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Author Topic: C1 and C2 DOUBTS HERE!!!!!  (Read 88596 times)

Offline astarmathsandphysics

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Re: C1 and C2 DOUBTS HERE!!!!!
« Reply #510 on: September 23, 2011, 10:56:37 am »
second one was tricky.
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Offline ~ Miss Relina ~

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Re: C1 and C2 DOUBTS HERE!!!!!
« Reply #511 on: September 23, 2011, 01:29:45 pm »
Quote from: astarmathsandphysics on September 23, 2011, 10:56:37 am
second one was tricky.
ya you know that I've finshed the papers in my home for doing it only  ;D
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Offline WARRIOR

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Re: C1 and C2 DOUBTS HERE!!!!!
« Reply #512 on: September 23, 2011, 03:58:56 pm »
how?
use algebra ro find the se of values for x for which x (x-5 ) > 36  [this one is easy i got the answer right x<-4,x>9]
use ur answer from part a , find the set of  values of y for which y^2 ( y^2 -5 ) > 36

how??

thanks in advanced. seems easy but not for me.


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Offline Arthur Bon Zavi

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Re: C1 and C2 DOUBTS HERE!!!!!
« Reply #513 on: September 23, 2011, 04:24:33 pm »
Quote from: Kimo Jesus on September 23, 2011, 03:58:56 pm
how?
use algebra ro find the se of values for x for which x (x-5 ) > 36  [this one is easy i got the answer right x<-4,x>9]
use ur answer from part a , find the set of  values of y for which y^2 ( y^2 -5 ) > 36

how??

thanks in advanced. seems easy but not for me.

Let y4 be x2.
Now try and get the values of x and then of y by :
x2 = y4
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Offline astarmathsandphysics

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Re: C1 and C2 DOUBTS HERE!!!!!
« Reply #514 on: September 27, 2011, 02:55:58 pm »
This is quite a trickly log question
Solve 6 log_4 x - 5x log_4 x +20x =24
Solution
(log_4 x) (6- 5x) =24-20x
(log_4 x) (6- 5x) -4(6-5x) =0
(log_4 x -4)(6- 5x) =0
Hence log_4 x -4 =0 \rightarrow log_4 x =4 \rightarrow x=4^4 =256 or 6-5x=0 \rightarrow x=6/5 .
« Last Edit: September 27, 2011, 03:00:53 pm by astarmathsandphysics »
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Offline MKh

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Re: C1 and C2 DOUBTS HERE!!!!!
« Reply #515 on: September 28, 2011, 01:17:14 pm »
Quote from: astarmathsandphysics on September 27, 2011, 02:55:58 pm
This is quite a trickly log question
Solve 6 log_4 x - 5x log_4 x +20x =24
Solution
(log_4 x) (6- 5x) =24-20x
(log_4 x) (6- 5x) -4(6-5x) =0
(log_4 x -4)(6- 5x) =0
Hence log_4 x -4 =0 \rightarrow log_4 x =4 \rightarrow x=4^4 =256 or 6-5x=0 \rightarrow x=6/5 .


Is it from the C2 specification? and which chapter?

Thanks.
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Offline astarmathsandphysics

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Re: C1 and C2 DOUBTS HERE!!!!!
« Reply #516 on: September 28, 2011, 09:31:33 pm »
I don't think it is c2 though it is logs. Someone sent it to me.
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Offline WARRIOR

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Re: C1 and C2 DOUBTS HERE!!!!!
« Reply #517 on: October 01, 2011, 01:30:29 pm »
In triangle PQR , PQ= (x+2 ) , PR = ( 5-x) and angle p = 30*. The area of the trangle is A.
a show that A = .25 ( 10 + 3x - x^2) --- easy.
b.use the method of completing the square or otherwise , to find the maximum value of A and give the corresponding value of x.
i really dont understand HOW THE ANSWER IS 3 1/16 WHEN X = 1/2

this is exactly the same.

