Question 10:
--- 10(i) ---
10(i) | m2 +2m+10=0 ⇒ m=−1±3i. | M1 | Finds complementary function.
x=e−t ( Acos3t+Bsin3t ) | A1
x= pcos3t+qsin3t | M1 | Finds particular integral.
⇒x(cid:5) =−3psin3t+3qcos3t ⇒x(cid:5)(cid:5)=−9pcos3t−9qsin3t | A1
( p+6q ) cos3t+( q−6p ) sin3t =37sin3t | M1
⇒ p=−6, q=1
x=e−t ( Acos3t+Bsin3t ) −6cos3t+sin3t | A1
x=3 when t =0 ⇒A−6=3⇒A=9 | B1 | Uses initial conditions.
x(cid:5)=e−t ( −3Asin3t+3Bcos3t ) −e−t ( Acos3t+Bsin3t )+18sin3t+3cos3t | M1 A1
x(cid:5)=0 when t=0⇒ 3B−A+3=0 ⇒ B=2
x=e−t ( 9cos3t+2sin3t ) −6cos3t+sin3t | A1 | Obtains solution, AEF.
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--- 10(ii) ---
10(ii) | As t→∞, e−t →0 ⇒ x≈ −6cos3t+sin3t | B1 | Obtains limit.
 1 6 
sin3t−6cos3t= 37 sin3t− cos3t
 
 37 37  | M1 | ( )
Converts to Rsin 3t−φ
( )
= 37sin 3t−tan−16 . | A1 | AG
3
Question | Answer | Marks | Guidance
11E(i) | ( 2r+1 )2 − ( 2r−1 )2 =8r | B1
n
⇒ 8∑r =−12 +( 2n+1 )2
r=1 | M1 | Sums both sides and uses method of differences.
⇒ ∑ n r = 1 n ( n+1 )
2
r=1 | A1 | AG.
3
11E(ii) | ( )( )
( 2r+1 )4 − ( 2r−1 )4 = ( 2r+1 )2 +( 2r−1 )2 ( 2r+1 )2 − ( 2r−1 )2 | M1 | Uses difference of squares or expands.
( ) ( )
= 8r2 +2 ( 8r )=16 4r3 +r | A1
n n  1 4  1 4
⇒4∑r3 +∑r =− 1− + n+
   
 2  2
r=1 r=1 | M1 | Sums both sides and uses method of differences.
4∑ n r3 =  n+ 1 4 −  1 4 − 1 n ( n+1 )
   
 2 2 2
r=1 | M1 | Uses formula for∑r
=    n+ 1  2 −   1  2       n+ 1  2 +   1  2   − 1 n ( n+1 )=n2 ( n+1 )2
  2 2   2 2  2
   | A1 | AG
5
11E(iii) | S = 1( 2N +1 )2( 2N +2 )2 =( 2N +1 )2( N +1 )2
4 | B1 | Uses formula for∑r3. AG.
1
Question | Answer | Marks | Guidance
11E(iv) | N
T =S−∑( 2r )3 =( 2N +1 )2( N +1 )2 −2N2 ( N +1 )2
r=1 | M1 | Eliminates even terms from S.
( )
( N +1 )2 2N2 +4N +1 | A1 | Accept ( N+1 )2 ( 2N ( N +2 )+1 ) .
2
11E(v) | S ( 2N +1 )2 4N2 +4N +1
= =
T 2N2 +4N +1 2N2 +4N +1 | M1 | Writes fraction as quadratic in N divided by quadratic
in N .
Converges to 2 as N →∞. | A1
2
11O(i) | ( )
x=−1⇒ y3 +2y−3=0⇒ ( y−1 ) y2 + y+3 =0 | M1 | Considers cubic polynomial in y.
There is only one real root (=1). | A1
2
Question | Answer | Marks | Guidance
11O(ii) |  dy 
2x+2 x + y
 
 dx  | B1 | Differentiates LHS correctly.
dy
=3y2
dx | B1 | Differentiates RHS correctly.
( )dy
⇒ 3y2 −2x = 2x+2y
dx
dy
x=−1, y=1⇒ =0.
dx | B1 | ( )
Substitutes −1,1 , AG.
3
11O(iii) | ( )d2y dy dy  dy
3y2 −2x + 6y −2 =2+2
 
dx2 dx dx  dx | M1 M1 | Differentiates equation again.
dy d2y 2
x=−1, y=1, =0 ⇒ =
dx dx2 5 | A1 | Substitutes ( −1,1 ) and dy =0, AG.
dx
3
11O(iv) | 0 dny  dn−1y 0 0 dn−1y
∫xn dx= xn  −n∫xn−1 dx
dxn dxn−1 dxn−1
 
−1 −1 −1 | M1 A1 | Integrates by parts.
=( )n+1
−1 A −nI .
n−1 n−1 | A1 | AG.
3
Question | Answer | Marks | Guidance
11O(v) | I =( −1 )4 A −3I = A −3 ( −A −2I )= 2 +6I
3 2 2 2 1 1 5 1 | M1 A1 | Uses reduction formulae.
2