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2 第2课时对数函数及其性质的应用(习题课) 纯答案

第2课时对数函数及其性质的应用(习题课)答案比较对数值的大小【解】 (1)因为函数y ＝ln x 是增函数，且0.3＜2，所以ln 0.3＜ln 2.(2)当a ＞1时，函数y ＝log a x 在(0，＋∞)上是增函数，又3.1＜5.2，所以log a 3.1＜log a 5.2；当0＜a ＜1时，函数y ＝log a x 在(0，＋∞)上是减函数．又3.1＜5.2，所以log a 3.1＞log a 5.2. (3)因为0＞log 0.23＞log 0.24，所以1log 0.23＜1log 0.24，即log 30.2＜log 40.2.(4)因为函数y ＝log 3x 是增函数，且π＞3，所以log 3π＞log 33＝1，同理，1＝log ππ＞log π3，即log 3π＞log π3.1．解析：选D.因为log 0.4x 为减函数，故log 0.44>log 0.46，故A 错；因为1.01x 为增函数，所以1.013.4<1.013.5，故B 错；由指数函数图象特点知，3.50.3>3.40.3，故C 错．2．解析：选A.因为a ＝30.5>1，b ＝log 312<0，0<c ＝log 32<1，所以a >c >b .解对数不等式【解】 (1)由题意可得⎩⎪⎨⎪⎧x ＞0，4－x ＞0，x ＜4－x ，解得0＜x ＜2.所以原不等式的解集为(0，2)． (2)当x ＞1时，log x 12＞1＝log x x ，解得x ＜12，此时不等式无解．当0＜x ＜1时，log x 12＞1＝log x x ，解得x ＞12，所以12＜x ＜1.综上所述，原不等式的解集为⎪⎭⎫⎝⎛1,21. (3)当a ＞1时，原不等式等价于⎩⎪⎨⎪⎧2x －5＞0，x －1＞0，2x －5＞x －1，解得x ＞4.当0＜a ＜1时，原不等式等价于 ⎩⎪⎨⎪⎧2x －5＞0，x －1＞0，2x －5＜x －1，解得52＜x ＜4.综上所述，当a ＞1时，原不等式的解集为{x |x ＞4}；当0＜a ＜1时，原不等式的解集为⎩⎨⎧⎭⎬⎫x |52＜x ＜4.1．解析：因为函数y ＝log 0.2x 在(0，＋∞)上是减函数，所以由log 0.22x <log 0.2(x －1)得⎩⎪⎨⎪⎧2x >0，x －1>0，2x >x －1，解得x >1，即x的取值范围为(1，＋∞)．答案：(1，＋∞)2．解：由题意知log a (3a －1)>0＝log a 1. 当a >1时，y ＝log a x 是增函数，所以⎩⎪⎨⎪⎧3a －1>1，3a －1>0，解得a >23，所以a >1；当0<a <1时，y ＝log a x 是减函数，所以⎩⎪⎨⎪⎧3a －1<1，3a －1>0，解得13<a <23.所以13<a <23.综上所述，a 的取值范围是⎪⎭⎫⎝⎛3231，∪(1，＋∞)．对数型函数的单调性【解】 (1)由4x －1>0，解得x >0，因此f (x )的定义域为(0，＋∞)． (2)设0<x 1<x 2，则0<4x 1－1<4x 2－1，因此log 4(4x 1－1)<log 4(4x 2－1)，即f (x 1)<f (x 2)，故f (x )在(0，＋∞)上单调递增．(3)因为f (x )在区间⎥⎦⎤⎢⎣⎡2,21上单调递增，又f ⎪⎭⎫⎝⎛21＝0，f (2)＝log 415，因此f (x )在区间⎥⎦⎤⎢⎣⎡2,21上的值域为[0，log 415]．解：因为1－2x ＞0，所以x ＜12.又设u ＝1－2x ，则y ＝f (u )是(0，＋∞)上的增函数．又u ＝1－2x ，则x ∈()⎝⎛⎭⎫－∞，12时，u (x )是减函数，所以函数f (x )＝log 2(1－2x )的单调递减区间是⎝⎛⎭⎫－∞，12，无单调递增区间．与对数函数有关的值域与最值问题【解】 (1)由题意得⎩⎪⎨⎪⎧1＋x >0，3－x >0，解得－1<x <3.所以f (x )的定义域为(－1，3)．(2)f (x )＝log a [(1＋x )(3－x )]＝log a (－x 2＋2x ＋3) ＝log a [－(x －1)2＋4]，－1<x <3，若0<a <1，则当x ＝1时，f (x )有最小值log a 4，所以log a 4＝－2，即a －2＝4，又0<a <1，所以a ＝12.若a >1，则当x ＝1时，f (x )有最大值log a 4，f (x )无最小值．综上可知，a ＝12.解：(1)由⎩⎪⎨⎪⎧f （1）＝1，f （2）＝log 212，得⎩⎪⎨⎪⎧log 2（a －b ）＝1，log 2（a 2－b 2）＝log 212，所以⎩⎪⎨⎪⎧a －b ＝2，a 2－b 2＝12，即⎩⎪⎨⎪⎧a －b ＝2，a ＋b ＝6，所以a ＝4，b ＝2.(2)由(1)知f (x )＝log 2(4x －2x )，设t ＝2x ，因为x ∈[1，3]，所以t ∈[2，8]．令u ＝4x－2x＝t 2－t ＝⎝⎛⎭⎫t －122－14，所以当t ＝8，即x ＝3时，u 最大，u max ＝56，故f (x )的最大值为log 256.1．解析：选C.因为x ≥2，所以log 2x ≥1，所以y ≥3. 2．解析：选B.易知函数y ＝lg|x |是偶函数．当x >0时，y ＝lg|x |＝lg x ，所以在区间(0，＋∞)上单调递增．由偶函数的性质知，函数在区间(－∞，0)上单调递减．3．解析：选C.由题意知，f (x )＝log a x (0<a <1)为减函数，则f (x )max ＝f (a )＝1，f (x )min ＝f (2a )＝1＋log a 2，所以1＝3(1＋log a 2)，即log a 2＝－23，解得a －23＝2，即a ＝24，故选C.4．解析：因为y ＝log 5x 与y ＝2x ＋1均为增函数，故函数f (x )＝log 5(2x ＋1)是其定义域上的增函数，所以函数f (x ) 的单调增区间是⎝⎛⎭⎫－12，＋∞. 答案：⎝⎛⎭⎫－12，＋∞ 5．解：设f (x )＝log a x (a >0且a ≠1)，因为f (4)＝2，所以log a 4＝2，所以a ＝2，所以f (x )＝log 2x ，所以f (2x －3)>f (x )⇒log 2(2x －3)>log 2x ⇒⎩⎪⎨⎪⎧2x －3>0，x >0，2x －3>x ⇒x >3，所以原不等式的解集为(3，＋∞)．[A 基础达标]1．解析：选C.由指数函数的性质可知，函数y ＝0.75x 为单调递减函数，又因为－0.1<0.1，所以0.75－0.1>0.750.1.2．解析：选D.f (x )的图象如图所示，由图象可知单调递增区间为[1，＋∞)．3．解析：选D.由函数f (x )的解析式知定义域为⎪⎭⎫ ⎝⎛∞+，61，设t ＝2x －13(t >0)，t 在⎪⎭⎫ ⎝⎛∞+，61上是增函数，y ＝21log t在(0，＋∞)上是减函数，由复合函数的单调性可知f (x )在⎪⎭⎫ ⎝⎛∞+，61上是减函数，故选D. 4．解析：选B.因为a x ≥1＝a 0的解集为{x |x ≤0}，所以0<a <1，所以x 2＋2≥2. 又因为函数y ＝log a (x 2＋2)的最大值为－1，则a ＝12.5．解析：选B.因为f (x )＝log 3x ，所以f (x )在(0，＋∞)上为增函数．又因为2>12>14，所以f (2)>f ⎪⎭⎫ ⎝⎛21>f ⎪⎭⎫ ⎝⎛41.6．解析：由y ＝log (2a －3)x 在(0，＋∞)上是增函数，所以2a －3>1，解得a >2. 答案：(2，＋∞)7．解析：由⎩⎪⎨⎪⎧2x ＋3＞0，5x －6＞0，2x ＋3＞5x －6，解得⎩⎪⎨⎪⎧x ＞－32，x ＞65，x ＜3，即65＜x ＜3，故不等式的解集为{x |65＜x ＜3}．答案：{x |65＜x ＜3}8．解析：因为a ＞1，所以f (x )＝log a x 在[a ，2a ]上递增，所以log a (2a )－log a a ＝12，即log a 2＝12，所以a 12＝2，a ＝4.答案：4 9．解：(1)因为y ＝log 3.1x 在(0，＋∞)上是增函数，所以log 3.10.5>log 3.10.2. (2)法一：因为y ＝21log x 在(0，＋∞)上是减函数，所以21log 8<21log 4.法二：21log 8＝－3，21log 4＝－2，由－3<－2知21log 8<21log 4.(3)因为log 56>log 55＝1，log 65<log 66＝1，所以log 56>log 65. 10．解：要使y ＝21log (1－x 2)有意义，则1－x 2＞0，所以x 2＜1，则－1＜x ＜1，因此函数的定义域为(－1，1)．令t ＝1－x 2，x ∈(－1，1)．当x ∈(－1，0]时，x 增大，t 增大，y ＝21log t 减小，所以当x ∈(－1，0]时，y ＝21log (1－x 2)是减函数；同理当x ∈[0，1)时，y ＝21log (1－x 2)是增函数．故函数y ＝21log (1－x 2)的单调增区间为[0，1)，且函数的最小值y min ＝21log (1－02)＝0.[B 能力提升]11．解析：选D.因为0<12<1，21log m <21log n <0，所以m >n >1，故选D.12．解析：选C.当－1<x <0时，0<x ＋1<1. 因为log a |x ＋1|>0，所以0<a <1，所以函数f (x )＝log a |x ＋1|在(－∞，－1)上递增，在(－1，＋∞)上递减． 13．解：(1)因为g (9)＝log a 9＝2，解得a ＝3，所以g (x )＝log 3x .因为函数y ＝f (x )的图象与g (x )＝log 3x 的图象关于x 轴对称，所以f (x )＝31log x .(2)因为f (3x －1)>f (－x ＋5)，所以31log (3x －1)>31log (－x ＋5)，则⎩⎪⎨⎪⎧3x －1>0，－x ＋5>0，3x －1<－x ＋5，解得13<x <32，即x 的取值范围为⎝⎛⎭⎫13，32.14．解：设t ＝x 2－2x ＋3＝(x －1)2＋2. 当x ∈R 时，t 有最小值2. 所以lg(x 2－2x ＋3)的最小值为lg 2.又因为y ＝a lg(x 2－2x ＋3)有最大值，所以0<a <1. 由f (x )＝log a (3－2x )，得其定义域为⎝⎛⎭⎫－∞，32. 设u (x )＝3－2x ，x ∈⎝⎛⎭⎫－∞，32，则f (x )＝log a u (x )．因为u (x )＝3－2x 在⎝⎛⎭⎫－∞，32上是减函数，所以f (x )＝log a u (x )在⎝⎛⎭⎫－∞，32上是增函数．所以f (x )＝log a (3－2x )的单调增区间为⎝⎛⎭⎫－∞，32. [C 拓展探究]15．解：(1)由题设，3－ax >0对x ∈[0，2]恒成立，且a >0，a ≠1.设g (x )＝3－ax ，则g (x )在[0，2]上为减函数，所以g (x )min ＝g (2)＝3－2a >0，所以a <32.所以实数a 的取值范围是(0，1)∪⎝⎛⎭⎫1，32. (2)假设存在这样的实数a ，则由题设知f (1)＝1，即log a (3－a )＝1，所以a ＝32.此时f (x )＝⎪⎭⎫ ⎝⎛-x 233log 23 但x ＝2时，f (x )＝0log 23无意义．故这样的实数a 不存在．。

