| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2009 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Product & Quotient Rules |
| Type | Find equation of tangent |
| Difficulty | Moderate -0.3 This is a straightforward multi-part question testing standard differentiation rules (product rule, quotient rule, chain rule) and finding a tangent line equation. Part (i) requires direct application of learned rules with no problem-solving, while part (ii) is a routine tangent calculation. Slightly easier than average due to the mechanical nature of all parts. |
| Spec | 1.07k Differentiate trig: sin(kx), cos(kx), tan(kx)1.07l Derivative of ln(x): and related functions1.07m Tangents and normals: gradient and equations1.07q Product and quotient rules: differentiation1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(u = x^2\), \(v = \cos 3x\); \(\frac{du}{dx} = 2x\), \(\frac{dv}{dx} = -3\sin 3x\) | M1 | Applies \(vu' + uv'\) correctly for their \(u, u', v, v'\) AND gives expression of form \(\alpha x\cos 3x \pm \beta x^2 \sin 3x\) |
| \(\frac{dy}{dx} = 2x\cos 3x - 3x^2\sin 3x\) | A1 | Any one term correct |
| A1 | Both terms correct and no further simplification to terms in \(\cos\alpha x^2\) or \(\sin\beta x^3\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(u = \ln(x^2+1) \Rightarrow \frac{du}{dx} = \frac{2x}{x^2+1}\) | M1 | \(\ln(x^2+1) \rightarrow \frac{\text{something}}{x^2+1}\) |
| A1 | \(\ln(x^2+1) \rightarrow \frac{2x}{x^2+1}\) | |
| \(\frac{dy}{dx} = \frac{\left(\frac{2x}{x^2+1}\right)(x^2+1) - 2x\ln(x^2+1)}{(x^2+1)^2}\) | M1 | Applying \(\frac{vu' - uv'}{v^2}\) |
| \(\frac{dy}{dx} = \frac{2x - 2x\ln(x^2+1)}{(x^2+1)^2}\) | A1 | Correct differentiation with correct bracketing but allow recovery |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| At \(P\), \(y = \sqrt{4(2)+1} = \sqrt{9} = 3\) | B1 | At \(P\), \(y = \sqrt{9}\) or \(3\) |
| \(\frac{dy}{dx} = \frac{1}{2}(4x+1)^{-\frac{1}{2}}(4)\) | M1* | \(\pm k(4x+1)^{-\frac{1}{2}}\) |
| \(\frac{dy}{dx} = \frac{2}{(4x+1)^{\frac{1}{2}}}\) | A1 aef | \(2(4x+1)^{-\frac{1}{2}}\) |
| At \(P\), \(\frac{dy}{dx} = \frac{2}{(4(2)+1)^{\frac{1}{2}}}\); \(m(\mathbf{T}) = \frac{2}{3}\) | M1 | Substituting \(x = 2\) into an equation involving \(\frac{dy}{dx}\) |
| \(y - 3 = \frac{2}{3}(x-2)\) | dM1* | \(y - y_1 = m(x-2)\) or \(y - y_1 = m(x - \text{their } x)\) with their tangent gradient and their \(y_1\); or uses \(y = mx + c\) with their tangent gradient, their \(x\) and their \(y_1\) |
| \(\mathbf{T}: 2x - 3y + 5 = 0\) | A1 | \(2x - 3y + 5 = 0\). Tangent must be in form \(ax + by + c = 0\) where \(a\), \(b\) and \(c\) are integers |
# Question 4:
## Part (i)(a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $u = x^2$, $v = \cos 3x$; $\frac{du}{dx} = 2x$, $\frac{dv}{dx} = -3\sin 3x$ | M1 | Applies $vu' + uv'$ correctly for their $u, u', v, v'$ AND gives expression of form $\alpha x\cos 3x \pm \beta x^2 \sin 3x$ |
| $\frac{dy}{dx} = 2x\cos 3x - 3x^2\sin 3x$ | A1 | Any one term correct |
| | A1 | Both terms correct and no further simplification to terms in $\cos\alpha x^2$ or $\sin\beta x^3$ |
## Part (i)(b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $u = \ln(x^2+1) \Rightarrow \frac{du}{dx} = \frac{2x}{x^2+1}$ | M1 | $\ln(x^2+1) \rightarrow \frac{\text{something}}{x^2+1}$ |
| | A1 | $\ln(x^2+1) \rightarrow \frac{2x}{x^2+1}$ |
| $\frac{dy}{dx} = \frac{\left(\frac{2x}{x^2+1}\right)(x^2+1) - 2x\ln(x^2+1)}{(x^2+1)^2}$ | M1 | Applying $\frac{vu' - uv'}{v^2}$ |
| $\frac{dy}{dx} = \frac{2x - 2x\ln(x^2+1)}{(x^2+1)^2}$ | A1 | Correct differentiation with correct bracketing but allow recovery |
## Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| At $P$, $y = \sqrt{4(2)+1} = \sqrt{9} = 3$ | B1 | At $P$, $y = \sqrt{9}$ or $3$ |
| $\frac{dy}{dx} = \frac{1}{2}(4x+1)^{-\frac{1}{2}}(4)$ | M1* | $\pm k(4x+1)^{-\frac{1}{2}}$ |
| $\frac{dy}{dx} = \frac{2}{(4x+1)^{\frac{1}{2}}}$ | A1 aef | $2(4x+1)^{-\frac{1}{2}}$ |
| At $P$, $\frac{dy}{dx} = \frac{2}{(4(2)+1)^{\frac{1}{2}}}$; $m(\mathbf{T}) = \frac{2}{3}$ | M1 | Substituting $x = 2$ into an equation involving $\frac{dy}{dx}$ |
| $y - 3 = \frac{2}{3}(x-2)$ | dM1* | $y - y_1 = m(x-2)$ or $y - y_1 = m(x - \text{their } x)$ with their tangent gradient and their $y_1$; or uses $y = mx + c$ with their tangent gradient, their $x$ and their $y_1$ |
| $\mathbf{T}: 2x - 3y + 5 = 0$ | A1 | $2x - 3y + 5 = 0$. Tangent must be in form $ax + by + c = 0$ where $a$, $b$ and $c$ are integers |4. (i) Differentiate with respect to $x$
\begin{enumerate}[label=(\alph*)]
\item $x ^ { 2 } \cos 3 x$
\item $\frac { \ln \left( x ^ { 2 } + 1 \right) } { x ^ { 2 } + 1 }$\\
(ii) A curve $C$ has the equation
$$y = \sqrt { } ( 4 x + 1 ) , \quad x > - \frac { 1 } { 4 } , \quad y > 0$$
The point $P$ on the curve has $x$-coordinate 2 . Find an equation of the tangent to $C$ at $P$ in the form $a x + b y + c = 0$, where $a , b$ and $c$ are integers.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C3 2009 Q4 [13]}}