OCR MEI C1 — Question 2 12 marks

Exam BoardOCR MEI
ModuleC1 (Core Mathematics 1)
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCurve Sketching
TypeRational curve intersections
DifficultyStandard +0.3 This is a standard C1 question combining graphical work, algebraic manipulation, and discriminant application. Part (i) is routine graph sketching, part (ii) involves straightforward equation solving by equating functions and using the quadratic formula, and part (iii) applies the tangency condition (discriminant = 0) in a predictable way. All techniques are core C1 content with no novel insight required, making it slightly easier than average.
Spec1.02d Quadratic functions: graphs and discriminant conditions1.02o Sketch reciprocal curves: y=a/x and y=a/x^21.02q Use intersection points: of graphs to solve equations
\includegraphics{figure_2} Fig. 12 shows the graph of \(y = \frac{1}{x-2}\).
  1. Draw accurately the graph of \(y = 2x + 3\) on the copy of Fig. 12 and use it to estimate the coordinates of the points of intersection of \(y = \frac{1}{x-2}\) and \(y = 2x + 3\). [3]
  2. Show algebraically that the \(x\)-coordinates of the points of intersection of \(y = \frac{1}{x-2}\) and \(y = 2x + 3\) satisfy the equation \(2x^2 - x - 7 = 0\). Hence find the exact values of the \(x\)-coordinates of the points of intersection. [5]
  3. Find the quadratic equation satisfied by the \(x\)-coordinates of the points of intersection of \(y = \frac{1}{x-2}\) and \(y = -x + k\). Hence find the exact values of \(k\) for which \(y = -x + k\) is a tangent to \(y = \frac{1}{x-2}\). [4]
Question 2:
AnswerMarks Guidance
2(i y = 2x + 3 drawn accurately
(1.6 to 1.7, 0.2 to 0.3)
AnswerMarks
(2.1 to 2.2, 7.2 to 7.4)M1
B1
B1
AnswerMarks
[3]at least as far as intersecting curve twice
intersections may be in form x = ..., y = ...ruled straight line and within 2mm of
(2, 7) and (1, 1)
if marking by parts and you see work
relevant to (ii), put a yellow line here
and in (ii) to alert you to look
AnswerMarks Guidance
2(ii 1
2x3
x2
1 = (2x + 3)(x  2)
1 = 2x2  x  6 oe
1 12 427
oe
22
1 57
isw
AnswerMarks
4M1
M1
A1
M1
AA11
AnswerMarks
[5]or attempt at elimination of x by
rearrangement and substitution
condone lack of brackets
for correct expansion; need not be simplified;
NB A0 for 2x2  x  7 = 0 without expansion
seen [given answer]
use of formula or completing square on given
equation, with at most one error
is eg coordinates;
1 57
after completing square, accept  or
4 16
AnswerMarks
bettermay be seen in (i) – allow marks; the
part (i) work appears at the foot of the
image for (ii) so show marks there
rather than in (i)
implies first M1 if that step not seen
implies second M1 if that step not seen
1
after 2x3 seen
x2
completing square attempt must reach
at least [2](x  a)2 = b or (2x  c)2 = d
stage oe with at most one error
AnswerMarks Guidance
2(iii 1
xk and attempt at rearrangement
x2
x2 k2x2k10
b2  4ac = 0 oe seen or used
AnswerMarks
[k =] 0 or 4 as final answer, both requiredM1
M1
M1
A1
AnswerMarks
[4]for simplifying and rearranging to zero;
condone one error;
collection of x terms with bracket not required
SC1 for 0 and 4 found if 3rd M1 not earned
AnswerMarks
(may or may not have earned first two Ms)eg M1 bod for x2 k2x2k
or M1 for x2 2kx2k10
= 0 may not be seen, but may be
implied by their final values of k
eg obtained graphically or using
calculus and/or final answer given as a
range
Question 2:
2 | (i | y = 2x + 3 drawn accurately
(1.6 to 1.7, 0.2 to 0.3)
(2.1 to 2.2, 7.2 to 7.4) | M1
B1
B1
[3] | at least as far as intersecting curve twice
intersections may be in form x = ..., y = ... | ruled straight line and within 2mm of
(2, 7) and (1, 1)
if marking by parts and you see work
relevant to (ii), put a yellow line here
and in (ii) to alert you to look
2 | (ii | 1
2x3
x2
1 = (2x + 3)(x  2)
1 = 2x2  x  6 oe
1 12 427
oe
22
1 57
isw
4 | M1
M1
A1
M1
AA11
[5] | or attempt at elimination of x by
rearrangement and substitution
condone lack of brackets
for correct expansion; need not be simplified;
NB A0 for 2x2  x  7 = 0 without expansion
seen [given answer]
use of formula or completing square on given
equation, with at most one error
is eg coordinates;
1 57
after completing square, accept  or
4 16
better | may be seen in (i) – allow marks; the
part (i) work appears at the foot of the
image for (ii) so show marks there
rather than in (i)
implies first M1 if that step not seen
implies second M1 if that step not seen
1
after 2x3 seen
x2
completing square attempt must reach
at least [2](x  a)2 = b or (2x  c)2 = d
stage oe with at most one error
2 | (iii | 1
xk and attempt at rearrangement
x2
x2 k2x2k10
b2  4ac = 0 oe seen or used
[k =] 0 or 4 as final answer, both required | M1
M1
M1
A1
[4] | for simplifying and rearranging to zero;
condone one error;
collection of x terms with bracket not required
SC1 for 0 and 4 found if 3rd M1 not earned
(may or may not have earned first two Ms) | eg M1 bod for x2 k2x2k
or M1 for x2 2kx2k10
= 0 may not be seen, but may be
implied by their final values of k
eg obtained graphically or using
calculus and/or final answer given as a
range
\includegraphics{figure_2}

Fig. 12 shows the graph of $y = \frac{1}{x-2}$.

\begin{enumerate}[label=(\roman*)]
\item Draw accurately the graph of $y = 2x + 3$ on the copy of Fig. 12 and use it to estimate the coordinates of the points of intersection of $y = \frac{1}{x-2}$ and $y = 2x + 3$. [3]

\item Show algebraically that the $x$-coordinates of the points of intersection of $y = \frac{1}{x-2}$ and $y = 2x + 3$ satisfy the equation $2x^2 - x - 7 = 0$. Hence find the exact values of the $x$-coordinates of the points of intersection. [5]

\item Find the quadratic equation satisfied by the $x$-coordinates of the points of intersection of $y = \frac{1}{x-2}$ and $y = -x + k$. Hence find the exact values of $k$ for which $y = -x + k$ is a tangent to $y = \frac{1}{x-2}$. [4]
\end{enumerate}

\hfill \mbox{\textit{OCR MEI C1  Q2 [12]}}
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Q1 12 Q2 12 Q3 13 Q4 11 Q5 12 Q6 12