Review Markscheme (5.1 - 5.3 and Past Material)
Review Markscheme (5.1 - 5.3 and Past Material)
Review Markscheme (5.1 - 5.3 and Past Material)
1a. [2 marks]
Markscheme
f ' ( x )=x 2+2 x−15 (M1)A1
[2 marks]
1b. [3 marks]
Markscheme
correct reasoning that f ' ( x )=0 (seen anywhere) (M1)
2
x + 2 x−15=0
valid approach to solve quadratic M1
( x−3 ) ( x+ 5 ), quadratic formula
correct values for x
3, −5
correct values for a and b
a = −5 and b = 3 A1
[3 marks]
1c. [1 mark]
Markscheme
A1
[1 mark]
1d. [5 marks]
Markscheme
normal to f at x=a is x = −5 (seen anywhere) (A1)
attempt to find y -coordinate at their value of b (M1)
f ( 3 )=¿ −10 (A1)
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences
in marking or structure. It appeared in a paper that permitted the use of a calculator, and so
might not be suitable for all forms of practice.
k
2 x+ 2 (A1)(A1)(A1) (C3)
x
1
Note: Award (A1) for 2 x, (A1) for + k , and (A1) for x−2 or 2.
x
Award at most (A1)(A1)(A0) if additional terms are seen.
[3 marks]
2b. [1 mark]
Markscheme
−2.5 ( −52 ) (A1) (C1)
[1 mark]
2c. [2 marks]
Markscheme
k
−2.5=2× (−2 ) + (M1)
(−2 )2
Note: Award (M1) for equating their gradient from part (b) to their substituted derivative
from part (a).
¿ (A1)(ft) (C2)
[2 marks]
3. [5 marks]
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences
in marking or structure.
( x=0 ⇒ ) y =1 (A1)
dy
appreciate the need to find (M1)
dx
¿ A1
dy
( x=0 ⇒ ) =−1 A1
dx
y−1
=−1 ( y=1−x ) A1
x−0
[5 marks]
4a. [2 marks]
Markscheme
attempt to form composite (in any order) (M1)
Markscheme
recognizing that the gradient of the tangent is the derivative (M1)
eg h'
correct derivative (seen anywhere) (A1)
h' ( x ) =4 x 3
correct value for gradient of f (seen anywhere) (A1)
'
f ( x )=1, m=1
setting their derivative equal to 1 (M1)
3
4 x =1
0.629960
x=
√
3 1
4
(exact), 0.630 A1 N3
[5 marks]
5. [2 marks]
Markscheme
valid method (M1)
f ' ( x )=0
eg , ,
Markscheme
(y − (−2)) = 12 (x − (−1)) (M1)
OR
−2 = 12(−1) + c (M1)
Note: Award (M1) for point and their gradient substituted into the equation of a line.
y = 12x + 10 (A1)(ft) (C2)
Note: Follow through from part (b).
[2 marks]
7a. [2 marks]
Markscheme
correct approach A1
π 2π
eg = (or equivalent)
6 period
period = 12 A1
[2 marks]
7b. [2 marks]
Markscheme
valid approach (M1)
max+min
eg b=max−amplitude
2
21.8+10.2
, or equivalent
2
b = 16 A1
[2 marks]
7c. [2 marks]
Markscheme
attempt to substitute into their function (M1)
7d. [5 marks]
Markscheme
valid attempt to set up a system of equations (M1)
two correct equations A1
p sin ( 2π
9 ) (
( 3−3.75 ) +q=2.5, p sin
2π
9 )
( 6−3.75 ) +q=15.1
7e. [2 marks]
Markscheme
attempt to use |f ( x)−g (x)| to find maximum difference (M1)
x = 1.64 A1
[2 marks]
8. [4 marks]
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences
in marking or structure.
A=P
use of the correct formula for area and arc length (M1)
perimeter is rθ+ 2r (A1)
Note: A1 independent of previous M1.
1 2
r (1 )=r ( 1 )+ 2r A1
2
r 2−6 r=0
r =6 (as r > 0) A1
Note: Do not award final A1 if r =0 is included.
[4 marks]
9. [7 marks]
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences
in marking or structure.
evidence of correctly substituting into circle formula (may be seen later) A1A1
1 2
eg θ r =12 , rθ=6
2
attempt to eliminate one variable (M1)
1 2
θr
eg 6 1 2 12
r = , θ= , =
θ r rθ 6
correct elimination (A1)
()
2 2
1 6 2 1 6 1 2 l r
eg × × r =12 , θ × =12 , A= ×r × , =2
2 r 2 θ 2 r 2r
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences
in marking or structure.
−0.394791,13
A(−0.395, 13) A1A1 N2
[2 marks]
10b. [1 mark]
Markscheme
13 A1 N1
[1 mark]
10c. [1 mark]
Markscheme
2 π , 6.28 A1 N1
[1 mark]
10d. [3 marks]
Markscheme
valid approach (M1)
eg recognizing that amplitude is p or shift is r
f ( x )=13 cos ( x +0.395 ) (accept p = 13, r = 0.395) A1A1 N3
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences
in marking or structure.
Markscheme
correct interval A2 N2
eg−1 ≤ x ≤3 , [−1 , 3], from −1 to 3
[2 marks]
13.
14.
15.