| Exam Board | OCR |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2011 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Line intersection: show lines are skew |
| Difficulty | Standard +0.3 This is a standard C4 vectors question with routine techniques: showing lines are skew (check non-parallel and non-intersecting), finding angle between direction vectors using dot product, and finding perpendicular from origin to line using scalar product condition. All are textbook methods requiring no novel insight, though part (iii) involves setting up and solving a perpendicular condition which adds slight complexity beyond pure recall. |
| Spec | 1.04c Extend binomial expansion: rational n, |x|<11.10b Vectors in 3D: i,j,k notation1.10c Magnitude and direction: of vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \((1+ax)^{1/2} = 1+\frac{1}{2}ax+\ldots+\frac{\frac{1}{2}\cdot\frac{-1}{2}}{2}(ax)^2\) | B1,B1 | N.B. third term \(= -\frac{1}{8}a^2x^2\) |
| Change \((4-x)^{-1/2}\) into \(k\left(1-\frac{x}{4}\right)^{-1/2}\), where \(k\) is likely to be \(\frac{1}{2}/2/4/-2\), & work out expansion of \(\left(1-\frac{x}{4}\right)^{-1/2}\) | ||
| \(\left(1-\frac{x}{4}\right)^{-1/2} = 1+\frac{1}{8}x+\ldots+\frac{\frac{-1}{2}\cdot\frac{-3}{2}}{2}\left(\frac{(-)x}{4}\right)^2\) | B1,B1 | N.B. third term \(= \frac{3}{128}x^2\) |
| OR change \(\{4-x\}^{1/2}\) into \(l\left(1-\frac{x}{4}\right)^{1/2}\), where \(l\) is likely to be \(\frac{1}{2}/2/4/-2\), & work out expansion of \(\left(1-\frac{x}{4}\right)^{1/2}\) | ||
| \(\left(1-\frac{x}{4}\right)^{1/2} = 1-\frac{1}{8}x-\frac{1}{128}x^2\) | B1 | (for all 3 terms simplified) |
| \(k=\frac{1}{2}\) (with possibility of M1+A1+A1 to follow) | B1 | \(l=2\) (with no further marks available) |
| Multiply \((1+ax)^{1/2}\) by \((4-x)^{-1/2}\) or \(\left(1-\frac{x}{4}\right)^{-1/2}\) | M1 | Ignore irrelevant products |
| The required three terms (with/without \(x^2\)) identified as \(-\frac{1}{16}a^2+\frac{1}{32}a+\frac{3}{256}\) or \(\frac{-16a^2+8a+3}{256}\) AEF ISW | A1+A1 8 | A1 for one correct term + A1 for other two |
| SC: B1 for \(\frac{1}{4}(1-\frac{x}{4})^{-1}\); B1 for \((1-\frac{x}{4})^{-1}=1+\frac{x}{4}+\frac{x^2}{16}\); M1 for multiplying \((1+ax)\) by their \((4-x)^{-1}\) | ||
| If result is \(p+qx+rx^2\), then to find \((p+qx+rx^2)^{1/2}\) award B1 for \(p^{1/2}(\ldots)\), B1 correct 1st & 2nd terms of expansion, B1 correct 3rd term; A1,A1 as before |
# Question 6:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $(1+ax)^{1/2} = 1+\frac{1}{2}ax+\ldots+\frac{\frac{1}{2}\cdot\frac{-1}{2}}{2}(ax)^2$ | B1,B1 | N.B. third term $= -\frac{1}{8}a^2x^2$ |
| Change $(4-x)^{-1/2}$ into $k\left(1-\frac{x}{4}\right)^{-1/2}$, where $k$ is likely to be $\frac{1}{2}/2/4/-2$, & work out expansion of $\left(1-\frac{x}{4}\right)^{-1/2}$ | | |
| $\left(1-\frac{x}{4}\right)^{-1/2} = 1+\frac{1}{8}x+\ldots+\frac{\frac{-1}{2}\cdot\frac{-3}{2}}{2}\left(\frac{(-)x}{4}\right)^2$ | B1,B1 | N.B. third term $= \frac{3}{128}x^2$ |
| OR change $\{4-x\}^{1/2}$ into $l\left(1-\frac{x}{4}\right)^{1/2}$, where $l$ is likely to be $\frac{1}{2}/2/4/-2$, & work out expansion of $\left(1-\frac{x}{4}\right)^{1/2}$ | | |
| $\left(1-\frac{x}{4}\right)^{1/2} = 1-\frac{1}{8}x-\frac{1}{128}x^2$ | B1 | (for all 3 terms simplified) |
| $k=\frac{1}{2}$ (with possibility of M1+A1+A1 to follow) | B1 | $l=2$ (with no further marks available) |
| Multiply $(1+ax)^{1/2}$ by $(4-x)^{-1/2}$ or $\left(1-\frac{x}{4}\right)^{-1/2}$ | M1 | Ignore irrelevant products |
| The required three terms (with/without $x^2$) identified as $-\frac{1}{16}a^2+\frac{1}{32}a+\frac{3}{256}$ or $\frac{-16a^2+8a+3}{256}$ AEF ISW | A1+A1 **8** | A1 for one correct term + A1 for other two |
| SC: B1 for $\frac{1}{4}(1-\frac{x}{4})^{-1}$; B1 for $(1-\frac{x}{4})^{-1}=1+\frac{x}{4}+\frac{x^2}{16}$; M1 for multiplying $(1+ax)$ by their $(4-x)^{-1}$ | | |
| If result is $p+qx+rx^2$, then to find $(p+qx+rx^2)^{1/2}$ award B1 for $p^{1/2}(\ldots)$, B1 correct 1st & 2nd terms of expansion, B1 correct 3rd term; A1,A1 as before | | |
---+ s \left( \begin{array} { l }
3 \\
2 \\
1
\end{array} \right) \quad \text { and } \quad \mathbf { r } = \left( \begin{array} { l }
1 \\
0 \\
0
\end{array} \right) + t \left( \begin{array} { r }
0 \\
1 \\
- 1
\end{array} \right)$$
respectively.\\
(i) Show that $l _ { 1 }$ and $l _ { 2 }$ are skew.\\
(ii) Find the acute angle between $l _ { 1 }$ and $l _ { 2 }$.\\
(iii) The point $A$ lies on $l _ { 1 }$ and $O A$ is perpendicular to $l _ { 1 }$, where $O$ is the origin. Find the position vector of $A$.
6 Find the coefficient of $x ^ { 2 }$ in the expansion in ascending powers of $x$ of
$$\sqrt { \frac { 1 + a x } { 4 - x } } ,$$
giving your answer in terms of $a$.
\hfill \mbox{\textit{OCR C4 2011 Q6 [8]}}