廣西柳鐵一中2013屆高三模擬數(shù)學(xué)文試題(二)

廣西柳鐵一中2013屆高三模擬（數(shù)學(xué)文)試題(二)一、選擇題要求：在每小題給出的四個(gè)選項(xiàng)中，只有一項(xiàng)是符合題目要求的。1.設(shè)集合A={x|-1≤x≤2}，B={x|1≤x≤3}，則下列命題正確的是（）A.A∩B=BB.A∪B={x|-1≤x≤3}C.A∩B=AD.A∪B={x|-1≤x≤2}2.函數(shù)f(x)=ax^2+bx+c（a≠0）的圖象開口向下，且f(-1)=2，f(1)=-2，則下列關(guān)于a，b，c的結(jié)論中正確的是（）A.a＜0，b=0，c=0B.a＜0，b=0，c≠0C.a＞0，b=0，c=0D.a＞0，b=0，c≠0二、填空題要求：把答案填在題目的橫線上。3.已知等差數(shù)列{an}的首項(xiàng)為1，公差為2，則第10項(xiàng)an=__________。4.若復(fù)數(shù)z=（1+i）^2013，則|z|=__________。三、解答題要求：解答應(yīng)寫出文字說(shuō)明、證明過(guò)程或演算步驟。5.（1）已知函數(shù)f(x)=x^3-3x+1，求函數(shù)f(x)的單調(diào)區(qū)間。（2）設(shè)a，b，c∈R，且a+b+c=0，證明：a^3+b^3+c^3=3abc。6.已知數(shù)列{an}是等比數(shù)列，且a1=2，公比為q（q≠1），求證：數(shù)列{an^2}也是等比數(shù)列，并求出其公比。四、解答題要求：解答應(yīng)寫出文字說(shuō)明、證明過(guò)程或演算步驟。7.（1）已知函數(shù)f(x)=x^2-4x+4，求函數(shù)f(x)的極值。（2）設(shè)A，B是兩個(gè)等價(jià)矩陣，且矩陣A的行列式值為-1，求矩陣B的行列式值。五、證明題要求：證明以下命題。8.證明：對(duì)于任意正整數(shù)n，都有1^2+2^2+3^2+...+n^2=n(n+1)(2n+1)/6。六、應(yīng)用題要求：解答應(yīng)寫出必要的文字說(shuō)明、計(jì)算過(guò)程或步驟。9.設(shè)某商品的成本為每件100元，銷售價(jià)格為每件150元，銷售量為1000件?，F(xiàn)進(jìn)行促銷活動(dòng)，每件商品降價(jià)10元，求在促銷活動(dòng)中該商品的利潤(rùn)。本次試卷答案如下：一、選擇題1.答案：B解析：集合A∪B包含A和B中的所有元素，因此A∪B={x|-1≤x≤3}。2.答案：B解析：函數(shù)f(x)=ax^2+bx+c的圖象開口向下，說(shuō)明a＜0。由于f(-1)=2和f(1)=-2，可以得到兩個(gè)方程：a(-1)^2+b(-1)+c=2a(1)^2+b(1)+c=-2解這個(gè)方程組，得到a=0，b=0，c≠0。二、填空題3.答案：21解析：等差數(shù)列的通項(xiàng)公式為an=a1+(n-1)d，其中a1是首項(xiàng)，d是公差。代入a1=1，d=2，n=10，得到an=1+(10-1)×2=21。4.答案：1解析：復(fù)數(shù)z=（1+i）^2013可以寫成（cos(π/4)+isin(π/4))^2013。由于（cos(π/4)+isin(π/4)）的模是1，所以|z|=1。三、解答題5.（1）答案：?jiǎn)握{(diào)遞增區(qū)間為(-∞,1)，單調(diào)遞減區(qū)間為(1,+∞)。解析：函數(shù)f(x)=x^3-3x+1的導(dǎo)數(shù)為f'(x)=3x^2-3。令f'(x)=0，解得x=±1。在x=-1和x=1處，函數(shù)的導(dǎo)數(shù)變號(hào)，因此x=-1是極大值點(diǎn)，x=1是極小值點(diǎn)。由于導(dǎo)數(shù)在x=-1左側(cè)為正，在x=-1右側(cè)為負(fù)，所以x=-1是單調(diào)遞增區(qū)間的右端點(diǎn)，x=1是單調(diào)遞減區(qū)間的左端點(diǎn)。5.（2）答案：a^3+b^3+c^3=3abc解析：由a+b+c=0，可以得到a+b=-c。將a+b=-c代入a^3+b^3+c^3中，得到：a^3+b^3+c^3=(a+b)^3-3ab(a+b)+c^3=(-c)^3-3ab(-c)=-c^3+3abc=3abc。6.答案：數(shù)列{an^2}是等比數(shù)列，公比為q^2。解析：由于{an}是等比數(shù)列，an=a1q^(n-1)。那么an^2=(a1q^(n-1))^2=a1^2q^(2n-2)。因此，數(shù)列{an^2}的首項(xiàng)是a1^2，公比是q^2。四、解答題7.（1）答案：極大值為f(2)=0，極小值為f(1)=0。解析：函數(shù)f(x)=x^2-4x+4可以寫成f(x)=(x-2)^2。這是一個(gè)完全平方的形式，因此它的頂點(diǎn)是(2,0)，所以x=2是極大值點(diǎn)，極大值為0。由于函數(shù)的導(dǎo)數(shù)在x=2兩側(cè)變號(hào)，x=2是極小值點(diǎn)，極小值也為0。7.（2）答案：-1解析：由于A和B是等價(jià)矩陣，它們的行列式值相等。已知A的行列式值為-1，所以B的行列式值也為-1。五、證明題8.答案：證明見下文。解析：使用數(shù)學(xué)歸納法證明。（1）當(dāng)n=1時(shí)，1^2=1，等式成立。（2）假設(shè)當(dāng)n=k時(shí)等式成立，即1^2+2^2+3^2+...+k^2=k(k+1)(2k+1)/6。（3）當(dāng)n=k+1時(shí)，需要證明1^2+2^2+3^2+...+k^2+(k+1)^2=[k(k+1)(2k+1)/6]+(k+1)^2?；?jiǎn)上述表達(dá)式，得到：1^2+2^2+3^2+...+k^2+(k+1)^2=[k(k+1)(2k+1)+6(k+1)^2]/6=[(k+1)(2k^2+3k+6k+6)]/6=[(k+1)(2k^2+9k+6)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(k+2)(2(k+1)+1)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(k+2)(2k+3)]/6=[(k+1)(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