Question

已知向量x=(1，t²-3),y=(-k，t)(其中實(shí)數(shù)k和t不同時(shí)為零)，當(dāng)|t|≤2時(shí),有x⊥y,當(dāng)|t|＞2時(shí),有x∥y.

(1)求函數(shù)關(guān)系式k=f(t);

(2)求函數(shù)f(t)的單調(diào)遞減區(qū)間;

(3)求函數(shù)f(t)的最大值和最小值.

Answer 1

解:(1)當(dāng)|t|≤2時(shí),由x⊥y得x·y=-k+(t²-3)t=0,得k=f(t)=t³-3t(|t|≤2).

當(dāng)|t|＞2時(shí),由x∥y得k=.所以k=f(t)=

(2)當(dāng)|t|≤2時(shí),f′(t)=3t²-3,由f′(t)＜0,得3t²-3＜0,解得-1＜t＜1.

當(dāng)|t|＞2時(shí),f′(t)==＞0.

∴函數(shù)f(t)的單調(diào)遞減區(qū)間是(-1,1).

(3)當(dāng)|t|≤2時(shí),由f′(t)=3t²-3=0得t=1或t=-1.∵1＜|t|≤2時(shí),f′(t)＞0,

∴f(t)_極大值=f(-1)=2，f(t)_極小值=f(1)=-2.又f(2)=8-6=2,f(-2)=-8+6=-2,當(dāng)t＞2時(shí),f(t)=＜0.

又由f′(t)＞0知f(t)單調(diào)遞增,∴f(t)＞f(2)=-2,即當(dāng)t＞2時(shí),-2＜f(t)＜0.

同理可求,當(dāng)t＜-2時(shí),有0＜f(t)＜2,綜合上述得,當(dāng)t=-1或t=2時(shí),f(t)取最大值2;

當(dāng)t=1或t=-2時(shí),f(t)取最小值-2.