The Student Room Group

pure 3

hi ay sorry im a lil stuck on these Qs, if u can help me il b very grateful!!

1) i) for the curve 2x^2 +xy+ y^2 = 14, find dy/dx in terms of x and y.

ii) deduce that there are 2 points on the curve 2x^2 + xy + y^2 = 14 at which the tangenst are parallel to the x-axis and find their coordinates.


2) i) the formula for tan2x = 2tanx/(1-tan^2x)

ii) By letting tanx= t show that the equation,

4tan2x + 3cotx sec^2x = 0

becomes

3t^4 - 8t^2 - 3= 0.

iii) find all the solutions of the equation,
4tan2x + 3cotx sec^2x = 0
which lie in the range 0<= x<=2pi.
Reply 1
:eek: what exam board is this from ? ( hope its not aqa !)
Reply 2
2) ii) 4tan2x + 3cotx sec2x = 0
split this into parts and change everything to tanx;
i first used the tan2x formula to and replaced the tanx with t,
so 4tan2x = 4(2t/1-t2)
3cotx = 3/t

for sec2x, use the identity, tan2x + 1 = sec2x

therefore sec2x = t2 + 1

therefore the whole equation can be replaced with;
4(2t/1-t2) + 3(1/t x t2 + 1) = 0

this simplifies to
8t2 + (3t2 + 3)(1- t2)

and then 3t4 - 8t2 - 3 = 0

to find the solutions, factorise the above equation to get
(3t2 + 1)(t2 - 3)
Reply 3
gemm
2) ii) 4tan2x + 3cotx sec2x = 0
split this into parts and change everything to tanx;
i first used the tan2x formula to and replaced the tanx with t,
so 4tan2x = 4(2t/1-t2)
3cotx = 3/t

for sec2x, use the identity, tan2x + 1 = sec2x

therefore sec2x = t2 + 1

therefore the whole equation can be replaced with;
4(2t/1-t2) + 3(1/t x t2 + 1) = 0

this simplifies to
8t2 + (3t2 + 3)(1- t2)

and then 3t4 - 8t2 - 3 = 0

to find the solutions, factorise the above equation to get
(3t2 + 1)(t2 - 3)


ok so that may be a bit confusing to read, but i got the right answer
Reply 4
1i) Diff. wrt x gives:
4x+y+x.dy/dx+2y.dy/dx=0
dy/dx=(-4x-y)/(x+2y)

ii) Tangent // to x-axis => grad.=0
dy/dx=0
(-4x-y)/(x+2y)=0
-4x-y=0................(1)
Also satisfies the eqn:
2x^2 +xy+ y^2 = 14.........(2)
Solving simultaneously gives 2 solutions.

2i) Can be proven using the sin2x and cos2x formulae:
tan2x=sin2x/cos2x=2sinxcosx/(1-2sin2x)
Dividing numerator and denominator by cos2x:
=> 2tanx/(sec2x-2tan2x)
= 2tanx/(1-tan2x)

ii) LHS
= 4tan2x + 3cotxsec2x = 0
=> 8t/(1-t2)+3t-1(1+t2) = 0
=> 8t+3t-1(1+t2)(1-t2) = 0
=> 8t+3(1/t-t3) = 0
=> 8t2+3-3t4 = 0
=> 3t4-8t2-3 = 0. =RHS

iii) (3t2+1)(t2-3) = 0
t2 = 3,-1/3
=> tan2x = 3,-1/3
=> tanx = ±&#8730;3 (reject t2= -1/3)
=> x = pi/3, -2pi/3
Reply 5
sumitk87
:eek: what exam board is this from ? ( hope its not aqa !)



hey, its ocr!! take it ur not doin dat board!! the amount of chapters in ocr pure 3 is soo much, theyv made it soo much easier with the new core modules theyv brought out instead of pure. so ridiculous!!!
Reply 6
gemm
ok so that may be a bit confusing to read, but i got the right answer

hey thank u very much!!! that was prob the most helpful answer iv had, i liked the lil comments 2 explain wot u were doin. thats where i tend 2 get lost most of the time wen ppl skip stages and then i get confuzzled!! thank u very much!!