CAIE P1 2019 November — Question 3 5 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2019
SessionNovember
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicChain Rule
TypeTangent with specified gradient
DifficultyModerate -0.3 This is a straightforward tangent problem requiring the chain rule to find the derivative, substitution to find the gradient at x=2, then using point-slope form. While it involves multiple steps (finding c, finding a via differentiation, finding b), each step is routine and the problem type is standard practice for AS-level students. Slightly easier than average due to clear structure and no conceptual challenges.
Spec1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations
3 The line \(y = a x + b\) is a tangent to the curve \(y = 2 x ^ { 3 } - 5 x ^ { 2 } - 3 x + c\) at the point \(( 2,6 )\). Find the values of the constants \(a , b\) and \(c\).
Question 3:
AnswerMarks Guidance
AnswerMarks Guidance
\(\frac{dy}{dx} = 6x^2 - 10x - 3\)B1
At \(x = 2\), \(\frac{dy}{dx} = 24 - 20 - 3 = 1 \rightarrow a = 1\)M1 A1
\(6 = 2 + b \rightarrow b = 4\)B1FT Substitute \(x = 2\), \(y = 6\) in \(y = (\text{their } a)x + b\)
\(6 = 16 - 20 - 6 + c \rightarrow c = 16\)B1 Substitute \(x = 2\), \(y = 6\) into equation of curve
Total: 5 
**Question 3:**

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{dy}{dx} = 6x^2 - 10x - 3$ | B1 | |
| At $x = 2$, $\frac{dy}{dx} = 24 - 20 - 3 = 1 \rightarrow a = 1$ | M1 A1 | |
| $6 = 2 + b \rightarrow b = 4$ | B1FT | Substitute $x = 2$, $y = 6$ in $y = (\text{their } a)x + b$ |
| $6 = 16 - 20 - 6 + c \rightarrow c = 16$ | B1 | Substitute $x = 2$, $y = 6$ into equation of curve |
| **Total: 5** | | |

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3 The line $y = a x + b$ is a tangent to the curve $y = 2 x ^ { 3 } - 5 x ^ { 2 } - 3 x + c$ at the point $( 2,6 )$. Find the values of the constants $a , b$ and $c$.\\

\hfill \mbox{\textit{CAIE P1 2019 Q3 [5]}}
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