In triangle ABC , AB= (2-X) BC = (X+1) ANGLE B = 120

show that AC^2 = x^2 - x + 7---easy
find the value of x for which AC has a minium value. The answer is 1/2

thanks
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Offline astarmathsandphysics

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Re: C1 and C2 DOUBTS HERE!!!!!
« Reply #518 on: October 01, 2011, 02:41:56 pm »
b. complete the square
0.25(10+3x-x^2)=0.25(10+(-1.5^2+3x-x^2)+1.5^2)=0.25(10-(x-1.5)^2-2.25)=0.25(12.25-(x-1.5)^2)
Put x=1.5 then x-1.5=0 and 0.25(12.25-(x-1.5)^2)=0.25*12.25=3 1/16
For the second question
x^2-x+7=(x-0.5)^2-0.5^2+7=(x-0.5)^2+6.75
min when (x-0.5)^2=0 so x=1/2
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Offline WARRIOR

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Re: C1 and C2 DOUBTS HERE!!!!!
« Reply #519 on: October 05, 2011, 06:18:38 am »

Good morning !
When drawing SKETCH graphs for logarithms (EXPONENTIAL EQUATIONS ) do we make a simple table of values? because they always if us a range of values of x?

PS : SORRY I JUST REALISED IT SAID DRAW AN ACCURATE GRAPH.
« Last Edit: October 05, 2011, 06:25:34 am by Kimo Jesus »
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Offline ~ Miss Relina ~

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Re: C1 and C2 DOUBTS HERE!!!!!
« Reply #520 on: October 05, 2011, 02:42:43 pm »
here we go  :D
a question to sove plz

click on the image for larger view
« Last Edit: October 05, 2011, 04:57:10 pm by ~Relina~ »
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Offline MKh

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Re: C1 and C2 DOUBTS HERE!!!!!
« Reply #521 on: October 05, 2011, 03:16:23 pm »
Salaam,
I have a few doubts in the Edexcel C1 textbook.

Edexcel C1 textbook - Chapter 5: Coordinate Geometry in the (x,y) plane - Mixed Exercise (5F) :-

Page 88, Question no. 8, part (c) --> Calculate the area of triangle ACD.
Page 89, question 12, part (b) --> find the area of triangle OAB.

Both triangles are on the (x,y) plane and the coordinates of all three vertices of each of them is known.

We have already studied the formula for finding the area of a triangle provided the length of two of its sides and the angle between them. We will also meet the same formula in the C2 module. Here, I don't think we have to use the same formula for finding the area of a triangle that we came across in IGCSE math.

I will look through the step-by-step solutions available with me but, for the moment, it would be great if someone who have solved this could help out.

Thank you.
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Offline astarmathsandphysics

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Re: C1 and C2 DOUBTS HERE!!!!!
« Reply #522 on: October 05, 2011, 03:26:55 pm »
I don't have the textbook. can you scan the questions
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Offline WARRIOR

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Re: C1 and C2 DOUBTS HERE!!!!!
« Reply #523 on: October 05, 2011, 03:44:21 pm »
hi i need help in a log question

LOG(base 2)X + LOG (base 4 ) = 2 Answer is 2.52 how?
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Offline WARRIOR

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Re: C1 and C2 DOUBTS HERE!!!!!
« Reply #524 on: October 05, 2011, 03:50:58 pm »
Quote from: astarmathsandphysics on October 05, 2011, 03:26:55 pm
I don't have the textbook. can you scan the questions

Q8 =  The straight line L1 passes through the points A and B with coordinates ( 0,-2 ) and (6,7) respectively.
a. find the equation of L1 in the form of y= mx + c
The straight line L2 with equation x + y = 8 cuts the y-axis at point C. The lines L1 and L2 intersect t the point D.
b. Calculate the coordinated of point D
c. Calculate the area of ACD

Q 12
Find an equation of a straight line passing through the points with coordinates  (-1,5) (4,-2) , giving your answer in the form of ax+ by +c=0
where a , b , c are integers.
The line crosses the x-axis  at the point A and the y -axis at the point B , and O is the origin
b.find the area of OAB.

thse are mkH's questions.
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But it ain't how hard you hit; it's about how hard you can get hit, and keep moving forward-Balboa
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