《工程流体力学(水力学)》第二版禹华谦课后习题答案西南交通大学出版社

《工程流体力学(水力学)》第二版禹华谦课后习题答案西南交通大学出版社欢迎光临阳光大学生网, 提供最全面的大学生课后习题答案和复习试题免费下载,////0>.阳光大学生网我们希望呵护您的眼睛,关注您的成长,给您一片绿色的环境,欢迎加入我们,一起分享大学里的学习和生活感悟,免费提供:大学生课后答案 ,大学考试题及答案 ,大学生励志书籍。

《水力学》李炜徐孝平主编 2000 年 6 月武汉水利电力大学出版社共 12 章全部习题的解答第一章1-1 解:3 3 3ρ 1.03g cm 1030kg m , 比重s 1.03, γ 10.094kN m1-2 解:2γ9789N /m3ρ 998.88kg m ,g 9.8?3 2μ gμ9.8 ×1.002 ×10 N ?S /m?6 2ν 1.003 ×10 m /sργ 9789?4γ11.82 × 0.15 ×10?5 2以上为水,以下为空气μρνν 1.089 ×10 N ?S /m g 9.81-3 解:d ν9 7dp ?K ?2.19 ×10 × ?1% 2.19 ×10 Pav1-4 解:3 3γ G v 0.678 /10 678kgf /m①用工程单位制:2 4ργ g 678 / 9.8 69.18kgfs /mγγ ×9.8N kgf 6644.4N m②用国单位制: (SI 制) :3ργ g 678kg m1-5 解:du u 1.531流速梯度 3.75 ×10 3sdy δ 0.4 ×10u3 2切应力τμ 0.1 ×3.75 ×10 3.75 ×10 Paδ2活塞所受的摩擦阻力 F τ A τπdl 3.75 ×10 ×3.14 ×0.14 ×0.16 26.38N1-6 解:作用在侧壁上粘性切力产生的力矩du r 0.2M A μr 2 πr h μω+1 2 ×3.14 × 0.2 × 0.4 × μ×101 + 68.3 μdy δ 0.003M 4.905∴μ 0.072Pa ?S68.3 68.31-7 解:2设u Ay +By +c; ①根据实际流体的无滑移现象,当 y0 时 u0∴C 0 (第三个常数项为零); ②∵y0.04m 时,u1m/sdu2则有 1A ×0.04 +B ×0.04; ③E 点的流体切应力为零,有 2Ay +B 0 , dy10.0016A + 0.04B 1 A ?625?则由联立方程求得解得:0.08A +B 0 B 50?du du-3?6τμυρ 1.0 ×10 ×1000 × 2 Ay+B )1 ×10 (-1250y+50 )dy dy-2当y0 处,τ 5 ×10 Pa-2当y0.02 处,τ 2.5 ×10 Pa当 y0.04 处,τ0 Pa由此可见均匀流横断面上切应力是呈直线分布的。

Linear Algebra_彭国华_第五章课后答案

Chapter 5 Linear Transformations 3. Proof. 因为Ak = 0, 我们有 (A − E )(Ak−1 + Ak−1 + · · · + E ) = −E. 所以(A − E )−1 = −(Ak−1 + · · · + E ). 4. Proof. Since f (x) and g (x) are relatively prime, there exists u(x) and v(x) such that u(x)f (x) + v (x)g (x) = 1. We have u(A)f (A) + v (A)g (A) = E . Suppose that there exists a non zero vector α in V such that f (A)α = g (A)α = 0. We get α = u(A)f (A)α + v (A)g (A)α = 0. Contradiction. 5. Proof. Since f (x) and g (x) are relatively prime, there exists u(x) and v(x) such that u(x)f (x) + v (x)g (x) = 1. In particular, we have f (A)u(A) + g (A)v (A) = E . For any vector α ∈ V , f (A)u(A)α + g (A)v (A)α = α, where f (A)u(A)α ∈ im f (A) and g (A)v (A)α ∈ im g (A). Thus, we get V = im f (A) + im g (A). 6. Proof. (1) First, we show that V = ker T + ker(T − 2E ). For any v ∈ V , v = 1 2 (T v − (T v − 2v )). We have (T − 2E )(T v ) = 0 which implies T v ∈ ker(T − 2E ). Similarly, we have T v − 2v ∈ ker T . Thus, V ⊂ ker T + ker(T − 2E ) ⊂ V . Next, we show that ker T ∩ ker(T − 2E ) = {0}. Suppose β ∈ ker T ∩ ker(T − 2E ), we have T β = (T − 2E )β = 0 which implies β = 0. We have proved that V = ker T ⊕ ker(T − 2E ). (2) Let α1 , · · · , αr be a basis of ker T and β1 , · · · , βs a basis of ker(T − 2E ). We have Tαi = 0, Tβj = 2βj . Thus T(α1 , · · · , αr , β1 , · · · , βs ) = (α1 , · · · , αr , β1 , · · · , βs ) In particular, T is similar to the diagonal matrix above. 7. Proof. 3 0 4 0 0 0 (1)T = 3 0 0 ; (2)T = −1.2 5.6 2.4 . 4 0 0 −1.2 0.6 −2.6 0 0 0 Es .

