| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2015 |
| Session | November |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Volumes of Revolution |
| Type | Multi-part: volume and area |
| Difficulty | Moderate -0.3 This is a straightforward multi-part question testing standard calculus techniques: differentiation (including chain rule), second derivative test, distance formula, and definite integration. All parts are routine applications of well-practiced methods with no novel problem-solving required, making it slightly easier than average for A-level. |
| Spec | 1.07e Second derivative: as rate of change of gradient1.07i Differentiate x^n: for rational n and sums1.07n Stationary points: find maxima, minima using derivatives1.08b Integrate x^n: where n != -1 and sums1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(f'(x)=2-2(x+1)^{-3}\) | B1 | |
| \(f''(x)=6(x+1)^{-4}\) | B1 | |
| \(f'0=0\) hence stationary at \(x=0\) | B1 | AG |
| \(f''0=6>0\) hence minimum | B1 [4] | www. Dependent on correct \(f''(x)\) except \(-6(x+1)^{-4}\rightarrow<0\) MAX scores SC1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(AB^2=(\frac{3}{2})^2+(\frac{3}{4})^2\) | M1 | |
| \(AB=1.68\) or \(\sqrt{45/4}\) oe | A1 [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Area under curve \(=\int f(x)=x^2-(x+1)^{-1}\) | B1 | Ignore \(+c\) even if evaluated. Do not penalise reversed limits |
| \(=\left(1-\frac{1}{2}\right)-\left(\frac{1}{4}-2\right)=\frac{9}{4}\) | ||
| (Apply limits \(-\frac{1}{2}\rightarrow1\)) | M1A1 | Allow reversed subtn if final ans positive |
| Area trap. \(=\frac{1}{2}(3+\frac{9}{4})\times\frac{3}{2}\) | M1 | |
| \(=\frac{63}{16}\) or \(3.94\) | A1 | |
| Shaded area \(\frac{63}{16}-\frac{9}{4}+\frac{27}{16}\) or \(1.69\) | A1 [6] | |
| ALT eqn \(AB\) is \(y=-\frac{1}{2}x+\frac{11}{4}\) | B1 | |
| Area \(=\int-\frac{1}{2}x+\frac{11}{4}-\int2x+(x+1)^{-2}\) | M1 | Attempt integration of at least one |
| \(=\left[-\frac{1}{4}x^2+\frac{11}{4}x\right]-\left[x^2-(x+1)^{-1}\right]\) | A1A1 | Ignore \(+c\) even if evaluated. Dep. on integration having taken place |
| Apply limits \(-\frac{1}{2}\rightarrow1\) to both integrals | M1 | Allow reversed subtn if final ans positive |
| \(\frac{27}{16}\) or \(1.69\) | A1 |
# Question 10:
## Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f'(x)=2-2(x+1)^{-3}$ | B1 | |
| $f''(x)=6(x+1)^{-4}$ | B1 | |
| $f'0=0$ hence stationary at $x=0$ | B1 | AG |
| $f''0=6>0$ hence minimum | B1 [4] | www. Dependent on correct $f''(x)$ except $-6(x+1)^{-4}\rightarrow<0$ MAX scores SC1 |
## Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $AB^2=(\frac{3}{2})^2+(\frac{3}{4})^2$ | M1 | |
| $AB=1.68$ or $\sqrt{45/4}$ oe | A1 [2] | |
## Part (iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Area under curve $=\int f(x)=x^2-(x+1)^{-1}$ | B1 | Ignore $+c$ even if evaluated. Do not penalise reversed limits |
| $=\left(1-\frac{1}{2}\right)-\left(\frac{1}{4}-2\right)=\frac{9}{4}$ | | |
| (Apply limits $-\frac{1}{2}\rightarrow1$) | M1A1 | Allow reversed subtn if final ans positive |
| Area trap. $=\frac{1}{2}(3+\frac{9}{4})\times\frac{3}{2}$ | M1 | |
| $=\frac{63}{16}$ or $3.94$ | A1 | |
| Shaded area $\frac{63}{16}-\frac{9}{4}+\frac{27}{16}$ or $1.69$ | A1 [6] | |
| **ALT** eqn $AB$ is $y=-\frac{1}{2}x+\frac{11}{4}$ | B1 | |
| Area $=\int-\frac{1}{2}x+\frac{11}{4}-\int2x+(x+1)^{-2}$ | M1 | Attempt integration of at least one |
| $=\left[-\frac{1}{4}x^2+\frac{11}{4}x\right]-\left[x^2-(x+1)^{-1}\right]$ | A1A1 | Ignore $+c$ even if evaluated. Dep. on integration having taken place |
| Apply limits $-\frac{1}{2}\rightarrow1$ to both integrals | M1 | Allow reversed subtn if final ans positive |
| $\frac{27}{16}$ or $1.69$ | A1 | |10 The function f is defined by $\mathrm { f } ( x ) = 2 x + ( x + 1 ) ^ { - 2 }$ for $x > - 1$.\\
(i) Find $\mathrm { f } ^ { \prime } ( x )$ and $\mathrm { f } ^ { \prime \prime } ( x )$ and hence verify that the function f has a minimum value at $x = 0$.\\
\includegraphics[max width=\textwidth, alt={}, center]{5c1ab2aa-3609-4245-b87a-98ecedc83a11-4_515_920_959_609}
The points $A \left( - \frac { 1 } { 2 } , 3 \right)$ and $B \left( 1,2 \frac { 1 } { 4 } \right)$ lie on the curve $y = 2 x + ( x + 1 ) ^ { - 2 }$, as shown in the diagram.\\
(ii) Find the distance $A B$.\\
(iii) Find, showing all necessary working, the area of the shaded region.
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\hfill \mbox{\textit{CAIE P1 2015 Q10 [12]}}