| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differentiating Transcendental Functions |
| Type | Calculate area under curve |
| Difficulty | Moderate -0.3 This is a multi-part question covering standard C3 techniques: quotient rule differentiation (routine), integration of sec²x (standard result), function transformations, and applying transformations to areas. All parts are textbook exercises requiring recall and direct application rather than problem-solving or insight. Slightly easier than average due to the guided structure and standard methods throughout. |
| Spec | 1.02w Graph transformations: simple transformations of f(x)1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.07q Product and quotient rules: differentiation1.08c Integrate e^(kx), 1/x, sin(kx), cos(kx) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{d}{dx}\left(\frac{\sin x}{\cos x}\right) = \frac{\cos x \cdot \cos x - \sin x \cdot (-\sin x)}{\cos^2 x}\) | M1A1 | Quotient (or product) rule (AG) |
| \(= \frac{\cos^2 x + \sin^2 x}{\cos^2 x} = \frac{1}{\cos^2 x}\) | A1 | |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Area \(= \int_0^{\pi/4} \frac{1}{\cos^2 x}\, dx\) | B1 | correct integral and limits (soi); condone no \(dx\) |
| \(= \left[\tan x\right]_0^{\pi/4}\) | M1 | \([\tan x]\) or \(\left[\frac{\sin x}{\cos x}\right]\) |
| \(= \tan(\pi/4) - \tan 0 = 1\) | A1 | unsupported scores M0 |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| \(f(0) = 1/\cos^2(0) = 1\) | B1 | must show evidence |
| \(g(x) = \frac{1}{2\cos^2(x + \pi/4)}\) | M1 | or \(f(\pi/4) = 1/\cos^2(\pi/4) = 2\) |
| \(g(0) = \frac{1}{2}\cos^2(\pi/4) = 1\) | A1 | so \(g(0) = \frac{1}{2}f(\pi/4) = 1\) |
| \((\Rightarrow\) f and g meet at \((0,1))\) | ||
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Translation in \(x\)-direction through \(-\pi/4\) | M1 | must be in \(x\)-direction, or \(\begin{pmatrix}-\pi/4\\0\end{pmatrix}\) |
| A1 | \(\begin{pmatrix}-\pi/4\\0\end{pmatrix}\) alone SC1 | |
| Stretch in \(y\)-direction scale factor \(\frac{1}{2}\) | M1 | must be in \(y\)-direction |
| A1 | ||
| asymptotes correct | B1ft | stated or on graph; condone no \(x = \ldots\), ft \(\pi/4\) to right only |
| min point \((-\pi/4,\ \frac{1}{2})\) | B1ft | stated or on graph |
| curves intersect on \(y\)-axis | B1 | |
| correct curve, dep B3, with asymptote lines and TP in correct position | B1dep | |
| [8] |
| Answer | Marks | Guidance |
|---|---|---|
| Same as area in (ii), but stretched by s.f. \(\frac{1}{2}\). So area \(= \frac{1}{2}\). | B1ft | \(\frac{1}{2}\) area in (ii); allow unsupported |
| [1] |
## Question 5:
### Part (i)
| $\frac{d}{dx}\left(\frac{\sin x}{\cos x}\right) = \frac{\cos x \cdot \cos x - \sin x \cdot (-\sin x)}{\cos^2 x}$ | M1A1 | Quotient (or product) rule **(AG)** |
|---|---|---|
| $= \frac{\cos^2 x + \sin^2 x}{\cos^2 x} = \frac{1}{\cos^2 x}$ | A1 | |
| **[3]** | | |
### Part (ii)
| Area $= \int_0^{\pi/4} \frac{1}{\cos^2 x}\, dx$ | B1 | correct integral and limits (soi); condone no $dx$ |
|---|---|---|
| $= \left[\tan x\right]_0^{\pi/4}$ | M1 | $[\tan x]$ or $\left[\frac{\sin x}{\cos x}\right]$ |
| $= \tan(\pi/4) - \tan 0 = 1$ | A1 | unsupported scores M0 |
| **[3]** | | |
### Part (iii)
| $f(0) = 1/\cos^2(0) = 1$ | B1 | must show evidence |
|---|---|---|
| $g(x) = \frac{1}{2\cos^2(x + \pi/4)}$ | M1 | or $f(\pi/4) = 1/\cos^2(\pi/4) = 2$ |
| $g(0) = \frac{1}{2}\cos^2(\pi/4) = 1$ | A1 | so $g(0) = \frac{1}{2}f(\pi/4) = 1$ |
| $(\Rightarrow$ f and g meet at $(0,1))$ | | |
| **[3]** | | |
### Part (iv)
| Translation in $x$-direction through $-\pi/4$ | M1 | must be in $x$-direction, or $\begin{pmatrix}-\pi/4\\0\end{pmatrix}$ |
|---|---|---|
| | A1 | $\begin{pmatrix}-\pi/4\\0\end{pmatrix}$ alone SC1 |
| Stretch in $y$-direction scale factor $\frac{1}{2}$ | M1 | must be in $y$-direction |
| | A1 | |
| asymptotes correct | B1ft | stated or on graph; condone no $x = \ldots$, ft $\pi/4$ to right only |
| min point $(-\pi/4,\ \frac{1}{2})$ | B1ft | stated or on graph |
| curves intersect on $y$-axis | B1 | |
| correct curve, dep B3, with asymptote lines and TP in correct position | B1dep | |
| **[8]** | | |
### Part (v)
| Same as area in (ii), but stretched by s.f. $\frac{1}{2}$. So area $= \frac{1}{2}$. | B1ft | $\frac{1}{2}$ area in (ii); allow unsupported |
|---|---|---|
| **[1]** | | |
---5 Fig. 9 shows the curve $y = \mathrm { f } ( x )$, where $\mathrm { f } ( x ) = \frac { 1 } { \cos ^ { 2 } x } , - \frac { 1 } { 2 } \pi < x < \frac { 1 } { 2 } \pi$, together with its asymptotes $x = \frac { 1 } { 2 } \pi$ and $x = - \frac { 1 } { 2 } \pi$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{431d496a-a606-4b92-9f5c-e12b074a7ba9-3_921_1398_538_414}
\captionsetup{labelformat=empty}
\caption{Fig. 9}
\end{center}
\end{figure}
(i) Use the quotient rule to show that the derivative of $\frac { \sin x } { \cos x }$ is $\frac { 1 } { \cos ^ { 2 } x }$.\\
(ii) Find the area bounded by the curve $y = \mathrm { f } ( x )$, the $x$-axis, the $y$-axis and the line $x = \frac { 1 } { 4 } \pi$.
The function $\mathrm { g } ( x )$ is defined by $\mathrm { g } ( x ) = \frac { 1 } { 2 } \mathrm { f } \left( x + \frac { 1 } { 4 } \pi \right)$.\\
(iii) Verify that the curves $y = \mathrm { f } ( x )$ and $y = \mathrm { g } ( x )$ cross at $( 0,1 )$.\\
(iv) State a sequence of two transformations such that the curve $y = \mathrm { f } ( x )$ is mapped to the curve $y = \mathrm { g } ( x )$.
On the copy of Fig. 9, sketch the curve $y = \mathrm { g } ( x )$, indicating clearly the coordinates of the minimum point and the equations of the asymptotes to the curve.\\
(v) Use your result from part (ii) to write down the area bounded by the curve $y = \mathrm { g } ( x )$, the $x$-axis, the $y$-axis and the line $x = - \frac { 1 } { 4 } \pi$.
\hfill \mbox{\textit{OCR MEI C3 Q5 [18]}}