2024-2025學(xué)年新教材高中數(shù)學(xué)課時素養(yǎng)評價二十四簡單冪函數(shù)的圖象和性質(zhì)含解析北師大版必修1

PAGE6-課時素養(yǎng)評價二十四簡潔冪函數(shù)的圖象和性質(zhì)(15分鐘35分)1.下列結(jié)論正確的是 ()A.冪函數(shù)圖象肯定過原點(diǎn)B.當(dāng)α<0時,冪函數(shù)y=xα是減函數(shù)C.當(dāng)α>1時,冪函數(shù)y=xα是增函數(shù)D.函數(shù)y=x2既是二次函數(shù),也是冪函數(shù)【解析】選D.函數(shù)y=x-1的圖象不過原點(diǎn),故A不正確;y=x-1在(-∞,0)及(0,+∞)上是減函數(shù),故B不正確;函數(shù)y=x2在(-∞,0)上是減函數(shù),在(0,+∞)上是增函數(shù),故C不正確.2.已知冪函數(shù)f(x)=kxα的圖象過點(diǎn)12,2,則A.12 B.1 C.32【解析】選A.因?yàn)閮绾瘮?shù)f(x)=kxα(k∈R,α∈R)的圖象過點(diǎn)12,2,所以k=1,f12=12α=2,即α=-3.在下列四個圖形中,y=x-1【解析】選D.函數(shù)y=x-124.冪函數(shù)的圖象過點(diǎn)(3,3),則它的單調(diào)遞增區(qū)間是 ()A.[-1,+∞) B.[0,+∞)C.(-∞,+∞) D.(-∞,0)【解析】選B.設(shè)冪函數(shù)為f(x)=xα,因?yàn)閮绾瘮?shù)的圖象過點(diǎn)(3,3),所以f(3)=3α=3=312,解得α=12,所以f(x)=x5.(2024·北京高一檢測)假如冪函數(shù)f(x)=xa的圖象經(jīng)過點(diǎn)(2,4),則f(x)在定義域內(nèi) ()A.為增函數(shù) B.為減函數(shù)C.有最小值 D.有最大值【解析】選C.因?yàn)閮绾瘮?shù)f(x)=xa的圖象經(jīng)過點(diǎn)(2,4),所以f(2)=2a=4,解得a=2,所以f(x)=x2所以f(x)在定義域先遞減再遞增,有最小值.【補(bǔ)償訓(xùn)練】已知2.4α>2.5α,則α的取值范圍是.

【解析】因?yàn)?<2.4<2.5,而2.4α>2.5α,所以y=xα在(0,+∞)上為減函數(shù),故α<0.答案:(-∞,0)6.已知冪函數(shù)f(x)=x-m2-2①在區(qū)間(0,+∞)上單調(diào)遞增;②對隨意的x∈R,都有f(-x)-f(x)=0.求冪函數(shù)f(x)的解析式,并求當(dāng)x∈0,4時,f(x)【解析】因?yàn)楹瘮?shù)在0,所以-m2-2m+3>0,解得:-3<m<1.因?yàn)?2<m<2,m∈Z,所以m=-1或m=0.又因?yàn)閒(-x)=f(x),所以f(x)是偶函數(shù),所以-m2-2m+3為偶數(shù).當(dāng)m=-1時,-m2-2m+3=4滿意題意,當(dāng)m=0時,-m2-2m+3=3不滿意題意,所以f(x)=x4,所以f(x)在0,所以f(x)min=f(0)=0,f(x)max=f(4)=256,所以值域是0,(20分鐘40分)一、單選題(每小題5分,共15分)1.(2024·瓊海高一檢測)若函數(shù)f(x)=(m2-6m+9)xm2-3m+1A.2 B.3 C.4 D.2或【解析】選D.因?yàn)楹瘮?shù)f(x)=(m2-6m+9)xm2-3m+1為冪函數(shù),所以m22.下列命題中,不正確的是 ()A.冪函數(shù)y=x-1是奇函數(shù)B.冪函數(shù)y=x2是偶函數(shù)C.冪函數(shù)y=x既是奇函數(shù)又是偶函數(shù)D.y=x12既不是奇函數(shù)【解析】選C.因?yàn)閤-1=1x,1-x=-1x,所以A正確;(-x)2=x2,所以B正確;-x=x不恒成立,所以C不正確;y=3.給出冪函數(shù):①f(x)=x;②f(x)=x2;③f(x)=x3;④f(x)=x;⑤f(x)=1x.f(x1+x22)>f(xA.1個 B.2個 C.3個 D.4個【解題指南】解決該題的關(guān)鍵是正確理解fx1+x22>f(x【解析】選A.①函數(shù)f(x)=x的圖象是一條直線,故當(dāng)x1>x2>0時,fx1+x②函數(shù)f(x)=x2的圖象是凹形曲線,故當(dāng)x1>x2>0時,fx1+x③在第一象限,函數(shù)f(x)=x3的圖象是凹形曲線,故當(dāng)x1>x2>0時,fx1+x④函數(shù)f(x)=x的圖象是凸形曲線,故當(dāng)x1>x2>0時,fx1+x⑤在第一象限,函數(shù)f(x)=1x的圖象是一條凹形曲線,故當(dāng)x1>x2fx1+x故僅有函數(shù)f(x)=x滿意當(dāng)x1>x2>0時,fx1+x二、多選題(共5分,全部選對的得5分,選對但不全的得3分,有選錯的得0分)4.下列函數(shù)中,其定義域和值域相同的函數(shù)是 ()A.y=x13 B.y=x-12 【解析】選A、B、C.A中y=x13=3x,定義域、值域都為R;B中y=x-12=1x定義域與值域都為(0,+∞);C中y=x三、填空題(每小題5分,共10分)5.已知函數(shù)f(x)=(m2-m-1)xm2-2m-2是冪函數(shù),且在【解析】在冪函數(shù)f(x)=(m2-m-1)xm令m2-m-1=1,得m2-m-2=0,解得m=2或m=-1;當(dāng)m=2時,m2-2m-2=-2,函數(shù)f(x)=x-2,在(0,+∞)上單調(diào)遞減,滿意題意;當(dāng)m=-1時,m2-2m-2=1,函數(shù)f(x)=x,在(0,+∞)上單調(diào)遞增,不滿意題意;所以實(shí)數(shù)m=2.答案:26.已知冪函數(shù)f(x)=x-m2-2m+3(m∈Z)為偶函數(shù),且在(0,+∞)【解析】因?yàn)閮绾瘮?shù)f(x)=x-m2-2m+3(m∈Z)為偶函數(shù),且在(0,+∞)上單調(diào)遞增,則指數(shù)是偶數(shù)且大于0,因?yàn)?m因此指數(shù)等于2或4,當(dāng)指數(shù)等于2時,求得m非整數(shù),所以m=-1,即f(x)=x4.所以f(2)=24=16.答案:16四、解答題7.(10分)已知冪函數(shù)f(x)=x(m2+m)-1(m∈N*)經(jīng)過點(diǎn)(2,2),試確定m【解析】因?yàn)閮绾瘮?shù)f(