# Can Anyone help me with these questions?

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• May 9th 2007, 08:05 AM
novadragon849
Can Anyone help me with these questions?
Hi I was wondering if anyone can help me with these questions so I know I got them right.

So can anyone go through these questions with step by step answers

I got these on the internet with no mark schemes and I need answers for all the questions since I self teach.

Thank You
• May 9th 2007, 08:20 AM
qbkr21
Re:
#1.
a.
f(x)=x^3+3x^2+5
f'(x)=3x^2+6x
f''(x)=6x+6

#2.

a. (1-2x)^5=1-10x+40x^2-80x^3+80x^4-32x^5

#4.

5^x=17
xln(5)=ln(17)

Divide ln(17) by ln(5)

Thus

x=ln(17)/ln(5)

#5.

b. f(x)=x^3+4x^2+x-6=(x-1)(x+2)(x+3)
c. f(x)=x^3+4x^2+x-6=(x-1)(x+2)(x+3)

Hence

x=1
x=-2
x=-3

This worksheet is a piece of cake. Hopefully this will get you started.

-qbkr21
• May 9th 2007, 08:33 AM
alinailiescu
#1.(a)
f'(x)=3x^2-6x
f"(x)=6x-6
(b)see attachment
• May 9th 2007, 08:43 AM
qbkr21
Re:
Quote:

Originally Posted by alinailiescu
#1.(a)
f'(x)=3x^2-6x
f"(x)=6x-6
(b)see attachment

Alinaliescu your differentiation and integration for part A & B are wrong

you were asked to integrate

f(x)=x^3+3x^2+5 NOT:eek: f(x)=x^3-3x^2+5
• May 9th 2007, 09:00 AM
alinailiescu
Yes, I realized that after I posted the replay. I tried to print out the problems, and the printer didn't work properly.
Anyway, you posted the right solutions. :)
Thanks
• May 9th 2007, 09:03 AM
qbkr21
Re:
I can see how you misread it. The vast majority of these PDF files people attach are very hard to read because of their clarity.
• May 9th 2007, 09:10 AM
alinailiescu
#3
Since the points given are the end points of the diameter, the midpoint would be the center of the circle.
((-1+3)/2, (4+6)/2)
thus, (1, 5) is the center.
You also need the radius,
r=sqrt[(3-1)^2+(6-5)^2]=sqrt(5)
So, the equation is:
(x-1)^2+(y-5)^2=5
• May 9th 2007, 09:25 AM
novadragon849
Thanks so is anyone able to do questions 6-10
• May 9th 2007, 09:36 AM
alinailiescu
#6
first ,use cos^2(x)=1-sin^2(x) to eliminate cosine function
That would give you:
2(1-sin^2(x))+1=5sinx
2-2sin^2(x)+1-5sinx=0
-2sin^2(x)-5sinx+3=0
This is a quadratic equation in sinx, so using the quadratic formula you have:
sinx=[5+sqrt(25-4*-2*3)]/[-2*2]=[5+7]/(-4)=12/-4=-3(which has no solution for x, since sinx should be between -1 and 1)
or
sinx==[5-sqrt(25-4*-2*3)]/[-2*2]=[5-7]/(-4)=1/2
so x=sin^-1(1/2)
x=pi/6
• May 9th 2007, 09:43 AM
r_maths
http://img406.imageshack.us/img406/5...t010ug9en7.png

This is a full past paper you are asking others to do.
Surely you can do at least one or two of them.
England exams are not far away.

@alinailiescu

You missed out 5(pi)/6
• May 9th 2007, 09:48 AM
novadragon849
Sorry if it seems your helping me do my past paper so far I have done must of them but I was checking if I got them right.
• May 9th 2007, 10:06 AM
r_maths
• May 9th 2007, 10:11 AM
alinailiescu
Pay attention,
6x^3/3=2x^3 and not 3x^3
So, the answer is -2. (which makes sense since the shaded area is under the x axis,)
Your area is 2.
• May 9th 2007, 11:18 AM
novadragon849
Anyone for 8 and 9, I finished 10 so don't worry about that.
• May 9th 2007, 11:50 AM
qbkr21
Re:
8.a

v=-70
v=70

Do you want to see the work?

8.b

C''=(2800/v^3)
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