dy dy − 2 y e−2 x = 2 + 2 y dx dx d dy y e−2 x ) = e −2 x − 2 y e−2 x ( dx dx ( e−2 x − 2 y ) ddxy = 2 + 2 y e−2 x
e−2 x
(a)
(b)
At P ,
Marks
A1 correct RHS
M1 A1 B1 M1
d y 2 + 2 y e−2 x = −2 x dx e − 2y
A1
d y 2 + 2 e0 = 0 = −4 dx e −2
M1
(5)
Using mm′ = −1 1 4 1 y −1 = ( x − 0) 4 x − 4y + 4 = 0 m′ =
M1 M1
or any integer multiple
A1
(4) [9]
Alternative for (a) differentiating implicitly with respect to y. dx dx e −2 x − 2 y e −2 x = 2 + 2y A1 correct RHS dy dy d dx y e −2 x ) = e−2 x − 2 y e−2 x ( dy dy ( 2 + 2 y e−2 x ) ddxy = e−2 x − 2 y
M1 A1 B1 M1
dx e −2 x − 2 y = d y 2 + 2 y e −2 x d y 2 + 2 y e −2 x = −2 x dx e − 2y
6666/01 GCE Mathematics June 2009
5
A1
(5)
Question Number
Q5
Scheme
Marks
dx dy = −4sin 2t , = 6 cos t dt dt
(a)
At t = (b)
π 3
,
Use of
dy 6 cos t ⎛ 3 ⎞ =− ⎜= − ⎟ dx 4sin 2t ⎝ 4sin t ⎠ 3 √3 m=− =− accept equivalents, awrt −0.87 √3 4× 2 2
cos 2t = 1 − 2sin 2 t x y cos 2t = , sin t = 2 6 2 x ⎛ y⎞ = 1− 2⎜ ⎟ 2 ⎝6⎠ y = √ (18 − 9 x )
M1 A1
(4)
M1
M1
( = 3 √ ( 2 − x ))
cao
A1
−2 ≤ x ≤ 2
k =2
B1
0 ≤ f ( x) ≤ 6
either 0 ≤ f ( x ) or f ( x ) ≤ 6
B1
Fully correct. Accept 0 ≤ y ≤ 6 , [ 0, 6]
B1
Leading to
(c)
B1, B1
(4)
(2) [10]
Alternatives to (a) where the parameter is eliminated 1
y = (18 − 9 x ) 2 1
1 dy 1 − = (18 − 9 x ) 2 × ( −9 ) dx 2 π 2π At t = , x = cos = −1 3 3 dy 1 1 √3 = × × −9 = − dx 2 √ ( 27 ) 2