OCR MEI C2 — Question 3 13 marks

Exam BoardOCR MEI
ModuleC2 (Core Mathematics 2)
Marks13
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicDifferentiation from First Principles
TypeFirst principles: x⁴ and higher power terms
DifficultyModerate -0.3 This is a standard C2 differentiation question that systematically builds understanding of derivatives through tangent lines, chord gradients, and first principles. Part (i) requires routine differentiation and point-slope form (4 marks suggests showing working). Parts (ii-iii) guide students through the limit definition of derivative with scaffolded steps. While pedagogically thorough with 13 total marks, each component uses standard techniques without requiring problem-solving insight, making it slightly easier than average.
Spec1.04a Binomial expansion: (a+b)^n for positive integer n1.07g Differentiation from first principles: for small positive integer powers of x1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations

  1. Find the equation of the tangent to the curve \(y = x^4\) at the point where \(x = 2\). Give your answer in the form \(y = mx + c\). [4]
  2. Calculate the gradient of the chord joining the points on the curve \(y = x^4\) where \(x = 2\) and \(x = 2.1\). [2]
    1. Expand \((2 + h)^4\). [3]
    2. Simplify \(\frac{(2 + h)^4 - 2^4}{h}\). [2]
    3. Show how your result in part (iii) (B) can be used to find the gradient of \(y = x^4\) at the point where \(x = 2\). [2]

Question 3:

AnswerMarks
3 (i)dy = 4x3
dx
dy
when x = 2, = 32 s.o.i.
dx
when x = 2, y = 16 s.o.i.
AnswerMarks
y = 32x – 48 c.a.o.M1
A1
B1
AnswerMarks
A1i.s.w.

AnswerMarks Guidance
3 (ii)34.481 2
0.1
3 (iii)
AnswerMarks Guidance
(A)16 + 32h + 24h2 + 8h3 + h4 c.a.o. 3
B1 for 3 terms correct
3 (iii)
AnswerMarks Guidance
(B)32 + 24h + 8h2 + h3 or ft 2
3 (iii)
AnswerMarks
(C)as h  0, result  their 32 from
(iii)(B)
gradient of tangent is limit of
AnswerMarks
gradient of chord1
1
Question 3:
--- 3 (i) ---
3 (i) | dy = 4x3
dx
dy
when x = 2, = 32 s.o.i.
dx
when x = 2, y = 16 s.o.i.
y = 32x – 48 c.a.o. | M1
A1
B1
A1 | i.s.w.
--- 3 (ii) ---
3 (ii) | 34.481 | 2 | M1 for 2.1424
0.1
3 (iii)
(A) | 16 + 32h + 24h2 + 8h3 + h4 c.a.o. | 3 | B2 for 4 terms correct
B1 for 3 terms correct
3 (iii)
(B) | 32 + 24h + 8h2 + h3 or ft | 2 | B1 if one error
3 (iii)
(C) | as h  0, result  their 32 from
(iii)(B)
gradient of tangent is limit of
gradient of chord | 1
1
\begin{enumerate}[label=(\roman*)]
\item Find the equation of the tangent to the curve $y = x^4$ at the point where $x = 2$. Give your answer in the form $y = mx + c$. [4]

\item Calculate the gradient of the chord joining the points on the curve $y = x^4$ where $x = 2$ and $x = 2.1$. [2]

\item \begin{enumerate}[label=(\Alph*)]
\item Expand $(2 + h)^4$. [3]

\item Simplify $\frac{(2 + h)^4 - 2^4}{h}$. [2]

\item Show how your result in part (iii) (B) can be used to find the gradient of $y = x^4$ at the point where $x = 2$. [2]
\end{enumerate}
\end{enumerate}

\hfill \mbox{\textit{OCR MEI C2  Q3 [13]}}