《材料力学》课后习题答案详细

N(x) F F x a
x (a,0]
轴力图如图所示。
[习题 2-2] 试求图示等直杆横截面 1-1、2-2 和平 3-3 上的轴力，并作轴
力图。若横截面面积 A 400mm2 ，试
求各横截面上的应力。
解：（1）求指定截面上的轴力
N11 20kN N 22 10 20 10(kN )
10000 100
0
100 100.0 0.0
10000 100
30
100 75.0 43.3
10000 100
45
100 50.0 50.0
10000 100
60
100 25.0 43.3
10000 100
90
100
0.0
0.0
[习题 2-7] 一根等直杆受力如图所示。已知杆的横截面面积 A 和材料的弹性模量 E。试作轴力图，并求杆端点 D 的位移。解：（1）作轴力图
N33 F 2F 2F F
轴力图如图所示。
1
（c）
解：（1）求指定截面上的轴力
N11 2F N22 F 2F F
（2）作轴力图
N33 2F F 2F 3F
轴力图如图所示。
（d）
解：（1）求指定截面上的轴力
N11 F
N 22
2F
qa
F
2F
F a
a
F
2F
（2）作轴力图
中间段的轴力方程为：
解：墩身底面的轴力为：
N (F G) F Alg
1000 (3 2 3.14 12 ) 10 2.35 9.8 3104.942(kN )
1000 (3 2 3.14 12 ) 10 2.35 9.8

数学：3.2.1《对数及其运算》课件(新人教B版必修1)

变式训练 5 (1)log2
求下列各式的值：
1 1 ＋log212－ log26； 24 2
(2)log9 2· 43＋log83)． (log
分析：(1)由于24,12,6都可分解成2和3的这两个质因子的积或幂，所以可运用对数运算性质直接将原式转化为含log23的式子再化简即可．或利用题中各对数均为同底的对数，可逆用运算性质将之化为一个对数的计算问题． (2)所含对数底数不同，因此可考虑用对数换底公式化为同底对数的运算．
(1)解法一：log2
1 1 ＋log212－ log26 24 2
分析：根据对数式的定义求解．
1 解：(1)log3 ＝x； 27 1 1 (3)log5 ＝－； 2 5 (5)10－3＝0.001； (4)( 2)4＝4； (6)( 2－1)－1＝ 2＋1.
评析：明确对数的定义是解题的关键．指数式与对数式是互逆的，二者能够相互转化，熟练掌握二者的互化，能够加深对指数式和对数式的理解，为后面学习对数函数打下坚实的基础．
分析：利用对数的性质求值，首先要明确解题目标是化异为同，先使各项底数相同，才能使用性质，再找真数间的联系，对于复杂的真数，可以先化简再计算．
解：(1)原式＝2log32－(log332－log39)＋3log32－3 ＝2log32－5log32＋2＋3log32－3＝－1. 10 (2)原式＝2lg5＋2lg2＋lg · lg(2×10)＋(lg2)2 2 ＝2lg(5×2)＋(1－lg2)· (lg2＋1)＋(lg2)2 ＝2＋1－(lg2)2＋(lg2)2＝3. 1 log 2· 73 2log log5 2· 79 2 5 log (3)∵ ＝ 1 1 3 log5 · 7 4 －log53·log74 log 3 3 lg2 lg3 · lg5 lg7 3 ＝－＝－， lg3 1lg4 2 · lg5 3lg7

