第一部分題型專項(xiàng)練壓軸題提分練(一)
壓軸題提分練(一)1. (2018威海模擬)已知橢圓C:2 2x y孑+ b2= i> b> 0)的離心率為且過點(diǎn),,動(dòng)直線I: y= kx+ m交橢圓C于不同的兩點(diǎn)A, B,且OA OB= 0(O為坐標(biāo)原點(diǎn)).(1)求橢圓C的方程.(2)討論3m2 2k2是否為定值?若為定值,求出該定值,若不是請(qǐng)說明理由. 解析:(1)由題意可知C = ¥,所以a2= 2c2= 2(a2 b2),即a2 = 2b2,a 2又點(diǎn)p(¥,中)在橢圓上,所以有4+4b2=1,2 2由聯(lián)立,解得b = 1, a = 2,2故所求的橢圓方程為2+y2= 1.(2)設(shè) A(x1, y1),B(X2, y2),由 OA OB= 0,可知 X1X2 + yy2 = 0.y= kx+ m,聯(lián)立方程組x22G+y =1,2 2 2消去y化簡(jiǎn)整理得(1 + 2k )x + 4kmx+ 2m 2 = 0,.2 222/口22 十、4km由= 16k m 8(m 1)(1 + 2k )>0,得 1+ 2k >m ,所以 X1 + x2 = 2, X1x21+ 2k22m2 221 + 2k2又由題知X1X2 + y1y2= 0,即 X1X2 + (kx1 + m)(kx2 + m) = 0,22整理為(1+ k )x1x2 + km(x1 + X2)+ m = 0.22 2m 24km 2將代入上式,得(1 + k )2 km + m = 0.1 + 2k21+ 2k22 23m 2 2k2221+ 2k2化簡(jiǎn)整理得=0,從而得到3m2 2k2 = 2.222. (2018 南寧二中模擬)設(shè)函數(shù) f(x)= a lnx+x ax(a R).(1)試討論函數(shù)f(x)的單調(diào)性;設(shè)(Kx) = 2x+ (a2 a)ln R)有兩個(gè)不相等的實(shí)根xi , X2,證明 h'解析:(1)由 f(x) = a2lnx+ x2 ax,可知2 2 2 a2x 一 ax一 af' (x)= + 2x a=x2x+ a x ax .因?yàn)楹瘮?shù)f(x)的定義域?yàn)?0,+x),所以,若a>0,當(dāng)x0, a)時(shí),f' (x)v0函數(shù)f(x)單調(diào)遞減,當(dāng)xG(a, +)時(shí),>0,函數(shù)f(x)單調(diào)遞增;若a = 0時(shí),f' (x) = 2x>0在xq0,+8)內(nèi)恒成立,函數(shù)f(x)單調(diào)遞增;若av0,當(dāng)xq0,a,f' (x)v0,函數(shù)f(x)單調(diào)遞減,當(dāng)x(2時(shí),f' (x)>0,函數(shù)f(x)單調(diào)遞增.2(2)證明:由題可知 h(x) = f(x) + K(x) = x + (2 a)x aln x(x> 0),2a 2x + 2 ax a 2x a x+ 1 所以 h (x) = 2x+ (2 a) x=入所以當(dāng) x0, |)時(shí),h' (x)v0;當(dāng) xq|,+8)時(shí),h' (x)>0x="2時(shí),h' ©x1 + x2欲證h' (一 )>0,只需證h'(x1 + x2aa) >h',又 h(x)二 2+ 子>0,即 h' (x)x,記 h(x)= f(x) + K(x),當(dāng) a>0 時(shí),若方程 h(x) = m(mXi + X2 a單調(diào)遞增,故只需證明>|.設(shè)xi, X2是方程h(x) = m的兩個(gè)不相等的實(shí)根,不妨設(shè)為0<xi<X2,+ ,21 2Xf njjyn則aln xi= m,aln X2= m,22兩式相減并整理得 a(xi X2+ In xi In X2)= xi X2 + 2xi 2x2,2 2 Xi X2+ 2xi 2x2 從而a =xi x2+ In xi In x22 2故只需證明2 xi x2 + In xi In X2xi + x2xi X2+ 2xi 2x222X1 X2+ 2xi 2X2即 Xi + X2>.(*)xi X2 + In xi In X2因?yàn)?xi X2+ In xi In X2V0,2xi 2x2所以(*)式可化為In xi In X2VXi + X2即 InXi<xiLi因?yàn)?0<Xi<X2,所以 0<:< i,不妨令t=予,所以得到In t<22, t qo,i).X2t+ i記 R(t) = In t 2t, tqo,i),所以 R' (t) = f 匕=上 i 2>0,當(dāng)且僅當(dāng) tt+ it (t+ i)2 t(t + if=i時(shí),等號(hào)成立,因此R(t)在(0,i)單調(diào)遞增.又 R(i)= 0,因此 R(t)<0, tqo,i).2t 2故 Intv, tqo,i)得證,t+1X1 + X2 從而h' (2 )>0得證.