Question 3:
3 | (i) | (1, 6) (0,1) (1,2) (2,3) (3,2) (4, 1) (5,6)
seen plotted
smooth curve through all 7 points
(0.3 to 0.5, 0.3 to 0.5) and
(2.5 to 2.7, 2.5 to 2.7) and (4, 1) | B2
B1 dep
B2
[5] | or for a curve within 2 mm of these points;
B1 for 3 correct plots or for at least 3 of the
pairs of values seen eg in table
dep on correct points; tolerance 2 mm;
may be given in form x = ..., y = ...
B1 for two intersections correct or for all the
x values given correctly | use overlay; scroll down to spare copy
of graph to see if used [or click ‘fit
height’
also allow B1 for (2 3, 0) and
(2, 3) seen or plotted and curve not
through other correct points
condone some feathering/ doubling
(deleted work still may show in scans);
curve should not be flat-bottomed or
go to a point at min. or curve back in at
top;
3 | (ii) | 1
x2 4x1
x3
1 = (x  3)( x2  4x + 1)
at least one further correct interim step with
‘=1’ or ‘=0’ ,as appropriate, leading to given
answer, which must be stated correctly | M1
M1
A1
[3] | condone omission of brackets only if used
correctly afterwards, with at most one error;
there may also be a previous step of
expansion of terms without an equation, eg
in grid
if M0, allow SC1 for correct division of
given cubic by quadratic to gain (x  3) with
remainder 1, or vice-versa | condone omission of ‘=1’ for this M1
only if it reappears
allow for terms expanded correctly
with at most one error
NB mark method not answer -
given answer is x3  7x2 + 13x  4 = 0
Quueessttion | er | Marks | Guidance
3 | (iii) | quadratic factor is
x2  3x + 1
substitution into quadratic formula or for
completing the square used as far as
x 32  5
2 4
3 5
oe
2 | B2
M1
A2
[5] | found by division or inspection;
allow M1 for division by x  4 as far as
x3  4x2 in the working, or for inspection
with two terms correct
condone one error
A1 if one error in final numerical expression,
but only if roots are real | no ft from a wrong ‘factor’;
isw factors