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Mathematics: Post your doubts here!

Haya Ahmed

Haya Ahmed

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Hey guys !! How to solve
Q31 and Q36 !! thanks :)

Nqkdjc.jpg
 
ZaqZainab

ZaqZainab

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Haya Ahmed said:
Hey guys !! How to solve
Q31 and Q36 !! thanks :)

Nqkdjc.jpg
Click to expand...
I couldnt understand 31 yet
but here is 36
I use the normal division method (Explained in the P3 text book)

We have 4x^4 +0ax^3 -9a^2x^2 +2(a^2-7)x -18
and 2x-3a
Untitled.png

(EDITED) I just got 31
multiply (x-root of3) with (x+root of 3)
you will get the 1st quadratic factor
to find the next you have to carry out division just like i did in 36
 
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W

Wolfgangs

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Can someone please help me with this question? Explain?

Answers: i - 12
ii - 8.45m
iii - 13
iv - 5.41m
 

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CaptainDanger

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Can I get a step by step solution for both these parts? (Mentioning all the identities!)
For the 2nd one, just explain how do we get to the second step and how does under-root 2 become just 2 in the next step?
Thanks in advance!

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10707946_10201737626402153_474615748_n.jpg
 
ZaqZainab

ZaqZainab

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Wolfgangs said:
Can someone please help me with this question? Explain?

Answers: i - 12
ii - 8.45m
iii - 13
iv - 5.41m
Click to expand...

s=-1.25
u=?
v=?
a=10
t=2.5
(units omitted because I keep thinking the "s" in "m/s" is related to displacement)
(Positive direction is upwards, hence negative acceleration)

i) The initial speed upwards

s=ut+0.5at^2
-1.25=2.5u-31.25
2.5u=30
u=12

ii) The greatest height reached

Here, we find the position where the ball reached zero velocity due to gravity.

From the previous question, we can see that u=12, and in this case, v=0, and a=-10

v=u+at
0=12-10t
10t=12
t=1.2

Now, we find the distance travelled:

s=(u+v)t/2
s=12*1.2/2
s=7.2

Don't forget to add 1.25, as this was the starting point. So, the ball actually reaches 8.45m into the air.

iii) The speed of the ball upon first impact with the ground

We know at the top of the flight, velocity is zero, so set u=0.
We also know that a=10 (this time, positive velocity is downwards).

The total time is 2.5, and 1.2 of that is spent travelling upwards. This means that the ball fell 1.3 seconds.

v=u+at
v=10*1.3
v=13

iv) The greatest height after the first bounce

The ball bounces up at the speed of 12.8. We know at the top of its bounce, v=0, and we also know that a=-10

So:

v=u+at
0=12.8-10t
10t=12.8
t=1.28

And with s=t(u+v)/2:

s=1.28(12.8)/2
s=8.192

(not done by me)
 
SitiPutri

SitiPutri

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jungly krinsky said:
please help with question 8 ii
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_qp_11.pdf
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_ms_11.pdf
Don't realy undertand what I am being asked to do here.
Click to expand...

To find the range you have to know the minimum and maximum value of f(x). See, (x-2)^2 for any x CAN'T be negative because it's a square. And since it can't be negative, so its least value must be 0, right?

From there we can get f(x) = -4 + k for the minimum value of f(x). And for the maximum value, it's just simply infinity because there's no other limitation.

Thus, the range is f(x) > -4 + k
 
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