【20套精选试卷合集】南京市钟英中学2019-2020学年高考数学模拟试卷含答案

数学卷 参考答案

一、选择题

1-5 DCACA 6-10 BADBB 二、填空题 11. 3,0; 12. i,1; 13. 40,2; 14. ?,(0,3]; 3?5??6?15. ?,1?; 16. 40 17. 9 三、解答题

18.解:(Ⅰ)由sinA?sin(B?C)?2sin2B,有

sin(B?C)?sin(B?C)?4sinBcosB,

展开化简得,cosBsinC?2sinBcosB, 又因为B??2,所以sinC?2sinB,

由正弦定理得,c?2b;

(Ⅱ)因为?ABC的面积S?5b2?a2,所以有由(Ⅰ)知c?2b,代入上式得

1bccosA?5b2?4b2cosA, 2b2sinA?5b2?a2,①

又由余弦定理有a2?b2?c2?2bccosA?5b2?4b2cosA, 代入①得b2sinA?4b2cosA, ∴tanA?4.

19.解:(Ⅰ)取BC中点G,连接FG,AG, 又∵F为BD的中点,CD?2EA,CD//AE, ∴FG?1CD?EA,且FG//AE, 2∴四边形AGFE是平行四边形, ∴EF//AG,

而且EF?平面ABC,AG?平面ABC, ∴EF//平面ABC;

(Ⅱ)∵?EAC?90,平面EACD?平面ABC,且交于AC, ∴平EA?面ABC,

由(Ⅰ)知FG//AE,∴FG?平面ABC, 又∵AB?AC,G为BC中点, ∴AG?BC,

如图,以GA,GB,GF所在直线为x,y,z轴建立空间直角坐标系,

o则A(1,0,0),B(0,3,0),D(0,?3,2),E(1,0,1),

uuuruuuruuur∴AB?(?1,3,0),BD?(0,?23,2),BE?(1,?3,1),

rn设平面BDE的法向量为?(x,y,z),则

ruuur???n?BD?0?z?3y?0,即, ruuur?????n?BE?0?x?3y?z?0r令y?1,得n?(0,1,3),

uuurrAB?n3∴直线AB与平面BDE所成角的正弦值为uuu. rr?4AB?n220.解(Ⅰ)f'(x)?3x?a,

①当a?0时,f'(x)?0恒成立,此时函数f(x)在R上单调递增; ②当a?0时,令f'(x)?0,得x???a, 3??a??a∴x????,???U??,???时,f'(x)?0;

???3?3?????aa?时,f'(x)?0, x????,????33?????a??aaa?∴函数f(x)的递增区间有???,???,??,???,递减区间有???,??.

????3?33?3??????(Ⅱ)由(Ⅰ)知:

①当a?0时,函数f(x)在[?1,1]上单调递增,此时M(a)?f(1)?1; ②当??aaa??1即a??3时,[?1,1]????,??,∴f(x)在[?1,1]单调递减,

?333???∴M(a)?f(?1)??1?2a,∵a??3,∴1?2a?5,即M(a)?5;

?aa?③当?3?a?0时,???,???[?1,1],

?33???而f(x)在??1,??????a??a?aa?,递增,在上递减, ?,1??,???????????3??3?33???a?????,1?????. 3??????∴M(a)?max?f?????a?????,f(1)?max????f??3???????2aaa???2由f?????1,得?a??a?1,令t??,则a??3t,

3333????32232∴2t?3t?1?0,即2(t?1)?3(t?1)?0?(t?1)(2t?1)?0,∴t?13,∴a??. 243∴当?3?a??时,

4当???a?a?????f?????1,∴M(a)?f????;

3?3????????a?3???a?0时,f?????1,∴M(a)?f(1)?1.

3?4?????3??. 4?综合①②③得:若M(a)?1,则实数a的取值范围为???,?21.解:(Ⅰ)抛物线的焦点为F?则切线PB的方程为:y?k?x??1?,0?,设切线PB的斜率为k, 4????1?1,即kx?y?k?0. ?4?41k?(?1)?1?0?k4?1,解得:k??4. ∴3k2?1∵P(x0,y0)(y0?1),∴k?4. 3y0?m① x0(Ⅱ)设切线方程为y?kx?m,由点P在直线上得:k?圆心C到切线的距离?k?mk2?1?1,整理得:m2?2km?1?0②

2将①代入②得:(x0?2)m?2y0m?x0?0③

设方程的两个根分别为m1,m2,由韦达定理得:m1?m2?2y0x0,m1m2??, x0?2x0?2x02?3x0从而AB?m1?m2?(m1?m2)?4m1m2?2,

(x0?2)22S?ABPx02?3x01?ABx0?x0?22(x0?2)x02(x02?3x0)(x0?1). 2(x0?2)x2(x2?3x)x2(2x2?11x?18)(x?1),则g'(x)??0, 记函数g(x)?(x?2)2(x?2)3g(x)min?g(1)?42,S?ABP的最小值为,当x0?1取得等号. 931.用数学归纳法证明如下: 222.解:(Ⅰ)猜想:0?an?(i)当n?1时,a1?1,结论成立; 221111?1?1(ii)假设n?k时结论成立,即0?ak?,则ak?1?ak2?ak??ak???,

2242?4?8∴0?ak?1?1,则n?k?1时,结论成立. 4(iii)由(i)(ii)可得,对任意n?N*,0?an?∴

1成立. 2an?1111?an??. an242(Ⅱ)易求得a2?1357,a3?,a4?,于是f(1)?2,f(2)?4,f(3)?10,f(4)?35, 4322048∴b1?a1,b2?a2,b3?a3,b4??a4,

f(n)an,所以?an?bn?an. ∵bn?(?1)∴Tn?a1?a2?a3?a4?b5?????bn?a1?a2?a3?a4?a5?????an. ∵

an?111?,有an?an?1, an222n?31?1??1?∴a3?a4?a5?????an?a3?a3????a3???????2?2??2?∴Tn?a1?a2??1??a3?a3????2?n?3?0,

3. 4又Tn?a1?a2?a3?a4?b5?????bn?a1?a2?a3?a4?a5?????an,

1?1??1?而?a4?a5?????an??a4?a4????a4???????2?2??2?∴Tn?a1?a2?a3?综上,当n?3时,

2n?4?1??a4??a4????2?n?4?0,

27. 32327. ?Tn?432

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