Question 8 - A-Level Maths

CAIE P1 2012 November — Question 8 9 marks

Exam Board	CAIE
Module	P1 (Pure Mathematics 1)
Year	2012
Session	November
Marks	9
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Chain Rule
Type	Find curve equation from derivative
Difficulty	Moderate -0.3 This is a straightforward multi-part calculus question requiring chain rule differentiation, finding/classifying a stationary point, and integration with a constant. All techniques are standard P1 procedures with no novel problem-solving required, making it slightly easier than average but not trivial due to the fractional power requiring careful application of the chain rule.
Spec	1.07e Second derivative: as rate of change of gradient 1.07i Differentiate x^n: for rational n and sums 1.07n Stationary points: find maxima, minima using derivatives 1.08b Integrate x^n: where n != -1 and sums

8 A curve is such that $$\frac { \mathrm { d } y } { \mathrm {~d} x } = 2 ( 3 x + 4 ) ^ { \frac { 3 } { 2 } } - 6 x - 8 .$$

Find $\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }$.
Verify that the curve has a stationary point when $x = - 1$ and determine its nature.
It is now given that the stationary point on the curve has coordinates $( - 1,5 )$. Find the equation of the curve.

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Answer	Marks	Guidance
(i) $f''(x) = [9]\times\left[(3x+4)^{\frac{1}{2}}\right] - [6]$	B2, 1, 0	1 each error [2]
(ii) $f'(-1) = 2 + 6 - 8 = 0$ hence stat value at $x = -1$; $f''(-1) = 9 - 6 = 3 > 0$ hence minimum	B1, B1	AG [2]
(iii) $y = \left[\frac{4}{15}\right] \circ \left[(3x+4)^{\frac{3}{2}}\right] - [3x^2 + 8x] + (c)$; Sub $(-1, 5)$: $-\frac{4}{15} - 3 + 8 + c = 5 \rightarrow c = -\frac{4}{15}$	B1, B1, B1, M1, A1	allow unsimplified $\frac{4}{...}$; Dependent on $c$ present cao [5]

(i) $f''(x) = [9]\times\left[(3x+4)^{\frac{1}{2}}\right] - [6]$ | B2, 1, 0 | 1 each error [2]

(ii) $f'(-1) = 2 + 6 - 8 = 0$ hence stat value at $x = -1$; $f''(-1) = 9 - 6 = 3 > 0$ hence minimum | B1, B1 | AG [2]

(iii) $y = \left[\frac{4}{15}\right] \circ \left[(3x+4)^{\frac{3}{2}}\right] - [3x^2 + 8x] + (c)$; Sub $(-1, 5)$: $-\frac{4}{15} - 3 + 8 + c = 5 \rightarrow c = -\frac{4}{15}$ | B1, B1, B1, M1, A1 | allow unsimplified $\frac{4}{...}$; Dependent on $c$ present cao [5]

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8 A curve is such that

$$\frac { \mathrm { d } y } { \mathrm {~d} x } = 2 ( 3 x + 4 ) ^ { \frac { 3 } { 2 } } - 6 x - 8 .$$

(i) Find $\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }$.\\
(ii) Verify that the curve has a stationary point when $x = - 1$ and determine its nature.\\
(iii) It is now given that the stationary point on the curve has coordinates $( - 1,5 )$. Find the equation of the curve.

\hfill \mbox{\textit{CAIE P1 2012 Q8 [9]}}

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Q1 3 Q2 3 Q3 4 Q4 4 Q5 5 Q6 6 Q7 8 Q8 9 Q9 10 Q10 10 Q11 13

Answer	Marks	Guidance
(i) \(f''(x) = [9]\times\left[(3x+4)^{\frac{1}{2}}\right] - [6]\)	B2, 1, 0	1 each error [2]
(ii) \(f'(-1) = 2 + 6 - 8 = 0\) hence stat value at \(x = -1\); \(f''(-1) = 9 - 6 = 3 > 0\) hence minimum	B1, B1	AG [2]
(iii) \(y = \left[\frac{4}{15}\right] \circ \left[(3x+4)^{\frac{3}{2}}\right] - [3x^2 + 8x] + (c)\); Sub \((-1, 5)\): \(-\frac{4}{15} - 3 + 8 + c = 5 \rightarrow c = -\frac{4}{15}\)	B1, B1, B1, M1, A1	allow unsimplified \(\frac{4}{...}\); Dependent on \(c\) present cao [5]