CAIE P1 2014 June — Question 6 7 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2014
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicImplicit equations and differentiation
TypeFind curve equation from dy/dx
DifficultyModerate -0.3 This is a straightforward integration problem requiring students to find a constant using the normal condition, then integrate a simple power function and apply a boundary condition. The steps are standard: use perpendicular gradient property, integrate using substitution or recognition of chain rule reverse, and find the integration constant. While it involves multiple techniques, each is routine for A-level and the question structure is typical.
Spec1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations1.08b Integrate x^n: where n != -1 and sums

6 A curve is such that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 12 } { \sqrt { } ( 4 x + a ) }\), where \(a\) is a constant. The point \(P ( 2,14 )\) lies on the curve and the normal to the curve at \(P\) is \(3 y + x = 5\).
  1. Show that \(a = 8\).
  2. Find the equation of the curve.

AnswerMarks Guidance
\(\frac{dy}{dx} = \frac{12}{\sqrt{4x+a}}\), \(P(2, 14)\) Normal \(3y + x = 44\)
(i) \(m\) of normal \(= -\frac{1}{3}\)B1 co
\(\frac{dy}{dx} = 3 = \frac{12}{\sqrt{4x+a}} \rightarrow a = 8\)M1 A1 Use of \(m_1m_2 = -1\). AG.
[3]
(ii) \(\int y = 12(4x+a)^{\frac{1}{2}} \div 2 \div 4\) \((+c)\)B1 B1 Correct without "\(\div 4\)", for "\(\div 4\)".
Uses \((2, 14)\)M1 Uses in an integral only. Dep 'c'.
\(c = -10\)A1 co All 4 marks can be given in (i)
[4]
$\frac{dy}{dx} = \frac{12}{\sqrt{4x+a}}$, $P(2, 14)$ Normal $3y + x = 44$ | | |
(i) $m$ of normal $= -\frac{1}{3}$ | B1 | co |
$\frac{dy}{dx} = 3 = \frac{12}{\sqrt{4x+a}} \rightarrow a = 8$ | M1 A1 | Use of $m_1m_2 = -1$. AG. |
| | [3] |
(ii) $\int y = 12(4x+a)^{\frac{1}{2}} \div 2 \div 4$ $(+c)$ | B1 B1 | Correct without "$\div 4$", for "$\div 4$". |
Uses $(2, 14)$ | M1 | Uses in an integral only. Dep 'c'. |
$c = -10$ | A1 | co All 4 marks can be given in (i) |
| | [4] |
6 A curve is such that $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 12 } { \sqrt { } ( 4 x + a ) }$, where $a$ is a constant. The point $P ( 2,14 )$ lies on the curve and the normal to the curve at $P$ is $3 y + x = 5$.\\
(i) Show that $a = 8$.\\
(ii) Find the equation of the curve.

\hfill \mbox{\textit{CAIE P1 2014 Q6 [7]}}