第一试(试题见本刊第5期) 一选择题 1.(B); 2.(C); 3.(D); 4.(B); 5.(A); 6.(C); 7.(B); 8.(D); 9.(A); 二解:y=|x-1|+|x-3|+(4x~2+4x+1)~(1/2)=|x-1|+|x-3|+|2x+1|=-4x+3 (x<-1/2)5 (-1/2≤x≤1)2x+3 (1≤x<3)4x-3 (x≥3) ∴当-1/2≤x≤1时y=|x-1|+|x-3|+(4x~2+4x+1)~(1/2)恒等于常数5。三、证明∵ABCD为圆外切四边形∴AB+CD=BC+DA(见下图) 两边平方:AB~2+2AB·CD+CD~2=BC~2+2BC·DA+DA~2 (1) 又∵AC⊥BD ∴AB~2-AB~2+BE~2,BC~2=BE~2+CE~2, The first test (for questions, see the 5th issue of the journal) A multiple-choice question 1.(B); 2.(C); 3.(D); 4.(B); 5.(A); 6.(C) ; 7.(B); 8.(D); 9.(A); Two solutions: y=|x-1|+|x-3|+(4x~2+4x+1)~(1/2) )=|x-1|+|x-3|+|2x+1|=-4x+3 (x<-1/2)5 (-1/2≤x≤1) 2x+3 (1≤x <3) 4x-3 (x≥3) When -1/2≤x≤1, y=|x-1|+|x-3|+(4x~2+4x+1)~(1/2) ) Constant is equal to constant 5. 3. Prove that ∵ ABCD is a round quadrilateral ∴ AB+CD=BC+DA (see the figure below) Two-sided square: AB~2+2AB·CD+CD~2=BC~2+2BC·DA+DA~2 ( 1) Also ∵AC⊥BD ∴AB~2-AB~2+BE~2,BC~2=BE~2+CE~2,