e56，··············································································································2分=9，所以公差d=2，···························································································4分16.解1）:正三棱柱ABC一A1B1C1，點E，G分別為棱AA1，CC1的中點，又:EG面EB1G,AC丈面EB1G,:AC//面EB1G.又:AC∩CF=C,且AC面ACF，CF面ACF,:面AFC//面EB1G.··································································································7分（2）法1：取AC的中點H，連接BH，B1H，AB1，CB1，:正三棱柱ABC一A1B1C1的體積為4，且AB=2，:H為AC的中點，正三角形ABC，:B1H丄AC,BH丄AC，:上BHB1為二面角B一AC一B1的平面角.··········································································12分法2：取BC的中點O，連接AO，過點O在平面BB1C1C中作OM//CC1,:正三棱柱ABC一A1B1C1，:CC1丄面ABC，:AO面ABC，BC面ABC，:CC1丄AO,CC1丄BC，:OM丄AO,OM丄BC，:O為BC的中點，正三角形ABC，:AO丄BC，設面ACB1的法向量n1=(x,y,z)，:n1.AC=0且n1.AB1=0,:CC1丄面ABC，:面ABC的法向量n2=CC1=(0,4,0)，分17.解1）記盲盒的外層包裝A型為事件A，盲盒的外層包裝B型為事件B，盲盒中含限量版商品為事件C，則P(C)=P(C|A).P(A)+P(C|B).P(B)····························································2分（2）小王抽中含限量版商品的盲盒數量為隨機變量X，X~B(5,)，··································6分則隨機變量X的概率分布為：(2,(2,(2,(2,(2,(2,X012345P1 1（3）若單個盲盒含限量版商品，該盲盒外層包裝為A型的概率為條件概率··········································································12分2518.解1）:橢圓的離心率為e==，又經過點(1，)，得又:a2=b2+c2，解得a=2,b=:橢圓方程為.···································4分4又點A，B在橢圓上，:{()24l4443此時直線l的斜率為±.·····································································8分②設A(x1,y1)，B(x2,y2)，當直線l斜率不為0時，設直線l的方程：x=my+4，與橢圓E：聯立，72m72m3(x21)所以f所以f(x)在(0,)上單調遞減；在(,+∞)上單調遞增.····················································4分22所以由與（1）同理可得f(x)在(0,)上單調遞減；在(,+∞)上單調遞增，所以···············································7分令只需證g(a)≥0即可.于是，類似可得φ(t)在(0,1)上單調遞減；在(1,+∞)上單調遞增，（3）不等式f(x)≥sinx恒成立，即ax2—lnx≥sinx恒成立，類似可得m(x)≥m(1)=0，所以x2—lnx≥x；······························································14分又令h(x)=xsinx(x>0)，所以h(x)>h(0)=0，所以x