CAIE P1 2019 November — Question 11 10 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2019
SessionNovember
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVolumes of Revolution
TypeRotation about y-axis, standard curve
DifficultyStandard +0.3 This is a straightforward volumes of revolution question requiring completing the square (routine A-level skill), rearranging to make x the subject, and applying the standard formula V = π∫x²dy. All steps are procedural with no novel insight required, making it slightly easier than average.
Spec1.02e Complete the square: quadratic polynomials and turning points1.08a Fundamental theorem of calculus: integration as reverse of differentiation4.08d Volumes of revolution: about x and y axes

11 \includegraphics[max width=\textwidth, alt={}, center]{0e4a249a-9e6a-49d4-996c-fe07b7730f59-18_650_611_260_762} The diagram shows a shaded region bounded by the \(y\)-axis, the line \(y = - 1\) and the part of the curve \(y = x ^ { 2 } + 4 x + 3\) for which \(x \geqslant - 2\).
  1. Express \(y = x ^ { 2 } + 4 x + 3\) in the form \(y = ( x + a ) ^ { 2 } + b\), where \(a\) and \(b\) are constants. Hence, for \(x \geqslant - 2\), express \(x\) in terms of \(y\).
  2. Hence, showing all necessary working, find the volume obtained when the shaded region is rotated through \(360 ^ { \circ }\) about the \(\boldsymbol { y }\)-axis.
    If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.

Question 11:
Part 11(i):
AnswerMarks Guidance
AnswerMarks Guidance
\((y =)(x+2)^2 - 1\)B1 DB1 2nd B1 dependent on 2 in bracket
\(x + 2 = (\pm)(y+1)^{1/2}\)M1
\(x = -2 + (y+1)^{1/2}\)A1
Part 11(ii):
AnswerMarks Guidance
AnswerMarks Guidance
\(x^2 = 4 + (y+1) - /+ 4(y+1)^{\frac{1}{2}}\)*M1A1 SOI. Attempt to find \(x^2\). The last term can be \(-\) or \(+\) at this stage
\((\pi)\int x^2 \, dy = (\pi)\left[5y + \frac{y^2}{2} - \frac{4(y+1)^{\frac{3}{2}}}{\frac{3}{2}}\right]\)A2,1,0
\((\pi)\left[15 + \frac{9}{2} - \frac{64}{3} - \left(-5 + \frac{1}{2}\right)\right]\)DM1 Apply \(y\) limits
\(\frac{8\pi}{3}\) or \(8.38\)A1
## Question 11:

### Part 11(i):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $(y =)(x+2)^2 - 1$ | B1 DB1 | 2nd B1 dependent on 2 in bracket |
| $x + 2 = (\pm)(y+1)^{1/2}$ | M1 | |
| $x = -2 + (y+1)^{1/2}$ | A1 | |

### Part 11(ii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $x^2 = 4 + (y+1) - /+ 4(y+1)^{\frac{1}{2}}$ | *M1A1 | SOI. Attempt to find $x^2$. The last term can be $-$ or $+$ at this stage |
| $(\pi)\int x^2 \, dy = (\pi)\left[5y + \frac{y^2}{2} - \frac{4(y+1)^{\frac{3}{2}}}{\frac{3}{2}}\right]$ | A2,1,0 | |
| $(\pi)\left[15 + \frac{9}{2} - \frac{64}{3} - \left(-5 + \frac{1}{2}\right)\right]$ | DM1 | Apply $y$ limits |
| $\frac{8\pi}{3}$ or $8.38$ | A1 | |
11\\
\includegraphics[max width=\textwidth, alt={}, center]{0e4a249a-9e6a-49d4-996c-fe07b7730f59-18_650_611_260_762}

The diagram shows a shaded region bounded by the $y$-axis, the line $y = - 1$ and the part of the curve $y = x ^ { 2 } + 4 x + 3$ for which $x \geqslant - 2$.\\
(i) Express $y = x ^ { 2 } + 4 x + 3$ in the form $y = ( x + a ) ^ { 2 } + b$, where $a$ and $b$ are constants. Hence, for $x \geqslant - 2$, express $x$ in terms of $y$.\\

(ii) Hence, showing all necessary working, find the volume obtained when the shaded region is rotated through $360 ^ { \circ }$ about the $\boldsymbol { y }$-axis.\\

If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.\\

\hfill \mbox{\textit{CAIE P1 2019 Q11 [10]}}