有d步驟我係飛左的
1. Let f(x)=x^2+(1-m)x+2m-5 , where m is a constant . Find the Discriminant of the equ. f(x)=0.
Hence find the range of values of m so that f(x)>0 , for all real values of x.
ans:
discr. = (1-m)^2 - 4(2m-5)
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=(m-3)(m-7)
hence part .
f(x)>0 , discr < 0
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點解當f(x)>0果陣 discr 會細過0
p.s(discr. = 判別式個三角形)
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2.
considerind case1 : x<= -1 case x>-1 (細過等於係 <=)
case 1 :
丨x+1丨=x+1
x=-1
case 2:
丨x+1丨=x+1
0=0
問題是ans是 x => -1 而不是 x <= -1
為啥??
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3.
Find the range of real values of m for which the wxpression x^2-mx+(m+3) is
always positive for all real values of x.
x^2-mx+(m+3) > 0
by completing method
x^2-mx+(m+3)
=(x-m/2)^2 - m^2/4 +m+3
which will be always positive if
-(m^2/4) + m +3 > 0
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點解-(m^2/4) + m +3 會係太過\0的
唔明係因為always positive指成條equ.每parth都係>0 定係 成條++埋埋先>0
i.e每parth都係>0 (x^2)>0 ,(-mx)>0 同埋 (m+3)>0
定係 (x^2-mx+(m+3))成條咁先可以>0
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由於不等式太過於煩的開系~長期沒有理會過
所以不清楚的concept太多
麻煩各位了
另外集中解釋便可以~不需要完成題目
幾條關於不等式的a.maths問題
2008-09-03 8:21 am
回答 (1)
2008-09-03 3:16 pm
✔ 最佳答案
Q1.For a quadratic function to be always positive, that means the function will never cut the x -axis. If it never cuts the x - axis, the equation ax^2 + bx + c = 0 will never have real roots, for the equation never have real roots, its delta must be smaller then 0. (Same argument applies to a function to be always negative if the coefficient of x^2 term is negative.)
Q2.
When you draw the graph of y = x+1 and y = absolute(x + 1), you will find that the 2 graphs overlaps from x = -1 to + infinity. That means x> and equal to -1 satisfy the equation x + 1 = abs( x + 1).
Q3.
Based on the principle of completing square method, y is a minimum when x = m/2 and y minimum = -(m^2/4) + m + 3. So for y to be always positive, y minimum must be always positive ( that is always above the x - axis). Thererfore, -(m^2/4) + m + 3 must be greater than 0.
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