)性分析人口變化情況，制定相關(guān)政策。3.若通項(xiàng)含指數(shù)、分式等，可嘗試作商法；若通項(xiàng)是多項(xiàng)式，作差法較合適；也可結(jié)合函數(shù)性質(zhì)，將數(shù)列看作函數(shù)上的離散點(diǎn)，利用函數(shù)單調(diào)性判斷。還可先列舉前幾項(xiàng)觀察規(guī)律再判斷。4.\(a_n+b_n\)一定遞增，因?yàn)閈((a_{n+1}