Question 10:
--- 10(i) ---
10(i) | 1 5 1  1 5 1 
   
1 −2 −2 → 0 −7 −3
   
 2 3 θ   0 0 θ+1 
    | M1 A1 | Reduces to echelon form. At least one row operation for M1.
( )=3
r A if θ≠−1 | A1
r ( A )=2 if θ=−1 | B1
4
--- 10(ii) ---
10(ii) | x +5y +z = −1
−7y −3z = 1
(θ+1)z = (θ+1) | M1 | Uses reduced form of augmented matrix or eliminates
variables from scratch.
4 6
z =1, y =− , x=
7 7 | A1
A1 | One correct.
All three correct.
3
--- 10(iii) ---
10(iii) | x +5y +z = −1
−7y −3z = 1
(θ+1)z = (θ+1)
z =t | M1 | Uses parameter.
3t+1 8t−2
y =− , x=
7 7 | A1 A1
3
Question | Answer | Marks | Guidance
--- 10(iv) ---
10(iv) | x+5y+ z =−1,
−7y −3z =1,
( θ+1 ) z =φ+1 | M1 | Uses reduced form of augmented matrix or eliminates
variables from scratch
θ=−1⇒φ=−1
so no solution (inconsistent). | A1
2
Question | Answer | Marks | Guidance
11E(i) | dw dy
w=cosy⇒ =−sin y
dx dx | B1
d2w d2y dy 2
=−sin y −cosy
 
dx2 dx2 dx | B1
d2w dw d2y dy 2 dy
+2 +w=−sin y −cosy −2sin y +cosy
 
dx2 dx dx2 dx dx | M1 | Uses substitution to obtain w-x equation, AG.
( )
=−cosy e−2xsecy =−e−2x | A1
4
Question | Answer | Marks | Guidance
11E(ii) | m2 +2m+1=0⇒m=−1 | M1 | Finds CF.
CF: w=( Ax+B ) e−x | A1
PI: w=ke−2x ⇒ w′=−2ke−2x ⇒ w′′=4ke−2x | M1 | Forms PI and differentiates.
4k−4k+k =−1⇒k =−1 | A1
w=( Ax+B ) e−x −e−2x | A1 | States general solution.
y = 1π w= 1 B= 3
x=0, 3 , 2 ⇒ 2 | B1 | Uses initial conditions to find constants.
w′=− ( Ax+B ) e−x + Ae−x +2e−2x | M1 | Differentiates general solution.
x=0, y = 1π, y'= 3, w'=−1 ⇒−1 =−3 +A+2⇒ A=−1
3 3 2 2 2 | M1 A1 | Substitutes initial conditions.
3  
y =cos−1 −x e−x −e−2x
  
2   | A1 | States particular solution for y in terms of x.
10
Question | Answer | Marks | Guidance
11O(i) | ( )
e2α−e−2α=2 eα +e−α ⇒eα−e−α=2 | M1 | eα+e−α.
Sets equations equal and divides by
e2α−2eα−1=0⇒ eα =1+ 2 | M1 A1 | Forms quadratic in eα, AG.
( )
α=ln 1+ 2 | A1 | Must be exact.
( )
r =2 1+ 2+ 2−1 =4 2 | M1 A1 | Substitutes to find r.
6
11O(ii) | B1 | C has correct shape.
1
B1 | C has correct shape.
2
B1 | Intersection points positioned correctly.
3
Question | Answer | Marks | Guidance
11O(iii) | ( ) ( )
ln1+√2 ln1+√2
2 ∫ ( eθ+e−θ ) 2 dθ− 1 ∫ ( e2θ−e−2θ ) 2 dθ
2
0 0
( )
ln1+√2
1 1
= ∫ 5+2e2θ+2e−2θ− e4θ− e−4θ dθ
2 2
0 | M1 A1 | 1
Uses ∫r2dθ to formulate correct area.
2
( )
ln1+√2
 1 1 
= 5θ+e2θ−e−2θ− e4θ+ e−4θ
 
 8 8 
0 | M1 A1 | Expands and integrates.
( ) 2 ( ) −2 1 ( ) 4 ( ) −4
=5ln(1+√2)+ 1+ 2 − 1+ 2 −  1+ 2 − 1+ 2  =5.82
8  | A1
5