Richard组合数学第5版-第5章课后习题答案（英文版）

Richard组合数学第5版-第5章课后习题答案（英⽂版）Math475Text:Brualdi,Introductory Combinatorics5th Ed. Prof:Paul TerwilligerSelected solutions for Chapter51.For an integer k and a real number n,we shown k=n?1k?1+n?1k.First assume k≤?1.Then each side equals0.Next assume k=0.Then each side equals 1.Next assume k≥1.RecallP(n,k)=n(n?1)(n?2)···(n?k+1).We haven k=P(n,k)k!=nP(n?1,k?1)k!.n?1 k?1=P(n?1,k?1)(k?1)!=kP(n?1,k?1)k!.n?1k(n?k)P(n?1,k?1)k!.The result follows.2.Pascal’s triangle begins111121133114641151010511615201561172135352171182856705628811936841261268436911104512021025221012045101···13.Let Z denote the set of integers.For nonnegative n∈Z de?ne F(n)=k∈Zn?kk.The sum is well de?ned since?nitely many summands are nonzero.We have F(0)=1and F(1)=1.We show F(n)=F(n?1)+F(n?2)for n≥2.Let n be /doc/6215673729.htmling Pascal’s formula and a change of variables k=h+1,F(n)=k∈Zn?kk=k∈Zn?k?1k?1=k∈Zn?k?1k+h∈Zn?h?2h=F(n?1)+F(n?2).Thus F(n)is the n th Fibonacci number.4.We have(x+y)5=x5+5x4y+10x3y2+10x2y3+5xy4+y5and(x+y)6=x6+6x5y+15x4y2+20x3y3+15x2y4+6xy5+y6.5.We have(2x?y)7=7k=07k27?k(?1)k x7?k y k.6.The coe?cient of x5y13is35(?2)13 185.The coe?cient of x8y9is0since8+9=18./doc/6215673729.htmling the binomial theorem,3n=(1+2)n=nk=0nSimilarly,for any real number r,(1+r)n=nk=0nkr k./doc/6215673729.htmling the binomial theorem,2n=(3?1)n=nk=0(?1)knk3n?k.29.We haven k =0(?1)k nk 10k =(?1)n n k =0(?1)n ?k n k 10k =(?1)n (10?1)n =(?1)n 9n .The sum is 9n for n even and ?9n for n odd.10.Given integers 1≤k ≤n we showk n k =n n ?1k ?1.Let S denote the set of ordered pairs (x,y )such that x is a k -subset of {1,2,...,n }and yis an element of x .We compute |S |in two ways.(i)To obtain an element (x,y )of S there are n k choices for x ,and for each x there are k choices for y .Therefore |S |=k n k .(ii)Toobtain an element (x,y )of S there are n choices for y ,and for each y there are n ?1k ?1 choices for x .Therefore |S |=n n ?1k ?1.The result follows.11.Given integers n ≥3and 1≤k ≤n .We shown k ? n ?3k = n ?1k ?1 + n ?2k ?1 + n ?3k ?1.Let S denote the set of k -subsets of {1,2,...,n }.Let S 1consist of the elements in S thatcontain 1.Let S 2consist of the elements in S that contain 2but not 1.Let S 3consist of the elements in S that contain 3but not 1or 2.Let S 4consist of the elements in S that do|S |= n k ,|S 1|= n ?1k ?1 ,|S 2|= n ?2k ?1 ,|S 3|= n ?3k ?1 ,|S 4|= n ?3k .The result follows.12.We evaluate the sumnk =0(?1)k nk 2.First assume that n =2m +1is odd.Then for 0≤k ≤m the k -summand and the (n ?k )-summand are opposite.Therefore the sum equals 0.Next assume that n =2m is even.Toevaluate the sum in this case we compute in two ways the the coe?cient of x n in (1?x 2)n .(i)By the binomial theorem this coe?cient is (?1)m 2m m .(ii)Observe (1?x 2)=(1+x )(1?x ).We have(1+x )n =n k =0n k x k,(1?x )n =n k =0nk (?1)k x k .3By these comments the coe?cient of x n in(1?x2)n isn k=0nn?k(?1)knk=nk=0(?1)knk2.2=(?1)m2mm.13.We show that the given sum is equal ton+3k .The above binomial coe?cient is in row n+3of Pascal’s /doc/6215673729.htmling Pascal’s formula, write the above binomial coe?cient as a sum of two binomial coe?ents in row n+2of Pascal’s triangle.Write each of these as a sum of two binomial coe?ents in row n+1of Pascal’s triangle.Write each of these as a sum of two binomial coe?ents in row n of Pascal’s triangle.The resulting sum isn k+3nk?1+3nk?2+nk?3.14.Given a real number r and integer k such that r=k.We showr k=rr?kr?1k.First assume that k≤?1.Then each side is0.Next assume that k=0.Then each side is 1.Next assume that k≥1.ObserverP(r?1,k?1)k!,andr?1k=P(r?1,k)k!=(r?k)P(r?1,k?1)k!.The result follows.15.For a variable x consider(1?x)n=nk=0nk(?1)k x k.4Take the derivative with respect to x and obtain n(1x)n1=nk=0nk(?1)k kx k?1.Now set x=1to get(?1)k k.The result follows.16.For a variable x consider(1+x)n=nk=0nkx k.Integrate with respect to x and obtain(1+x)n+1 n+1=nk=0nkx k+1k+1+Cfor a constant C.Set x=0to?nd C=1/(n+1).Thus (1+x)n+1?1n+1=nk=0nkx k+1k+1.Now set x=1to get2n+1?1 n+1=k+1.17.Routine.18.For a variable x consider(x?1)n=nk=0nk(?1)n?k x k.Integrate with respect to x and obtain(x?1)n+1 n+1=nk=0nk(?1)n?kx k+1k+1+Cfor a constant C.Set x=0to?nd C=(?1)n+1/(n+1).Thus (x?1)n+1?(?1)n+1n+1=nk=0nk(?1)n?kx k+1k+1Now set x =1to get(?1)n n +1=n k =0n k(?1)n ?k 1k +1.Therefore1n +1=n k =0 n k (?1)k 1k +1 .19.One readily checks2 m 2 + m 1=m (m ?1)+m =m 2.Therefore n k =1k 2=nk =0k 2=2nk =0 k 2 +n k =0k1=2 n +13 +n +12 =(n +1)n (2n +1)6.20.One readily checksm 3=6 m 3 +6 m 2 + m1.Thereforen k =1k3=n=6nk =0 k3+6n k =0 k2 +n k =0k1 =6 n +14 +6 n +13 +n +12 =(n +1)2n 24= n +12 2.621.Given a real number r and an integer k .We showrk=(?1)kr +k ?1k .First assume that k <0.Then each side is zero.Next assume that k ≥0.Observe r k =(r )(r 1)···(r k +1)k !=(?1)kr (r +1)···(r +k ?1)k !=(?1)kr +k ?1k.22.Given a real number r and integers k,m .We showr m m k = r k r ?km ?k.First assume that mObserver m m k =r (r ?1)···(r ?m +1)m !m !k !