CAIE P1 2014 June — Question 10 8 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2014
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicAreas Between Curves
TypeCurve-Line Intersection Area
DifficultyModerate -0.3 This is a standard area-between-curves question requiring finding intersection points by solving a quadratic equation, then integrating the difference of functions. While it involves multiple steps (finding intersections, setting up integral, integrating polynomials, evaluating), these are all routine techniques for P1 level with no conceptual challenges or novel insights required. Slightly easier than average due to straightforward algebra and integration.
Spec1.08e Area between curve and x-axis: using definite integrals1.08f Area between two curves: using integration
10 \includegraphics[max width=\textwidth, alt={}, center]{0b047754-84f2-46ea-b441-7c68cef47641-3_812_720_1484_715} The diagram shows the curve \(y = - x ^ { 2 } + 12 x - 20\) and the line \(y = 2 x + 1\). Find, showing all necessary working, the area of the shaded region.
AnswerMarks Guidance
pts of intersection \(2x + 1 = -x^2 + 12x - 20\)M1A1 Attempt at soln of sim eqns. co
\(\rightarrow x = 3, 7\)
Area of trapezium \(= \frac{1}{4}(7)(7+15) = 44\)M1A1 Either method ok. co
(or \(\int(2x+1)dx\) from 3 to \(7 = 44\))
Area under curve \(= -\frac{1}{3}x^3 + 6x^2 - 20x\)B2,1 –1 each term incorrect
Uses 3 to 7 \(\rightarrow (54\frac{2}{3})\)DM1 Correct use of limits (Dep 1st M1)
Shaded area \(= 10\frac{2}{3}\)A1 co
[8]
OR
\(\int\left[-x^2 + 10x - 21\right] = -\frac{x^3}{3} + 5x^2 - 21x\) Functions subtracted before integration
M1 subtraction, A1A1A1 for integrated terms, DM1 correct use of limits, A1 Subtraction reversed allow A3A0. Limits reversed allow DM1A0 
pts of intersection $2x + 1 = -x^2 + 12x - 20$ | M1A1 | Attempt at soln of sim eqns. co |
$\rightarrow x = 3, 7$ | | |
Area of trapezium $= \frac{1}{4}(7)(7+15) = 44$ | M1A1 | Either method ok. co |
(or $\int(2x+1)dx$ from 3 to $7 = 44$) | | |
Area under curve $= -\frac{1}{3}x^3 + 6x^2 - 20x$ | B2,1 | –1 each term incorrect |
Uses 3 to 7 $\rightarrow (54\frac{2}{3})$ | DM1 | Correct use of limits (Dep 1st M1) |
Shaded area $= 10\frac{2}{3}$ | A1 | co |
| | [8] |
**OR** | | |
$\int\left[-x^2 + 10x - 21\right] = -\frac{x^3}{3} + 5x^2 - 21x$ | | Functions subtracted before integration |
M1 subtraction, A1A1A1 for integrated terms, DM1 correct use of limits, A1 | | Subtraction reversed allow A3A0. Limits reversed allow DM1A0 |
10\\
\includegraphics[max width=\textwidth, alt={}, center]{0b047754-84f2-46ea-b441-7c68cef47641-3_812_720_1484_715}

The diagram shows the curve $y = - x ^ { 2 } + 12 x - 20$ and the line $y = 2 x + 1$. Find, showing all necessary working, the area of the shaded region.

\hfill \mbox{\textit{CAIE P1 2014 Q10 [8]}}
This paper (11 questions)
View full paper
Q1 3 Q2 5 Q3 6 Q4 6 Q5 7 Q6 7 Q7 8 Q8 8 Q9 8 Q10 8 Q11 9