| Exam Board | OCR MEI |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Curve Sketching |
| Type | Graphical equation solving with auxiliary line |
| Difficulty | Moderate -0.8 This is a straightforward C1 question testing basic polynomial manipulation and solving. Part (i) requires writing factors from given roots, (ii) is routine expansion, (iii) is graphical reading, and (iv) involves standard algebraic division and quadratic formula application. All techniques are standard with no problem-solving insight required, making it easier than average. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.02n Sketch curves: simple equations including polynomials1.02q Use intersection points: of graphs to solve equations |
| Answer | Marks | Guidance |
|---|---|---|
| 1 | (i) | y = (x + 5)(x + 2)(2x 3) or |
| y = 2(x + 5)(x + 2)( x 3/2) | 2 | |
| [2] | M1 for y = (x + 5)(x + 2)(x 3/2) or |
| Answer | Marks |
|---|---|
| (x + 5)(x + 2)(2x 3) = 30 etc | allow ‘f(x) =’ instead of ‘y = ‘ |
| Answer | Marks | Guidance |
|---|---|---|
| 1 | (ii) | correct expansion of a pair of their linear two- |
| Answer | Marks |
|---|---|
| answer y = 2x3 + 11x2 x 30 | M1 |
| Answer | Marks |
|---|---|
| [2] | ft their factors from (i); need not be |
| Answer | Marks |
|---|---|
| were there | allow only first M1 for expansion if their |
| Answer | Marks | Guidance |
|---|---|---|
| 1 | (iii) | ruled line drawn through ( 2, 0) and (0, 10) |
| Answer | Marks |
|---|---|
| 5.3 to 5.4 and 1.8 to 1.9 | B1 |
| Answer | Marks |
|---|---|
| [3] | tolerance half a small square on grid at |
| Answer | Marks |
|---|---|
| ‘coordinate’ form such as (1.8, 5.3) | insert BP on spare copy of graph if not |
| Answer | Marks | Guidance |
|---|---|---|
| 1 | (iv) | 2x3 + 11x2 x 30 = 5x + 10 |
| Answer | Marks |
|---|---|
| 4 | M1 |
| Answer | Marks |
|---|---|
| [5] | for equating curve and line; correct eqns |
| Answer | Marks |
|---|---|
| dependent only on 4th M1 | annotate this question if partially correct |
Question 1:
1 | (i) | y = (x + 5)(x + 2)(2x 3) or
y = 2(x + 5)(x + 2)( x 3/2) | 2
[2] | M1 for y = (x + 5)(x + 2)(x 3/2) or
(x + 5)(x + 2)(2x 3) with no equation or
(x + 5)(x + 2)(2x 3) = 0
but M0 for y = (x + 5)(x + 2)(2x 3) 30 or
(x + 5)(x + 2)(2x 3) = 30 etc | allow ‘f(x) =’ instead of ‘y = ‘
ignore further work towards (ii)
but do not award marks for (i) in (ii)
1 | (ii) | correct expansion of a pair of their linear two-
term factors ft isw
correct expansion of the correct linear and
quadratic factors and completion to given
answer y = 2x3 + 11x2 x 30 | M1
M1
[2] | ft their factors from (i); need not be
simplified; may be seen in a grid
must be working for this step before given
answer
or for direct expansion of all three factors,
allow M2 for
2x3 + 10x2 + 4x2 3x2 + 20x 15x 6x 30
oe (M1 if one error)
or M1M0 for a correct direct expansion of
(x + 5)(x + 2)(x 3/2)
condone lack of brackets if used as if they
were there | allow only first M1 for expansion if their
(i)has an extra 30 etc
do not award 2nd mark if only had
( x 3/2) in (i) and suddenly doubles
RHS at this stage
condone omission of ‘y =’ or inclusion of
‘= 0’ for this second mark (some cands
have already lost a mark for that in (i))
allow marks if this work has been done
in part (i) – mark the copy of part (i) that
appears below the image for part (ii)
1 | (iii) | ruled line drawn through ( 2, 0) and (0, 10)
and long enough to intersect curve at least
twice
5.3 to 5.4 and 1.8 to 1.9 | B1
B2
[3] | tolerance half a small square on grid at
( 2, 0) and (0, 10)
B1 for one correct
ignore the solution 2 but allow B1 for both
values correct but one extra or for wrong
‘coordinate’ form such as (1.8, 5.3) | insert BP on spare copy of graph if not
used, to indicate seen – this is included
as part of image, so scroll down to see it
accept in coordinate form ignoring any y
coordinates given;
1 | (iv) | 2x3 + 11x2 x 30 = 5x + 10
2x3 + 11x2 6x 40 [= 0]
division by (x + 2) and correctly obtaining 2x2
+ x 20
substitution into quadratic formula or for
completing the square used as far as
x 7 2 209 oe
4 16
7 209
[x ] oe isw
4 | M1
M1
M1
M1
A1
[5] | for equating curve and line; correct eqns
only
for rearrangement to zero, condoning one
error
or showing that (x + 2)( 2x2 + 7x 20) = 2x3
+1 x2 6x 40, with supporting working
condone one error eg a used as 1 not 2, or
one error in the formula, using given
2x2 + 7x 20 = 0
dependent only on 4th M1 | annotate this question if partially correct\includegraphics{figure_1}
Fig. 12 shows the graph of a cubic curve. It intersects the axes at $(-5, 0)$, $(-2, 0)$, $(1.5, 0)$ and $(0, -30)$.
\begin{enumerate}[label=(\roman*)]
\item Use the intersections with both axes to express the equation of the curve in a factorised form. [2]
\item Hence show that the equation of the curve may be written as $y = 2x^3 + 11x^2 - x - 30$. [2]
\item Draw the line $y = 5x + 10$ accurately on the graph. The curve and this line intersect at $(-2, 0)$; find graphically the $x$-coordinates of the other points of intersection. [3]
\item Show algebraically that the $x$-coordinates of the other points of intersection satisfy the equation
$$2x^2 + 7x - 20 = 0.$$
Hence find the exact values of the $x$-coordinates of the other points of intersection. [5]
\end{enumerate}
\hfill \mbox{\textit{OCR MEI C1 Q1 [12]}}