(m ?k )!=r (r ?1)···(r ?k +1)k !(r ?k )(r ?k ?1)···(r ?m +1)(m ?k )!= r k r ?k m ?k .23.(a) 2410.(b) 94 156.(c) 949363.(d)94156949363.24.The number of walks of length 45is equal to the number of words of length 45involving10x ’s,15y ’s,and 20z ’s.This number is45!10!×15!×20!.725.Given integers m 1,m 2,n ≥0.Shown k =0m 1k m 2n ?k = m 1+m 2n .Let A denote a set with cardinality m 1+m 2.Partition A into subsets A 1,A 2with cardinalitiesm 1and m 2respectively.Let S denote the set of n -subsets of A .We compute |S |in two ways.(i)By construction|S |= m 1+m 2n .(ii)For 0≤k ≤n let the set S k consist of the elements in S whose intersection with A 1has cardinality k .The sets {S k }n k =0partition S ,so |S |= nk =0|S k |.For 0≤k ≤n we now compute |S k |.To do this we construct an element x ∈S k via the following 2-stage procedure: stage to do #choices 1pick x ∩A 1 m 1k2The number |S k |is the product of the entries in the right-most column above,which comes to m 1k m 2n ?k .By these comments |S |=n k =0m 1k m 2n ?k .The result follows.26.For an integer n ≥1shown k =1 n k n k ?1 =12 2n +2n +1 ? 2n n .Using Problem 25,n k =1 n k nk ?1 =n k =0n k n k ?1 =n k =0n k nn +1?k =2n n +1 =12 2n n ?1 +12 2n n +1.8It remains to show12 2nn ?1 +12 2n n +1 =12 2n +2n +1 ? 2n n.This holds since2n n ?1 +2 2n n + 2n n +1 = 2n +1n +2n +1n +1= 2n +2n +1.27.Given an integer n ≥1.We shown (n +1)2n ?2=nk =1Let S denote the set of 3-tuples (s,x,y )such that s is a nonempty subset of {1,2,...,n }and x,y are elements (not necessarily distinct)in s .We compute |S |in two ways.(i)Call an element (s,x,y )of S degenerate whenever x =y .Partition S into subsets S +,S ?with S +(resp.S ?)consisting of the degenerate (resp.nondegenerate)elements of S .So |S |=|S +|+|S ?|.We compute |S +|.To obtain an element (s,x,x )of S +there are n choices for x ,and given x there are 2n ?1choices for s .Therefore |S +|=n 2n ?1.We compute |S ?|.To obtain an element (s,x,y )of S ?there are n choices for x,and given x there are n ?1choices for y ,and given x,y there are 2n ?2choices for s .Therefore |S ?|=n (n ?1)2n ?2.By these comments|S |=n 2n ?1+n (n ?1)2n ?2=n (n +1)2n ?2.(ii)For 1≤k ≤n let S k denote the set of elements (s,x,y )in S such that |s |=k .Thesets {S k }nk =1give a partition of S ,so |S |= n k =1|S k |.For 1≤k ≤n we compute |S k |.To obtain an element (s,x,y )of S k there are n k choices for s ,and given s there are k 2ways to choose the pair x,y .Therefore |S k |=k 2 nk .By these comments|S |=n k =1k 2 n k .The result follows.28.Given an integer n ≥1.We shown k =1k n k 2=n 2n ?1n ?1 .Let S denote the set of ordered pairs (s,x )such that s is a subset of {±1,±2,...,±n }andx is a positive element of s .We compute |S |in two ways.(i)To obtain an element (s,x )of S There are n choices for x ,and given x there are 2n ?1n ?1 choices for s .Therefore|S |=n 2n ?1n ?1.9(ii)For1≤k≤n let S k denote the set of elements(s,x)in S such that s contains exactlyk positive elements.The sets{S k}nk=1partition S,so|S|=nk=1|S k|.For1≤k≤nwe compute|S k|.To obtain an element(s,x)of S k there are nkways to pick the positiveelements of s and nn?kways to pick the negative elements of s.Given s there are kways to pick x.Therefore|S k|=k nk2.By these comments |S|=nk=1knk2.The result follows.29.The given sum is equal tom2+m2+m3n .To see this,compute the coe?cient of x n in each side of(1+x)m1(1+x)m2(1+x)m3=(1+x)m1+m2+m3.In this computation use the binomial theorem.30,31,32.We refer to the proof of Theorem5.3.3in the text.Let A denote an antichain such that|A|=nn/2.For0≤k≤n letαk denote the number of elements in A that have size k.Sonk=0αk=|A|=nn/2.As shown in the proof of Theorem5.3.3,≤1,with equality if and only if each maximal chain contains an element of A.By the above commentsnk=0αknn/2nknk≤0,with equality if and only if each maximal chain contains an element of A.The above sum is nonpositive but each summand is nonnegative.Therefore each summand is zero and the sum is zero.Consequently(a)each maximal chain contains an element of A;(b)for0≤k≤n eitherαk is zero or its coe?cient is zero.We now consider two cases.10Case:n is even.We show that for0≤k≤n,αk=0if k=n/2.Observe that for0≤k≤n, if k=n/2then the coe?cient ofαk isnonzero,soαk=0.Case:n is odd.We show that for0≤k≤n,eitherαk=0if k=(n?1)/2orαk=0 if k=(n+1)/2.Observe that for0≤k≤n,if k=(n±1)/2then the coe?cient ofαk is nonzero,soαk=0.We now show thatαk=0for k=(n?1)/2or k=(n+1)/2. To do this,we assume thatαk=0for both k=(n±1)/2and get a contradiction.By assumption A contains an element x of size(n+1)/2and an element y of size(n?1)/2. De? ne s=|x∩y|.Choose x,y such that s is maximal.By construction0≤s≤(n?1)/2. Suppose s=(n?1)/2.Then y=x∩y?x,contradicting the fact that x,y are incomparable. So s≤(n?3)/2.Let y denote a subset of x that contains x∩y and has size(n?1)/2. Let x denote a subset of y ∪y that contains y and has size(n+1)/2.By construction |x ∩y|=s+1.Observe y is not in A since x,y are comparable.Also x is not in A by the maximality of s.By construction x covers y so they are together contained in a maximal chain.This chain does not contain an element of A,for a contradiction.33.De?ne a poset(X,≤)as follows.The set X consists of the subsets of{1,2,...,n}. For x,y∈X de?ne x≤y whenever x?y.Forn=3,4,5we display a symmetric chain decomposition of this poset.We use the inductive procedure from the text.For n=3,,1,12,1232,233,13.For n=4,,1,12,123,12344,14,1242,23,23424,For n=5,,1,12,123,1234,123455,15,125,12354,14,124,124545,1452,23,234,234525,23524,2453,13,134,134535,13534,345.1134.For 0≤k ≤ n/2 there are exactlyn kn k ?1symmetric chains of length n ?2k +1.35.Let S denote the set of 10jokes.Each night the talk show host picks a subset of S for his repertoire.It is required that these subsets form an antichain.By Corollary 5.3.2each antichain has size at most 105 ,which is equal to 252.Therefore the talk show host can continue for 252nights./doc/6215673729.htmlpute the coe?cient of x n in either side of(1+x )m 1(1+x )m 2=(1+x )m 1+m 2,In this computation use the binomial theorem.37.In the multinomial theorem (Theorem 5.4.1)set x i =1for 1≤i ≤t .38.(x 1+x 2+x 3)4is equal tox 41+x 42+x 43+4(x 31x 2+x 31x 3+x 1x 32+x 32x 3+x 1x 33+x 2x 33)+6(x 21x 22+x 21x 23+x 22x 23)+12(x 21x 2x 3+x 1x 22x 3+x 1x 2x 23).39.The coe?cient is10!3!×1!×4!×0!×2!which comes to 12600.40.The coe?cient is9!3!×3!×1!×2!41.One routinely obtains the multinomial theorem (Theorem 5.4.1)with t =3.42.Given an integer t ≥2and positive integers n 1,n 2,...,n t .De?ne n = ti =1n i .We shownn 1n 2···n t=t k =1n ?1n 1···n k ?1n k ?1n k +1···n t.Consider the multiset{n 1·x 1,n 2·x 2,...,n t ·x t }.Let P denote the set of permutations of this multiset.We compute |P |in two ways.(i)We saw earlier that |P |=n !n 1!×n 2!×···×n t != n n 1n 2···n t.12(ii)For1≤k≤t let P k denote the set of elements in P that have?rst coordinate x k.Thesets{P k}tk=1partition P,so|P|=tk=1|P k|.For1≤k≤t we compute|P k|.Observe that|P k|is the number of permutations of the multiset{n1·x1,...,n k?1·x k?1,(n k?1)·x k,n k+1·x k+1,...,n t·x t}. Therefore|P k|=n?1n1···n k?1n k?1n k+1···n t.By these comments|P|=tn1···n k?1n k?1n k+1···n t.The result follows.43.Given an integer n≥1.Show by induction on n that1 (1?z)n =∞k=0n+k?1kz k,|z|<1.The base case n=1is assumed to hold.We show that the above identity holds with n replaced by n+1,provided that it holds for n.Thus we show1(1?z)n+1=∞=0n+z ,|z|<1.Observe1(1?z)n+1=1(1?z)n11?z=∞k=0n+k?1kz k∞h=0z h=0c zwherec =n?1+n1+n+12+···+n+ ?1=n+.The result follows.1344.(Problem statement contains typo)The given sum is equal to (?3)n .Observe (?3)n =(?1?1?1)n=n 1+n 2+n 3=nnn 1n 2n 3(?1)n 1+n 2+n 3=n 1+n 2+n 3=nnn 1+n 2+n 3=nnn 1n 2n 3(?1)n 2.45.(Problem statement contains typo)The given sum is equal to (?4)n .Observe (?4)n =(?1?1?1?1)n=n 1+n 2+n 3+n 4=nnn 1n 2n 3n 4(?1)n 1+n 2+n 3+n 4=n 1+n 2+n 3+n 4=nnn 1n 2n 3n 4(?1)n 1?n 2+n 3?n 4.Also0=(1?1+1?1)n= n 1+n 2+n 3+n 4=nnn 1n 2n 3n 4(?1)n 2+n 4.46.Observe√30=5=5∞ k =01/2k z k.For n =0,1,2,...the n th approximation to √30isa n =5n k =0 1/2k 5?k.We have14n a n051 5.52 5.4753 5.47754 5.47718755 5.477231256 5.4772246887 5.4772257198 5.4772255519 5.477225579 47.Observe101/3=21081/3=2(1+z)1/3z=1/4,=2∞k=01/3kz k.For n=0,1,2,...the n th approximation to101/3isnk=01/3k4?k.We haven a n021 2.1666666672 2.1527777783 2.1547067904 2.1543852885 2.1544442306 2.1544327697 2.1544350898 2.1544346059 2.15443470848.We show that a poset with mn+1elements has a chain of size m+1or an antichain of size n+1.Our strategy is to assume the result is false,and get a contradiction.By assumption each chain has size at most m and each antichain has size at most n.Let r denote the size of the longest chain.So r≤m.By Theorem5.6.1the elements of the posetcan be partitioned into r antichains{A i}ri=1.We have|A i|≤n for1≤i≤r.Thereforemn+1=ri=1|A i|≤rn≤mn, 15for a contradiction.Therefore,the poset has a chain of size m+1or an antichain of size n+1.49.We are given a sequence of mn+1real numbers,denoted{a i}mni=0.Let X denote the setof ordered pairs{(i,a i)|0≤i≤mn}.Observe|X|=mn+1.De?ne a partial order≤on X as follows:for distinct x=(i,a i)and y=(j,a j)in X,declare xof{a i}mni=0,and the antichains correspond to the(strictly)decreasing subsequences of{a i}mni=0sequence{a i}mni=0has a(weakly)increasing subsequence of size m+1or a(strictly)decreasingsubsequence of size n+1.50.(i)Here is a chain of size four:1,2,4,8.Here is a partition of X into four antichains:8,124,6,9,102,3,5,7,111Therefore four is both the largest size of a chain,and the smallest number of antichains that partition X. (ii)Here is an antichain of size six:7,8,9,10,11,12.Here is a partition of X into six chains:1,2,4,83,6,1295,10711Therefore six is both the largest size of an antichain,and the smallest number of chains that partition X.51.There exists a chain x116。

