| Exam Board | AQA |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2013 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tangents, normals and gradients |
| Type | Find tangent given derivative expression |
| Difficulty | Moderate -0.5 This is a straightforward C1 question requiring basic integration to find the curve equation and substitution to find the tangent. The gradient at P is found by direct substitution (giving 9), then using point-slope form. Part (b) requires polynomial integration and using the given point to find the constant. All techniques are routine with no problem-solving insight needed, making it slightly easier than average. |
| Spec | 1.07m Tangents and normals: gradient and equations1.08a Fundamental theorem of calculus: integration as reverse of differentiation1.08b Integrate x^n: where n != -1 and sums |
| Answer | Marks | Guidance |
|---|---|---|
| (Gradient \(= 10 - 6 + 5) = 9\) | B1 | correct gradient from sub \(x=1\) into \(\frac{dy}{dx}\) |
| \(y - 4 =\) 'their 9' \((x - 1)\) or \(y =\) 'their 9' \(x + c\) and attempt to find \(c\) using \(x = 1\) and \(y = 4\) | M1 | must attempt to use given expression for \(\frac{dy}{dx}\) and must be attempting tangent and not normal and coordinates must be correct |
| \(y = 9x - 5\) | A1 | 3 marks |
| Answer | Marks | Guidance |
|---|---|---|
| \((y =) \frac{10}{5}x^5 - \frac{6}{3}x^3 + 5x + C\) | M1 | one term correct |
| A1 | another term correct | |
| \(4 = 2 - 2 + 5 + C \Rightarrow C = -1\) | m1 | substituting both \(x = 1\) and \(y = 4\) and attempting to find \(C\) |
| \(y = 2x^5 - 2x^3 + 5x - 1\) | A1 cso | 5 marks |
**6(a)**
(Gradient $= 10 - 6 + 5) = 9$ | B1 | correct gradient from sub $x=1$ into $\frac{dy}{dx}$
$y - 4 =$ 'their 9' $(x - 1)$ or $y =$ 'their 9' $x + c$ and attempt to find $c$ using $x = 1$ and $y = 4$ | M1 | must attempt to use given expression for $\frac{dy}{dx}$ and must be attempting tangent and not normal and coordinates must be correct
$y = 9x - 5$ | A1 | 3 marks | condone $y = 9x + c,...$ $c = -5$
**6(b)**
$(y =) \frac{10}{5}x^5 - \frac{6}{3}x^3 + 5x + C$ | M1 | one term correct
| A1 | another term correct
$4 = 2 - 2 + 5 + C \Rightarrow C = -1$ | m1 | substituting both $x = 1$ and $y = 4$ and attempting to find $C$
$y = 2x^5 - 2x^3 + 5x - 1$ | A1 cso | 5 marks | must have $y = ...$ and coefficients simplified
---6 The gradient, $\frac { \mathrm { d } y } { \mathrm {~d} x }$, of a curve at the point $( x , y )$ is given by
$$\frac { \mathrm { d } y } { \mathrm {~d} x } = 10 x ^ { 4 } - 6 x ^ { 2 } + 5$$
The curve passes through the point $P ( 1,4 )$.
\begin{enumerate}[label=(\alph*)]
\item Find the equation of the tangent to the curve at the point $P$, giving your answer in the form $y = m x + c$.
\item Find the equation of the curve.
\end{enumerate}
\hfill \mbox{\textit{AQA C1 2013 Q6 [8]}}