OCR PURE — Question 4 5 marks

Exam BoardOCR
ModulePURE
Marks5
PaperDownload PDF ↗
TopicDifferentiation from First Principles
TypeFirst principles: polynomial (find gradient)
DifficultyModerate -0.5 This is a straightforward application of the first principles definition to find the derivative of a simple quadratic at a specific point. While it requires careful algebraic manipulation, it's a standard textbook exercise with no conceptual challenges beyond applying the definition correctly.
Spec1.07g Differentiation from first principles: for small positive integer powers of x
4 The quadratic polynomial \(2 x ^ { 2 } - 3\) is denoted by \(\mathrm { f } ( x )\).
Use differentiation from first principles to determine the value of \(\mathrm { f } ^ { \prime } ( 2 )\).
Question 4:
AnswerMarks Guidance
Considers \(\dfrac{f(2+h) - f(2)}{h}\)B1 (AO 2.1) Or considers \(\dfrac{f(x+h)-f(x)}{h}\) with \(x=2\) substituted at some point
\(f(2+h) = 2(2+h)^2 - 3 = 2h^2 + 8h + 8 - 3\)M1 (AO 1.1) Considers \(f(2+h)\) and attempts to expand; or considers \(f(x+h)\) and attempts to expand
\(f(2+h) - f(2) = (2h^2 + 8h + 5) - 5 = 2h^2 + 8h\)A1 (AO 1.1) Correct simplified expression for \(f(2+h) - f(2)\); correct simplified expression for \(f(x+h) - f(x)\)
\(\dfrac{f(2+h) - f(2)}{h} = 2h + 8\)A1 (AO 1.1) Correct simplified expression for \(\dfrac{f(2+h)-f(2)}{h}\); correct simplified expression for \(\dfrac{f(x+h)-f(x)}{h}\)
\(f'(2) = \lim_{h \to 0} \dfrac{f(2+h) - f(2)}{h} = 8\)A1 (AO 2.2a) cao – must be explicit that the limit (and not simply \(h = 0\)) is considered
Total: [5]
## Question 4:

Considers $\dfrac{f(2+h) - f(2)}{h}$ | B1 (AO 2.1) | Or considers $\dfrac{f(x+h)-f(x)}{h}$ with $x=2$ substituted at some point

$f(2+h) = 2(2+h)^2 - 3 = 2h^2 + 8h + 8 - 3$ | M1 (AO 1.1) | Considers $f(2+h)$ and attempts to expand; or considers $f(x+h)$ and attempts to expand

$f(2+h) - f(2) = (2h^2 + 8h + 5) - 5 = 2h^2 + 8h$ | A1 (AO 1.1) | Correct simplified expression for $f(2+h) - f(2)$; correct simplified expression for $f(x+h) - f(x)$

$\dfrac{f(2+h) - f(2)}{h} = 2h + 8$ | A1 (AO 1.1) | Correct simplified expression for $\dfrac{f(2+h)-f(2)}{h}$; correct simplified expression for $\dfrac{f(x+h)-f(x)}{h}$

$f'(2) = \lim_{h \to 0} \dfrac{f(2+h) - f(2)}{h} = 8$ | A1 (AO 2.2a) | cao – must be explicit that the limit (and not simply $h = 0$) is considered

**Total: [5]**

---
4 The quadratic polynomial $2 x ^ { 2 } - 3$ is denoted by $\mathrm { f } ( x )$.\\
Use differentiation from first principles to determine the value of $\mathrm { f } ^ { \prime } ( 2 )$.

\hfill \mbox{\textit{OCR PURE  Q4 [5]}}
This paper (12 questions)
View full paper
Q1 2 Q2 3 Q3 5 Q4 5 Q5 9 Q6 6 Q7 9 Q8 11 Q9 3 Q10 8 Q11 7 Q12 7