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OrlandoBloom<3 said:
OrlandoBloom<3 said:I am having a huggeee problem with graphs !
Especially ones involving curves !
I cannot figure out what we are supposed to know for the exam !
Please help !
Jazakallah =)
Plus this question is giving me a headache
http://www.xtremepapers.net/CIE/Cambrid ... _qp_43.pdf
Question 6 part C & D
Oh okaaayyy I did get the D part very clearly =)Math_angel said:OrlandoBloom<3 said:
Assalamoalaikum!!
I am extremely sorry for not replying....i'll help u out now!!
OrlandoBloom<3 said:I am having a huggeee problem with graphs !
Especially ones involving curves !
I cannot figure out what we are supposed to know for the exam !
Please help !
Jazakallah =)
Plus this question is giving me a headache
http://www.xtremepapers.net/CIE/Cambrid ... _qp_43.pdf
Question 6 part C & D
ok so cud u mind telling what kind of problem do u have with the graphs...
anyways...for Q:6
here u go..... they r asking to find the value of k....in such questions involving solutions of equations using graphs....what u have to do make LHS of eqn....same as the eqn. of curve/graph..now draw another line with the eqn u got on RHS!! and then the solution will be the x coordinates of the point which intersect the graph... OK?
now in this case LHS is already same as the eqn of the curve....so the solutions for the eqn in part (c) is where the line y=k intersects with the graph....now what u r asked to find is the k value where this eqn. doesnt has any solutions...now check at which value it isnt possible to get any solutions....remember these shud be integer values!! that's the answer...so after plotting the graph notice that in the eqn y = k when k is 1 , 0 , -1.....it cannot intersect the curve....right??
so these are the values of k where the eqn. ..... has no solutions.....
i hope u understood....if not let me know....bcoz i personally feel i havent given a clearer explanation....so if u still have problem let me know...i can give a bit clearer explanation than this....or tell me which area are u still confused with
ok for part (d) : u know that to find gradient of curve at any point we draw a tangent....and the gradient of the tangent at that point is equal to the gradient of the curve!!!
now remember lines like this ==> / have positive gradient and lines like this ==> \ have negative gradient
so find out at which points will the graph has tangents like this => / .....write the range of these values....u'll probably have two sets ...i cud have given the answer but i havent drawn the graph so u work it out urself please...and ask me if u r stuck at any point
i hope i have cleared ur doubt and sorry once again....
Math_angel said:Assalamoalaikum!!
Rahma Magdy said:
I Know this might sound silly..
But Right know in our schools we are having a 2-weeks Christmas break,, After it directly, we have Pre-Mock examinations..
I have maths first day, I cant seem to be in contact with my teacher,,,
I need to know how to find F-1 (x )
the F inverse x !!!
..
Kindly..
oh so all u have to do is write the eqn of f(x) in terms of x and y
then make x the subject ....now replace y with x,,,,,this is f-1(x)
Eg: f(x) = 3x - 2 ...find f-1 (x)
---solution---
y = 3x - 2
y + 2 = 3x
x = (y + 2)/3
f-1(x) = (x + 2)/3
hope u got it now
OrlandoBloom<3 said:
:? f-1 x from a graph?? i dont think ever heard of it?????? whatever it is atleast now i cant remember...may be i forgot...u sure u have seen a question like that?? show me one if possible...plzz!Rahma Magdy said:
Rahma Magdy said:
Farhana.Crystalaire said:
Farhana.Crystalaire said:
JmCullen said:
JmCullen said:
Math_angel said:hmm...sorry..couldnt manage to find time to give u explanation of the chapter!!
anyways..if u have doubts at specific places..i'd be glad to help..
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