数学Paper 1讨论专区

walrein_lim88

回复 620# idontwant2b


    (a)很简单而已，
       To be quadratic equation, coeficient of X^2 cannot = 0
      K + 1 cannot = 0
      K cannot = -1
     ( K is the element of real number ,K cannot be -1)

无头猫

今年的maths1算是很容易的说 >v<

Log

回复 616# lonely_world

b). let the roots be a ( a>0）and b ( b<0),then,
P.O.R , ab<0
              c/a <0
              (k+2)/ (k+1）<0
              ∴ -2< k < -1

但是，这问题有没有出错？有problem

~HeBe~_@

Show that the equation f(x)=(k+1)x^2+(2k+3)x+(k+2)=0 has real roots
[f(x)=0 has real roots 意思是b^2 - 4ac ≥ 0 就是
equals to 0 or greater than 0 i.e.
equal real roots or distinct real roots]
for
all real values of k. find the set of values of k for which

f(x)=0 is a quadratic equation and has one positive & one negative real roots.
[意思是f(x) has two distinct real roots with one 1 +ve and 1 -ve real roots]

Sketch the graph of f(x)=(k+1)x^2+(2k+3)x+(k+2) for k=-1 1/2.


Solution：

f(x) = (k+1)x^2 + (k+3)x + (k+2) = 0

b^2 - 4ac = (k+3)^2 - 4(k+1)(k+2)
= 1 (Since b^2 - 4ac > 0,
then it has two distinct real roots)

f(x) = 0 => (k+1)x^2 + (k+3)x + (k+2) = 0

x = [-b ± √(b^2 - 4ac)]/2a
= [-(2k+3) ± √(1)]/2(k+1)
= -1 or -2(K+2)/2(k+1)
= -1 or -(K+2)/(k+1)

The two distinct real roots are -1 and -(K+2)/(k+1).
Since f(x) = 0 has one positive and one negative roots, then
x=-1 is negative root but x=-(K+2)/(k+1) is positive root.

Therefore, -(K+2)/(k+1)≥0
(K+2)/(k+1)≤0
-2 < k < -1

Sketch the graph of f(x)=(k+1)x^2+(2k+3)x+(k+2) for k=-1 1/2.

Solution:


Substitute k=-1 1/2 into f(x), then f(x) = -1/2 x^2 + 1/2

When f(x) = 0, then the distinct roots of f(x)= 0 are x = ±√1.
When x= 0 , then f(0) = 1/2 [It is y-intercept]


Find the turning point,
f(x) = -1/2 x^2 + 1/2
f'(x) = - x = 0 => x=0, and y = 1/2

f "(x) = -1 < 0 (It is a max. point)
Therefore, (0,1/2) is a max. point)


The graph passes through points (1,0), (-1,0), and (0,1/2).
Sketch the graph based on those points with 'n' shape.

lonely_world

show that 1 is a zero of the polynomial f(x)=x^3 -7x^2 +7x -11. hence, find the polynomial g(x) such that f(x)=(x-1)g(x).
show that g(x) is always positive.hence, determine the set of values of x for which x^2 +17  ≤ 7x + (11/x)

[0<x ≤1]

麻烦教我这题..

walrein_lim88

show that 1 is a zero of the polynomial f(x)=x^3 -7x^2 +7x -11. hence, find the polynomial g(x) such ...
lonely_world 发表于 3-1-2010 10:57 PM

walrein_lim88

walrein_lim88 发表于 3-1-2010 11:21 PM

    我的答案有点错误
因为那个x是分母,x不可以=0
所以那个range of x shud be { 0<x<1 }
If the question ask find the set of value x, got the word SET. must put this { }

白羊座aries

lim   (√ x -√ 3 ) / [(x )(x-3)]
x->3

ans:√ 3/18

lim
x->1-  |x^2 -1|/(x-1)

ans: -2


fx=(x-1)(x-3), x<3
find value of x such that f(x)=f^-1(x)
f^-1(x) is inverse of fx

我找到f^-1
怎样找value of x =.=

四月一日的小皮

can show ur working for f(x) = f^-1(x)?

四月一日的小皮

是x=０，３，和两个complex value？

四月一日的小皮

lim   (√ x -√ 3 ) / [(x )(x-3)]
x->3

其实（x＾２ － y＾２）可以是（x － y）（x + y）。这个应该是common了。用这个就可以消除掉（x － ３）这一组了。

lim
x->1-  |x^2 -1|/(x-1)


这个因为lim－＞１－的关系，做法也比较复杂点。记得别忽略modulus的function和application。（x － １）可以被cancel掉，别忽略sign。

白羊座aries

是x=０，３，和两个complex value？
四月一日的小皮 发表于 5-1-2010 10:19 PM

   
我也是算到2个complex values,但是答案给其中一个。
..
(x^2 -4x +1)^2 = x+1
这么做

白羊座aries

lim   (√ x -√ 3 ) / [(x )(x-3)]
x->3

其实（x＾２ － y＾２）可以是（x － y）（x + y）。这个应该 ...
四月一日的小皮 发表于 5-1-2010 10:35 PM

可以不要只是解释吗？能帮忙做吗？如果你这么解释我就会，那么我早就会了

Log

fx=(x-1)(x-3), x<3
find value of x such that f(x)=f^-1(x)
f^-1(x) is inverse of fx
是不是x=0.6972？

白羊座aries

fx=(x-1)(x-3), x
Log 发表于 5-1-2010 11:41 PM

   
对啦，0.6972...
还有一个complex number为什么拒绝？

Log

for function and graph , we must always take REAL x values

Log

这题这样做，f(x）=f-1(x),if you sketch ,you will find that this two curves intersect at y=x .
thus,f(x)=f-1(x)=x
        f(x)=x
        x^2-4x+3=x
.....

walrein_lim88

可以不要只是解释吗？能帮忙做吗？如果你这么解释我就会，那么我早就会了
白羊座aries 发表于 5-1-2010 10:54 PM

   

白羊座aries

这题这样做，f(x）=f-1(x),if you sketch ,you will find that this two curves intersect at y=x .
thus, ...
Log 发表于 5-1-2010 11:54 PM

   
f-1x = surd x+1
fx=f-1x
then,
x^2 -4x +1 = surd x+1
(x^2-4x+1)^2 = x+1 .
应该是这样才对阿

walrein_lim88

f-1x = surd x+1
fx=f-1x
then,
x^2 -4x +1 = surd x+1
(x^2-4x+1)^2 = x+1 .
应该是这样 ...
白羊座aries 发表于 6-1-2010 12:14 AM

    是什么答案？？