| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2013 |
| Session | January |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Standard trigonometric equations |
| Type | Function properties and inverses |
| Difficulty | Standard +0.3 This is a multi-part question testing transformations, differentiation using quotient rule, and integration by substitution. While it has several parts (17 marks total), each component is relatively standard: (i) identifying transformations is routine, (ii) quotient rule differentiation is straightforward C3 content, (iii) the substitution is guided and follows a standard pattern, and (iv) requires recognizing the relationship between the two functions. The question is slightly easier than average because it's heavily scaffolded with clear instructions at each step and no novel problem-solving required. |
| Spec | 1.02w Graph transformations: simple transformations of f(x)1.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^21.07q Product and quotient rules: differentiation1.08h Integration by substitution |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int_0^{\pi/4} f(x)dx = \int_0^{\pi/4} \frac{\sin x}{\cos x}dx\) | M1 [7] | substituting to get \(\int \frac{-1}{u}(du)\); ignore limits here, condone no \(du\) but not \(dx\); allow \(\int 1u \cdot -du\); but for A1 must deal correctly with the \(-ve\) sign by interchanging limits |
| let \(u = \cos x, du = -\sin x \, dx\) | ||
| when \(x = 0, u = 1\), when \(x = \pi/4, u = 1/\sqrt{2}\) | ||
| \(= \int_1^{1/\sqrt{2}} \frac{-1}{u}du\) | A1 | NB AG |
| \(= \int_{1/\sqrt{2}}^1 \frac{1}{u}du\) * | ||
| \(= [\ln u]_{1/\sqrt{2}}\) | M1 | \([\ln u]\) |
| \(= \ln 1 - \ln(1/\sqrt{2})\) | A1 | |
| \(= \ln \sqrt{2} = \ln 2^{1/2} = \frac{1}{2}\ln 2\) | A1 | mark final answer |
| Answer | Marks | Guidance |
|---|---|---|
| Area = area in part (iii) translated up 1 unit. | M1 [2] | soi from \(\pi/4\) added; A1cao; o.e. (as above) |
| So = \(\frac{1}{2}\ln 2 + 1 \times \pi/4 = \frac{1}{2}\ln 2 + \pi/4\). | A1cao |
### (iii)
$\int_0^{\pi/4} f(x)dx = \int_0^{\pi/4} \frac{\sin x}{\cos x}dx$ | M1 [7] | substituting to get $\int \frac{-1}{u}(du)$; ignore limits here, condone no $du$ but not $dx$; allow $\int 1u \cdot -du$; but for A1 must deal correctly with the $-ve$ sign by interchanging limits
let $u = \cos x, du = -\sin x \, dx$ | |
when $x = 0, u = 1$, when $x = \pi/4, u = 1/\sqrt{2}$ | |
$= \int_1^{1/\sqrt{2}} \frac{-1}{u}du$ | A1 | NB AG
$= \int_{1/\sqrt{2}}^1 \frac{1}{u}du$ * | |
$= [\ln u]_{1/\sqrt{2}}$ | M1 | $[\ln u]$
$= \ln 1 - \ln(1/\sqrt{2})$ | A1 |
$= \ln \sqrt{2} = \ln 2^{1/2} = \frac{1}{2}\ln 2$ | A1 | mark final answer
### (iv)
Area = area in part (iii) translated up 1 unit. | M1 [2] | soi from $\pi/4$ added; A1cao; o.e. (as above)
So = $\frac{1}{2}\ln 2 + 1 \times \pi/4 = \frac{1}{2}\ln 2 + \pi/4$. | A1cao |
---Fig. 8 shows parts of the curves $y = f(x)$ and $y = g(x)$, where $f(x) = \tan x$ and $g(x) = 1 + f(x - \frac{1}{4}\pi)$.
\includegraphics{figure_8}
\begin{enumerate}[label=(\roman*)]
\item Describe a sequence of two transformations which maps the curve $y = f(x)$ to the curve $y = g(x)$. [4]
\end{enumerate}
It can be shown that $g(x) = \frac{2\sin x}{\sin x + \cos x}$.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item Show that $g'(x) = \frac{2}{(\sin x + \cos x)^2}$. Hence verify that the gradient of $y = g(x)$ at the point $(\frac{1}{4}\pi, 1)$ is the same as that of $y = f(x)$ at the origin. [7]
\item By writing $\tan x = \frac{\sin x}{\cos x}$ and using the substitution $u = \cos x$, show that $\int_0^{\frac{1}{4}\pi} f(x)dx = \int_{\frac{1}{\sqrt{2}}}^1 \frac{1}{u}du$.
Evaluate this integral exactly. [4]
\item Hence find the exact area of the region enclosed by the curve $y = g(x)$, the $x$-axis and the lines $x = \frac{1}{4}\pi$ and $x = \frac{1}{2}\pi$. [2]
\end{enumerate}
\hfill \mbox{\textit{OCR MEI C3 2013 Q8 [17]}}