Question 1:
1 | (a) | ∂z
=12+6y
∂x
∂z
=−30+6x
∂y
∂z
=0⇒12+6y=0⇒ y=−2
∂x
∂z
=0⇒−30+6x=0⇒x=5
∂y
y = –2, x = 5 ⇒ z = 60 (so the only SP is
(5, –2, 60)) | B1
B1
M1
M1
A1
[5] | 1.1
1.1
1.1
1.1
1.1 | ∂f
For both B marks accept eg or z
x
∂x
or f .
x
Both could be embedded in a vector.
Setting one partial derivative to 0 and
finding the value of one variable
Setting the other partial derivative to
0 and finding the value of the other
variable
SP found and no others. Do not ISW.
Must be coordinates not vector. | For both B marks accept d or δ but
must be correct indication of both
parameters (ie ∂x/∂z = or ∂z = is B0).
Ignore z component.
This mark can only be awarded if it is
clear that both derivatives are 0 at the
SP (eg stated or by giving one SP with
the found co-ordinates).
Answer in format x =..., y =..., z = ... is
sufficient if not invalidated.
1 | (b) | 24x + b ≡ 12x – 30a + 6ax = (12 + 6a)x – 30a
so 12 + 6a = 24 (or b = –30a)
a = 2
a = 2 ⇒ b = –60 | M1
A1
A1FT
[3] | 3.1a
2.2a
2.2a | Substituting y = a into the equation of
the surface, equating to 24x + b and
correctly comparing constants or x
coefficients (soi by correct answers).
FT –30×their value of a.
SCB1 (after M0 or M1) if a and b
not properly identified (eg so 2, –60). | Condone 12x + 6ax = 24x for M1.
Could find a correct equation by
choosing a value of x (eg x = 0).
Can be embedded so f(x, 2) = 24x – 60
is sufficient for A1A1.
Ignore answers at “a =” and “b =” in
PAB unless no answers given or a and/
or b not identified in main working.
1 | (c) | z = 12 ⇒ 12 = 12x – 30y + 6xy
12x−12 2x−2 8
y= = =−2+
30−6x 5−x 5−x
(so vertical asymptote is) eg x = 5
(...and horizontal asymptote is) y = –2 www | M1
A1
A1
[3] | 3.1a
2.2a
2.2a | or 2 = 2x – 5y + xy oe
Substituting z = 12 and rearranging to
a useful form eg y = g(x).
Either asymptote obtained.
Both asymptotes obtained with no
errors in working. | 12−12x
May see y= oe or
6x−30
12+30y 2+5y 8
x= = =5− oe
12+6y 2+ y 2+ y
Condone if asymptotes not correctly
labelled.
1 | (d) | x = 3, y = 2 ⇒ z = 12
  12+6y   24   24 
      
 ∇g=  −30+6x  =  −12  ⇒ eg r.  −12  =...
    −1      −1     −1  
 3   24 
   
r.∇g= 2 . −12 =72−24−12=36
   
   
12  −1 
 24 
 
∴equation is r. −12 =36 oe
 
 
 −1  | B1
M1
A1FT
[3] | 1.1
1.1
2.5 | Clear statement of value of z or sight
3
 
of eg  2  .
 
12
Forming grad g to find normal at S
using x = 3 and y = 2 in the
expressions for f and f from (a), and
x y
using it.
Condone use of f rather than g. Can
be implied by 2 correct coordinates.
Must be used in equation of plane,
not line.
FT 48–“z” miscalculated from f(3, 2).
Not z = –1 unless clearly from f(3, 2).
Or any non-zero multiple. Must be
in vector form; r.n = p, (r – a).n = 0
 3   0   1 
     
or r= 2 +λ −1 +µ 0 oe.
     
     
12 12 24
Any ‘free’ scalar products must be
evaluated but ISW once acceptable
form reached. | z must be evaluated.
grad g must be formed as a vector;
stating the components is not
sufficient. z co-ordinate must be –1.
Must realise that grad g gives the
normal to S for M1 so do not ignore
subsequent work on ∇g.
If (B1)M0 then SC1 for using
∂f ∂f
z−c= (x−a)+ (y−b)
to derive
∂x ∂y
the correct answer.
Or eg
x  24    3  24 
        
y . −12 =36 or r− 2 . −12 =0
      
  z    −1       12      −1  
1 | (e) |  36  36
   
∇g= −30 ⇒(r=) a+λ −30
   
−1 −1
 
 0  36  3
     
4 +λ −30 . 3 =52 or
     
 −120 −1  −2
   
3(36λ)+3(4 – 30λ) – 2(–120 – λ) = 52
∴12+240+λ(108−90+2)=52
 0   36  −360
     
λ=−10⇒r= 4 −10 −30 = 304
     
     
−120 −1 −110
   
(so point is (−360, 304, −110)) | M1
M1
A1
[3] | 1.1
2.1
2.2a | Forming grad g using x = 0 and y = 4
in the expressions for f and f from
x y
(a) and proper attempt to form the
equation of the normal line. Attempt
at grad can be implied by two correct
coordinates. Not grad g from (d).
Condone use of f rather than g.
Intersecting (r =) a + λb (with
 0 
 
a= 4 and b = their new grad g)
 
 
−120
with the given plane. Need not be
seen in vector form.
Not grad g from (d). Not [3, 3, –2].
AG Correct parameter for their line
must be derived from cartesian
equation and substituted in.
Condone proof completed by finding
position vector.
AG so must see correctly derived
value of λ found and correctly
substituted in.
Expect to see 20λ + 252 = 52 =>
λ = –200/20 => λ = –10 | z co-ordinate must be –1. Must realise
that grad g gives the normal at A for
M1 so do not ignore subsequent work
on ∇g.
grad must be direction vector of line.
If "∇g=" not seen, M1 can be
awarded but see guidance for A1.
If M1M0 then SCB1 for both
verifying that (–360, 304, –110) lies on
the plane and using λ = –10 (however
derived) to verify that it also lies on the
normal line. Both verifications need to
be complete.
Accept statement that parameter is to
be substituted as evidence of
substitution.
Derivation of correct position vector or
“x = ..., y = ..., z = ...” is sufficient.
Do not award A1 if "∇g="never seen
in any form or described or identified
in some way as normal to the surface.