高二级2013—2014学年度第二学期第二次阶段测试数学试卷(文科)答案一、选择题:二、填空题11、;12、;13、;14、三、解答题15.解:在中,由可知,是锐角,所以,………………………2分由正弦定理···························……5分(2)………………………………………………8分……12分1题号12345678910答案ADBAAABCBB1216.【答案】(1)设{a}的公差为d,则S=.………………………2分由已知可得…………………………4分…………………………6分(2)由(I)知…………………9分从而数列………13分17.(Ⅰ)因故由于在点处取得极值故有··············3分即················4分化简得解得·················7分(Ⅱ)由(Ⅰ)知,·················8分令,得·················9分(-∞,-2)-2(-2,2)2(2,+∞)+0-0+↗极大值↘极小值↗由题设条件知,得·············11分此时,,,,·················12分2nn1(1)2nnnad111330,1,1.5105,adadad解得n=2-.naan故的通项公式为212111111(),(32)(12)22321nnaannnn21211nnnaa的前项和为1111111-+-++)2-1113232112nnnn(因此上的最小值为,最大值为·················13分18.解:(Ⅰ)每套丛书售价定为100元时,销售量为万套,此时每套供货价格为元,书商所获得的总利润为万元.·················5分(Ⅱ)每套丛书售价定为元时,由得,,依题意,单套丛书利润,···············8分∴,···························10分∵,∴,由,··················12分当且仅当,即时等号成立,此时,.··················14分19.解:(1)由题意可知an+1n+1=ann⋅12,所以数列nan是以12为首项,12为公比的等比数列.……………………3分(2)由(1)可知an=n2n,所以Sn=1⋅12+2⋅(12)2+3⋅(12)3++n⋅(12)n∴12Sn=1⋅(12)2+2⋅(12)3++n⋅(12)n+1相减整理得Sn=2−12n−1−n2n.……………………8分(3)不等式a⋅2n+1