2024-2025学年江苏南京六校联合体高二下学期3月调研数学试题含答案

2024-2025学年高二第二学期六校联合体3月调研测试项中，只有一项是符合题目要求的，请把答案填涂在答题卡相应位置上.⊥α,则λ的值是()A2B12CA．y=±2xB．y=±33xC．y=±33x或y=±3xD．y=±12x或y=±2x8．已知函数y＝ax与y＝ex有两条公共切线，则实数a的取值范围是项中，有多项符合题目要求，请把答案填涂在答题卡相应位置上．全部选对得6分，部分选对得部分分，不选或有错选的得0分.9．已知函数f(x)＝13x3－2x2＋3x，下列说法正确的有A．函数f(x)在x＝0处的切线方程为y＝3xD．若方程f(x)＝a仅有1个解，则11．已知在平行六面体ABCD－A1B1C1D1中，AB＝1，AD＝AA1＝2，且∠A1AB=∠A1AD=A．BD1＝AD－AB＋AA1相应位置上.高依次递减，则不同的站法有▲种.14．已知F是抛物线C：y2＝2px(p＞0)的焦点，M是抛物线的准线与x轴的交点，且过点M应写出必要的文字说明，证明过程或演算步骤.已知(1－2x)10＝a0＋a1x＋a2x2+···+a10x10.(2)求a0＋a1+···+a10的值；(3)求|a1|＋|a2|+···+|a10|的值.PM(1)求证：CM//平面PAB；APQqBCmmm02024-2025学年高二第二学期六校联合体3月调研测试1－8ABBBCDCA9－11ADBCDABD＝－由S6＝126得b1(1－26)1－2＝126，所以b1＝2，则bn＝2n，·······················…·9分··················································…·11分所以T2n＝(a1＋a3+···+a2n－1)＋(b2＋b4+···+b2n)=n＋n(n－1)2×4＋4(1－4n)1－4=2n2－n＋4(4n－1)3·······…·…···············…·…·········…·…··········…·15分,→→→→因为n1·=0，所以n1⊥CM，3分∥又因为BC=12AD，所以ME＝又因为BCzPMAQDyBxC····················…·…··········…·8分所以，平面PAB与平面PCD所成二面角的余弦值为34.··························…·9分(3)设BQBD=λ,则=λ,=+λ=(3－3λ,2λ,0)(0，0，1)，设平面PAQ的法向量为n3＝(x0，y0，z0)，则n3·可得(\r(3)λz0＝0，令y0＝1，所以n3＝(－2λ3，1，0).·····→所以d＝3PD3|n·||n|＝4λ23(1-λ)2解得λ=12．14分Δ=36m2＋36(3m2＋4)＝144(m2＋1)，∴|PQ|＝1＋m2·|y3－y4|＝1＋m2·16m2＋16＝4(m2＋1)，························…·12分∴μ|AB|+λ|PQ|＝(3m2＋4)μ12(m2＋1)+λ4(m2＋1)＝(3m2＋4)μ+3λ12(m2＋1)＝C(C为定值).··············································································…·15分∴3m2(4C-μ)＋(12C－3λ-4μ)＝0，任意的实数m恒成立方法二：设l倾斜角为θ,∴|AB|＝2ab2a2－c2cos2θ=2×2×34－cos2θ=124－cos2θ,························…·8分|PQ|＝2psin2θ=4sin2θ,·······················································…·12分∴μ|AB|+λ|PQ|＝(4－cos2θ)μ12+λsin2θ4＝4μ+3λsin2θ-μcos2θ12为定值，…·…·15分∴3λ=-μ时即λμ=-13时，μ|AB|+λ|PQ|为定值.·······························…·17分6＝4×3×2×1×0×(－1)6!＝0················································…·2分2＝0.5×(－0.5)2!＝－18(写－0.125也对)·········································…·4分＝1＋C1.8·0.1＋C1.8·0.12＋C1.8·0.13＋…·…