Expert: Paul Klarreich - 8/30/2009

Question
QUESTION: hello there..hopefully you can help me to answer such questions..im doing Ib..n found its higher level math quite a tough 1..here the questions
1.   Show that x-1 is a factor of p(x)=x2k –(m+1)x2 + m where k and m are integers
2.   |1/5x3 -5x + 2 | and k(x)= logx 3 …find the set of values of x such that h(x)<k(x)
3.   solve for x if 1- 2b/(x-a) = (a2-b2)/(x2-2ax+a2),
4.   Find the set value of a such that ax2 +ax +3+ a ≤ 0 for all real value of x

p/s; some question..2 is considered as square actually..3..cube..


ANSWER: Questioner: belle
Country: Malaysia
Category: Advanced Math
Private: No
Subject: higher level math for IB
Question: hello there..hopefully you can help me to answer such questions..im doing Ib..n found its higher level math quite a tough 1..here the questions

1. Show that x-1 is a factor of p(x)=x2k –(m+1)x2 + m where k and m are integers

2. |1/5x3 -5x + 2 | and k(x)= logx 3 …find the set of values of x such that h(x)<k(x)

3. solve for x if 1- 2b/(x-a) = (a2-b2)/(x2-2ax+a2),

4. Find the set value of a such that ax2 +ax +3+ a ≤ 0 for all real value of x

p/s; some question..2 is considered as square actually..3..cube..
...................................................................
Hi, Belle,

p/s. I ask questioners to tell me what topic they are studying; 'IB' does not mean anything here.

1. Show that x-1 is a factor of p(x)=x2k –(m+1)x2 + m where k and m are integers

The Factor Theorem says: x-1 is a factor of p(x) if p(1) = 0.

That should be enough.
....................................
2. |1/5x3 -5x + 2 | and k(x)= logx 3 …find the set of values of x such that
h(x) < k(x)

h(x) ?????????????????????????  What is  logx 3?

Please check your notation.  Look at other answers to see how to write math.
....................................
3. solve for x if 1- 2b/(x-a) = (a2-b2)/(x2-2ax+a2)

Here is your equation:

     2b       a^2 - b^2
1 - ------- = -----------
   x - a      (x - a)^2

x - a - 2b    a^2 - b^2
---------- = -----------   << combine fractions on the left.
 x - a      (x - a)^2

(x - a)(x - a - 2b) = a^2 - b^2   << multiply by  (x-a)^2

This is going to be a quadratic.  Simplify this way:

Let Y = x - a:   << temporary change of variable.

Y(Y - 2b) = a^2 - b^2  << temporary change of variable.

Y^2 - 2bY  + b^2  = a^2  << multiply out and bring over b^2

(Y - b)^2 = a^2   << factor left side.

Y - b = +-a    << take roots.

y     = b +- a

x - a = b +- a  << put back  x - a for Y.

Solutions:

x - a = b - a  << use the minus
x = b          << one solution

x - a = b + a  << use the plus.
x = b + 2a     << other solution.

Check:  

Oh, I think I will leave that to you.  YOU SHOULD DO IT.

..........................................
4. Find the set value of a such that ax2 +ax +3+ a ≤ 0 for all real value of x.

Since  ax^2 + ax + 3 + a  is a quadratic in x, its graph is a parabola.  If it has to be <= 0, it must be a 'vertex on top' parabola, with  a < 0.  If the vertex is above the x-axis, it would have two real roots, but it cannot.

If it had two real roots, the discriminant (look that word up, please) would be positive, but it cannot.

So the discriminant must be negative or zero:

D = a^2 - 4(a)(3 + a) <= 0

a^2 - 12a - 4a^2 <= 0

- 12a - 3a^2 <= 0

-3a(a + 4) <= 0

a(a + 4) >= 0

Intersections at  a = 0,  a = -4.

If  a > 0, both are positive.  That is ok, but we need  a < 0.  No good.
If  -4 < a < 0, a is neg, a+4 is pos.  No good.
If  a < -4, a is neg, a+4 is neg.  GOOD.

That is it!  The set of values is  a <= - 4


---------- FOLLOW-UP ----------

QUESTION: sorry  for the mistake...I've corrected it
h(x)=|1/5x3 -5x + 2 | and k(x)= logx 3 …find the set of values of x such that
h(x) < k(x)..
p/s; actually, logx 3 is just like log10 4<<(example)..x and 10 is the 'footer' of log..i don't know how to write it in correct way..again, i do apologize..

of your explanation, some of them..i cant really understand it..

ax^2+ax+3+a..<<<how could you simplify it as >>>D = a^2 - 4(a)(3 + a) <= 0
how does + symbol change into multiply 4(a)(3 + a)?..

Answer
Questioner: belle
Private: no
Subject:  

QUESTION: sorry  for the mistake...I've corrected it
h(x)=|1/5x3 -5x + 2 | and k(x)= logx 3 …find the set of values of x such that
h(x) < k(x)..
p/s; actually, logx 3 is just like log10 4<<(example)..x and 10 is the 'footer' of log..i don't know how to write it in correct way..again, i do apologize..

of your explanation, some of them..i cant really understand it..

ax^2+ax+3+a..<<<how could you simplify it as >>>

D = a^2 - 4(a)(3 + a) <= 0

>> I did not 'simplify it' -- this is the discriminant.  Look up your work on solving quadratic equations.

how does + symbol change into multiply 4(a)(3 + a)?..

>> It doesn't.  See above comment.

You could also do an analysis like this:

Find the vertex V of the parabola:

y = ax^2 + ax + 3 + a

Complete the square:

y = ax^2 + ax          + 3 + a
y = a(x^2 + x       )  + 3 + a
y = a(x^2 + x + 1/4 )  + 3 + a - a/4
y = a(x + 1/2)^2  + 3 + 3a/4

Now you must have the y-coordinate of the vertex <= 0.

3 + 3a/4 <= 0

3a/4 <= -3

a <= -4

That is it.


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I don't have much help for you on the last one:

k(x) = log(base x) (3) = y

means  x^y = 3

y ln x = ln 3

So:  k(x) = ln 3/ln x

Now you could write:

|1/5x^3 -5x + 2 | < ln 3/ln x

But that does not help much.