OCR MEI C2 — Question 3 12 marks

Exam BoardOCR MEI
ModuleC2 (Core Mathematics 2)
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicTangents, normals and gradients
TypeNormal meets curve/axis — further geometry
DifficultyModerate -0.3 This is a standard C2 calculus question covering routine techniques: sketching a quadratic, finding a tangent using differentiation, and finding where a normal intersects the curve. All parts follow textbook procedures with no novel problem-solving required. The algebraic method in part (iii) involves substituting the normal equation into the curve and solving a quadratic, which is straightforward. Slightly easier than average due to the predictable structure and standard techniques, though the multi-part nature and 12 total marks provide some substance.
Spec1.02f Solve quadratic equations: including in a function of unknown1.02n Sketch curves: simple equations including polynomials1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations
The point A has \(x\)-coordinate 5 and lies on the curve \(y = x^2 - 4x + 3\).
  1. Sketch the curve. [2]
  2. Use calculus to find the equation of the tangent to the curve at A. [4]
  3. Show that the equation of the normal to the curve at A is \(x + 6y = 53\). Find also, using an algebraic method, the \(x\)-coordinate of the point at which this normal crosses the curve again. [6]
Question 3:
AnswerMarks Guidance
3(i) sketch of parabola the right way up
cutting y-axis at 3 and either x-axis at 1 and
AnswerMarks
3 only or minimum value at (2, -1)B1
B1
AnswerMarks
[2]intersections must be marked on graph or
shown worked out next to sketch
AnswerMarks Guidance
3(ii) y´ = 2x  4
at A y´ = 6
at A y = 8 soi
y  their 8 = 6(x  5) or substitution of
(5, their 8) into y = 6x + c and evaluation of
AnswerMarks
cM1*
A1
B1
M1dep*
AnswerMarks
[4]must be obtained by calculus
implied by y = 6x  22;
M0 if value of y´ not y used
AnswerMarks Guidance
3(iii) 1
m 
their6
y  8 = 1/6 (x  5) oe and interim step
completing to given answer
53x
 x2 4x3 oe
6
x2  23x350oe
6 6
(x- 5)(6x + 7)
x = 7 oe isw (accept 1.17 or better)
AnswerMarks
6M1
A1
M1*
A1
M1dep*
A1
AnswerMarks
[6]NB answer x + 6y = 53 given
must be three terms
or correct substitution in quadratic formula
or correct completion of square
AnswerMarks
previous M1 implied by correct answerM0 if clearly obtained from
x + 6y = 53
325
if quadratic in y, then B2 for y 
36
= 9.0277…
B2 for x = 7 oe obtained from
6
correct value for y
Question 3:
3 | (i) | sketch of parabola the right way up
cutting y-axis at 3 and either x-axis at 1 and
3 only or minimum value at (2, -1) | B1
B1
[2] | intersections must be marked on graph or
shown worked out next to sketch
3 | (ii) | y´ = 2x  4
at A y´ = 6
at A y = 8 soi
y  their 8 = 6(x  5) or substitution of
(5, their 8) into y = 6x + c and evaluation of
c | M1*
A1
B1
M1dep*
[4] | must be obtained by calculus
implied by y = 6x  22;
M0 if value of y´ not y used
3 | (iii) | 1
m 
their6
y  8 = 1/6 (x  5) oe and interim step
completing to given answer
53x
 x2 4x3 oe
6
x2  23x350oe
6 6
(x- 5)(6x + 7)
x = 7 oe isw (accept 1.17 or better)
6 | M1
A1
M1*
A1
M1dep*
A1
[6] | NB answer x + 6y = 53 given
must be three terms
or correct substitution in quadratic formula
or correct completion of square
previous M1 implied by correct answer | M0 if clearly obtained from
x + 6y = 53
325
if quadratic in y, then B2 for y 
36
= 9.0277…
B2 for x = 7 oe obtained from
6
correct value for y
The point A has $x$-coordinate 5 and lies on the curve $y = x^2 - 4x + 3$.

\begin{enumerate}[label=(\roman*)]
\item Sketch the curve. [2]

\item Use calculus to find the equation of the tangent to the curve at A. [4]

\item Show that the equation of the normal to the curve at A is $x + 6y = 53$. Find also, using an algebraic method, the $x$-coordinate of the point at which this normal crosses the curve again. [6]
\end{enumerate}

\hfill \mbox{\textit{OCR MEI C2  Q3 [12]}}
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Q1 13 Q2 11 Q3 12 Q4 12 Q5 5