计算机系统结构习题课(2012)-万继光

• 计算机系统有三个部件可以改进，这三个部件的加速比如下：部件加速比1＝30；部件加速比2＝20；部件加速比3＝10； • (1) 如果部件1和部件2的可改进比例为30％，那么当部件3的可改进比例为多少时，系统的加速比才可以达到10？ • (2) 如果三个部件的可改进比例为30％、30％和20％，三个部件同时改进，那么系统中不可加速部分的执行时间在总执行时间中占的比例是多少？
M W B S S S S S S E X M W B S S S E X I D I F M W B E X I D I F M W B S S S S E X I D I F M W B S S S E M W X B S S S I F
CPU时间 = IC ×CPI×时钟周期时间 n = （CPIi×ICi）×时钟周期时间
i=1
n （CPIi×ICi） n ICi 时钟周期数 i=1 CPI = = = （CPIi× IC ） IC IC i=1
f MIPS速率 CPI 10 6
▲ 2/101
习题1.7
对于一台400MHz计算机执行标准测试程序，程序中指令类型，执行数量和平均时钟周期数如下：
习题2.14
IEEE754
e N m rm
为便于软件的移植，浮点数的表示格式应该有统一标准（定义）。1985年IEEE提出了IEEE754标准。该标准规定基数为2，阶码E用移码表示，尾数M用原码表示，根据原码的规格化方法，最高数字位总是1，该标准将这个1缺省存储，使得尾数表示范围比实际存储的多一位。
采用定向技术消除数据相关
习题3.11（1）
1 LOOP: LW R1 0(R2) IF ADDI R1 R1 #1 SW 0(R2) R1 ADDI R2 R2 #4 SUB R4 R3 R2 BNZ R4 LOOP 2 ID IF 3 EX ID IF 4 5 6 7 8 9 1 0 1 1 1 2 1 3 1 4 1 1 1 1 1 5 6 7 8 9

