《同济大学第六版高等数学上下册课后答案全集》.doc

同济第六版高等数学课后答案全集 第一章 习题1(1 1( 设A((((( (5)((5( (()( B([(10( 3)( 写出A(B( A(B( A\B及A\(A\B)的表达式( 解 A(B((((( 3)((5( (()( A(B([(10( (5)( A\B((((( (10)((5( (()( A\(A\B)([(10( (5)( 2( 设A、B是任意两个集合( 证明对偶律( (A(B)C(AC (BC ( 证明 因为 x((A(B)C(x(A(B( x(A或x(B( x(AC或x(BC ( x(AC (BC( 所以 (A(B)C(AC (BC ( 3( 设映射f ( X (Y( A(X( B(X ( 证明 (1)f(A(B)(f(A)(f(B)( (2)f(A(B)(f(A)(f(B)( 证明 因为 y(f(A(B)((x(A(B( 使f(x)(y ((因为x(A或x(B) y(f(A)或y(f(B) ( y(f(A)(f(B)( 所以 f(A(B)(f(A)(f(B)( (2)因为 y(f(A(B)((x(A(B( 使f(x)(y((因为x(A且x(B) y(f(A)且y(f(B)( y( f(A)(f(B)( 所以 f(A(B)(f(A)(f(B)( 4( 设映射f ( X(Y( 若存在一个映射g( Y(X( 使( ( 其中IX、IY分别是X、Y上的恒等映射( 即对于每一个x(X( 有IX x(x( 对于每一个y(Y( 有IY y(y( 证明( f是双射( 且g是f的逆映射( g(f (1( 证明 因为对于任意的y(Y( 有x(g(y)(X( 且f(x)(f[g(y)](Iy y(y( 即Y中任意元素都是X中某元素的像( 所以f为X到Y的满射( 又因为对于任意的x1(x2( 必有f(x1)(f(x2)( 否则若f(x1)(f(x2)(g[ f(x1)](g[f(x2)] ( x1(x2( 因此f既是单射( 又是满射( 即f是双射( 对于映射g( Y(X( 因为对每个y(Y( 有g(y)(x(X( 且满足f(x)(f[g(y)](Iy y(y( 按逆映射的定义( g是f的逆映射( 5( 设映射f ( X(Y( A(X ( 证明( (1)f (1(f(A))(A( (2)当f是单射时( 有f (1(f(A))(A ( 证明 (1)因为x(A ( f(x)(y(f(A) ( f (1(y)(x(f (1(f(A))( 所以 f (1(f(A))(A( (2)由(1)知f (1(f(A))(A( 另一方面( 对于任意的x(f (1(f(A))(存在y(f(A)( 使f (1(y)(x(f(x)(y ( 因为y(f(A)且f是单射( 所以x(A( 这就证明了f (1(f(A))(A( 因此f (1(f(A))(A ( 6( 求下列函数的自然定义域( (1)( 解 由3x(2(0得( 函数的定义域为( (2)( 解 由1(x2(0得x((1( 定义域为((( (1)(((1( 1)((1( (()( (3)( 解 由x(0且1(x2(0得函数的定义域D((1( 0)((0( 1]( (4)( 解 由4(x2(0得 |x|(2( 定义域为((2( 2)( (5)( 解 由x(0得函数的定义D([0( (()( (6) y(tan(x(1)( 解 由(k(0( (1( (2( ( ( ()得函数的定义域为(k(0( (1( (2( ( ( ()( (7) y(arcsin(x(3)( 解 由|x(3|(1得函数的定义域D([2( 4]( (8)( 解 由3(x(0且x(0得函数的定义域D((((( 0)((0( 3)( (9) y(ln(x(1)( 解 由x(1(0得函数的定义域D(((1( (()( (10)( 解 由x(0得函数的定义域D((((( 0)((0( (()( 7( 下列各题中( 函数f(x)和g(x)是否相同？为什么？ (1)f(x)(lg x2( g(x)(2lg x( (2) f(x