Question 4:
4 | (i) | h = 20, stops growing | B1
[1] | AG need interpretation
(ii) | h = 20 – 20et/10
dh/dt = 2et/10
20 et/10= 20 – 20(1 – et/10) = 20 – h
=10dh/dt
when t = 0, h = 20(1 – 1 ) = 0
………………………………………………………
OR verifying by integration
dh dt
 
20h 10
ln(20h)0.1tc
h0,t0,cln20
ln(20h)0.1tln20
20h
ln( )0.1t
20
20h20e0.1t
h20(1e0.1t) | M1A1
M1
A1
B1
M1
A1
B1
M1
A1
[5] | differentiation (for M1 need ket/10, k const)
oe eg 20 – h = 20 – 20(1 – et/10) = 20et/10
= 10dh/dt (showing sides equivalent)
initial conditions
……………………………………………………………………………..
sep correctly and intending to integrate
correct result (condone omission of c, although no further marks are possible)
condone ln (h – 20) as part of the solution at this stage
constant found from expression of correct form (at any stage) but B0 if say
c = ln (–20) (found using ln (h – 20))
combining logs and anti-logging (correct rules)
correct form (do not award if B0 above)
4 | (iii) | 200 A B
 
(20h)(20h) 20h 20h
 200 = A(20  h) + B(20 + h)
h = 20  200 = 40 B, B = 5
h = 20  200 = 40A , A = 5
200 dh/dt = 400  h2
200
  dhdt
400h2
5 5
 (  )dhdt
20h 20h
 5ln(20 + h) 5ln(20  h) = t + c
When t = 0, h = 0  0 = 0 + c  c = 0
20h
 5ln t
20h
20h
 et/5
20h
 20 + h ===  h)et/5 = 2200 t/5  h et/5
 h + h et/5 = 20 et/5  20
 h(et/5+ 1) = 20(et/5  1)
20(et/51)
 h
et/51
20(1et/5)
 h *
1et/5 | M1
A1
A1
M1
A1
B1
M1
DDMM11
A1
[9] | cover up, substitution or equating coeffs
separating variables and intending to integrate (condone sign error)
substituting partial fractions
ft their A,B, condone absence of c, Do not allow ln(h-20) for A1.
cao need to show this. c can be found at any stage. NB c = ln(–1) (from ln(h –20))
or similar scores B0.
anti-logging an equation of the correct form . Allow if c = 0 clearly stated
(provided that c = 0) even if B mark is not awarded, but do not allow if c omitted.
Can ft their c.
maki h the subject, dependent on previous mark
NB method marks can be in either order, in which case the dependence is the other
way around.(In which case, 20 + h is divided by 20 – h first to isolate h).
AG must have obtained B1 (for c) in order to obtain final A1.
4 | (iv) | As t  ∞, h  20. So long-term height is 20m. | B1
[1] | www
4 | (v) | 1st model h = 20(1  e01) = 1.90..
2nd model h = 20(e1/5 1)/(e1/5 + 1)= 1.99..
so 2nd model fits data better | BB11
B1
B1 dep
[3] | O 1st model h = 2 gives t = 1.05..
2nd model h = 2 gives t = 1.003..
dep previous B1s correct
5 (i) 1 =4 − 1 2(1− 1 x2) − 1 2
4−x2 4
1 3
(− )(− )
= 1 [1+(− 1 )(− 1 x2)+ 2 2 (− 1 x2)2+...]
2 2 4 2! 4
1 1 3
= + x2+ x4+...
2 16 256
(ii) ∫ 1 1 dx≈∫ 1 ( 1 + 1 x2+ 3 x4)dx
0 4−x2 0 2 16 256
= ⎡1 1 3 ⎤1
x+ x3+ x5
⎢ ⎥
⎣2 48 1280 ⎦
0
1 1 3
= + +
2 48 1280
= 0.5232 (to 4 s.f.)
(iii) 1 1 ⎡ x⎤ 1
∫ dx= arcsin
⎢ ⎥
0 4−x2 ⎣ 2⎦
0
= π/6 = 0.5236 | M1
M1
A1
A1
M1ft
A1
B1
[7] | Binomial coeffs correct
Complete correct expression inside
bracket
cao