NOIP第五章习题一答案

第 3 页共 9 页
/* 4.数组逆序重【1.6 编程基础之一维数组 04】将一个数组中的值按逆序重新存放。例如，原来的顺序为 8,6,5,4,1。要求改为 1,4,5,6,8。输入: 输入为两行：第一行数组中元素的个数 n（1<n<100)，第二行是 n 个整数，每两个整数之间用空格分隔。输出: 输出为一行：输出逆序后数组的整数，每两个整数之间用空格分隔。样例输入: 5 86541 样例输出: 14568 */ #include "iostream" #include "iomanip" using namespace std; int main() { int count; int num[100]; cin>>count; for(int i=0;i<count;i++) { cin>>num[i]; } int n=count; int temp = count-1; if(count%2==0) count=count/2; else count=(count-1)/2; for(int i=0;i<count;i++) { int m; m=num[i]; num[i]=num[temp]; num[temp]=m; temp--; }
//
} /* 7.向量点积计算【1.6 编程基础之一维数组 07】在线性代数、计算几何中，向量点积是一种十分重要的运算。给定两个 n 维向量 a=(a1,a2,...,an)和 b=(b1,b2,...,bn)，求点积 a·b=a1b1+a2b2+...+anbn。输入: 第一行是一个整数 n(1<=n<=1000)。第二行包含 n 个整数 a1,a2,...,an。第三行包含 n 个整数 b1,b2,...,bn。相邻整数之间用单个空格隔开。每个整数的绝对值都不超过 1000。输出: 一个整数，即两个向量的点积结果。样例输入: 3 146 215

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(1) 矩阵的特征值及其相似对角矩阵; 矩阵B
( 2) 行列式 B 及 A2 − 3 E 的值 .
2 −2 0 4. (7分)判断矩阵 A = − 2 1 − 2 可否对角化 ? 0 −2 0 若可对角化 , 求出可逆矩阵 U使U − 1 AU为对角矩阵 .
2.(10分 ) 设矩阵 A与B相似 , 其中 −1 − 2 2 y 0 0 A= 0 1 0 , B = 0 1 0 0 0 0 1 0 x
(1)求x和y的值; ( 2) 求可逆阵 P , 使得 P − 1 AP = B .
3 . (10 分 )已知三阶矩阵 A 的特征值为 1, 2 , − 1, 设矩阵 B = A − 2 A 2 + 3 A 3 , 试求
6. A− 1Y ;
2 2 2 7. x1 + 2 x 2 + 3 x 3 + 4 x1 x2 + 8 x1 x3 − 2 x 2 x3 ; 8. t < −1.
0 ( 2) k 1 ; 二、1.(1)t = 8; − 2 1 0 1 2. (1) x = 1, y = −1; ( 2) P = 0 1 0 . 0 1 1 0 2 0 ( 2) A2 − 3 E = 4. 3. (1) 0 18 0 ; 0 0 − 6 1 − 2 2 4. 可对角化 , U = 2 − 1 − 2 . 2 2 1
8. 当t满足
2 2 时, 二次型 f ( x1 , x 2 , x 3 ) = tx1 + tx 2 + 2 tx 3 + 2 x1 x 2 + 2 x1 x 3 − 2 x 2 x 3是负定的 . 计算题（二、计算题（共40分）．分 3 0 1 1. (7分)设 2是矩阵 A = 1 t 3 的特征值 , 求 1 2 3 (1)t的值; ( 2)对应于2的所有特征向量 .
T T
量, Q是三阶正交矩阵 , 试求常数 α , β .
测试题答案
一、 1.2 ( n重 ) , 任意 n维非零向量; 3. 2,1 ( 二重 ) ; 2. −1, −5, 4; 4. x = 0, y = 1;
1 2 5. 0 0
2 0 1 0
0 1 3 0
x 2 , x 3 ) = x12 + x 22 + x 32 + 5. ( 6分 )将二次型 f
4 x1 x 2 + 4 x1 x 3 + 4 x 2 x 3化为标准型 .
三、证明题（共20分）．证明题（ 20分
1.(5分) 设A为n阶实对称矩阵, 且满足 A3 + A2 + A = 3 E , 证明A是正定的矩阵.
值为2和1(二重 ), 那么B的特征值为
2 0 4. 已知矩阵 A = 0 0 0 1 则x = ,y= 0 2 1 与 B = 0 0 x 0 y 0 相似 , − 1 0 0
2 ( x1 , x 2 , x 3 , x4 ) = x12 + 2 x 22 + 3 x 3 + 4 x1 x 2 5. 二次型 f
2.(5分 )设A与B是正定矩阵 , 证明 : AB是正定矩阵的充要条件是 A与B可交换 .
四、（8分）设二次型、（8
2 2 2 f = x1 + x2 + x3 + 2ax1 x2 + 2 βx2 x3 + 2 x1 x3
经正交变换 X = QY 化成
2 2 f = y1 + 2 y3
其中 X = ( x1 , x 2 , x 3 ) , Y = ( y1 , y 2 , y 3 ) 是三维列向
− 2 0 0 −1 U AU = 0 1 0 . 0 0 4
2 2 2 5. f = − y1 − y2 + 5 y3 .
四、 α = β = 0.
+ 2 x 3 x 2的矩阵是 6. 当
2 ( x1 , x2 , x3 ) = x12 + x22 + 5 x3 + 时, 实二次型 f
2tx1 x 2 − 2 x1 x 3 + 4 x 2 x3是正定的.
4 1 2 7. 矩阵A = 2 2 − 1 对应的二次型是 4 −1 3
第五章
∗
测试题
一、填空题(每小题4分，共32分)．填空题(每小题4 32分
1. 设A是n阶方阵, A 是A的伴随矩阵 , A = 2, 则方阵 B = AA 的特征值是
∗
, 特征向量是
3 2
2. 三阶方阵 A的特征值为 1,−1,2, 则 B = 2 A − 3 A 的特征值为
− 1 1 0 − 1 − 4 1 3. 设A = − 4 3 0 , B = 1 3 0 且A的特征 1 0 2 0 0 2