How was AS Math P1?

[yash1972007]

The answers of Q.2 (i) was 60 and (ii) was 40 na? -_-

-40

 

[Lostsoul]

I
wrote decreasing because when u increased x, the gradient decreases.....:S

And I wrote neither for the same reason..

 

[Lostsoul]

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

I have got just lik etill here. How much do I get?
Also, the exoansion stuff, my first one was, so automnatically second will be wriong, the first coefficient of x^2 i did a silly calc mistake, it was 60 I got 120, just a small error.. and the second part there was a 20, where we had to do 60-20, I got the twenty right. How much overall would I loose?

also, the functions domain, range, i left it -__- How much do I loose?

GP, I think I just got the ratio.. I did not manage to finish it. How much do I loose? Please help

 

[pakiboy]

do you remeber the question ?
as far as i remember the domain of f(x) was <=1 and the x value was in denominator
so how can the domain of f inverse x be 0 because if the x in F(x) is 0 and in denominator the answer would be infinte
i might be wrong but just tell me the question if you remeber it

the domain was >= 1
i dnt remember the exact quest.

 

[pakiboy]

what was it about?

as far as I know decreasing means the discriminant is negative, and since it was (x-something)^2 it could never be negative.. :/ though I don't know when is it going to be "neither"

why do you say that :L

The question wasn't about reflection.. If you saw the answer which someone wrote above you can see its all co-ordinate geometry. The "reflection" word in the question was just figured out by common sense.
I don't think reflections are listed in the syllabus, and most people didn't study it either

just to correct u guyz!

 

[Lostsoul]

Yeah the threshold will be around 58 or 59 !

Naaaahiiiiiiiiin :'( :'(
Okay I become dramatics over some matter
That's over estimating the paper
The GT should be 50 as for penalizing the examiner for setting a question out of the syllabus. Let us all protest!

 

[Mayedah]

Naaaahiiiiiiiiin :'( :'(
Okay I become dramatics over some matter
That's over estimating the paper
The GT should be 50 as for penalizing the examiner for setting a question out of the syllabus. Let us all protest!

nothing was out of syllabus i think ..it was just using ur basic knowledge of coordinate geometry !

 

[Lostsoul]

nothing was out of syllabus i think ..it was just using ur basic knowledge of coordinate geometry !

LOL! Yess you are right.. A reflection question came after loong for me, so I prolly required more time to think over it and solve it.. Never mind, lets hope for the best

 

[0Louis0]

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

I have got just lik etill here. How much do I get?

The question has 7 marks and I never saw it before in the past papers so I don't think anyone can help you estimating.. but I would guess at around 3

Also, the exoansion stuff, my first one was, so automnatically second will be wriong, the first coefficient of x^2 i did a silly calc mistake, it was 60 I got 120, just a small error.. and the second part there was a 20, where we had to do 60-20, I got the twenty right. How much overall would I loose?

If everything else is correct you lose 1 mark for the final answer of the first part, and nothing for the second

also, the functions domain, range, i left it -__- How much do I loose?

I can't remember what did the question say and how many marks it had in total, did it ask for the inverse and its domain?

GP, I think I just got the ratio.. I did not manage to finish it. How much do I loose? Please help

I'd say 1 since you only needed to substitute

 

[Lostsoul]

The question has 7 marks and I never saw it before in the past papers so I don't think anyone can help you estimating.. but I would guess at around 3

If everything else is correct you lose 1 mark for the final answer of the first part, and nothing for the second

I can't remember what did the question say and how many marks it had in total, did it ask for the inverse and its domain?

I'd say 1 since you only needed to substitute

The function question asked for the inverse..

 

[Arslanjaved]

the domain was >= 1
i dnt remember the exact quest.

well i remember the domain and it was x<=1

 

[NouranAyman]

answer to Q 7 guys please!

 

[Saadman]

Grade Thresholds!? And was that function decreasing or increasing!? Thanks!! =)

 

[A star]

Grade Thresholds!? And was that function decreasing or increasing!? Thanks!! =)

increasing hoping for <60

 

[Saadman]

increasing hoping for <60

Yesss!! I gave increasing too!! Thanks! =)

 

[Ahmed Khalel]

Are you guys kidding ?
no matter how hard paper comes I have never seen GT of P1 less than 57,58
And this was relatively easy paper. Q7 was also easy it's just that very few people were able to pick it

Most of the question were easy ,put personally i needed more time to finish it.

 

[shafayat]

messed up! :/

 

[LeQuavina]

I
wrote decreasing because when u increased x, the gradient decreases.....:S

I wrote the same exact thing.

 

[istudent]

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

got same answer but I did half question on answer sheet and half on extra sheet would it cause any problem?????

 

[jysim123]

Hi fellow members,

Just want to ask you guys what the answers/solutions are for the quadratic equation in cos^2x (or was it cos^2 theta??)

Thanks!
(Were there four solutions?)