Question 7:
7 | (a) | 2 2
dy 1 dy  1   1 
= x2 − ⇒ 1 +   = 1 + x4 − 1 +  or x2 + 
dx 4x2 dx  2 16x4   4x2 
x3 1  1  x5 x 1 
A = 2π∫   +  x2 + dx or 2π∫   + +  dx
 3 4x 4x2   3 3 16x3 
3
 x6 x2 1  2 1145 −95
= 2π + −  = 2π −  = 155 π
 18 6 32x2  576 576  72
 1
2 | 1.1
1.1a
1.1
1.1 | M1
A1
A1
A1
[4] | Not necessarily as a perfect square
Correct integrand soi at any stage (any correct form, including
involving a square-root)
BC or from correct integration
(b) | Line is y = 3 x+ 1 and cuts the x-axis at (− 2, 0)
4 6 9
DR Distance (− 2, 0) to ( 3, 31 ) is ( 31 )2 + ( 31 )2 = 155
9 2 24 18 24 72
Distance (− 2, 0) to ( 1, 13 ) is ( 13 )2 + ( 13 )2 = 65
9 2 24 18 24 72
Curved SA required is π RL−π rl = π ( 31.155 – 13 . 65 ) = 55 π
24 72 24 72 24 | 3.3
3.4
1.1
1.1 | B1
M1
A1
A1 | Cutting-point on x-axis BC
Attempt at either of the two slant heights (or use of calculus)
Both correct
Cao (possibly BC)
Alt.
Line is y = 3 x+ 1
4 6 | B1
SA = 2π∫ ( 3x+ 1 )( 1+(3)2 ) dx
4 6 4 | M1 | Attempt at surface area integral with y and “ds”
A1 | Correct integrand in any form (ignore limits for now)
= 55 π
24 | A1 | Cao (possibly BC)
[4]
(c) | 1 1.5x3 1 
Mean value of y is M = ∫   +  dx
3 − 1  3 4x
2 2 0.5
 x4 1 1.5
=  + lnx = 5 + 1ln3 or 0.691 …
12 4  12 4
 0.5
Curved SA cylinder = 2πrh (with r = y , h = 1)
m
( )
= 2π 5 + 1ln3 = 4.3437 to 4 d.p.
12 4 | 3.3
1.1
3.4 1.1 | M1
A1
M1
A1
[4] | Attempt to find the mean value of y over this interval
BC
Or by integration
Allow exact answer and condone awrt 4.344 (i.e. to 4 s.f.)
(a) | DR 2(p−2)p − 2 ≡ 2(–2)p – 2 (mod p) = – (–2)p – 1
≡ – 1 (mod p) by FLT
since hcf((–) 2, p) = 1 | 2.1 1.2
2.4
2.2a | M1 A1
A1
E1
Alt. I DR | E1
Since p and (p – 2) are consecutive odds, hcf(p – 2, p) = 1
and so (p−2)p − 1 ≡ 1 (mod p) by FLT | B1
⇒ (p−2)(p−2)p − 2 ≡ 1 (mod p) | M1
⇒ (−2)(p−2)p − 2 ≡ 1 (mod p) ⇒ 2(p−2)p − 2 ≡ –1 (mod p) | A1 | AG correctly reasoned
Alt. II DR | M1
A1 | By the Binomial Theorem
(since each term apart from the last has a factor of p)
Statement only is fine
( ) p−2 ( )
(p−2)p − 2 = pp − 2 + (p – 2) pp − 3 (−2) +   pp − 4 (–2)2 + …
 2 
p−2 ( )
… +   p2 (−2)p − 4 + (p – 2)p(–2)p – 3 + (–2)p – 2 ≡ (–2)p – 2
p−4
(mod p)
Then 2(p−2)p − 2≡ –(–2)p – 1 (mod p) and (–2)p – 1 ≡ 1(mod p) by | A1
FLT
⇒ result since hcf(– 2, p) = 1 | E1 | Can also argue via (p – 2) but there’s more to justify
[4]
DR mod 3, N ≡ 2×(1)34 −215
≡ 2×1−2=0 since 2odd ≡ 2 (mod 3) and 2even ≡ 1 (mod 3)
( )
2×3434 −215 = 235 × 1734 – 215 = 215 220×1734 −1
( ) ( )( )
220×1734 −1 = 210 ×1717 −1 210 ×1717 +1
Using (a), 2 × 1717 ≡ –1 (mod 19)
so first bracket ≡ –29 – 1 = –513 = – 19 × 27 ≡ 0 (mod 19)
and N is a multiple of 19 | 3.1a
2.4
3.1a
2.5
3.1a
2.4
3.2a | M1
A1
B1
M1
M1
M1
A1 | Working mod 3
Fully justified that 3 | N
Indices work to factor out the 215
Use of the difference-of-two-squares factorisation
Using (a)’s result
Working mod 19 in one or both brackets
Valid conclusion
By the Binomial Theorem
(since each term apart from the last has a factor of p)
Statement only is fine
M1
A1
Alt. (for factor 19)
( )
2×3434 −215 = 235 × 1734 – 215 = 215 220×1734 −1 = 215B | B1 | Indices work to factor out the 215
mod 19, 218 ≡ 1 by FLT | M1 | (since 19 is prime and hcf(2, 19) = 1)
B ≡ 22 × 1734 – 1 = (2 × 1717 – 1)( 2 × 1717 + 1) | M1 | Use of the difference-of-two-squares factorisation
Using (a), second bracket is divisible by 19 | M1 | Using (a)’s result
and N is a multiple of 19 | A1 | Valid conclusion
[7]