1.选择题- (共5题)
①集合A={x∈Z|x=2k﹣1,k∈Z}与集合B={x∈z|x=2k+3,k∈Z}是相等集合;
②若函数f(x)的定义域为[0,2],则函数f(2x)的定义域为[0,4];
③函数y= {#mathml#}{#/mathml#} 的单调减区间是(﹣∞,0)∪(0,+∞);
④不存在实数m,使f(x)=x2+mx+1为奇函数;
⑤若f(x+y)=f(x)f(y),且f(1)=2,则 {#mathml#}{#/mathml#} + {#mathml#}{#/mathml#} +…+ {#mathml#}{#/mathml#} =2016.
其中正确说法的序号是( )
①集合A={x∈Z|x=2k﹣1,k∈Z}与集合B={x∈z|x=2k+3,k∈Z}是相等集合;
②若函数f(x)的定义域为[0,2],则函数f(2x)的定义域为[0,4];
③函数y= {#mathml#}{#/mathml#} 的单调减区间是(﹣∞,0)∪(0,+∞);
④不存在实数m,使f(x)=x2+mx+1为奇函数;
⑤若f(x+y)=f(x)f(y),且f(1)=2,则 {#mathml#}{#/mathml#} + {#mathml#}{#/mathml#} +…+ {#mathml#}{#/mathml#} =2016.
其中正确说法的序号是( )
①集合A={x∈Z|x=2k﹣1,k∈Z}与集合B={x∈z|x=2k+3,k∈Z}是相等集合;
②若函数f(x)的定义域为[0,2],则函数f(2x)的定义域为[0,4];
③函数y= {#mathml#}{#/mathml#} 的单调减区间是(﹣∞,0)∪(0,+∞);
④不存在实数m,使f(x)=x2+mx+1为奇函数;
⑤若f(x+y)=f(x)f(y),且f(1)=2,则 {#mathml#}{#/mathml#} + {#mathml#}{#/mathml#} +…+ {#mathml#}{#/mathml#} =2016.
其中正确说法的序号是( )
( I)判断f(x)的奇偶性;
( II)求证:f(x)+f( {#mathml#}{#/mathml#} )为定值;
(III)求 {#mathml#}{#/mathml#} + {#mathml#}{#/mathml#} + {#mathml#}{#/mathml#} +f(1)+f(2015)+f(2016)+f(2017)的值.
2.单项选择- (共7题)
A.I walking | B.was walking | C.I am walking | D.walking |
A.learning | B.to learn | C.learn | D.to learning |
A.speaking | B.speak | C.spoken | D.is spoken |
A.for example | B.such as | C.including | D.liking |
— Tom asked me ______ _____ _______ go to Xiamen ________.
A.that; I ; would ; the next week. |
B.if ; you ; would ; the next week |
C.if ; I; would; next week |
D.if; I; would ; the next week |
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【1】题量占比
选择题:(5道)
单项选择:(7道)
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【2】:难度分析
1星难题:0
2星难题:0
3星难题:0
4星难题:3
5星难题:0
6星难题:3
7星难题:0
8星难题:0
9星难题:0