AFTER 24 hr .. P1 discussion here !

[ali khan saaed]

Wait, what?
You're saying it was tan^-1(x) = -7?
Doesn't that make it x = tan(-7)
?

yes

 

[ahmed abdulla]

foR ALL WHO GOT THE K VALUE AS 6 3 AND -23
FOR POINT OF REFLECTION
ANY EXPAINATION?

 

[yousef]

if some one is sure of his answers ...can u explain us all te touhg questions like refletion and stuff??

 

[ali khan saaed]

The question gives point R (-1.3) and line of 3y + 2x = 33 as the mirror. We are looking for the image of R, (let's name it as P)

Rearranging the equation of the mirror gives us y = -2x/3 + 11

Therefore gradient (m) of mirror is -2/3.

Since we understand that the line PR will be PERPENDICULAR to mirror (properties of reflection), therefore m of PR = 1.5.

Work out the equation of PR from the point (-1,3) and m of 1.5

y - 3 = 1.5 (x+1)

y = 1.5 x + 4.5

Another properties of mirror is that distance from object to mirror = distance from image to mirror. Therefore point of intersection of line PR and the mirror will be the MIDPOINT for line PR.

Working out simultaneous equation on PR and mirror,

y = 1.5 x + 4.5
y = -2x/3 + 11

1.5 x + 4.5 = -2x/3 + 11

9x + 27 = -4x + 66

13x = 39
x = 3

Subs x = 3 to any of the 2 equation above, we will get y = 9

Since we know R, and the midpoint, we can work out P.

(-1 + x)/2 = 3 for the x-coordinate and (3+y)/2 = 9 for the y-coordinate

Therefore x = 7 and y = 15

 

[Beaconite007]

yes

So then which one was correct?

 

[Kangxidi]

foR ALL WHO GOT THE K VALUE AS 6 3 AND -23
FOR POINT OF REFLECTION
ANY EXPAINATION?

For K values in sequence,

The two values of K are 63 and -21. I couldn't exactly remember the question, but I think it was ar^2 = 4a and ak = [a(1-r^6)]/1 - r

Solving ar^2 = 4a

r^2 = 4a/a, cancelling a will give you r^2 = 4

Remember that since it is square, your answer will come in BOTH POSITIVE AND NEGATIVE, a mistake which I suppose many of us have done.

As a result r will either be 2 or -2.

Substituting r =2 to your second equation will give you K = 63
Substituting r = -2 to your second equation will give you K = -21.

 

[Zuhair Rayyes]

Wait, what?
You're saying it was tan^-1(x) = -7?
Doesn't that make it x = tan(-7)
?

ohh sry, i meant tan x = -7

 

[Lujain M.]

hey guys for the binomial question second part was the answer 104 or something like that?

 

[Obsidian Fl1ght]

hey guys for the binomial question second part was the answer 104 or something like that?

Well, the coefficient I got was 40.

 

[Khan Boi]

It was 40

 

[Kangxidi]

40 for the second one.

 

[dragoninja98]

no kidding, for me the battle of M1, S1 and P3 all remain to be fought !
Come what may, either we rise or we fall !

In Sha Allah.
The battle of M1 still remains to be fought.

 

[Kangxidi]

All the best for the other maths' papers guys.

 

[Numnum]

The question gives point R (-1.3) and line of 3y + 2x = 33 as the mirror. We are looking for the image of R, (let's name it as P)

Rearranging the equation of the mirror gives us y = -2x/3 + 11

Therefore gradient (m) of mirror is -2/3.

Since we understand that the line PR will be PERPENDICULAR to mirror (properties of reflection), therefore m of PR = 1.5.

Work out the equation of PR from the point (-1,3) and m of 1.5

y - 3 = 1.5 (x+1)

y = 1.5 x + 4.5

Another properties of mirror is that distance from object to mirror = distance from image to mirror. Therefore point of intersection of line PR and the mirror will be the MIDPOINT for line PR.

Working out simultaneous equation on PR and mirror,

y = 1.5 x + 4.5
y = -2x/3 + 11

1.5 x + 4.5 = -2x/3 + 11

9x + 27 = -4x + 66

13x = 39
x = 3

Subs x = 3 to any of the 2 equation above, we will get y = 9

Since we know R, and the midpoint, we can work out P.

(-1 + x)/2 = 3 for the x-coordinate and (3+y)/2 = 9 for the y-coordinate

Therefore x = 7 and y = 15

I did everything except the last part and wrote R as (3,9). How many marks do you think I'll get?

 

[AdeelRox]

i wrote x<=25 as well...

almost every1 in my centre wrote that except for only one of my big nerd fellow

 

[AdeelRox]

bcz its -2.5<=x<=0

it is not -2.5<=x<=0 it is -2.5<=x<0. X cannot be zero
anyways i did it wrong by omitting the zero inequality

 

[pakiboy]

it is not -2.5<=x<=0 it is -2.5<=x<0. X cannot be zero
anyways i did it wrong by omitting the zero inequality

thnks for correcting!

 

[A star]

it is not -2.5<=x<=0 it is -2.5<=x<0. X cannot be zero
anyways i did it wrong by omitting the zero inequality

if that is right then if it wasnt for the Q7 i would have achieved 72 darn it :/

 

[Numnum]

if that is right then if it wasnt for the Q7 i would have achieved 72 darn it :/

You seem pretty sure about your grades.

 

[A star]

thats cause i and a friend went and discussed our answers i got this whole question iguess wrong and i forgot one value of K rest was ok . well u can have an estimation

You seem pretty sure about your grades.