| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 20 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Product & Quotient Rules |
| Type | Find derivative of product |
| Difficulty | Standard +0.3 This is a structured multi-part question on product rule differentiation with standard follow-up applications. While it covers several techniques (product rule, odd functions, turning points, integration by parts), each part is routine for C3 level with clear signposting. The algebraic manipulations are straightforward, and no novel problem-solving insight is required—slightly easier than average due to its guided nature. |
| Spec | 1.05a Sine, cosine, tangent: definitions for all arguments1.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^21.07e Second derivative: as rate of change of gradient1.07q Product and quotient rules: differentiation1.08i Integration by parts |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(x\cos 2x = 0\) when \(x=0\) or \(\cos 2x = 0\) | M1 | \(\cos 2x = 0\) |
| \(2x = \frac{\pi}{2}\) | M1 | or \(x = \frac{1}{2}\cos^{-1}0\) |
| \(x = \frac{1}{4}\pi\) | A1 | \(x = 0.785..\) or \(45\) is M1 M1 A0 |
| P is \(\left(\frac{\pi}{4}, 0\right)\) | [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(f(-x) = -x\cos(-2x)\) | M1 | \(-x\cos(-2x)\) |
| \(= -x\cos 2x\) | E1 | \(= -x\cos 2x\) |
| \(= -f(x)\), Half turn symmetry about O | B1 [3] | Must have two of: rotational, order 2, about O (half turn = rotational order 2) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(f'(x) = \cos 2x - 2x\sin 2x\) | M1 | Product rule |
| A1 [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(f'(x) = 0 \Rightarrow \cos 2x = 2x\sin 2x\) | ||
| \(2x\frac{\sin 2x}{\cos 2x} = 1\) | M1 | \(\frac{\sin}{\cos} = \tan\) |
| \(x\tan 2x = \frac{1}{2}\) | E1 [2] | www |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(f'(0) = \cos 0 - 2.0.\sin 0 = 1\) | B1ft | Allow ft on (their) product rule expression |
| \(f''(x) = -2\sin 2x - 2\sin 2x - 4x\cos 2x\) | M1 | Product rule on \(2x\sin 2x\) |
| \(= -4\sin 2x - 4x\cos 2x\) | A1 | Correct expression – mark final expression |
| \(f''(0) = -4\sin 0 - 4.0.\cos 0 = 0\) | E1 [4] | www |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Let \(u = x\), \(\frac{dv}{dx} = \cos 2x \Rightarrow v = \frac{1}{2}\sin 2x\) | M1 | Integration by parts with \(u=x\), \(\frac{dv}{dx} = \cos 2x\) |
| \(\int_0^{\pi/4} x\cos 2x\,dx = \left[\frac{1}{2}x\sin 2x\right]_0^{\pi/4} - \int_0^{\pi/4}\frac{1}{2}\sin 2x\,dx\) | A1 | |
| \(= \frac{\pi}{8} + \left[\frac{1}{4}\cos 2x\right]_0^{\pi/4}\) | A1 | \(\left[\frac{1}{4}\cos 2x\right]\) – sign consistent with their previous line |
| M1 | Substituting limits – dep using parts | |
| \(= \frac{\pi}{8} - \frac{1}{4}\) | A1 | www |
| Area of region enclosed by curve and \(x\)-axis between \(x=0\) and \(x=\frac{\pi}{4}\) | B1 [6] | Or graph showing correct area – condone P for \(\frac{\pi}{4}\) |
## Question 3:
### Part (i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $x\cos 2x = 0$ when $x=0$ or $\cos 2x = 0$ | M1 | $\cos 2x = 0$ |
| $2x = \frac{\pi}{2}$ | M1 | or $x = \frac{1}{2}\cos^{-1}0$ |
| $x = \frac{1}{4}\pi$ | A1 | $x = 0.785..$ or $45$ is M1 M1 A0 |
| P is $\left(\frac{\pi}{4}, 0\right)$ | [3] | |
### Part (ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $f(-x) = -x\cos(-2x)$ | M1 | $-x\cos(-2x)$ |
| $= -x\cos 2x$ | E1 | $= -x\cos 2x$ |
| $= -f(x)$, Half turn symmetry about O | B1 [3] | Must have two of: rotational, order 2, about O (half turn = rotational order 2) |
### Part (iii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $f'(x) = \cos 2x - 2x\sin 2x$ | M1 | Product rule |
| | A1 [2] | |
### Part (iv)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $f'(x) = 0 \Rightarrow \cos 2x = 2x\sin 2x$ | | |
| $2x\frac{\sin 2x}{\cos 2x} = 1$ | M1 | $\frac{\sin}{\cos} = \tan$ |
| $x\tan 2x = \frac{1}{2}$ | E1 [2] | www |
### Part (v)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $f'(0) = \cos 0 - 2.0.\sin 0 = 1$ | B1ft | Allow ft on (their) product rule expression |
| $f''(x) = -2\sin 2x - 2\sin 2x - 4x\cos 2x$ | M1 | Product rule on $2x\sin 2x$ |
| $= -4\sin 2x - 4x\cos 2x$ | A1 | Correct expression – mark final expression |
| $f''(0) = -4\sin 0 - 4.0.\cos 0 = 0$ | E1 [4] | www |
### Part (vi)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Let $u = x$, $\frac{dv}{dx} = \cos 2x \Rightarrow v = \frac{1}{2}\sin 2x$ | M1 | Integration by parts with $u=x$, $\frac{dv}{dx} = \cos 2x$ |
| $\int_0^{\pi/4} x\cos 2x\,dx = \left[\frac{1}{2}x\sin 2x\right]_0^{\pi/4} - \int_0^{\pi/4}\frac{1}{2}\sin 2x\,dx$ | A1 | |
| $= \frac{\pi}{8} + \left[\frac{1}{4}\cos 2x\right]_0^{\pi/4}$ | A1 | $\left[\frac{1}{4}\cos 2x\right]$ – sign consistent with their previous line |
| | M1 | Substituting limits – dep using parts |
| $= \frac{\pi}{8} - \frac{1}{4}$ | A1 | www |
| Area of region enclosed by curve and $x$-axis between $x=0$ and $x=\frac{\pi}{4}$ | B1 [6] | Or graph showing correct area – condone P for $\frac{\pi}{4}$ |3 Fig. 8 shows part of the curve $y = x \cos 2 x$, together with a point P at which the curve crosses the $x$-axis.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{00c12cc4-f7ee-4219-8d34-a1854284f65d-2_425_974_478_591}
\captionsetup{labelformat=empty}
\caption{Fig. 8}
\end{center}
\end{figure}
(i) Find the exact coordinates of P .\\
(ii) Show algebraically that $x \cos 2 x$ is an odd function, and interpret this result graphically.\\
(iii) Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$.\\
(iv) Show that turning points occur on the curve for values of $x$ which satisfy the equation $x \tan 2 x = \frac { 1 } { 2 }$.\\
(v) Find the gradient of the curve at the origin.
Show that the second derivative of $x \cos 2 x$ is zero when $x = 0$.\\
(vi) Evaluate $\int _ { 0 } ^ { \frac { 1 } { 4 } \pi } x \cos 2 x \mathrm {~d} x$, giving your answer in terms of $\pi$. Interpret this result graphically.
\hfill \mbox{\textit{OCR MEI C3 Q3 [20]}}