Maths P1 13th/oct .. how was it?

[RatJumper]

the paper wasnt too difficult but a litle tricky...domain of f inverse was the range of F(x)..so it was more than 3.h(x) was x^2+3(i dont remember th exact answer...may be -3)...wat was ur answer of vector's question.....at my school ..every had different answer execpt few people

yeah i got 69.6 degrees. you??

and btw, so r u sayin ur range was >3 or it was >=3.

 

[khalid]

yes i got my angle in 60's and my teacher told me that range of function is the domain of its inverse.. so i wrote exactly the same .still i am not sure

 

[RatJumper]

yeah but did u rite >3 or >=3

 

[Islu_jf]

i wrote >3 bcuz the x was >0 so i thot its ganna be the same!!

 

[RatJumper]

oh ok.. i rote >= 3 for my range :/

for the inverse domain, did u rite <3?

 

[grimreaper_123]

Aoa guyz..
overall my paper went great..im just unsure bout the volume question...my ans was 0.917pi..can anyone cunfirm that?
thnx

 

[khalid]

nai sir....maximun students( including me)except few did the volume question wrong.answer was 2.88 pie some thing. i made silly mistake.. didnt put the correct integrals.

 

[Islu_jf]

yeah bcoz in all the past papers .. its a rule ..

the question was x(2)-4x+7
and when u do the complete square form it will be.. (x-2)square+3
so therefore in the nxt part it says.
Obtain an expression for f-1 bla bla and state the domain..

so if the range of the first one was y>3 so it will be the same x>3

 

[RatJumper]

yeah bcoz in all the past papers .. its a rule ..

the question was x(2)-4x+7
and when u do the complete square form it will be.. (x-2)square+3
so therefore in the nxt part it says.
Obtain an expression for f-1 bla bla and state the domain..

so if the range of the first one was y>3 so it will be the same x>3

oh ok so the domain was also >3 damm i rote <3 . oh well.. how many marks did i loose for that?

 

[Islu_jf]

one mark bcuz the whole one had 3/// 2 for the f inverse and 1 for the domain..
no worries.. =D

 

[RatJumper]

thank god lol hey dude, also, wt ws ur answer for the k question??

 

[Islu_jf]

i had k=4 ..
and the coordinates = (0.5 ,15/2) nt really sure.. u ?
i will try to attach the paper sumwhere if um allowd to!

 

[RatJumper]

i thought they asked for the range of k.. thus my anser was 0 < k < 4 :/ and my coordinates were (0.5,2)

 

[Islu_jf]

k here u go the question..

A curve has eq. y=kx*2 +1 and a line has eq. y=kx, where k is a non-zero constant.

(i) find the set of values for which the curve and the line have no common points. [3]
(ii)state the value of k for which the line is a tangent to the curve and, for this case, find the coordinates of the point where the line touches the curve.[4]

 

[RatJumper]

Thankx for that bro
and since it said set of values i think we needed to find the range and not just state the value of k. since the equation would look like k^2 - kx + 1 < 0 when we use b^2 - 4ac, ...when we solve for k, we get two values, 0 and 4. those would be our two limits for k wouldnt they?

 

[Islu_jf]

yeah .. my anser was for both parts

 

[RatJumper]

awesum what was ur guys's answer for the rate of change question part 2 where it sed 5 years..

 

[khalid]

mine was 7/30.

 

[RatJumper]

how did u do it? wt ws ur equation for the dx/dt for part 1

 

[khalid]

dont remeber the exact but in first we just had to differentiate and then in part 2 put 5=t