STPM-学校功课討論区

mathlim

原帖由 darksider 于 22-7-2008 09:33 PM 发表
15a) Find the range of values of the expression 5x^2 -200x + 1969 as x changes between 0 and 50 inclusive.

5x² - 200x + 1969 = 5(x - 20)² - 31
0 ≤ x ≤ 50
- 20 ≤ x - 20 ≤ 30
0 ≤ (x - 20)² ≤ 900
0 ≤ 5(x - 20)² ≤ 4500
- 31 ≤ 5(x - 20)² - 31 ≤ 4469
∴ - 31 ≤ 5x² - 200x + 1969 ≤ 4469

mathlim

原帖由 darksider 于 22-7-2008 09:33 PM 发表
b) Find , in terms of a and b , the range set of values for t such that 3t^2 + 6at >= 4b^2 - 3a^2 where a,b are given positive numbers.

3t² + 6at ≥ 4b² - 3a²
3t² + 6at + 3a² - 4b² ≥ 0
(√3t + √3a)² - (2b)² ≥ 0
(√3t + √3a + 2b)(√3t + √3a - 2b) ≥ 0
t ≥ (- √3a + 2b)/√3 或 t ≤ (- √3a - 2b)/√3

darksider

谢谢你！请问我可以有你们的msn contact吗？谢谢！

darksider

原帖由 mathlim 于 2008/7/23 07:14 AM 发表


5x² - 200x + 1969 = 5(x - 20)² - 31
0 ≤ x ≤ 50
- 20 ≤ x - 20 ≤ 30
0 ≤ (x - 20)² ≤ 900
0 ≤ 5(x - 20)² ≤ 4500
- 31 ≤ 5(x - 20)² - 31 ≤ 4469
∴ - 31 ≤ 5x&sup ...

这个- 20 ≤ x - 20 ≤ 30 怎样变成0 ≤ (x - 20)² ≤ 900？

sqr 全部 = (-20)^2 ≤  (x-20)^2 ≤ (30)^2
                      400 ≤  (x-20)^2 ≤  900

那个400 怎么会变成0呢？

mathlim

我有msn，但是都没有在用。

mathlim

原帖由 darksider 于 23-7-2008 02:28 PM 发表
这个- 20 ≤ x - 20 ≤ 30 怎样变成0 ≤ (x - 20)² ≤ 900？

sqr 全部 = (-20)^2 ≤  (x-20)^2 ≤ (30)^2
                      400 ≤  (x-20)^2 ≤  900

那个400 怎么会变成0呢？

a ＞ 3
a^2 ＞ 9

但是
a ＞ -3
a^2 ≯ 9
而是
a ＞ -3
a^2 ≥ 0

darksider

Given the equation y = ax^2 -bx + c where a,b andn c are positive coefficients , show that when x changes , the smallest value of y is -p/4a where p = b^2 - 4ac and find the corresponding value of x.
Sketch its graph for the following cases:
i) p>0 , in case i) the points where the graph intercept the x-axis are P and Q , and L is a point on the graph where y has the smallest value . The midpoint of PQ is M, show that  a.MP^2 = LM

P是不是只有用Sum of roots and product of roots才能找出来？
那个MP是不是利用distance的formula来找？

darksider

原帖由 mathlim 于 2008/7/24 07:53 AM 发表


a ＞ 3
a^2 ＞ 9

但是
a ＞ -3
a^2 ≯ 9
而是
a ＞ -3
a^2 ≥ 0

恩，明白了，谢谢

darksider

show that tan 3x =   3tanx - tan^3 x
                               -------------------
                                 1 - 3 tan^2 x

hamilan911

tan 3x = tan(x+2x) = (tanx + tan2x) / (1-tanxtan2x) = [tanx + 2tanx/1-tan^2 x] / [ 1-tanx(2tanx/1-tan^2 x] = [3tanx - tan^3 x] / [1 - 3tan^2 x]

darksider

prove that cos^2 A/2 = [(b^2 + c^2 - a^2)(b^2 +c^2 +a^2)] / 4bc

mathlim

原帖由 darksider 于 5-8-2008 12:02 AM 发表
prove that cos^2 A/2 = [(b^2 + c^2 - a^2)(b^2 +c^2 +a^2)] / 4bc

你的题目都错了！

cos²A/2
= (1 + cosA) / 2
= [ 1 + ( b² + c² - a² ) / 2bc ] / 2
= ( 2bc + b² + c² - a² ) / 4bc
= [ (b + c)² - a² ] / 4bc
= (b + c + a)(b + c - a) / 4bc

Allison

我不会polynomial 那课的 graphical inequalities。。。有谁能教我吗？
我知道有下面这几种题目，我只会1和2 的而已。。。
1。y = x^2 + 4
2。y = x^3 + 1
3。y = square root of x
4。y = 1/lxl
5。y = 12/（3-x）
6。y = l 1/（x+1）l
7。y = 3/x

[ 本帖最后由 Allison 于 9-8-2008 06:29 PM 编辑 ]

darksider

Find the exact value of x if 2- ln (3-x) =0
                   ---- 2 = ln (3-x)
                         e^2 = 3-x
                          x = 3 - e^2 , 对吗？

Solve the equation 3x^4 . e^(-2 lnx) + 2x - 1 = 0

~HeBe~_@

i)   对。

ii)     3x^4 . e^(-2 lnx) + 2x - 1 = 0
=>    3x^4 . e^( lnx^ (-2) ) + 2x - 1 = 0
=>    3x^4 . x^ (-2) + 2x - 1 = 0
=>    3x^2  + 2x - 1 = 0
=>  (3x - 1)(x +1) = 0
=>  x =1/3   or   x = -1

darksider

原帖由 Allison 于 2008/8/9 06:25 PM 发表
我不会polynomial 那课的 graphical inequalities。。。有谁能教我吗？
我知道有下面这几种题目，我只会1和2 的而已。。。
1。y = x^2 + 4
2。y = x^3 + 1
3。y = square root of x
4。y = 1/lxl
5。y = 1 ...

你有msn吗？add 了，比较容易教。

darksider

A circle is divided by a chord into two segments such that the areas of the segments are in the ratio 3:1. If the chord subtendes an angle of 2dita at the centre, where dita is acute, show that if the angle is measured in degrees and if sigma = 2dita - 90degree, then cos sigma = pie sigma / 180

by drawing the graphs of cos sigma and pie sigma / 180 for values of sigma between 30 deg and 50 deg, estimate the value of sigma to the nearest half degree, that statisfies the above equation, then obtain an approximation of the value of dita.

不死的ProG

cos3X=sinX

找X, 0<2X<2Pi

darksider

complementary angle formulae = cos (3x) = sin (x) ==> cos ( 3x) = cos ( 90 - x )

Case 1 : 3x = 90 - x + 360n where n is an integer
Solve for x and substitute appropriate value of n to get all the values of x satisfying the integer.

Case 2 : 3x = - (90 - x) + 360n where n is an integer
Solve for x and substitute appropriate value of n to get all the values of x satisfying the integer.

不死的ProG

原帖由 darksider 于 6-9-2008 06:06 PM 发表
complementary angle formulae = cos (3x) = sin (x) ==> cos ( 3x) = cos ( 90 - x )

Case 1 : 3x = 90 - x + 360n where n is an integer
Solve for x and substitute appropriate value of n to get all the ...

只有这种方法吗?