[咸阳三模]咸阳市2024年高考模拟检测(三)数学.考卷答案

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[咸阳三模]咸阳市2024年高考模拟检测(三)数学.考卷答案试卷答案

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分析(1)利用列举法确定基本事件,即可求函数y=f(x)在区间[1,+∞)上是增函数的概率;
(2)方程$\frac{{x}^{2}}{{a}^{2}}$+$\frac{{y}^{2}}{{b}^{2}}$=1表示焦点在x轴上且离心率小于$\frac{\sqrt{3}}{2}$的椭圆,故$\left\{\begin{array}{l}{{a}^{2}>{b}^{2}}\\{\frac{\sqrt{{a}^{2}-{b}^{2}}}{a}<\frac{\sqrt{3}}{2}}\end{array}\right.$,化简得$\left\{\begin{array}{l}{a>b}\\{a<2b}\end{array}\right.$,又a∈[1,5],b∈[2,4],画出满足不等式组的平面区域,利用面积比,即可求方程$\frac{{x}^{2}}{{a}^{2}}$+$\frac{{y}^{2}}{{b}^{2}}$=1表示焦点在x轴上且离心率小于$\frac{\sqrt{3}}{2}$的椭圆的概率.

解答解:(1)∵函数f(x)=ax2-4bx+1的图象的对称轴为直线x=$\frac{2b}{a}$,要使f(x)=ax2-4bx+1在区间[1,+∞)上为增函数,当且仅当a>0且$\frac{2b}{a}$≤1,即2b≤a.…(2分)
若a=1,则b=-1;
若a=2,则b=-1或1;
若a=3,则b=-1或1.
∴事件包含基本事件的个数是1+2+2=5.…(4分)
而满足条件的数对(a,b)共有3×5=15个
∴所求事件的概率为$\frac{5}{15}$=$\frac{1}{3}$.…(6分)
(2)方程$\frac{{x}^{2}}{{a}^{2}}$+$\frac{{y}^{2}}{{b}^{2}}$=1表示焦点在x轴上且离心率小于$\frac{\sqrt{3}}{2}$的椭圆,故$\left\{\begin{array}{l}{{a}^{2}>{b}^{2}}\\{\frac{\sqrt{{a}^{2}-{b}^{2}}}{a}<\frac{\sqrt{3}}{2}}\end{array}\right.$…(8分)
化简得$\left\{\begin{array}{l}{a>b}\\{a<2b}\end{array}\right.$
又a∈[1,5],b∈[2,4],画出满足不等式组的平面区域,如图阴影部分所示,
…(10分)
阴影部分的面积为$\frac{15}{4}$,故所求的概率P=$\frac{15}{32}$.…(12分)

点评本题考查概率的计算,考查学生的计算能力,区分两种类型是关键.