Question 4:
--- 4(a) ---
4(a) | Substitutes x = 6 into p(x) | 1.1a | M1 | p(6)=4×63−15×62−48×6−36
=0
( )
∴x−6is a factor of p x
Completes reasoned proof by
stating p(6)=0 and clearly states
that this implies that x - 6 is a
factor | 2.1 | R1
--- 4(b)(i) ---
4(b)(i) | Factorises with at least two
terms correct
or
differentiates p(x) with at least
two terms correct | 3.1a | M1 | 4x3−15x2−48x−36=(x−6)( 4x2+9x+6 )
b2 −4ac=92 −4×4×6
=−15
<0
∴4x2 +9x+6=0has no real roots
( )=0
Hence p x has exactly one
real root.
Obtains fully correct quadratic
factor
or
obtains fully correct derivative | 1.1b | A1
Calculates their discriminant
or
sets their quadratic = 0 or
sketches their quadratic
PI concluding no real root
or
sets their derivative = 0 and
obtains their turning points (x, y)
OE | 1.1a | M1
States -15 < 0 and concludes
reasoned argument OE
or
states roots are non-real roots
and concludes reasoned
argument
or
states correct turning points and
concludes reasoned argument | 2.1 | R1
--- 4(b)(ii) ---
4(b)(ii) | States coordinates of point of
intersection | 1.1b | B1 | (6, 0)
Total | 7
Q
5(a) | Marking instructions | AO | Marks | Typical solution
Substitutes t = 15.9 hours and
N
N = 0 in the model to find k
2
OE
PI by correct value of k | 3.4 | M1 | N
0 = N e−15.9k
2 0
k =0.0436
0.1=e−0.0436t
t =52.8hours
t = 2.2 days
Obtains correct k
AWFW [0.043, 0.044] or
𝑙𝑙𝑙𝑙(0.5) | 1.1b | A1
− 15.9
Substitutes their value of k and
N =0.1N in the model to find t
0
OE | 3.4 | M1
Solves their equation correctly
to find their t
AWFW [52.3, 53.6] if correct k
used | 1.1a | M1
Obtains correct t in days
AWRT 2.2 days
Accept [2 days 4 hours, 2 days
6 hours] or 3 days
Condone 2 days if 2.2 days
seen | 3.2a | A1