【四川卷】四川省宜賓市普通高中2022級（2025屆）高三第一次診斷性考試（宜賓一診）（11.19-11.21）數(shù)學(xué)試卷答案

一診數(shù)學(xué)參考答案第1頁共5頁宜賓市普通高中2022級第一次診斷性測試一、選擇題題號123456789選項(xiàng)ABCCACDBABCACBCD 四、解答題：本題共5小題，共77分。解答應(yīng)寫出文字說明，證明過程或演算步驟。1513分）B1M同理可證B1M丄AM，又MA∩MC=M所以B1M丄平面AMC，又B1M平面B1MC，（2）解：取AC的中點(diǎn)O，連結(jié)OM，OB1， 1615分）解1）該射手恰好命中一次的概率··········································6分（2）該射手的總得分X的所有可能取值為：0，1，2，3，4····················7分所以2于是，X的分布列為X01234P 1715分）所以f(A)=2sin(2A+)=1即，··········································································4分所以，··························································9分 π所以4a22c=32sin(B+3)=33+2·····························10分sin2BsinBtan2BtanB2t分一診數(shù)學(xué)參考答案第2頁共5頁一診數(shù)學(xué)參考答案第3頁共5頁1817分）2故雙曲線方程為x2-y=1.................................................................................4分解2故雙曲線方程為x2-y=1.................................................................................4分4由于直線l1與雙曲線的左右兩支相交，所以方程(*)有兩個(gè)同號的實(shí)根2124m2-12124m2-1由O,Q,M三點(diǎn)共線得=4m............=-1...........(ii)由解得：M............................................................................9分 顯然點(diǎn)M的橫坐標(biāo)為定值，縱坐標(biāo)隨m變化而變化 ②由MN=MA+MB可知，四邊形MANB是平行四邊形，M-l1AB，dM-l1 2+2+m2.................................................................................13分32t3 2,m2 tt3故f(t)max=f((|,)=..............................................................................15分 當(dāng)且僅當(dāng)t=,即m=±時(shí)取等號..................1917分）所以u(x)在(0,1)單調(diào)遞增，在(1,+∞)單調(diào)遞減.................................................................4分（2）①f(x)=u(x)+v(x)=（2x2f'(x)=4xlnx+(2x記g(x)=2lnx+1+a，則f'(x)=2x.g(x)，g'(x)== f'(x)=2xg(x)≥0→f(x)單調(diào)遞增，無極值點(diǎn). 注意到當(dāng)x→+∞時(shí)f'(x)→+∞,故f'(x)=0在(1,+∞)有一個(gè)所以f'(x)=0在(,1)上有一個(gè)根x1，故f(x)恰有兩個(gè)極值點(diǎn)，符合題意.一診數(shù)學(xué)參考答案第4頁共5頁綜上實(shí)數(shù)a的取值范圍為(2,+∞).........................................................................................10分方法二：可用參變分離法求解，(閱卷時(shí)酌情給分.)且f(x)在(0,x1)，(x2,+∞)單調(diào)遞增，在(x1,x2)單調(diào)遞減，因?yàn)閒(1)=0從而f(x1)>f(1)>0，同理可得f(x2)<f(1)<0..............................12分所以f(x)在(0,x1)和(x2,+∞)上各有一個(gè)零點(diǎn)，結(jié)合f(1)=0可知f(x)共有三個(gè)零點(diǎn)..................................................................................14分所以若=0，則f0故f(x)的三個(gè)零點(diǎn)可以表示為：x0,1,............................................................